Power-plant-engineering.docx

SOLUTIONS TO PROBLEMS IN

POWER PLANT ENGINEERING MKS UNITS

PEDRO S. DE LEON

Submitted by: BSME 5-4 PUP-MANILA

Submitted to: Engr. Robel Nomorosa PPE Instructor

Chapter 1

Introduction

Problem 1-1 A train weighing 1600 tons is pulled up a 2% grade by 4476 kw. Train resistance, 8727.273 kg. What is the speed, MPH?

1000 kg =1,600.00 kg. ton

Train weight in kg = 1600 tons x Work done by 4476 kw = 4476

kn−m 1000 n x s kn

= 4476000 n-m/s

Force exerted along the incline (F) F = 1600000 sin( tan−10.02) + 8727.273 kg F = 40720.875 kg Let V = constant velocity of train up to the incline in m/s FV = 40720.875 x V This is equal to the work done by 4476 kw Equating the two works 40720.875 V =4476000 n−m/s x

kg 9.81 n

Solving for V: V=

4476000 kg−m/s = 11.205 m/s 40720.875 kg( 9.81)

V = 11.205 m/s x km/1000 m x miles/1.6kfm x 3600 s/hr = 25.211 miles/hr Problem I-2 A pump is lifting water through 5.486m to fill a 566.254 m3 tank. The overall energy efficiency is 80%. Calculate the length of time that 7.46 kw applied to the pump will require to complete the job. Solution: Total kgs. of water pumped to tank at 1000kg/m3 = 566.254 m3 (1000 kg/m 3 ¿=¿ 11.205 m/s Total work input to pump at 80% efficiency =

( kg water )(head) Efficiency

=

566254 kg x 5.486 m 0.80

= 3883086.8 kg-m

Work output of 7.46 kw = 7.46

kn−m 60 s kn−m x = 447.6 s min min

Time required to fill the tank

¿ = total work input ¿ pump rate of doing work =

3883086.8 kg−m x 9.81 n /kg 447.6 knm/min x 1000m/kn

= 85.105 min. ans

Problem I-4 The difference in tension between the sides of belt running over a 0.762 m diameter pulley is 22.727 kg. Pulley is 500 rpm. What kw is transmitted? Is the direction of the belt leaving the pulley a factor? Solution: Net tension = 22.727 kg Torque

= Net tension x Radius of Pulley = 22.727 kg x

0.762 m 2

= 8.659 kg – m a.) Kw transmitted = 2 π 2 =

nt ; kw 1000

π rev n (500 )(8.659 kg−m x .81 ) rev min kg n s 1000 x 60 kn min

m Kw x s kn−m = s = 4.448 kw. Ans 4.448 kn−

b.) The direction of the belt leaving the pulley is a factor. Installations with the slack side above the tight side below, will have greater capacity than those where the slack side is below. Also installation where the belts are horizontal will transmit greater kw than those where the belts are vertical.

Problem I-6

What is the power of a steam jet 0.015 m in diameter moving at 761.936 m/s? kg Steam condition, 1 406 dry and saturated. cm2 Solution: The weight of the steam leaving the nozzle will be found by the following formula W=

Eqn. (1)

AVo ; m/s V1

Where: A = Area of jet =

π (0.015)2 m2 = 0.00018 m 2 4

Vo = Velocity of jet = 761.963 m/s Given V1 = Specific volume of steam at 1.4 g = 0.19

kg dry and saturated cm2

m3 from steam table kg

Substituting these values in equation (1) above and solving for W. W=

0.0018 m2 (761.963

m ) s

0.19 m3 /kg W = 0.722 kg/s Kinetic energy of the jet KE = 0.722 =

WVs2 2g

2 kg 2 2 (761.963) m / s s m 2(9.81 2 ) s

= 21365.151 kg-m/s KE in kw = 21365.151 kg – m/s x 9.81 n/kg x kn/1000n x kw/ = 209.592 kw. ans

kn m s

Problem 1-7 Forty percent of the electrical input to a motor driven pump is converted into a hydraulic jet 0.013 m in diameter, for the purposes of washing down ashes. Find the jet velocity in m/s. The motor has a 3-phase, 220 V, 7.5 amp. rating. Power factor 85%.

Solution: = √ 3 EI cos θ

Watts input to motor

= √ 3 x 220 x 7.5 x 0.85 = 2420 W = 2.42 Kw Since 40% of this is converted into a hydraulic jet, energy converted to jet = 2.42 x 0.40 = 0.968 Kw = 968 W = 968 N-m/s Jet energy is also = (AVsW)

Vs 2 2g

Where: A = Area of jet = π/4 D2 W = Density of water: 1000 kg/m3

Vs = Velocity of jet, m/s

Equating jet energy equation to 968, and substituting known values in the equation, we have

π Vs2 2 2 3 4 (0.013) m x Vs x 1000kg/m x 2¿ ¿ = 981 N-m/s  Vs = 52.214. m/s. ans. Problem 1-10 A turbo generator rotating mass has a moment of inertia of 555 hyls/ m 2. It is delivering 2500 kw at 1800 rpm. The load then suddenly increases to 2550 kw, the developed steam power remaining unchanged. What is the resulting speed in rpm after 10 seconds?

Solution: Since the developed steam power remains the same, the increase in energy required must come from the rotating mass. This will cause the rotating mass to have a decrease in kinetic energy, hence, a decrease in speed. Kinetic energy of rotating mass;

KE =

1 I W 21 2

Where: I - Moment of inertia = 555 hyls/m2 W1 - Angular velocity = RPS x 2π = 1800/60 x 2π RAD /s

 KE =

1 1800 (555)( x 2π)2 2 60

KE = 9860130 kg - m x 9.81

n kn x kg 1000 n

KE = 96727.875 kn - m

Change in load = 2550 - 2500 = 50 kw Total change of energy in 10 seconds.

= 50 kn - m/s x 10 /s = 500 kn - m This is also the total KE Energy change of the rotating mass KE of Rotating Mass = 96727.875 - 500 = 96227.875 kn -m Where W2 = Angular velocity after 10 seconds Potential Kw capacity at 90% electric efficiency = 762.472 x 0.90 = 686.22 Kw ans.

Problem 1-16 An electric heater is to heat 11 kg of oil per min. from 4.5°C to 65.5°C. Specific heat of the oil is 2.1 kJ/kg °C. How many watts should this heater consume?

Solution: To find the heat absorbed by the oil use the equation: Q=WCΔT

Where: Q = Total heat absorbed; kJ/min. W = Weight of substance; kg/min. C = Specific heat of substance; kJ/kg°C ΔT = Change of temp. of substance.

Substituting the known values in this equation; Q = 11 kg/min (2.1 kj/kg°C) (65.5 - 4.5)°C

Q = 1409.1 kJ/min. This will also be the heat consumed by the heater assuming no losses. Converted to Kw. Consumption of heater = 1409.1 Kj/min x min/60s x kw/kj/s = 23.485 Kw ans.

Problem 1-18

An electric motor converted 700 W of electrical input into work at 58.186 kg-m/s. The speed was 1750 RPM. Find the kg-m of driven torque and the motor efficiency.

Solving for W2 ;

W 22

W

2 2

=

=

2 x 96227.875 555

RAD s 2 π ( RPS)

346.767

RPS =

346.767/2π =

55.189

RPM =

60 x 55.189 = 3311.381 RPM ans. resulting speed after 10 seconds

Problem 1-12 A hoist is to raise a 11147.727 N mine cage at the rate of 4.57 m/s. Mechanical efficiency of the hoist is 92%. What is the Kw required to drive at this speed?

Solution: Work done by the hoist = 11147.727 x 4.57 = 50945.114 n-m/s

Kw required at 92% Eff. of hoist =

50945.114 = 55.375 Kw 1000 x 0.92

Problem 1-14 The flow of a river is 4.25 m/s at a site where a 22.86 m hydrostatic head can be created by the erection of a dam. What is the potential capacity of a hydro-electric power plant if installed at this site? Consider that the hydraulic efficiency of energy conversion can be 80%, and that the electric efficiency can be 90%.

Solution: Kw that can be developed from the water flow = Qwh x hydraulic efficiency = 4.25 m3/s (9.81 kn/m3) (22.86 m)(0.80) = 762.472 KW

Problem 1-16 An electric heater is to heat 11 kg of oil per min. from 4.5ºC to 65.5ºC. The specific heat of the oil is 2.1kJ/kg ºC. How many watts should this heater consume? Solution: To find the heat absorbed by the oil use the equation: Q=WC ∆ T Where:

Q = Total heat absorbed; kJ/min.

W =Weight of substance; kg/min. C = Specific heat of substance; kJ/kg ºC ∆ T = Change of temp. substance. Substituting the known values in this equation; Q=11

kg kJ 2.1 (65.5−4.5)℃ min kg ℃

Q=1409.1

(

)

kJ min

This will also be the heat consumed by the heater assuming no losses. Converted to Kw. Consumption of heater

= 1409.1 kJ/min x min/60s x kw/kJ/s = 23.485 Kw ans.

Problem 1-18 An electric motor converted 700 W of electrical into work at 58.186 kg-m/s. The speed was 1750 RPM. Find the kg-m of driven torque and the motor efficiency. Solution: Kw output of motor = 58.186 kg-m/s (9.81 m/s2) = 570.80 n-m/s = 0.571 kn-m/s = 0.571 kw

Since Kw =

2 πNt 1000

And Substituting 0.571 Kw kn−m n ( 1000 ) kw(1000) s kn Torque= = 2π N 2π rev min x 1750 x rev min 60 s 0.571

Torque=3.115 n−m x

kg 9.81 n

Torque = 0.317 kg-m ans. Problem 1-20 In a change over from steam to electric heating, a unit which had been condensing 4.5 kg dry and saturated steam at 2.109 kg/cm 2 absolute each 15 min. is to be electrically heated from a 220 V Circuit. How many 12-Ohm resistor's in parallel would be needed to supply the same heat?

Solution: Latent heat of evaporation at 2.109 kg/cm 2 hfg = 2198.535 kJ/kg from steam tables Heat given off by 4.5 kg/min dry and sat. steam at 2.109kg/cm 2 = hfg x 4.5 = 2198.535 x 4.5 = 9893.408 kJ/15 min. Heat given off in 1 hr = 9893.408 x 60/15 = 39573.63 kJ/hr If the same heat is to come from an electric source ¿ 39573.63

kJ hr kw x x h r 3600 s kJ s

= 10.993 Kw = 10993 W Since watts =

E2 R

Total resistance required

= E2/watts = (220)2/10993 = 4.403 Ohms

If this total resistance is to come from 12 Ohm resistors in parallel; No. of resistors required

¿

12 =2.725 4.403

Say 3 ans. Actual total resistance

¿

12 =4 O h ms 3

Problem 1-26 A power plant serves a factory having two 22-Kw motors and ten 3.7-Kw motors. Assume the efficiency of motors 80%, of transmission line 95%, of generator 92%. a) What should the rated capacity of the generator be if it is assumed that all motors might be delivering their rated power simultaneously? b) What should be the rated capacity of the engine? 1.

Maximum total Kw output = 2 x 22 +10 x 3.7 = 81 Kw

Kw capacity of generator to take care of this load at 80% motor efficiency and 95% transmission eff. ¿ 2.

81 = 106.579 Kw ans. 0.8 x 0.95

Rated capacity of engine at 92% generator eff. ¿

106.579 Kw = 115.848 Kw ans. 0.92

Problem 1-28 The overall efficiency of an electric system, coal pile to lights, is 11%. What fraction of a kg coal containing 30250 kJ/kg must be consumed to light a 100 W lamp for 45 min? Solution: kJ equivalent of 1-100 W lamp lighted for 45 min. ¿ 100

J s kJ x 45 min x 60 =27000 J x s min 1000 J

= 270 kJ Required heat input from fuel for 11% efficiency. ¿

270 kJ =2454.546 kJ 0.11

Fraction of 30250 kJ/kg coal to supply 2545.546 kJ ¿

2545.546 kJ 30250 kJ /kg

= 0.081 kg ans.

Chapter 2 The Variable Load Problem

Problem 2-1 A central station is supplying energy to a community through two sub-stations. One substation feeds four distributor circuits; the other six. The maximum daily recorded demands are Power Station

……………………………………………………. 12,000Kw

Substation

A

……………………………………………………. 6,000Kw

Feeder

1

……………………………………………………. 1,700

Feeder

2

……………………………………………………. 1,800

Feeder

3

……………………………………………………. 2,800

Feeder

4

……………………………………………………. 600

Substation

B

……………………………………………………. 9,000Kw

Feeder

1

……………………………………………………. 620

Feeder

2

……………………………………………………. 1,500

Feeder

3

……………………………………………………. 1,000

Feeder

4

……………………………………………………. 2,900

Feeder

5

……………………………………………………. 2,200

Feeder

6

……………………………………………………. 3,000

Calculate the diversity factor between: a)

Substation

b)

Feeders on Substation A

c)

Feeders on Substation B

Solutions: a) Diversity Factor Between Substations = =

= 1.25 ans. b) Diversity Factor between Feeders on Substation A = = =

1.15 ans.

c) Diversity Factor Between Feeders on Substation B =

= = 1.247

Problem 2-6 The annual peak load on a 15,000kw power plant is 10,500kw. Two substations are supplied by this plant. Annual energy dispatched through Substation A is 27,500,000kw-hr with a peak load at 8,900kw, while 16,500,000 are sent through substation B with a peak load at 6,650kw Neglect line losses. Find: a)

Diversity Factor between Substations and

b)

Capacity Factor of the Power Plant

Solution a) Diversity Factor between Substations. =

=

=1.45 ans. b) Capacity Factor Actual energy produced annually = 27,500,000 =44,000,000

 Capacity Factor =

+16,500,000

= = 0.335 ans.

Problem 2-8 A distributing transformer supplies a group of general power customer having a connected load of 186kw. Demand factor = 0.75. If the load factor for the group will average 45% and the energy sells at 3½ cents per kw-hr, what will be the monthly (30day) income from energy delivered through this transformer? Assume average motor efficiency is 75%

Solution: Demand Factor (DF) =

Actual Max. Demand = Connected Load x Demand Factor = 186kw x 0.75 = 139.5kw Simultaneous Maximum Demand (SMD) =

From table 2-2 p42, PPD by Morse, Diversity Factor for general power service between consumer = 1.5 SMD =

= 93kw

With 75% Motor Efficiency, the Simultaneous Max. Demand on distributing Transformer is =

= Peak load

= = 124 kw  Average Load on Transformer, = Load Factor x Peak Load = 0.45 x 124 kw = 55.8 kw kw – hrs delivered 30 days in one month =

=

55.8kw

= 40,176

 Income from Energy Delivered = P0.035

X 40,176

= P1,406.16/month

ans.

Problem 2-11 A power plant is said to have had a use factor of 48.5% and a capacity factor of 42.4%. How many hours did it operate during the year?

Solution: From the definitions of use factor and capacity factor we can get the following equations: Annual Kw −hrs Produced 1. Plant Use Factor ¿ kw Capacity × No . of hrs of Operation

Actual Energy Production hrs 2. Plant Capacity Factor kw Capacity ×8760 yr ¿

Let:

PUF – Plant Use Factor PCF – Plant Capacity Factor TEP – Total Energy Produced in one year KWC – Kilowatt Capacity NHO – Number of Hrs of Operation TEP – Annual Kw-Hrs produced = Actual Energy Production

Divide Eq. 1 by Eq. 2 1 2 P UF=

TEP KWC × NHO

TEP

P CF=

KWC × 8760

hrs yr

TEP PUF KWC × NHO = PCF TEP hrs KWC ×8760 yr PUF 8760 hrs / yr hrs PCF = ∴ NHO=8760 PCF NHO yr PUF

(

PUF hrs 0.424 =8760 PCF yr 0.485

(

∴ NHO=7658.23

)

hrs ans . yr

)

Problem 2-12 A central Station has annual factors as follows: Load Factor 58.5%, Capacity Factor 40.9%, Use factor, 45.2%. The reversed carried over and above peak load is 8,900 kW. Find (a) installed capacity, (b) annual energy production, (c) hours per year not on service.

Solution: Eq .1 Capacity Factor=

¿

Ave . Load × 8760 hrs/ yr hrs Installed Capacity × 8760 yr

Ave . Load Installed Capacity

But: Ave. Load = Load Factor × Peak Load Installed Capacity = Peak Load + Reserve

∴ 0.409= 0.409=

Load Factor × Peak Load Peak Load + Reserve

0.585 × Peak Load Peak Load +8900

0.409 ( Peak Load +8900 )=0.585× Peak Load 0.409 ¿

( 0.585−0.409 ) Peak Load=0.409( 8900) ( 0.176 ) Peak Load=3640.1 ∴ Peak Load=

3640.1 =20,682.38kW 0.176

(a) Installed capacity or kW consumed: = Peak Load + 8900 = 20, 682.38 + 8900

= 29,532.38 kW ans.

(b) Average Load = Load Factor x Peak Load = 0.585 x 20, 638.38 Annual Production = Ave. Load x 8760 hrs/yr = (0.585 x 20,682.38) 8760 = 105, 988,924.5

kW −hrs yr

(c) Load Factor (LF) = ¿

∴ 0.452= Hrs .∈Use=

hrs yr ans.

Ave . Load ×8760 Installed Capacity × Hrs .∈Use

Annual Production Installed Capacity × Hrs .∈Use

105,988,924.5 kW −hr / yr 29,582.38 kW × Hrs .∈Use

105,988,924.5 kW−hr / yr ( 0.452 ) 29,582.38 kW

¿ 7926.63 hrs / yr Hrs per yr not ∈service=8760

hrs hrs −7926.63 yr yr

= 833.37 hrs ans. Problem 2-15 A 50, 000 kW steam plant delivers an annual output of 238,000 kW-hr with a peak load of 42,860 kW. (a) What is the annual load factor? (b) What is the capacity factor?

Solution: (a) Load Factor =

Average Load( AL) Peak Load(PL)

TEP = Total kw-hrs Produced per Yr. = 238,000,000 kw-hr/yr AL = =

TEP No . of Hrs∈OneYr . 238 ,000,000 kw−hr / yr 8760 hrs / yr

AL = 27,168.94 kW

Annual Load Factor =

Average Load Peak Load

=

27,168.94 kW 42,860 kW

= 0.634

(b) Capacity Factor = =

ans.

TEP kw Capacity × 8760hrs/ yr 238,000,000 kW−hr / yr 50000 kW ×8760 hrs / yr

= 0.543 hrs.

ans.

Problem 2-17 Given load factor, 0.48, installed capacity, 35,000 kW, reserve over peak, 3000 kW, hours out of service per year, 410. Find the capacity and use factor and use factor.

Solution: Installed Capacity = Peak Load + reserve over peak 35,000 = Peak Load + 3000 kW Peak Load

= 35,000 kW – 3,000 kW = 32,000 kW

∴ Average Load =Load Factor × Peak Load = 0.48 × 32, 000 kW = 15,360 kW

(a) Capacity Factor, (CF) = (CF) = =

Annual Production( AP) Installed Capacity ×8760 hrs / yr

AP=Average Load ×8760 hrs / yr Installed Capacity ×8760 hrs/ yr 15 ,360 kW × 8760 hrs / yr 35 ,000 kW ×8760 hrs / yr

CF = 0.438

(b) Use Factor =

ans.

Annual Production Installed Capacity × Hrs .∈Service

15,360 kw ×8760 hrs / yr = 35,000 kw × ( 8760−410 ) hrs/ yr UF = 0.460

ans.

Problem 2-21 The system shown in Fig. 2-7 consists in part of a transformer serving customers e, f, and g. Estimate the peak load on the transformer. e. store building with 5 kw lighting, 25 kw small motor power. f. store building with 18 kw lighting, 35 kw small motor power. g. store building with 55 kw lighting, 80 kw small motor power. Solution:

From Table 2.1, page 42. PPD by Morse: Typical Demand Factors: Commercial Lighting of Stores & Offices = 0.70 Typical Demand Factors: For General Power Service 15 kw – 75 kw = 0.55 Over 75 kw = 0.5 Demand Factor, (DF) = Thus,

Maximum Demand ,(MD) Connected Load ,(CL)

MD = (DF) (CL)

For Customer e: MD = (5 kw) (0.70) + (25 kw) (0.55) = 17.25 kw For Customer f: MD = (18 kw) (0.70) + (35 kw) (0.55) = 31.85 kw For Customer g: MD = (55 kw) (0.70) + (80 kw) (0.50) = 78.5 kw Total MD = MDe + MDf + MDg

= 17.25 kw + 31.85 kw + 78.5 kw = 127.6 kw From table 2.1 p. 42 PPD by Morse, diversity factor for commercial lighting and general power service = 1.5 Total Max Demands Diversity Factor, (DF) = Simultaneous Max Demand ¿ Peak Load on the Transformer Peak Load on the transformer =

127.6 kw 1.5

= 85.97 kw ans. Problem 2-25 Steam flow meters on an industrial process line recorded maximum flow rates of 26,258.5 kg/hr for cycles of 15 seconds duration. To keep this fluctuation off the boiler and allow them to steam evenly at an average rate of 3401.36 kg/hr, a Ruths steam accumulator was installed, working between 3.2 and 2.5 kg/cm 2 gage. A water surface area of sufficient size to limit rate of steam disengagement to 0.305 m 3/sec/m2, was considered necessary to ensure dry steam production. Find suitable dimensions for the accumulator tank, (i.e., diameter x length), note as actually installed the tank was 1.83 x 7.32 m long. Solution: Maximum Steam Flow Rate Above the Average = 26,258.5 kg/hr – 3401.36 kg/hr = 22,855.14 kg/hr This corresponds to a flow rate as follows: 22,857.14 kg/hr x 1hr/3600 sec = 6.35 kg/sec Amount of steam in 15 secs. = 6.35 kg/sec x 15 sec = 96.25 kg steam Enthalpy of liquid at 3.2 kg/cm2 = 2737 KJ/kg Enthalpy of liquid at 2.5 kg/cm2 = 2730 KJ/kg

Average Steam Enthalpy: =

2737+2730 = 2733.5 KJ/kg 2

Enthalpy of liquid at 3.2 kg/cm2 = 609 KJ/kg Enthalpy of liquid at 2.5 kg/cm2 = 581 KJ/kg Let: w = kg water in the accumulator at 3.2 kg/cm2 w= 95.25kg = kg water in the accumulator at 2.5 kg/cm 2 Energy Discharged Per Cycle By Accumulator = Kg Steam Representing Max. Flow Above Average = 96.25 kg (2733.5 KJ/kg) = 260,365.875 KJ/cycle This is also equal to w(609) – (w - 95.25) 581, the heat given off by the water in the accumulator, equating the two values. w(609) – (w - 95.25) 581 = 260,365.875 609 w – 581 w + 55,340.25 = 260,365.875 w=

260,365.875−55,340.25 28

w = 7,322 kg water in tank when full charged to 3.2 kg/cm 2 For maximum steam flow, consider the maximum flow area, i.e. the diametral area (area through the center). Specific Volume of Steam (Vapor) at 3.2 kg/cm 2 = 0.5325 m3/kg Specific Volume of Steam (Vapor) at 2.5 kg/cm 2 = 0.4486 m3/kg Average Specific Volume =

0.5325+0.4486 2

= 0.4905 m3/kg Thus, Steam Flow Rate: = 0.4905 m3/kg (6.35kg/sec) = 3.115 m3/sec

m3 sec Flow Area = = 10.21 m2 3 m 0.305 2 sec−m 3.115

Finally: If diameter is 1.83 m, Length =

10.21m2 = 5.58 m 1.83 m

say = 6 m ans. Check: Vol. when half full =

π 2 d l 8

=

π ( 1.83 m )2 (6 m) = 7.9 m3 8

Vol. of water (7,322 kg at 3.2 kg/cm2) Specific Volume = 0.00108 m3/kg V = 0.00108 m3/kg (7.322 kg) = 7.9 m3 Problem 2-28 A hydraulic storage plant has a unit similar to that shown in Figure 2-15. This efficiency of the generator-motor is 96%, of the turbine 80%, of the pump 75%. Average elevation between upper and lower pools = 30m. Assume 2% loss of head in pipe

friction. This unit was installed to carry a daily peak load of 1400 kw-hr. There is a daily peak load of 1400 kw-hr. There is a daily evaporation loss of stored water amounting to 907 tons. Calculate the overall efficiency of conversion. Solution: Combined Efficiency = Energy Input, Eit =

Eot = (Ngm)(Nt) Eit

Eot QwHnet = k (Ngm)(Nt )

Net Head, Hnet = (1 – 2%) (30 m) = 29.4 m Mass Flow: = Mt = Qw Thus, Mt =

k Eot ; k = Constant (Ngm)( Nt ) Hnet = 6116.3

=

(

6116.3

kg−m kw−min

kg−m min ( 1400 kw−hr ) 60 kw−min hr ( 96 % )( 80 % ) (29.4 m)

)

Mt = 22,754,092.26 kg (The amount of water required by turbine for the generator motor to generate 1400 kw – hr)

Eop = (Ngm)(Np) Eip Energy Output for Pumping: Eop =

Mp Hp ; where k = Constant, k = 6116.3

kg−m kw−min

Mp = Amount of H20 Pumped Hp = Pump Head Pump Head: Hp = 30 m (1 + 2%) = 30.6 m Solve for Energy Input to Gen-Motor: Eip =

Eop Mp Hp = ( Ngm ) ( Np) k ( Ngm ) (Np)

Amount of Water Pumped, = Amount used by the turbine + Amount lost by evaporation = 22,754,092.26 kg + 907 tons (1000kg/ton) = 22,661,092.26 kg Thus, (23,661,092.26 kg)(30.6 m) Eip = kg−m min 6116.3 (60 )(96 % )( 75 %) kw−min hr

(

)

Eip = 2,740.20 kw – hrs Efficiency of Conversion: Nc =

EnergyOutput of System as GeneratingUnit System as PumpingUnit ¿ Energy Input ¿

Nc =

1400 kw−hr x 100% 2740.20 kw−hr

Nc = 51.09%

Chapter 3 Power Plant Economics

Problem 3-1 A power plant costs $375,800 to build. Its life is assumed 20 yrs, salvage, 15%. Find the sinking fund payment. Interest compounded annually at 5%. Solution: Sinking Fund Factor = =

r n (1+r ) −1 0.05 0.05 = 20 (1+0.05) −1 2.65−1

= 0.0303 Sinking fund payment = (principal – salvage) (sinking fund factor) = (375,800 – 1.5 x 375,800) (0.0303) = $9,658.06 ans. Problem 3-2 Find the production cost per 1000 kg steam in a steam plant when the evaporation rate is 3.3 kg steam per kg coal; initial cost of plant, $150,000; annual operation cost exclusive of coal, $15,000. Assume life of 20 years; no final value;

interest on borrowed capital, 4%, on sinking fund, 3%. Average steam production, 14,400 kg/hr; cost of coal, $8,00 per ton. Solution: Sinking fund payment =( P−S ) ×

r ( 1−r )n−1

= ( 150000−0 ) ×

0.03 ( 1+0.03 )20−1

= $5,580.00 Interest on capital at 4% = 150,000 x 0.04

$6,000.00

Annual operation cost exclusive of coal(given)

$15,000.00

Annual cost of production exclusive of coal

= $ 26,580.00

Steam production per year = average steam production/hr x 8760 = 14,500 x 8760 = 127,020,000 kgs kgs coal used per year at steaming rate of 3.3 kgs steam/kgs coal

=

127,020,000 =38,409,909kgs 3.3

Annual cost of coal at $8.00 per ton =

38,409,909 x=$ 307,507.27 1000

Grand Total Annual Production Cost =$ 26,580.00+$ 307,927.27 = $334,507.27 Cost per 453.5 kg steam =

Annual Cost x 1000 Annual Steam Production

=

334. 507.27 x 1000 kgs 127,020,000 kg

= $ 2.63 ans. Problem 3-4 The bonds issued to build certain power plant have a face value of $2,500,00 and bear interest at 4.5%. They are to be retired at the end of the 20 years by an accumulated sinking fund which will yield 4% compounded annually. Find the annual payment to the account of capital investment. Solution: Let r = rate of interest on sinking fund = 0.04 given n = term = 20 years given The sinking fund factor =

=

r ( 1−r )n−1 0.04 0.04 = =0.0336 20 ( 1.04 ) −1 2.19−1

Sinking fund payments = Principal x Sinking Fund Factor = 2,500,000 x 0.0336 = $84,000.00 Interest on bonds at 4.5 = 2,500,000 x 0.045 = $112,500.00 Total Annual Payment = $196,500.00

Problem 3-6 An ash disposal system of a steam plant cost $30,000 when new. It is now 4 years old. The annual maintenance costs for the four years have been $2,000, $2,500, $2,675, and $3,000. Interest rate = 6%. A news system is guaranteed to have an

equated annual maintenance and operation cost not exceeding $1,500. Its costs is $47,000 installed. Life of each system, 7 years, salvage value, 5% of first cost. Present value of old system is same as salvage value. Would it be profitable to install the new system? Solution: The old and the new system will be compared in the basis of costs actually prevented or incurred by replacement.

Preliminary Calculations: Total maintenance cost of old plant in four years = 2,000 + 2,500 + 2,6750 + 3,000 = $9,925.00 Average annual maintenance cost =

9,925 =$ 2,481.25 4

Prevented replacement Depreciation of old plant =

Sale Value−Salvage Value Remaining Life

=

0.05 x 30,000−0.5 x 30,00 7−4

=0 Average interest on (Sale Value – Salvage Value)

=0

Interest on Salvage Value at 5% = 0.05(0.05 x 30,000) = $ 750.00 Maintenance Cost Total Prevented by Replacement Incurred by Replacement Depreciation =

Principal −Salvage Life

= $2,481.25 = $3,231.25

=

44,650 47,000−(0.5 x 47,000) =, 7 = $6,378.57 7

r n+1 Average Interest = ( 2 )( n )(Principal – Salvage) 0.06 8 = ( 2 )( 7 ¿(44,650)

= $1,530.00

Interest of Salvage= 0.06 x 0.04(47,000)

= $ 141.00

Maintenance and Operation Cost

= $1,500.00

Total Incurred by Replacement

= $9,459.57

Replacement is unprofitable by (9,549.57 – 3,231.25) or $6,318.32 a year Problem 3-7 Ten years ago, a small steam plant of 2,000 kw capacity costing $125 per kw was erected. The life was estimated at 15 years and the salvage value at 5%. At present, abandoning the old plant in favor of a new diesel plant is being considered. A market has been found for the old engine and boiler equipment at $15,000. The remainder of the old plant can be utilized in the new and is given a valuation of $8,500. Depreciation has been figured on the straight line basis. What is the difference between the depreciation book value of the old plant and its sale value? How would the difference be taken cared of in the reconstruction? Solution: Initial cost of the plant

= $125 x 2,000 = $250,000

Estimated Salvage Value

= 5% of initial cost = 0.05 x 250,000 = $ 12,500

Depreciation, (Straight Line) =

Initial Cost−Salvage Value Life

=

250,000−12,500 15

= $5,833.33 per year

Total depreciation in 10 years = 15,833.33 x 10 = $158,333.3 Depreciation Book Value after 10 years = Initial Cost - Depreciation = $250,000 - $158,333.30 Sale Value of the Old Plant (Given) Difference Between Book Value and Sale Value

= $91,666.70 = $15,000.00 = $76,666.70

This difference is to be ignored in the reconstruction. Refer to item 1, first line on page 74 of the text. Problem 3-9 The annual costs expected by a utility system in supplying a certain residential suburb of 45,000 customers are: Fixed Element, $345,000; Energy Element, $180,000; Customer Element, $300,000; Return in Investment $200,000. During the year, 17,050,000 kw will be registered on customer parameters and their maximum demand on the power plant will be 5,500 kw. Diversity factor from Table 2-2. (a) Form a straight line meter rate (b) Form a three charge rate, putting ¾ of the profit in the Energy Element, 1/44 in Fixed Element

(c) Form a room rate in which the Customer Element is a fixed monthly service charge and the Fixed Element is obtained in the first kw-hr per room. (Assume average home, 6 rooms) Energy Element is uniformly distributed. Solution: (a) Straight Line Meter Rate Fixed Element

=$

345,000.00 Given

Energy Element

=$

180,000.00 Given

Customer Element

=$

300,000.00 Given

Profit Element

=$

200,000.00 Given

Total Annual Production Cost

= $ 1,025,000.00 Annual Production Cost____

Straight Line Meter Rate =

Energy Delivered to Customer / yr Energy Delivered to Customer:

17,050,000 kw-hrs / yr

Straight Line Meter Rate =

= $0.06 per kw – hr.

1,025,000_ 17,050,000 (b) Three Charge Rates From Table 2-2, p.42, and for residential lighting. Diversity Factor Between Consumers, 3-4, Use 3.5 Diversity Factor Between Transformers

1.3

Diversity Factor Between Feeders

1.2

Diversity Factors Between Substations

1.1

From the above values of diversity factors, the diversity factorfor the power plant is the product of the 4 factors or 3.5 x 1.3 x 1.2 x 1.1 = 6 Maximum Demand on Power Plant = 5,500 kw, given Customer’s Peak = Maximum Demand x Diversity Factor\ = 5,500 x 6 = 3,300 kw Demand Charge = Fixed Element x ¼ of profit (Given) = $345,000 + ¼ x $200,000 = $395,000 per year Total Demand Charge The Unit Demand Charge = $395,000

=

Customer’s Peak

33,000

= $12.00 per kw per year or $1.00 per kw per month The Energy Charge = Energy Element + ¾ of profit (Given) = $180,000 + ¾ (200,000) = $330,000 per year Total Energy Charge per year___ Unit Energy Charge =

Total Energy Delivered to Customers $330,000.00______ =

$17,050,000.00 kw-hrs / yr = $0.0194 per kw-hr

Service Charge = Customer Element = $300,000 per year Total Service Charge / yr 12 x No. of Customers

Unit Service Charge Per Month =

$300,000.00_ =

12 x 45,000 =

$0.555 per month

Summarizing, final rate is as follows: Plus $1.00 per month per kw Max. Demand Plus 55.5 ¢ per kw-hr used. (c) Room Rate Total Service Charge / yr Service Charge =

12 x No. of Customers =

12 x 45,000

$300,000.00_ = $0.555 per month same as above

If we consider the profit as part of the Fixed Element, and it is desired to obtain this in the first 4 kw-hr/room, then annual sum to be collected this way = $345,000 + $200,000 = $545,000 per year Unit Charge per kw per month for 1st 4 kw-hr per room. Sum to be collected per yr_____________ =

No. of customers x No. of room / customer x 4 x 12 $545,000______ = $45,000 x 6 x 4 x 12

=

$0.0422/(kw-hr)(room) per month

Energy Remaining for Straight Energy Charge = Total Energy Delivered – Energy Accounted for Already

=17,050,000 – 45,000 x 4 x 12\ =17,050,000 – 12,900.000 = 41,150,000 kw-hrs / yr Energy Element_ Energy Remaining

Unit Energy =

$180,000 4,150,000 =

=

$0.0435

Summarizing room rate is as follows: $0.555 per month service charge Plus $0.0422 per kw-hr month for the first 4 kw-hr per counted room (Note: The energy charge may look too high compared to that of (b) at $0.0194 kw-hr) Alternately, the charge based on counted room may include the Fixed Elements, profit + 50%, profit of Energy Element, total sum obtained was = $345,000 + $200,000 + 180,000 2 = $635,000.00 Change for 1st 4 kw-hr per room is then $635,000_________ 45,000 x 6 x 4 x 12

= $0.0492 say $0.049

Amount collected this way = 0.049 = $634,000.00 Total annual cost = sum of all elements = $725,000 Energy left for straight energy charge = 4,150,000 same as before Straight energy charge for remaining energy =

$725,000 – 634,000 4,150,000 = $0.0129, say $0.022/ kw-hr

12,900,000 x

This is more attractive than thepreceding rates. Summarizing, room rate: $0.555 Service charge per month Plus $0.049 per kw-hr per month for the 1 st 4 kw-hr/counted room Plus $0.022 per kw-hr for all energy in excess of this.

Problem 3-13 Annual costs ni a certain power system are: For fixed costs:

Plant, $1,750,000; Primary Lines, $600,000; Secondary Lines, $1,250,000

For operating cost: Plant, $75,000 indirect and $950,000 direct. For distribution system, $500,000 Direct customer expense, $400,000; profit, 8% of fixed cost. Output, 1.2 x 108 kw-hr Assume 50,000 customers an 20% transmission loss. Find the straight line meter rate. Solution: Total Fixed Costs (Given) Plant

$1,750,000

Primary Lines Secondary Line

600,000 1,250,000 3,600,000

Operating Cost: Plant (Indirect)

$75,000

(Direct)

950,000

Distributing System

500,000 $ 1,525,000

Customer Expenses

400,000

Profit, 0.08 x Fixed Costs = 0.08 x $3,600,000

= $

Annual Production Cost

288,000

= $ 5,813,000

At 20% transmission loss, kw-hrs registered on customers meters = Annual Production x 0.8 = 1.2 x 108 x 0.8 = 0.96 x 108 kw-hrs Straight Line Meter Rate =

Annual Production Cost___ Energy Delivered to Customer

=

$5,813,000 0.96 x 108 =

$0.606 per kw/hr

Problem 3-15 An air preheater installation will cost $12,500. Its life is assumed to be 8 yrs. Salvage value is nothing. Annual maintenance and repair are estimated to average $150. Use compound interest at 6% and find the annual cost of the preheater.

Solution: r____ Sinking Fund Factor = ( 1 + r )n -1 =

0.06___ (1.06)8 – 1

= 0.101 Sinking Fund Payment = (P - S) x Sinking Fund Factor = (12,500 – 0) x 0.101 = $1,262.50 Interest on Capital

= 0.06 x $12,500

Maintenance and Repair (Given)

=

750.00

=

150.00

Total Annual Cost

= $2,162.50

Problem 3-16 What initial cost will an annual saving of $675 per yr. for ten yrs, amortize at 4% interest compounded annually? Solution: Value of principal, P in 10 yrs, at 4% compounded annually = P (1.04)10 = 1.48 P

Sinking Fund Factor =

r____ ( 1 + r )n -1

=

0.04___ (1.04)10 – 1

=

0.04___ 1.48 – 1

= 0.0834 Sum accumulated at the end of ten yrs. = =

Annual Savings__ Sinking Fund Factor $675_ 0.0834

= $8,100 This sum should also be equal to the value of the principle after ten yrs or 1.48 P = $8,100 Or

P = $8,100 1.48 = $5,470

Problem 3-17

Amt. amortized ans.

A 30 mhp condensate pump motor has been burned beyond repair. The plant superintendent has two replacement alternatives. Manufacturer “A” offers to replace the original (which was an “A” motor) for $510. Manufacturer “B” offers a cheaper motor at $400 but can only guarantee 87% efficiency whereas the “A” motor is guaranteed for 89%. The installation operates 25% of the time at full load, and 75% of the time at half load where the two efficiencies become 85% and 84%, respectively. Assume a motor comparison period of 5 years, interest rate 8%, equal maintenance costs. Electric energy is charged for at rate of 1 ½ ȼ per kw-hr. a) Which motor is more economical to buy? b) At what energy cost do they become equal alternatives? Solution: Assume that the motor will run continuously for the whole year, then. No. of operating hrs. at full load

= ¼ × 8760 = 2190 hrs.

No. of operating hrs. at half load = ¾ × 8760 =6570 hrs. Calculations for Energy Costs: 1) Motor “A” Kw-hrs consumption at full load at 89% eff. ¿

30×0.746×6570 0.85

¿ 86,500 Expected kw-hrs consumption/ year = 141,500 kw/hrs. Energy cost at $0.015/kw-hr

= 0.015 × 141,500 = $2,122.50/ year

2) Motor “B” Kw-hrs consumption at full load at 87% eff. ¿

30×0.746×2190 0.87

¿ 56,300 kw/hrs. Kw-hrs consumption at half load at 84% eff. 1 ( ×30)×0.746×6570 2 ¿ 0.84 ¿ 87,500 Expected kw-hrs consumption/ year = 143,800 kw-hrs Energy cost at $0.015/kw-hr

=0.015 × 143,800 = $2,157.00

The annual cost, excluding maintenance which is the same for both, will be computed for each motor and then compare. Depreciation will be based on sinking fund method. Sinking Fund Factor =

r ( 1+r )n -1

¿

0.08 (1+0.08 )5 -1

¿ 0.17 Annual cost computations for Motor “A”: Depreciation = Principal × Sinking Fund Factor =$510 × 0.17

$86.70

Interest on investment = 0.08 × 510

$40.80

Energy cost

$2,122.50

Annual cost excluding maintenance

$2,250.00

Annual cost computations for Motor “B”: Depreciation = Principal × Sinking Fund Factor =$400 × 0.17

$68.00

Interest on investment = 0.08 × 400

$32.00

Energy cost

$2,157.00

Annual cost excluding maintenance

$2,257.00

∴ Motor A is more favorable by $2,257.00-$2,250.00 or $7.00 per year Note: Straight line depreciation may also be used, and it can be shown that motor “A” is still more economical but the difference will be $7.22/yr. a) Both motors will be equal alternatives if their total annual cost are equal. Let C= Energy cost for which the two motors are equal alternatives. Then from the relation: Annual cost for Motor “A” = Annual cost for Motor “B” $86.70 + $40.80 + C (141,500) = $68.00+$32.00+C (143,800) C [143,800-141,500] = $86.70+ $40.80 - $68.00 - $32.00 C= $27.50/ 2,300 C= $0.0119

Therefore, the two motors are equal alternatives when energy cost is 1.19 ȼ

Problem 3-20 A 500-kw lighting plant cost $95 per kw installed. Fixed charges, 14%, operating costs, 1.34 per kw-hr. the plant averages 150 kw for 5,000 hr of the year, 420 kw for 1,000 hr, and 20 kw for the remainder. What is the unit cost of production of electric energy? Solution: Plant output per year: =150 × 5,000+ 420 × 1,000+ 20 × [8,760-(5,000+1,000)] =750,000+ 420,000+ 35,200 =1,205,200 kw-hrs. Installed cost of power plant: = cost per kw installed × kw installed = $95 × 500 = $47,500 Fixed Charges = 0.14 × $47,500

=$ 6,650.00

Operating Cost = 0.013 × 1,205,200

=$15,667.60

Annual Cost of Production Cost per kw-hr

=

$22,317 1,205,200

= $0.0184 ans.

=$22,317.60

Problem 3-22 A costumer’s meter reads 29,543 kw-hr on May 1, and 29,598 kw-hr on June 1. Find the amount of his electric bill for May based on the following rates: a) 7 ȼ per kw-hr. b) 10 ȼ per kw-hr for the first 35 kw-hr; 5 ȼ per kw-hr for the next 25 kw-hr; 3 ȼ per kw-hr for all in excess of 60 kw-hr. Solution: Reading on meter, June 1

29,598 kw-hrs.

Reading on meter, May 1

29, 543 kw-hrs

Energy consumption for May by difference

=

55 kw-hrs

=

$3.50

= 20 × $0.05

=

$1.00

Total bill for month of May

=

$4.50 ans.

(a) Bill at 7 ȼ per kw-hrs.

= 55 × $0.07 = 3.85

(b) 1st 35 kw-hrs at 10 ȼ = 35 × $0.10 Cost of energy above 35 kw-hrs but below 60 kw-hrs = (55-35) × $0.05

Problem 3-26

The rate for a commercial customer is $6.00 per kw per month for the first kw of maximum demand, plus $5.00 per kw per month for the next 6 kw of maximum demand, plus $4.00 per kw per month for all the maximum demand in excess of 7 kw, plus energy charge as follows: First 100 kw-hr. at 4 ȼ per kw- hr. all remaining energy at 1 ȼ per kw-hr. what type of rate is this? How much is the customer’s bill in a month when the registered 15-kw maximum demand and consumer 1,850 kw-hr? Solution: (a) This is blockhopkinson rate: (b) Computations for Costumer’s Bill: 1. Demand Charge; 1st kw

$6.00

Next kw @ $5.00

$30.00

In excess of 7 kw @ $4.00 =8×4

$32.00

2. Energy charge; 1st 100 kw-hrs @ 0.04 = 100 × 0.04

$4.00

All in excess of 100 kw-hrs @ 0.01 = 1750 × 0.01 Total Bill

= Sum of all charges

$17.50 =

$89.50 ans.

Chapter 4 The Power Plant Building

Problem 4-1 How high can a solid, unloaded brick wall (2.29 tons per cu. m) be carried without the compressive stress on the lower course exceeding (a) 12 Solution: (a) Compressive Stress, fc fc =

Wt. of Wall, kg Area under stress, cm 2

Weight of Wall: W = ρAH Where: ρ = density, kg/cm2 A = area, cm2 H = height, cm Thus, fc =

W ρAH = = ρH A A

kg kg . 2 ; (b) 17.6 cm cm 2

(

fc = 2.29

tons 1000 kg ( 1m H in m ) 3 ton 100cm cm

fc = 0.229 H

)(

)

(

2

)

kg cm 2

Therefore equating, 0.229 H

kg kg = 12 2 cm cm 2

H = 52.4 m (b) For 17.6

ans.

kg compressive stress, and using the same relations derived above. cm 2 kg cm 2 Height in m = kg 0.229 cm 2 17.6

H = 78.86 m

ans.

Problem 4-8 A 39.62 m high chimney of radial brick masonry is described by the following top and bottom dimensions: D2 = 1.905 m, d2 = 1.524 m, D1 = 3.2 m, d1 = 2.4 m. Uniform batter, weight 200 tons. Find the maximum compressive stress under 160.9 km per hour wind load. Will a base crack be opened to windward? Solution: (a) From eq. 4-3 p. 98 and for 160.9 kph wind Pwh =

100 H 2 ( 2 R2 + R 1 ) 3

Substituting the unknown values with all linear dimensions in m

Pwh =

100 H2 ( 2 R2 + R 1 ) 3

100 ( 39.62 )2 = ( 2 x 0.9525 + 1.6 ) 3 Pwh = 183,398.47 kg-m But, Wz = Pwh or, z =

=

183,398.47 kg -m kg 200 tons x 1000 ton

z = 0.917 m Figure:

Pwh w

From Eq. 4-7, p. 102 Kern Radius, K =

=

R21 + r 21 4 R1

( 1.60 )2 + ( 1.2 )2 4 ( 1.60 )

= 0.625 m From Eq. 4-9, p. 103 Limiting Displacement for z y=

=

R r 2 + 4 R

(

)

1.6 1.2 2 + 4 1.6

(

)

= 1.1 > z > k From Eq. 4-11, p. 103, when y > z > k Maximum Compressive Stress, z z - 1 k k

z-k y-k

2

[( ) ( )( ) ]

fz = fc 1 +

Where fc = compressive stress due to dead load fc =

w π ( R21 - r 21 ) 200 tons x 1000

=

kg ton

π [ ( 1.6 )2 - (1.2 )2 ]

= 56,841.05

kg m2

Then, z 0.917 = = 1.47 k 0.625 And, z-k y-k

2

0.917 - 0.625 2 = = 0.378 1.1 - 0.625

( ) (

)

Substituting these values in Eq. 4-11 above maximum compressive stress, fz = 58,814.05 [ (1 + 1.47) - (1 - 1.47)(0.378) ] = 150,495.77

kg m2

ans.

(b) Size z s greater than the Kern Radius k. A crack will open but chimney will be safe since z < y and fc < 40,000 psf Problem 4-10

Using Table 4-3, estimate the proportions of a radial brick chimney of 2.44 m inside diameter x 45.72 m high. Masonry weight, 1,926.54

kg , wind 160.9 kph. Test the m3

base at mid-height sections for maximum compressive stress. No lining. Solution: Table 4-3, p. 100 gives the top wall thickness as 180 to 230 mm. Depending on chimney diameter assume 205 mm. Inside diameter of top = d1 = 2.44 m (Given) Therefore, outside diameter of top = Inside diameter + 2 x Thickness

(

= 2.44 m + 2 x

205 m 1000

)

= 2.85 m Ratio

H 45.72 m = = 18.74 d2 2.44 m

Interpolating from Table 4-3 Batter = 17.984

mm m

Height

Outside radius R1 at Bottom =

D2 2.85 mm + 17.984 ( 45.72 m ) + Batter x H= 2 m 2

(

= 1.425 m + 822.285 mm x

1m 1000 mm

= 2.25 m Outside diameter D1 = 4.5 m Also from Table 4-3, min. wall thickness at base

)

= 9.25 H = (45.72)(9.5) = 422.91 mm say 423 mm or

= 0.423 mm

Inside Radius ar base, r1 = Outside Radius – Thickness of wall = 2.25 m – 0.423 m = 1.827 m Inside Diameter, d1 = 2r1 = 2(1.827) = 3.654 m or

I.

= 3654 mm

Analysis of base for max. compressive stress: First, calculate the volume of the material used, then the total weight. Volume of the material will be calculated by the finding the difference between the outside and the inner volumes of the chimney treating as the frusta of a cone.

Volume of a Frustum = Where:

1 H ( B + b + √ Bb ) , cu. m 3

H = Vertical height, m B = Area of base, sq. m b = Area of top, sq. m

Outside conical volume =

45.72π [ ( 2.25 )2 + ( 1.425 )2 + (2.25)(1.425) ] 3

=

45.72π [ 5.0625 + 2.030625 + 3.2063 ] 3

= 493.112 cu. m Inside conical volume =

45.72π [ ( 1.827 )2 + (1.22 )2 + (1.827)(1.22) ] 3

=

45.72π [ 3.337929 + 1.4884 + 2.22894 ] 3

= 337.79 cu. m Volume of the chimney material = Outside Volume – Inside Volume = 493.112 – 337.79 = 155.322 cu. m At 1926.54 kg per cu. m chimney weight W = 155.322 x 1926.54 = 299,234.05 kg From Eq. 4-3, p. 98, wind load moment

Pwh =

100 H 2 ( 2 R2 + R 1 ) 3

100 ( 45.72 )2 Pwh = ( 2 x 1.425 + 2.25 ) 3 = 355,354.13 kg-m But, Wz = Pwh or, z = =

Pwh w

355,354.13 kg -m 299,234.05 kg

= 1.188 m From Eq. 4-7, p. 102, Kern Radius R 21 + r 21 k= 4 R1

( 2.25 )2 + ( 1.827 )2 = 4 ( 2.25 ) = 0.9334 m From Eq. 4-9, p. 103, limiting displacement for z y=

=

R r 2 + 4 R

(

)

2.25 1.827 2 + 4 2.25

(

)

= 1.582 m > z > k Since y > z > k from Eq. 4-11, p. 103, maximum compressive stress,

z z - 1 k k

z-k y-k

2

[( ) ( )( ) ]

fz = fc 1 +

Where fc = unit dead load stress =

w π ( R21 - r 21 ) =

299,234.05 kg π [ ( 2.25 )2 - ( 1.827 )2 ]

= 55,230.64 kg per sq. m z 1.188 m = k 0.9334 m = 1.273 Then, z-k y-k

2

( ) ( =

1.188 - 0.9334 2 = 0.1541 1.582 - 0.9334

)

Substituting these values in Eq. 4-11 above fz = 55,230.64 [ (1 + 1.273) - (1 - 1.273)( 0.1541) ] fz = 123,215.73

kg kg or 12.322 2 m cm 2

Conclusion: Crack will open at base since z > k. But chimney will be safe since y > z and fz < 40,000 PSF or 19.511 II.

kg m2

Analysis at mid-height section for maximum compressive stress: Assuming smooth taper inside & outside of chimney

Dm = =

D2 + D1 2 4.5 + 2.85 2

= 3.675 m dm = =

d2 + d1 2 2.44 + 3.654 2

= 3.047 m

As in (I), Volume of a frustum of a cone =

1 H ( B + b + √ Bb ) 3

Where the symbol mean the same things, Then: Outside conical volume of upper half chimney, =

22.86π 3

3.675 2

2

3.047 2

2

[( )

2

+ (1.425 ) +

x 1.425)] (3.675 2

= 192.121 cu. m Inside Volume =

22.86π 3

[(

x 1.22) ] ) + (1.22 ) +(3.047 2 2

= 135.69 cu. m Volume of Chimney Material = Outside Volume – Inside Volume = 192.121 – 135.69 = 56.431 cu. m

At 1,926.54

kg , weight of upper half m3

w = 1,926.54 x 56.431 = 108,716.58 kg In the succeeding computations, all items with subscript 1 will have the subscript m to denote midsection. Pwh =

100 H 2 ( 2 R2 + Rm ) 3 100 ( 22.86 )2 (2 x 1.425 + 1.8375 ) 3

=

Pwh = 81,653.063 kg-m Pwh w

z = =

81,653.063 kg -m 108,716.58 kg

= 0.751 m Kern Radius, k R 2m + r 2m k= 4 Rm

( 1.8735 )2 + ( 1.5235 )2 = 4 ( 1.8735 ) = 0.775 m y=

Rm rm 2+ 4 Rm

=

(

)

1.8375 1.5235 2+ 4 1.8375

(

)

= 2.6 m > k > z Since z > k, there will be no cracking at midheight Also max. stress, from Eq. (4-8) p. 102

(

f = fc 1 +

=

z k

)

w

0.751 1+ ( 0.775 ) π(R - r ) 2 m

=

2 m

108,716.58 kg 0.751 1+ 2 2 0.775 π ( 1.8375 - 1.5235 )

[

f = 64,565.51

f in

]

kg m2

kg kg 1 m2 = 64,565.51 x cm 2 m2 ( 100 )2 cm 2 = 6.4566

kg cm 2

Problem 4-19 The following data refer to a 6-cycle, 450 kw, 650 bmhp diesel engine alternator unit having direct connected exciter. Weight of engine and flywheel: 63,636.364 kg Weight of alternator & exciter: 6,818.182 kg Bedplate of engine: 3.05 m x 7.3 m Length of whole unit: 10.67 m Width of generator bedplate: 3.05 m The subsoil is average sand. Design and detail a foundation for this engine. The foundation top is all in one horizontal plane. Calculate the amount of sand, stone, cement, and a form lumber requiring for its construction.

Solution: (a) From Table 4-4, p. 105, safe bearing capacity of sand is 19.5 to 58.5

ton m2

For average sand, bearing power will be 19.5 + 58.5 ton = 39 2 2 m ton 1 of this or 19.5 2 as allowable stress. 2 m

For machine foundation, take

Total weight of unit = wt. of engine + wt. of alternator = 63,636.364 + 6,818.182 = 70,454.546 kg From Table 4-5, p. 108, wt. of foundation for multicylinder diesel engine = 570

kg bmhp

At this ratio, wt. of foundation, = 570 x 650 = 370,500 kg (This is the max. value and not necessary at all times) Foundation maybe decreased when complete dampening is unnecessary, say 363,636.364 kg. Total weight of unit and foundation = 70,454.546 + 363,636.364 = 434,090.91 kg Since foundation is very much heavier than unit, we shall assume that total weight is more or less uniformly distributed to the soil. Total base area of unit = 3.05 x 10.67 = 32.54 m2

Maximum area required for foundation based on soil capacity

= =

Total Weight Allowable Soil Stress 434,090.91 kg tons kg 19.5 2 x 1000 ton m

= 22.26 m2 But area of foundation cannot be less than engine base area. Min. area required = 32.54 m2 Allowing 15 cm on all sides of engine, Desired Foundation Area = 3.35 x 10.97 = 36.75 m2 Corresponding Foundation Depth =

Volume of Foundation Area of Foundation

(Assuming Uniform Cross-Section) At 2,408.17

kg , volume concrete m3 =

363,636.364 2,408.17

= 151 m3 151 m 3 Corresponding Foundation Depth = 36.75 m 2 = 4.11 m As this is rather a deep foundation, we may reduce it to say, 3.05 m for which the foundation will have to be spread out. Bearing area is then not a governing factor. If depth = 3.05 m Area of Foundation =

151 m 3 2 = 49.51 m 3.05 m

If length of one side is assumed same as before as 10.97 m Width of Foundation =

49.51 m 2 = 4.5 m 10.97 m

Note: From the foregoing computations, it is clear that other satisfactory combinations can be arrived at. (b) From Table 4-1, p. 90, recommended mixture is 1 : 3 : 5 for which required material per cu. m. mixture (or 2 tons) is given as follows: Cement……………………………….. 6.2 sacks Sand………………………………….. 0.52 cu. m Stone…………………………………. 0.86 cu. m Since weight of foundation = 363,636.364 kg or 400 tons Required materials are: 400 2

Cement

=

6.2 x

= 1,240 sacks

Sand

=

0.52 x

400 2

= 104 cu. m

Stone

=

0.86 x

400 2

= 172 cu. m

Note: (1) Some text recommend 1 : 2 : 4 mixture for machine foundation (2) Of course, in the Philippines, sand and stone (gravel) are sold per cu. m

Problem 4-20 The condenser of a 10,000-kw turbine is rigidly carried by its own foundation and is connected to the turbine by a copper expansion joint whose dimensions are 2.13m x 2.82m, approximately rectangular. The water and steam in the condenser normally weigh 9,090.91 kg, condenser pressure 0.07 kg/cm 2 abs. what is the variation in weight on the condenser foundation between in-service and out-of-service conditions? Solution: From Eq. 4-15, p.111 Vacuum pull, F = A(1.03-Pc), kg Where:

A = Exhaust Nozzle Area (100 cm)2 = (2.13 x 2.82) m x 2 m 2

= 60,066 sq. cm. Pc = Condense Pressure = 0.07

kg abs cm 2

 F = 60, 066 cm2 (1.03 – 0.07)

kg cm 2

= 57,663.36 kg upward Force due to weight of water & steam = 9090.91 kg downward Net force upward = 57,663.36 – 9090.91 = 48, 572.45 kg When in-service, weight of the foundation is 48,572.49 kg less than when out-of-service. Problem 4-23 A 60-kw vertical, single-cylinder, steam engine-generator set has a bedplate 152.4cm x 228.6cm. Its weight if 5 tons. Design a foundation to rest on average sandy soil. Foundation mass in accordance to Table 4-5. Solution: From Table 4-4, p. 105, safe bearing power of sand is 19.5 to 58.5 tons/m 2 For average sand bearing power =

19.5 + 58.5 2

= 39 tons/m2

Since this is a machine foundation, Allowable Soil Stress =

39 or 19.5 tons / m2 2

From Table 4-5, p. 108 Foundation weight for single-cylinder steam engine = 320 kg / Bmhp Bmhp for 60 kw at approximately 1.5 hp / kw = 90 hp Foundation weight require = 320 kg/Bhp x 90 = 28, 800 kg Foundation volume at 2,408.17 kg per cu. M 28,800 kg = 2408.17 kg m3 = 11.96 m3 Total weight of engine and foundation = 5 ton x 100

kg + 28,800 kg ton

= 33,800 kg At 19572.66 kg/m2 safe soil stress, foundation area should not be less than, 33,800 kg 19, 572.66

kg m2

= 1.73 m2

Area of engine bedplate = (152.4 / 100) x (228.6 / 100) = 3.48 m2 This area will govern the minimum foundation area.

If allowance on the sides are given, so that the area is increased to 5 m 2, Depth of Foundation =

Volume of Foundation Area

11.96 m 3 = 5 m2 = 2.4 m Since this is a small unit, it will be desirable to lessen the foundation depth, say to 1.5 m, for which the area of foundation required =

11.96 m3 = 7.97 m2 1.5 m

A foundation 2.4 m x 3.2 m 1.5 m deep will satisfy this. Other proportions will also be satisfactory, provided the minimum requirements are fulfilled. Problem 4-27 Piles are driven in a quicksand (bearing = 5.38 ton/m 2) on 91.4 cm centers. They are driven until the penetration under the last hammer blow is 2.54 cm. the driver’s 1-ton hammer drops 2.6 m. piles are sawn off and surmounted by a concrete slab 91.4 cm thick. What average bearing power, in tsm, does this foundation provide? Solution: From Eq. 4-14, p. 107 Safe Load on Pile =

16.66 WH , kg s + 2.54

Where: W = wt. on file driver hammer = 100 kg (given) H = fall of hammer, ft = 2.6 m

S = penetration under last blow = 2.54 cm Safe Load on Pile =

16.66 (1000 kg)( 2.6 m) 2.54 cm + 2.54

= 8,526.77 kg Area of Soil/Pile = Area of Slab on Pile = 91.4 x 91.4 = 8353.96 m2 = 0.84 m2 Total Load Capacity of Soil Around Each Pile = 0.84 x (1/2) x 1000 = 420 kg Total Capacity of Soil and Pile = 420 + 8,526.77 = 8,946.77 lb/m Weight of Concrete Slab/Pile = Volume x Density = (0.914)3 x 2,408.17 kg/m3 = 1,838.76 kg Net Capacity per Pile surmounted by slab = 8,946.77 – 1,838.76 = 7,108.01 kg Since this is net capacity over an area of 0.836 m 2, Average Bearing Power in tsm 7,108.01 kg kg = 1000 x 0.836 m 2 ton

= 8.5 tons per sq. m (tsm) Chapter 5 Fuels and Combustion

Problem 5-1 Six hundred twenty-three cubic meters of a fuel gas are passed through a meter at 0.35 kg/cm2, 8.8 °C. Barometer, 756 mmHg. a) Find the commercial sales volume of this gas b) Tanks contain 214 m3 of fuel oil at 11.7 °C. S.G. = 0.945. Find the volume and weight of this quantity of oil measure at 15.6°/15.6°. Solution: a) Let: V1 be the commercial sales volume of the gas.

(

P1 = 756 mmHg 1.36 ×10 -3

kg/ cm2 mmHg

)

= 1.028 kg / cm2

Absolute pressure of the gas P2 = 0.35 kg/cm2 + 1.028 kg/cm2 P2 = 1.378 kg/cm2 Absolute temperature of gas T2 = 8.8°C + 273 = 281.8 K Absolute temperature of atmosphere (assumed standard): T1 = 15.6°C + 273 = 288.6 K

Volume of gas through meter V2 = 623 m3 (given) Volume of gas at atmospheric condition, V1 = to be determined Then using the general gas equation: P1 V 1 P2 V 2 = T1 T2 Substituting known values: 1.028 V 1 1.378 (623) = 288.6 K 281.8 K 1.378 (623) (288.6 Or V1 = 281.8 (1.028) = 855.26 cu.m, commercial sales volume ans. b) Volume at 11.7°C = 214 m3 From page 118, text for fuel oils, the volumetric coefficient of expansion is 0.0007 per °C Thus corrected volume = 214 m3 + 214 m3 (0.0007/°C)(15.6-11.7)°C = 214.58 m3 Weight of this volume of oil S.G. = 0.945 W = 214.58 m3 (0.945 x 1000 kg/m3) W = 203,000 kg ans.

Problem 5-6 Testing a certain West Virginia Pocahontas Coal by bomb calorimeter yielded these results: ∆T = 3.49°C. Fuel sample, 1.0535 g. Fuse wire consumed, 0.0073 g. Water charged 1855.68 g. Water equivalent of bomb, 470 g. Test coal had a 0.96% moisture. Calculate Qh for dry coal. (Q of fuse wire, 1600 cal per g). Solution: Apparent heat released, Q = W’C∆T Where:

W’ = weight of water bath and water equivalent of outfit = C=

1855.68 + 470 g = 2325.68 g Specific heat

=

1 calorie per gram per °C

∆T =

Temperature rise of water

=

3.49 °C

Substituting these values to the equation above, Q= =

2325.68 x 1 cal/g -°C x 3.49°C 8116.6 cal

Total heat from fuse wire, at 1600 cal/g =

0.0073 g (1600 cal/g) = 11.68 calories

Actual heat released by fuel =

apparent heat – heat from fuse wire

=

8116.6 – 11.68 = 8104.92 calories

Since the fuel sample is 1.0535 g Qh of fuel =

8104. 92 = 7693.32 calories per g 1.0535

Thus, heat value of fuel Qh =

Qh 8104. 92 = = 7767 cal/g ans. m 1−0.96 %

Problem 5-8 Fuel oil of 16° Be is stored in a tank to supply a boiler whose rated maximum steaming rate is 6804 kg per hr at 10.54 kg/cm 2. Feedwater, 48.9°C. Average thermal efficiency, 70%. Consider that plant’s capacity factor is 45% and that 1 ½ month’s supply is carried. Make any necessary assumptions and calculate the number of 1.8 m Dia x 4.6m long cylindrical tanks needed to hold this quantity. Solution: Assuming continuous operation, the average steaming rate at a capacity factor of 45% in 6804 kg/hr x 0.45 = 3062 kg/hr In 1 ½ months (assuming a 30-day month, 24-hour day) Total kg steam produced = 3060 kg/hr x 24 hrs/day x 1 ½ months x 30 day/month = 388,800 kgs Thermal Efficiency of Plant ¿ ηt = Energy output ¿ steam Energy input ¿ fuel ¿ ηt = Where:

Ws( hs−hf ) Wf x Qh Ws = Steam rate, kg/hr hs =

enthalpy of steam, kJ/kg

hf =

enthalpy of feed water, kJ/kg

Wf =

fuel rate, kg/hr

Qh =

heat value of fuel, kJ/kg

From the steam table For saturated steam at 10.54 kg/cm2 hs =

2782 kJ/kg

For water at 48.9°C hf =

204.5 kJ/kg

Heat value of the fuel, Qh =

42,450 – 43(Be + 10) kJ/kg

Qh =

42,450 – 43(16 + 10)

Qh =

40,032 kJ/kg Ws( hs−hf ) nt Qh

Thus, Wf =

Substitute values, we have Wf =

3062 ( 2782−204.5 ) kJ /kg kJ 0.70(40,032 ) kg

Wf =

281.64 kg/hr of fuel oil

Total amount of fuel required for 1 ½ months (assume 30 days per month, and 24hrs, per day) Wf =281.64 kg/hr x 24 hrs/day x 30 days/month x 1½ month Wf =

304,200 kg

Volume of fuel =

Wf Pf

Specific gravity of the fuel, SGf =

140 140 = = 0.96 Be+ 130 16+130

Thus, 304,200 kg kg = 317 m3 Vf = 0.96(1000 ) m3 Volume of one tank V=

π (1.8 m)2 (4.6 m) 4

V = 11.70 m3 per tank Number of tanks required: =

317 m3 11.70 m3 /tank

= 27 tanks

ans.

Problem 5-10 Reduce the analysis of a coal designated from Table 5-4 to a.) moisture-free basis b.) moisture and ash-free basis Solution: Arbitrarily picking coal No. 4, corresponding proximate analysis Moisture

=

2.89%

Fixed Carbon =

26.61%

Ash

2.32%

=

Ultimate analysis from the same table Sulfur

=

.55%

Hydrogen

=

4.99%

Carbon

=

84.11%

Nitrogen

=

1.64%

Oxygen

=

6.39%

Ash

=

2.32%

=

100%

Total Preliminary Computations:

It is clear from the above analysis that part of the hydrogen and oxygen in the coal exist as moisture and the rest in free form. Since, 1 kg water, 1/9 kg is hydrogen and 8/9 kg is oxygen In .0289 kg moisture, H2 =

.0289 9 = 0.0032 kg

O2 = .0289 (8/9) = 0.0257 kg

∴ Free H2 in 1 kg coal as fired = .0499 - .00322 = .04668 kg Free O2 in 1 kg coal as fired = .0639 - .0257 = .0382 kg a.) Analysis on moisture-free basis: =

Analysis on as received basis 1−moisture

1 – moisture = 1- .0289 = .9711 Corresponding proximate analysis, moisture-free basis: Fixed Carbon

=

69.18 = 0.9711

71.3%

Ash

=

2.32 = .9711

2.38%

Volatile Matter

=

25.61 = .9711

26.32%

Total =

100.00%

Corresponding ultimate analysis, moisture-free basis: Sulfur

=

.55 = .9711

.567

Hydrogen

=

4.688 = .9711

4.82

Carbon

=

84.11 = .9711

84.61

Nitrogen

=

1.64 = .9711

1.60

Ash

=

3.32 = .9711

2.38

Total =

100.00%

b.) Analysis on moisture and ash-free basis: =

Analysis on as fired basis 1−moisture−ash

1 – moisture – Ash = 1 – .0289 – .0232 = .9479 Thus, Proximate Analysis: Fixed Carbon = Volatile Matter =

69.18 = .9479 25.61 = .9479

Total =

72.9% 27.1% 100.00%

Ultimate Analysis: Sulfur

=

.55 = .9479

.582

Hydrogen

=

4.668 = .9479

4.93

Carbon

=

84.11 = .9479

88.728

Nitrogen

=

1.64 = .9479

1.73

Oxygen

=

3.82 = .9479

4.03

Total

=

100.00%

Problem 5-12 Given a coal with proximate analysis as follows, find heating value and kg air required per kg coal at 50% excess. Moisture, 4.47; volatile, 22.74; fixed carbon, 53.72; ash 19.07. Solution: Since the ultimate analysis of this coal is not given, the properties required for this coal will be determined using the empirical formula found in Table 5-5 p.129. In this connection, it will be necessary to convert the analysis from as fired basis, to combustible basis. Analysis on combustible basis =

As fired 1− Ash−moisture

1 – Ash – moisture = 1 – .1907 – .0447 = 7640 kg combustible per kg coal Volatile Matter, V

=

.2274 = .7646

298 kg per kg combustible

Fixed Carbon

=

.5372 = .7646

.7 kg per kg combustible

From Table 5-5, for a coal with a range of volatile matter in combustible of 16-36%. The heat value is HV = (16,160 – 2,250 v) (2,326 j/g) kg combustible Thus, HV = (16,160 – 2,250 x 22.74%) (2,326 j/g) HV = 36,398 j/g of combustible Since there are .7646 kg combustible/ kg coal HV = .7646 x 36,398 kg coal = 27,830 j/g

ans

a.) Also from Table 5-5, p. 129, for V in the range of 16 – 36% of the combustible: H = 0.0457 + 0.0206 V H = 0.0457 + 0.0206 (22.74%) ∴ H = 0.0504

g hydrogen g combustible

Total Carbon in combustible (with V from 0 to 36%) C = 0.943 – 0.242 V C = 0.943 – 0.242 (22.74%) ∴ C = 0.8879

g hydrogen g combustible

Theoretical air needed per kg coal

=

11.5C + 34.5 ( H –

O ) + 4.3S 8

Disregarding oxygen and sulfur Theoretical air per kg combustible =

11.5(0.8879) + 34.5(0.0504)

=

11.95 kg/kg combustible

Per kg coal; Theoretical air: = 11.95

kg air kg combustible

x 0.7646

kg combustible kg coal

kg air = 9.13 kg coal At 50% excess air, actual air required = 1.5 (9.13 kg air / kg coal) = 13.7 kg/kg coal

ans.

Problem 5-14 The as – fired proximate analysis is taken of a coal of West Virginia origin. Moisture, 1.75; volatile, 21.70; fixed carbon, 69.13; ash, 7.42. when burned with what is presumed to be sufficient air for complete combustion, the CO2 tests 10% by volume. What percent excess air is indicated?

Solution: The proximate analysis will be converted to combustible basis in order to be able to use Table 5-5. Analysis on Combustible basis =

As-fired basis 1-moisture-Ash

1 – moisture – Ash = 1 – 0.0175 – 0.0742 = 0.9083 kg Volatile Matter per kg combustible ¿ 0.217

kg V.M. kg coal

0.9083 ¿ 0.239

kg V.M. kg coal

kg V.M. kg coal

Fixed carbon 0.6913 = = 0.762 kg/kg combustible kg combustible 0.9083

From Table 5-5. p. 129 Carbon in Combustible = C (for V, 0 – 36%) = 0.943 – 0.242 V = 0.943 – 0.242 (0.239) = 0.885 kg/kg combustible Hydrogen in Combustible = H (for V, from 16% up) = 0.457 + 0.206 V = 0.457 + 0.206 (0.239) = 0.0506 kg/kg combustible From Eq. 5-18, p. 144 for complete combustion Excess air =

Where: R =

20.9 R R + 2.37 = R+3 CO 2( R+3) C  '  H

C1 = carbon burned per kg coal C1 = 0.885 x 0.9083 assuming all carbon is burned H = hydrogen burned by air/kg coal = 0.0506 x 0.9083 Therefore: R =

0.885 x 0.9083 = 17.5 0.0506 x 0.9083

Substituting all known values in Eq. 518 above, Excess air =

=

20.9(17.5) (17.5 + 2.37) 10(17.5+3) (17.5+3) 20.9 (17.5) 19.87 20.5 20.5 (10)

= 0.815% or 81.5% ans.

Problem 5-16 Predict the Orsat analysis resulting from the combustion of a coal deg. Ignited from Table 5-4, with 50% excess air 10% combustible in the refuse. Solution Arbitrarily picking coal no. 1, the corresponding ultimate analysis is as follows: S

=

.57%

H2

=

C

= 86.37

N

=

2.7

.91

O2

=

3.55

Ash =

5.9__ 100.00%

Theoretical air required = 11.5C + 34.5 (H -

0 ) + 4.3 S 8

= 11.5(0.8637) + 34.5 (0.0270 -

0.0355 ) + 4.3 (0.0057) 8

= 10.735 kg/kg coal Since 10% of refuse is combustible, 90% of the refuse is ash Total refuse/kg coal =

Ash 0.059 9 9

= 0.0655

kg refuse kg coal

Assuming all combustible in refuse is carbon Then unburned carbon = 0.1 x 0.0655 = 0.00655 kg/kg coal and carbon burned/lb coal = 0.8637 - 0.00655 = 0.85715 kg/kg coal Gaseous products of combustion at 50% excess air: CO2

=

0.857 x 3.67

= 3.14

H2O

=

0.027 x 9

= 0.243

SO2

=

0.0057 x 2

= 0.0114

O2

=

(0.5 x 10.73) x 0.232

= 1.245

N2

=

(1.5 x 10.73) x 0.768 + 0.0091

= 12.71

The probable Orsat analysis will be a volumetric analysis of the dry flue gas consisting of: CO2, O2, and N2 Mols CO2/kg coal =

3.14 = 0.0713 44

Mols O2/kg coal

=

1.245 = 0.0389 32

Mols N2/kg coal

=

12.71 = 0.444 28

Mols dry gas = 0.5542 Mols/kg coal Probable Orsat analysis: (Analysis by Volume) CO2 =

0.0713 x 100% = 12.9% ans. 0.5542

O2 =

0.0389 x 100% = 7.02% ans. 0.5542

N2 =

0.444 x 100% = 80.08% ans. 0.5542

Problem 5-19 From the analysis of a coal sample taken during a test, it is estimated that the asfired total carbon was 76.5% and free hydrogen was 4.7%. From other tests C 1 = 0.7263, ORSAT; CO2, 10.53; O2 3.74; CO, 0.49; N2, 85.24. Find A:F Ratio Solution: From Eq. 5-15, p. 139, A:F = 3.03 C1[

N2 ] CO 2+CO

Substituting the given values in the problem, for the corresponding terms in the equation above, we get A:F = 3.03 (.7263) [

=

85.24 ] ( 10.53 )+ 0.49

3. 03 x 0.7263 x 85.24 11.02

= 17 kg air/kg coal ans.

Problem 5-25 A coal as designated from Table 5-4 is burned in a pulverized coal furnace with 20% excess air. Preheated air temperature, 148.9°F; furnace outlet, temperature, 1010°C. No incomplete combustion. Calculate kcal absorbed by the gases leaving the furnace, per kg coal. Solution: Arbitrarily picking coal no. 1, the corresponding ultimate analysis is as follows: Sulfur

=

0.57%

Hydrogen

=

2.7%

Carbon

= 86.37%

Nitrogen

=

Oxygen

=

3.55%

Ash

=

5.9%

Theoretical Air required:

0.91%

= 11.5C + 34.5 (H -

0 ) + 4.3 S 8

= 11.5(0.8637) + 34.5 (0.027 -

0.0355 ) + 4.3 (0.0057) 8

= 9.93 + 0.78 + 0.0245 = 10.735 kg/kg coal

Assuming complete combustion and no combustible in the refuse, the gaseous products of combustion with 20% excess air will be as follows: CO2

=

3.67 x 0.8637

= 3.17

H2O

=

9 x 0.027

= 0.243

SO2

=

2 x 0.0057

= 0.0114

N2

=

1.2 x 10.7345 x 0.768 - 0.0091

= 9.902

O2

=

0.2 x 10.7345 x 0.232

= 0.498

Weight of dry flue gas (CO2, N2, and O2) = 3.17 + 9.902 + 0.498 = 13.57 kg per kg coal Assuming Cp. for water vapor at 1010°C, the specific heat for water is 0.51 kcal/kg °C Qs = sensible heat absorbed by dry flue gas + sensible heat absorbed by water vapor = (13.57 x 0.267) (1010 - 148.9°C) = (0.243 x 0.53) (1010 - 148.9°C)

= (3120 + 107) = 3227 kcal/kg coal Latent heat in the products: = mass of H2O vapor hfg hfg = latent heat of vaporization of water at 148.9°C hfg = 560 kcal/kg

Therefore: Latent heat = 0.243 (560) = 136.1 kcal/kg coal Therefore: Total heat absorbed by products of combustion is = 3227 + 136.1 = 3363.1 kcal/kg coal ans.

Problem 5-28 A coal of following composition is burned to an Orsat gas analysis of 9% CO2. Refuse analysis, 15% combustible. Flue gas at 232 °C. Find the density of this flue gas, C, 76.82; H2, 4.96: O2, 6.38: N2, 1.46: S, 1.39; Ash, 6.26, Moisture, 2.73. Solution Ultimate analysis is given as follows: C

76.82%

H2

4.96%

O2

6.38%

N2

1.46%

S

1.39%

Ash

6.26%

Moisture

2.73%

Ash coal kg refuse / kg coal = kg 0.85 =

0.0626 0.85

= 0.0737 kg/ kg coal Combustible in refuse

Carbon burned/ kg coal

=

Total refuse – Ash/kg coal

=

0.0737 – 0.0626

=

0.0111 kg/ kg coal

=

C’

= 0.7682 – 0.0111 = 0.7571 kg/kg coal

Assuming complete combustion, 0.7571 kg carbon will burn to (3.67 x 0.7571) kg CO2 or CO2 = 3.67 x 0.7571 = 2.71 kgs/kg coal

Mols CO2/kg coal

=

2.78 Molecular Wt .

=

2.78 44

= 0.063 mols/kg coal Given that CO2, is 9% flue gas Dry flue gas =

0.0632 = 0.702 mol/ kg coal 0.09

CO2 = 0.0632 mols N2 and O2 in flue gas by difference = 0.6388 mols/ kg coal Mols N2/ kg coal =

Wt of N 2∈flue gas 28

Mols O2/ kg coal =

Wt of O 2∈ flue gas 32

Solve for the last two items above as follows: Theoretical air required = 11.5 C + 34.5 (H-

O ) + 4.3 S 8

= 11.5 (0.7682) + 34.5 (0.0496-

00638 ) + 4.3 (0.0139) 8

= 10.345 kgs/kg coal Let e = Excess air Then actual air supplied = 10.345 (1 + e) N2 in flue gas = 10.345 (1 + e) x 0.768 + 0.0146 =7.9646 + 7.95 e kg/kg coal Mol. N2 in flue gas

=

7.9646+7.95 e 28

= 0.2845 + 0.284 e mols/kg coal Since Mols N2 + Mols Q2 in flue gas = 0.6388 mols/kg coal as previously computed.

Then, 0.2845 + 0.284 e + 0.0753 e = 0.6388 3593 e = 0.6388 – 0.2845 = 0.3543 Or Excess Air e =

0.3543 = 0.987 or 98.7% 0.3593

Kgs gaseous products of combustion: CO2

=

3.67 x 0.7571

=

2.78

H2O

=

9 x 0.0496 + 0.0273

=

0.4743

N2

=

10.345 (1.987) x 0.768 + 0.0146

=

15.8446

O2

=

10.345 (0.987) x 0.232

=

2.37

SO2

=

2 x 0.0137

=

0.0274

=

21.4963 kg/kg coal

Total

Mols product/kg coal Dry flue gas (As per previous computation) H2O

=

0.4733 18

= 0.0263

SO2

=

0.0274 64

= 0.00043 = 0.72873 mols

=

0.702 mols/kg

Chapter 6 Industrial Combustion Engine Power Plant

Problem 6-3 Using problem data, calculate (1) ideal thermal efficiency ( with n= 1.35 in place of δ), (2) mean effective pressure, (3) power from a 6-cylinder, 4 cycle, 327-RPM engine operating on this cycle. Solution: Data froom problem 6-2 P1

=

0.9 kg/cm2 ab

P2

=

38.7 kg/cm2 ab

t1

=

15.6 °C

R

=

2.5

n

=

1.35

DxL

=

25 cm x 38 cm

(1) Ideal Thermal Efficiency, 1 ni= 1 - n−1 nr

V1 Rn−1 ;r= ; P1V1 = P2V2 V2 R−1

[ ]

V1 P2 1n =( ) V2 P1 r=(

1 38.7 1.35 so; ) 0.9

ni = 1 -

1 1.35(16)0.35

[

2.51.35 −1 1.5

ni= 0.54 ans. (2) Mean Effective Pressure

[

Pmi = P1r

n r n−1 ( R−1 ) −( Rn −1) (r −1)(n−1)

= 0.9 Pmi

kg 1.35 ( 16 )0.35 ( 1.5 )−(2.51.35−1) abs 16 cm2 (15)(0.35)

= 7.95

[

kg | | ¿ cm2

(3) Power from 6-cylinders HP =

Pmep LANp 449.702

HP = ¿ ¿ HP = 323 Hp

Problem 6-6

]

]

]

The specific gravity of oil tested at 29.4 °C is 0.852. Find its API degrees. Solution °API =

141.5 −131.5 SG @15.6

From p.118 of the text, volumetric coefficient of expansion of oil is 0.0007 per °C. Volume at 15.6°C,: =

[1-0.0007 per °C (29.4 – 15.6) °C] volume at 29.4 °C

=

0.99 x volume at 29.4°C

=

ρ ρ H2 O

m

=

Mass of substance

V

=

Volume of substance

ρ

=

Density of H2O

S.G.

=

m VρH 2 O

Where:

Since mass is constant (S.G.)(v)(ρ H2O) = m = constant; Since H2O is also constant (S.G.)(v) @ 15.6°C = (S.G.)(v) @ 29.4°C (X) @ 29.4°C (S.G.)(v) @ 15.6°C = (S.G.)(v) @ 29.4°C (S.G.)(v) @ 15.6°C = (S.G.)@ 29.4°C (v)@ 29.4°C V @15.6°C

( 0.991 )

= (0.852) =0.86 Therefore: °API

141.5 −131.5 0.86

°API = 33.0 Ans

Problem

6-9 Using median data from Fig. 6-5, find the oil storage volume, m 3, needed for a two week supply of 25 °API fuel oil to operate 1014HP engine 70% of the time at full load , 10% at

3 load. It is idle 20% of the time. 4

Solution: From Figure 6-15, p.164 For full load, upper limit of Fuel rate = 0.19

kg BHP−hr

Lower Limit = 0.17 For

kg BHP−hr

3 load, 4 Upper Limit = 0.18

kg BHP−hr

Lower Limit = 0.17

kg BHP−hr

MEDIAN VALUES @ Full Load = 0.18 @

kg BHP−hr

3 kg Load = 0.17 4 BHP−hr

Fuel consumed in two weeks

(

7 days 24 hrs kg x 0.18 x 1014 hp = 42,930 kg wk days BHp−hr

) 3 7 days 24 hrs kg x 0.17 x 1014 hp = 5790 kg @ load = 10 % ( 2 wk x 4 wk days ) BHp−hr @ Full load = 7 % 2 wk x

Total fuel consumed = 42,930 + 5790 = 48,720 kg

SPECIFIC GRAVITY: S.G. =

141 141 = 131.5+ ° API 131.5+25

S.G. = 0.904 Volume required for storage : 48,720 kg = 0.904(1000 kg ) m3 = 53.9 gm3 ans.

Problem 6-12 An engine on test developed 154 BHP steadily 30 min. during which it consumed 15.34 kg fuel. The same fuel when tested at 23.9°C with a 15.68° hydrometer, showed S.G. of 0.905. Calculate thermal efficiency, based on BHP and Qh. Solution: Volume coefficient of expansion = 0.0007 per °C Specific gravity of fuel at 15.6/15.6° 0.905 1−0.0007( 23.9−15.6) = 0.91 For heat value, equation 6-8, p. 158, PPE by Morse, states that Qh = 51,716 - 8793.8 (S.G.)2

Qh = 51,716 –- 8793.8 (0.91)² :. Qh = 44,430

kj kg

J g

Brake thermal efficiency: amount of fuel consumed ŋtb=

2648 amount of fuel consumed ; but Wb = Wb Q Bhp x time of testing

15.34 kg kg 1 = = 0.199 154 bhp hr bhp−hr 2

Thus, ŋtb=

2648 x 100% (0.199)( 44,430)

ŋtb = 30% ans

Problem 6-14 Find the percent excess air represented by 32:1 A:F ratio with Cetane as the fuel. Solution Combustion equation for cetane C16H34 + 24.5 O2 = 16 CO2+ 17 H20 Equation indicates that 1mol C16H34+ 25.5 mol O2 = 16 mol CO2 + 17 mol H20 Since: 1 mol CARBON = 12 kg C 1 mol OXYGEN = 32 kg O2 1 mol HYDROGEN = 2 kg H2 %3D Then: 1 mol C16H34 = 16 (12 kg) + 34 (1 kg) = 226 kg

1 mol CO2= 12 kg + 32 kg = 44 kg 1 mol H2O = 2 kg +

1 (32 kg) = 18 kg 2

Thus: 226 kg C16H34 + 24.5 (32) kg O2 = 16 (44 kg) Co2+ 17(18 kg) H20 or 1 kg C16H34 + =

24.5 x 32 kg O2 226

16 x 44 17 x 18 kg CO2 + kg H2O 226 226

1 kg C16 H34 + 3.469 kg O2 = 3.115 CO2 + 1.354 kg H2 Since air is composed of approximately 23.2% oxygen and 76.8% nitrogen by weight. Then amount of oxygen in the air 0.232 kg oxygen kg air

Theoretical air required for each kilogram of cetane 3.469 kg O 2 kg O 2 = 0.232 kg air = 15 kg air Therefore, percentage air supplied in excess of theoretical requirement: % Excess =

32−15 x 100% 15

% Excess = 113.3% ans

Problem 6-18

A 39.37 cm x 55.88 cm x 327 - RPM 16 cylinder, 4-cycle stationary diesel engine is connected to a 3125 - KVA (80% power factor) generator. It also drives a 30kw exciter. Assume a generator efficiency of 92% and determine BMEP at rated load. Solution Rated Load = KVA x Power Factor =3125 KVA x 0.8 =2500 kw Exciter Load = 30 kw Total Load = 2530 kw ans

Brake Horsepower of Engine 2530 kw kw (0746 )(0.92) Hp

= 3686.327 SAY 3687 HP ans.

Also; BHP =

Bmep =

Bmep LANp 449,702

BHP 449,702 LANp

kg−cm ) Hp−min Bmep = π 327 strokes ( 16 cyl ) (5 5.88 cm ) (39.37 ) 2 ( ) 4 2 min

( 3687 ) (449702

[

Bmep = 9.32

kg ans cm

]

Problem 6-19 Specifications of a 4-cylinder, 4-cycle gas engine are 10.16 cm x 15.2 cm x 900 rpm. Mechanical efficiency, 86%. Find the kw output of a direct connected generator of 90% efficiency when the indicated mep is 7.73 kg

Solution: Pmep LANp Nc ; where : Nc – no of cylinders 449,702

Indicated HP =

(

IHP = 7.73

kg ¿¿ cm

)

IHP = 38.2 HP At 86% mechanical efficiency BHP = 0.86 (38.2 hp) BHP = 32.88 Hp Kilowatt output at 90% generator efficiency = 32.88 HP 0.746

kw 90 % HP

= 22.1 kw ans.

Problem 6-20 An engine-type generator with 30 poles generates 3-phase 60-cycle current at 600V. Its rated output is 400kW engine specifications 40.64 cm x 50.8 cm. Find the break MEP existing when line current is 360 Amp, power factor, 0.8 Solution: KW Output =

√ 3 EI cos Ө

= √ 3 ( 600v ) ( 360A ) ( 0.8 ) x

1kW 1000W

= 2993 kW Since rated output is 400 kW, the generator operates at Speed of rotation =

N

=

3 load 4

120f n 120 x 60 30

From Fig. A-16, p. 675, standard generator efficiency at

3 4

load = 92.2%

(For 400 – KW, 240 – RPM Generator) From table 6-3, p. 185 STD. Deduction for Generator Efficiency = 2.8% @ Thus, Actual Generator Efficiency; =

92.2% - 2.8%

=

89.4%

Brake Horsepower Engine: NGen =

BHP

Gen output Brake HP of Engine Gen output = N Gen 300 kW

BHP

=

89.4% 0.746

kW HP

3 4

Load

= 450 HP Assume 6-cylinders, 4 stroke Then; Bmep

=

BHP 449,702 LANp Nc 450 BHP 449,702

Bmep

=

Bmep

= 4.27

50.8 cm

kg - cm hp-min

π (40.64 cm) 2 ( 120 ) (6 cyl ) 4

kg cm 2

Problem 6-24 Find Ntb and Bmep of an 8-cylinder, 4-cycle, diesel engine which is direct connected to a 2300-V, 3-phase generator rated at 1250 kW. Engine specifications are 39.37 cm x 55.88 cm x 327 RPM. During a test, this unit used 259.5 kg fuel oil of 44,550

J heating g

value. Wattmeter readings at the start and finish of the one-hour test were 25,156 and 26,378 kW. For electrical efficiency use NEMA data, corrected by table 6-3.

Solution: Generator Output: =

(Wattmeter reading after test – reading before test) ÷ No. of hour test

=

(26,378 – 25,156) ÷ 1 = 1,222 kW

Output in % of rated capacity 1,222 kW 1,250 kW

=

=

0.978

From Fig. A-16, p. 657, NEMA efficiency at given RPM and Load = 95% From Table 6-3, deduction = 0.83% Net Efficiency = 94.17% Corresponding BHP required 1,222 kW kW = (0.764 ) (0.9417) HP

= 1740 HP

With fuel consumption of 259.5 kg in one hour kg hr Fuel Rate = 1740 HP 259.5

= 0.0149

kg BHP - Hr

a.) Brake Thermal Efficiency ɳtb =

2648 Wb Q

=

ɳtb = 0.398

Bmep =

ans

BHP ( 449,702) LANp Nc kg-cm ) HP-min π 327 8 (55.88 cm) [ ( 39.37 cm)2 ] ( )( ) 4 2 min (1740 HP) ( 449,702

=

2648 (0.0149) (44550)

∴ Bmep

= 8.97

kg cm 2

ans.

Problem 6-26 Estimate the fuel storage tank capacity for a diesel plant having 5000 kW installed capacity. Expected plant capacity factor, 55%. Fuel contract is to be made for semi-monthly delivery, but allow 100% extra for contingency. Also determine the gpm transfer pump capacity.

Solution: From Fig. 6-15, p. 164, the greatest fuel consumption to be expected at 55% capacity is 3.4

kW-hr liter

The minimum fuel rate = 2.3

kW - hr liter

Averege fuel value: =

3. 4 + 2.3 2

=

2.85

kW - hr liter

At 55% plant capacity factor and without allowance for emergencies, Quantity for 15-day storage

24hr ) ( 15-day ) (0.55) day kW - hr 2.85 liter

(5000 kW) ( =

=

347,400 liters

ans.

Allowing 100% for contingencies: Total storage capacity = 694,800 liters

ans.

Problem 6-29 Cooling water for a 507 HP diesel engine is pumped to cooling tower at 60˚C. It is desired to cool the water to a maximum temperature of 37.8˚C under an atmosphere condition of 32.2˚C dry bulb temperature, 27.8˚C wet bulb temperature. Find the required capacity and efficiency of the cooling tower.

Solution: If the brake thermal efficiency of the engine is taken as 30% and the loss to cooling water is 32%, then Eq. 6-16, p.178 may be used to find the amount of cooling water. W = 674.58

BHP t2 - t1

In which li hr

W

= cooling water,

BHP

= rated brake, HP

t2 – t1 = inlet, outlet water temperature, ˚C

W

= 674.58

500 (60-37.8)

= 15,200

li ans. hr

Efficiency of cooling tower Cooling efficiency =

ta - t b ta - tw

Where: ta =

temp. of H2O entering

tb

=

temp. of H2O leaving

tw

=

atmospheric wet bulb temp.

Thus; Cooling Efficiency =

60 - 37.8 60 - 27.8

= 0.689 x 100%

ans.

= 68.9%

Problem 6-30 A diesel plant has a cooling system employing a cooling tower that loses to the atmosphere approximately 5% of the water circulating. When the two 507 – hp engines are operated at full load on an average day, the tower cools the water form 54.4 to 35°. What should the capacity of a water softening plant for the make up be in Solution:

li ? min .

Assuming 30% brake thermal efficiency and 32% less to cooling water, then from Eq. 6-16: W =

Bhp li ; t 2−t 1 hr .

674.58

Then: W = (674058)

W = 294

507 li 1 hr . ( 54.4−35 )( hr . 60 min . )

li min .

With 5% loss of water, Minimum capacity of water softening plant =

5% of 294

= 14.7

li min .

li = (0.05) min . ans.

(294 minli . )

Chapter 7 Gas Turbine Power Plant

Problem 7-1 Calculate the work done per kg of gas expanding from 6.33 kg/ cm 2 ab to 1.05kg/ cm2 ab in a gas turbine of 82% internal efficiency. Initial temperature, 750°C; ȣ = 1.34; MW = 29. Solution: Eq. (1) ȣ =

Cp = 1.34 (given) Cv

Let Cp = 29 Cp Mol specific heat at constant pressure. Cv = 29 Cv Mol specific heat at

constant volume J ; kinetic theory of gasses mol

Cp – Cv = 8.345

29 Cp – 29 Cv = 8.345 Eq. (2) Cp – Cv =

8.345 = 0.288 29

From Eq. (1) above; Cv =

Cp 1.34

Substituting in Eq. (2) Cp 1 Cp - 1.34 = 0.288 = Cp 1− 1.34 = 0.288

(

or Cp =

0.288 = 1.135 for the process of expansion 3-4 0.252

Solve for temperature after expansionT 4: T3 P3 = T4 P4

[ ]

T3 = T4 T4 =

)

ȣ−1 ȣ

6.33 1.05

[ ]

=

P2 P1

[ ]

1.34−1 1.34

ȣ−1 ȣ

= 1.57

T3 ( 750+273 ) ° K = 1.57 1.57

T 4=651° K Ideal work of turbine = Cp ( T 3−T 4 ) = 1.135 [ ( 750+273 ) −651 ]

J = 422 g gas flow Actual work done at 82% turbine efficiency

(

= 0.82 422

= 346 or = 346

J g

)

J g kJ ans. kg

Problem 7-2 Products of combustion with ȣ of 1.35 =; 556° K; MW = 29, are moving within an exhaust pipe at 174 m/sec; 1.12 kg/cm2 ab static pressure. Find the total pressure and temperature. Solution: From Eq. 7-6 p. 195, PPE by Morse, M=

V ; where: V = gas velocity, m/s √ ȣ gRT ȣ = 1.35 (given) g = 9.8 m/ s2 R = gas constant

kJ kg mol ° K kJ 29 kg mol

8.32 R=

R = 0.287

M = Mach Number

kJ 1000 N . m 1kg kg−m x x = 29.26 kg ° K kJ 9.807 N kg ° K

Substituting these values in Eq. 7-6 above: 174 m/sec M=

√

(

1.35 9.8

m kg−m 29.26 556 ° K 2 kg ° K sec

)(

)

M = 0.375 From Eq. 7-4, p. 194 Total Pressure, ȣ−1 M2 2

[ ( ) ]

tp = p 1+

ȣ−1 ȣ

2 kg 0.35 ( ) tp = 1.22 1+ 0.375 2 cm2

[ ( )

tp = 1.127

kg cm2

ans.

Total Temperature

[

tT = T 1+

M ( ȣ−1 2 ) ] 2

0.35 (0.375)2 2

[ ( )

tT = 556° K 1+ tT = 570 ° K

ans.

]

]

0.35 1.35

Problem 7-4 Kerosene is the fuel o a gas turbine plant: f = 0.012 T 3

= 972 ° K;

pressure ratio, 4.5, exhaust to atmosphere. Find the available energy, kg – m per kg air flow. Solution: Determine probable T 4 by Trial solution First assuming for process 3-4 =1.34 T3 P3 = T4 P4

[ ]

ȣ−1 ȣ

0.34

= ( 4.5) 1.34 = 1.464 T4 =

T3 1.464

=

972° K 1.464

From Fig. 7-2, p. 193 for f = 0.012, Cp at 664° K = 1.10

J g°C

J Cp at 972° K = 1.16 g ° C Average Cp =

1.10+1.16 2

= 1.13

Assume MW = 29

(

Cp = 29 1.13

J J = 32.77 mol . g°C

)

J g°C

J Cv = 32.77 – 8.345 = 24.42 mol . ȣ =

Cp 32.77 = Cv 24.42

ȣ = 1.34

∴ no need for recalculation

Available energy = (1 + f) Cp (T 3−T 4 ¿

[

= ( 1.012 )( 1.13 ) J

J (972 - 664) ° C g° C

]

kJ

N .m

= 352 g = 352 kg x 1000 kJ kg . m

= 35,910 kg

1kg

x 9.807 N

ans.

Problem 7-8 Consider that Fig, 7-11 is characteristics of all gas turbine plants and determine the efficiency of the plant at half load. Plant has a pressure ratio of 6, other data same as in problem 7. Solution: Data from problem 7. Type of gas turbine cycle…. Open Fuel…… C12 H26

Intake air at P1 =1

kg ab c m2

T1 = 29.4˚C T3 = 750 ˚C Qc= 43,155

J g

Compressor efficiency….. 0.80 Internal turbine efficiency...0.84 Combustion efficiency……95% Also as given in the problem:

Pressure ratio,

P2 P3 = =6 P1 P4

For process 1-2, substance is pure air, therefore, assume ɣ = 1.4 For the ideal (Isentropic) compression, process, 1-2 T 2 P2 = T 1 P1

( )

=( ) 6

0.4 1.4 ¿ ¿

ɣ−1 ɣ ¿ ¿

= (6)

0.286

= 1.67

Ideal temp. T2 = 1.67 T1 = 1.67 (29.4 + 273)˚K = 505 ˚K Let T 21 = actual temperature at the end of the compression process Δ h1 = actual work on air by compressor Δ h1 =Cp ( T 21 −T 1 ) = Cp

(T 2−T 1) ηc

T 21 = T1 +

(T 2−T 1) ηc

= 302.4 +

505−302.4 0.80

= 555.65 ˚K Process 2-3 Calculations: Value of Cp for process 2-3 must first be determined as a preliminary trial. Assume fuel-air ratio. f = 0.0121 Referring to Fig. 7-2, p. 193, J Cp at 5550.65 ˚K =1.05 g ¿ ¿ ˚C

J Cp at 1023 ˚K = 1.17 g ¿ ¿ ˚C

J Average Cp = 1.11 g ¿ ¿ ˚C

Heat to riase products of combustion to 750˚ C J Qf = 1.11 g (1023 – 555.65)˚ C ¿ ¿ ˚C

J =518.7 g ¿ ¿ Fuel air ratio to supply this heat at 95% combustion efficiency

518.7 f= 43,155

J g

J ( 0.95 ) g

= 0.0126

This is close enough to assumed value of f = 0.0121 and therefore a recalculation is unnecessary. A new value for f and a recalculation must be made. If value of f previously assumed do not check out.

Process 3-4 Caculation: First determine the probable temperature after the isentropic process 3-4. As a preliminary trial. Assume ɣ = 1.34 Let T4 = Temperature after isentropic expansion T3 = 750 ˚ C = 1023 ˚K (as given)

Then, T 2 = P 2 T 1 P1

( )

ɣ−1 ɣ ¿ ¿

0.34 ¿ = 1.34 = 1.575 (6) ¿

T4 =

T3 1023 = =649.5 ˚ K 1.575 1.575

From Fig. 7-2, p. 193 and for f = 0.0121 Cp for 649.5˚ K =1.08 Cp for 1023˚ K = 1.17 1.08+1.17 J =1.125 ˚ C Average Cp = 2 g ¿ ¿ Assume molecular weight of products = 29

Cp = 29 (1.125) = 32.625 Cv = 32.625 – 8.345 = 24.28 and =

Cp 32.625 = =1.34 (close enough to assumed volume) Cv 24.28 Work calculation: Cp at (T1 = 302.4˚K) = 1.00 CP at (T2 = 505˚K) = 1.03 1.00+1.03 J =1.015 ˚ C Average Cp for process 1-2 = 2 g ¿ ¿ Work of compression per g air: J Wc =

˚C

g ( 505−302.4 ) ˚ C = 257 J 19 x 1.015 ¿ ¿ 0.80

Actual compression work for g air WT = (1 + f) Cp 3-4 (T3 – T4) ηt = (1.0120) (1.125) (1023 – 649.5) 0.84 = 357 J Plant Thermal efficiency excluding combustion losses: =

WT −Wc 357−257 = =0.19 Qf 518.7

Including combustion losses, plant thermal efficiency = 0.19 x 0.95 = 0.18 at rated load

Referring to 7-11 p. 205, for simple open cycles and at 50% rated load, operating efficiency is 73% of full load thermal efficiency. Therefore, for this plant, thermal efficiency at half load = 0.18 x 0.73 = 0.13 ans.

Problem 7-9 Find the full-load efficiency, air rate, and exhaust temperature of an open cycle gas turbine plant to be built for optimum pressure ratio with T 3 = 1028˚K, T1 = 288 ˚K, ηC = 0.79. Oil fuel with QL = 43155

J . Combustion efficiency, 94%. Standard sea level g

atmospehere. Specific heat data as in Fig. 7-2.

Solution: T21 = is actual temperature after compression T41 = is actual temperature at end of expansion

Optimum pressure ratio

P2 can be found by using the value of T2 P1

Obtained from Eq. 10, p 197 as follows: 2 (ηT ηc T3 T1) T2- [T1 – (1- ηc ηT) T3] T22 = ηT ηc T32 T1 Values given in the problem will be substituted in the above question. But first, it will be convenient to find ηT ηc: ηT ηc = 0.82 x 0.79 = 0.648 2 (0.648 x 1028 x 288) T2 – [288 – (1 – 0.648) 1028] T22 = 0.648 (1028)2 (288) 383,580 T2 + 74.06 T22 = 1.9716 x 108 T22 + 5179.3 T2 – 2, 662,166=0 By Pythagorean theorem: T2 = -517.3 ± (5179.3)22 – 4 (-2,662,166) T2 = 417 ˚K Optimum Pressure ratio, P 2 = T 2 P1 T 1

( )

P 2 471 = P 1 288

( )

ɣ−1 ɣ ¿ ¿

1.4 1.4−1 ¿ ¿

P2 P1

❑

( ) =5.6

therefore,

Actual temp. at end of compression T22 = T1 +

T 2−T 1 ηc

T22 = 288 +

417+288 0.79 = 519.6˚K

Calculation for process 2-3: Assume f = 0.0135 (trial value) From Fig. 7-2, p. 193 with f = 0.0135 Cp at 519.6 ˚K = 1.055 Cp at 1028 ˚K = 1.175 Average value =

1.055+ 1.175 J =1.115 2 g˚ C

Qf needed to raise products of 1028 ˚K = 1.115

= 567

J (1028−516.6)˚ C g˚ C

J g

Fuel – air ration: f=

Qf QL Comb . eff

J g f= = 0.0139 (close enough) J 43,155 0.94 g 567

(

)

Calculation for process 3-4: Determine probable T4. First assume ɣ = 1.34 T3 T4 =

( 5.6 )

0.34 1.34 ¿ ¿

1028˚ K T4 =

( 5.6 )

0.34 1.34 ¿ ¿

= 664˚K

From Fig. 7-2 and for f = 0.0135, Cp at 1028 ˚K = 1.175 Cp at 664 ˚K = 1.09 1.175+ 1.09 J =1.132 2 g˚C

Average Cp =

Assume MW = 29: Cp = 29 (1.132) = 32.84

J mol

Cv = 32.84 – 8.345 = 24.49

ɣ=

J mol

CP 32.84 = =1.341 (close enough) CV 24.49

Therefore, T4 = 664˚K as before Process 1-2 (f = 0), Fig. 7-2 p. 193 Cp at 288˚K = 1.0 Cp at 519.6 ˚K = 1.03 J Average Cp= 1.015 g ˚ c ¿ ¿

Actual work of compression (1-2) WC1 =

CP ( T 2−T 1 ) ηc J

WC1 = 1.015 g ˚ c ¿ ¿

J

WC1 = 235 g air ¿ ¿

(471−288)˚ C 0.79

Actual Turbine work: WT1 = Cp (T3 – T4) ηT (1 + f) J = 1.132 g ˚ c (1028 – 664)˚ C (0.82) (1 + 0.0135) ¿ ¿ J

WT1 = 342.4 g air ¿ ¿

Network of plant: Wnet = WT1 – Wc1 = 342.4 – 235 = 107.4

J Gair

Air rate: m = 2685 107.4

KJ hp−hr

KJ kg air

m = 25 kg air per hp-hr or

m = 33.5 kg air per kw-hr ans.

Thermal efficiency excluding combustion losses: J WT 1−Wc 1 g air = ηT = = 0.189 Qf J 567 g air ¿ ¿ ¿ 107.4

Considering combustion losses: ηt = 0.189 x 0.94 = 0.178 ans. Actual exhaust temp.

T41 = T3 – ηt (T3 – T4) = 1028 – 0.82 (1028 – 664) = 729.5˚K Therefore. T41 = 729.5 – 273 = 456.5 ˚C ans.

Problem 7-11 To the plant described in prob. 9 is added a regenerator of 50% effectiveness. Calculate the plant thermal efficiency.

Solution: From prob. 9 Actual temperature leaving the compressor, T21 = 519.6 ˚K Actual exhaust temperature,

T41 = 729.5 ˚K The condition entering and leaving the compressor and the turbines will remains unchanged, therefore, the net work of the cycle will remain unchanged Wnet = WT1 – WC1 = 107.4

J as in prob. 9 g air

Due to the regenerator, the temperature entering the combustor will increase to T 2a. This will decrease the required heat input to the plant for the same net work and should therefore result in an increased efficiency. Effectiveness = Temp. drop in heat realizing fluid Max. temp. differences between the two fluids Let: Tb = temp. of gases leaving the regenerator T 41−Tb Thus, Eex = T 4 1−T 21 ¿ ¿ Solve for Tb: Eex (T41 – T21) = T41 – Tb Therefore, Tb= T41 - Eex (T41 – T21) Substitute values: Tb = 729.5 – 50% (729.5 – 519.6) Tb = 624.5 ˚K For trial value Assume equal temperature drops for exhaust gases and air entering combustor: Ta – T21 = T41 – Tb Therefore, Ta = T21 + Ta1 – Tb

Ta = 519.6 + 729.5 – 624.5 = 624.6 ˚K

From Fig. 7-2 p. 193 for air at 624.6 ˚K Cpa = 1.05

J g˚C

For air at 519.6˚ K ; Cp21 =1.03

J g˚C

J

Average Cp = 1.04 g ˚ C For the exhaust gases f = 0.0135 Cp at 729.5˚K = 1.11

J g˚C

Cp at 624.5˚K = 1.08

J g˚C

Average Cp = 1.095

J g˚C

Solve for actual value of Ta:

Heat balance Cp air (Ta – T21) = ( 1 + f ) Cp gases (T41 – Tb)

Solve for Ta: Ta = T21 + (1 + f)

Cp gases (T41 – Tb) Cp air

Ta = 519.6 + 1.0135

(729.5 – 624.5) ( 1.095 1.04 )

Ta = 631.6˚K At 631.6 ˚K, Cp = 1.05

J same g˚C

Therefore, no need for recalculations, For the combustion process a-3, J g˚C

At Ta = 631.6 ˚K

Cp = 1.05

At T3 = 1028 ˚K

Cp = 1.175

J g˚C

Average Cp = 1.112

J g˚C

Heat required to raise air temp. from 631.6 ˚K to 1028 ˚K Qf = 1.112

J (1028 – 631.6)˚C g˚C

= 440

J g air

Therefore, plant thermal efficiency Wnet

ηT = Qf (combustor eff.) J g air = (94%) J 440 g air 107.4

ηT = 0.23 ans.

Problem 7-15 As open cycle gas turbine plant will exhaust into was heat boilers which receives 65.6˚C feed water. The plant is otherwise like that described in Ex. 1 sec. 7-3. The gas turbine plant produces 3550 hp. How much saturated steam at 3.52 kg/cm 2 gage can be produce per hour, assuming that the final gas temperature will be 37.8˚C above the steam temperature. Solution: From Ex. 1, Sec. 7-3, Air flow = 26.7 kg per hp-hr Fuel rate = 0.014

Kg fuel Kg air

Actual exhaust gas temperature = 756˚K From steam tables: Kg hg = enthalpy of steam at 3.56 Kgcm2 ¿ ¿ = 2743

J g

hf = enthalpy of feedwater at 65.6˚C = 274.3

J g

Saturation temp. at 37.8 ˚C above steam temperature : Final gas temp. = 147.2 + 37.8 = 185 ˚C

From Fig. 7-2 p. 193 for f = 0.014, At 185 ˚C or 458 ˚K = Cp = 1.05

At 756 ˚K: Cp = 1.11

Average Cp = 1.08

J g˚C

J g˚C

J g˚C

Total heat given off by exhaust gases: = (1 + f) (air rate) (hp) Cp ΔT

[

= ( 1.014 ) 26.7

kg J ( 3550 hp ) 1.08 ( 756−458 ) ˚ C hp−hr g˚ C

]

(

)

KJ

= 31 x 106 hr

Heat required to produce 1 kg of steam: = hg – hf = (2743 – 274.3)

= 2468.7

KJ kg

KJ kg

Kgs of steam that can be produced KJ hr = KJ 2468.7 kg 31 x 106

= 12, 600 kg per hr ans.

Chapter 8 Vapor Cycles

Problem: 8-2 A simple steam engine plant has a boiler feed water at 95.6 °C. Steam is supplied to the engine at 97% dry. Atmospheric exhaust, steam rate, 9.3 kg per ihp - hr, Nm=87%. Generator of 91.5% efficiency is directly connected to the engine. Find: a) Ng ; b) Nt based on the ihp and bhp ; c) Nc Solution:

From Steam tables at 95.6 °C, Enthalpy of feed water: hf31 = 400 kj/kg For steam @689.4 kPaa, 97% quality, hf

= 697.34 kj/kg

hfg = 2066.2 kj/kg ∴ h1 = hf + x hfg

= 697.34 + (0.97) 2066.2 = 2701.5

kj kg

From Mollier Chart: After isentropic expansion to atmospheric pressure, h = 2436

kj kg

Enthalpy of water at 101.325 kPa, Hf3 = 419

kj kg

Thus: Rankine Cycle efficiency ∴NR =

h 1−h 2 h 1−hf 3

Note: hf3 = enthalpy of exhaust pressure liquid water NR = b)

2700−2436 kj /kg = 0.116 ans. 2700−419 kj /kg

Actual thermal efficiency of cycle based on IHP kj kw−hr ihp ( 2700−400 ) kj /kg 0.746 kw 3600

Nu =

kg (9.3 ihp−hr )(

Nu = 0.126 ans.

)

Based on BHP: Nu = Nu x Nm Nu = 0.126 x 0.87 = 0.11 ans.

c) Combined Thermal and Electric Efficiency, kj kw−hr ihp ( 2700−400 ) kj /kg 0.746 kw 3600

Nc =

kg (9.3 ihp−hr )(

)

(0.87 bhp/ihp)(0.746 kw/hp)(0.915) Nc = 0.099

Problem: 8-6 A 75-kw turbine generator has a steam rate of 12.7 kg per kw-hr. Steam at 1206.3 kPa g, 55.6 °C superheat. Exhaust, 50.8 mm Hg abs. Find Nc of a simple vapour cycle incorporating this unit. Solution: Enthalpy of steam At 1206.3 kPag, 55.6 °C Sh, h1 = 2927

kj kg

Enthalpy of liquid at 50.8 mm Hg abs,

Hf3 = 160.5

kj kg

Combined Thermal Efficiency

3600 Nc =

kj kw−hr

kg (12.7 kw−hr ) ( 2927−1605 ) kj/kg

Nc = 0.1025 ans.

Problem: 8-8 A steam power plant has operating conditions shown in Fig. 8-8p. Determine all flows for a boiler output of 1 kg per min. and record the same on a copy of a flow diagram. Calculate the cross vapour cycle efficiency and the plant efficiency main exhaust at 0.211 kg/cm2 ab., 90% dry neglect pipe line friction and static head.

Solution: Enthalpy at 1: h1 = 2788 kj/kg

Enthalpy at 4: h4 = 2376 kj/kg Enthalpy at 7: hf = 391 kj/kg at 93.3 C hf = 254 kj/kg at 0.211 kg/cm2 ab

Work input to pump, Pw = Vft (P5 – P4) = 0.00101 kg/ cm2 (15.82-0.211)kg/cm2 (cm/m)2 = 157.6 kg-m Pump work: Pw = 157.6 kg-m/kg x

1 KJ 101.95 kg−m

kj = 1.54 kg condensate through pump Pump steam rate = 31.5

kg or hp−hr

kg hp−hr = 0.01173 kg/kj KJ 2685 hp−hr 31.5

Kg of pump steam per kg of boiler steam W1 = 1.54

KJ kg boiler steam

x 0.01173

kg pump steam W1 =0.018 kg boiler steam kgs. Steam through turbine,

kg pump steam KJ

1 - W1 = 1 - 0.018 = 0.982

kg turbine steam kg boiler steam

Indicated work of pump driver: =

pump work Nm

=

1.54 kj/kg 0.85

= 1.81

kj kg boiler steam

Since this is the total ΔH for 0.018 kg pump steam per kg boiler steam: ΔH = kg pump steam

kj kg boiler steam kg pump steam 0.018 kg boiler steam 1.81

kj = 100.5 kg pump steam Then: h2 = h1 – Δh h2 = 2788 – 100.5 h2 = 2687.5 Nc Set up heat Balance: Heat absorbed by feedwater = Heat given up by pump steam hfg

= (hf1 + Pw) = W1 (h1 – hf7)

∴hfg

= (hf1 + Pw) = W1 (h1 – hf7)

hf6

= (254

kj kj + 1.54 + 0.018kg/kg) kg kg

[2687.5-390.8] kj/kg hf6

= 297

kj kg

Heat output of cycle: = Indicated work of turbine = 0.982

kg (h1 - h41) kg boiler steam

= 0.982

kg (2786-2376)kj/kg kg boiler steam

= 404.6

kj kg

= h1 – hf2 = 2788-297 = 2491kj/kg Cross Cycle efficiency 404.6 = 249.1 = 0.162

Heat Output of Plant: = 404.6

kj kg

x 0.90 (turbine eff.) = 364.14

kj kg

Heat Input to Plant:

= Heat input to boiler KJ kg boiler = 0.75 boiler eff . 2491

= 3321.3

kj kg

Plant Efficiency 364.14 = 3321.3

:

kj kg kj = 0.11 or 11% ans. kg

Problem: 8-10 Draw a flow diagram of a rankine cycle vapour steam power plant. Steam engine drives 150-kw generator of 90% electrical efficiency. Steam rate, 6.7 kg per bhp – hr; steam pressure, 1034.6 kPa g; 55.6 Sh., exhaust to condenser at 15.2 cm Hg abs. No feedwater heating. Motor driven boiler feed pump. Find a) N g b) Ntb c) Ng

Solution:

BHP of steam engine 150 kw kw = = 223 bhp 0.746 x 0.90 hp Steam flow through engine: =223 bhp x 6.7 = 14.94

kg bhp−hr

kj kg

Enthalpy of steam at 1034.6 kPa g, 55.6 C h1 = 2919

kj kg

Enthalpy of steam at 15.2 cm hg abs (after isentropic expansion) kj h2 = 2252 kg Enthalpy of condensate at 15.2cm Hg abs, hf2 = 252.8

kj kg

m3 Volume of condensate: 0.001019 kg Pump Work: Wp = Vf3 (P1 – P3) = 0.001019

m3 101.325 kPaa [(1034.6 + 101.325) kPaa – 15.2 cm Hg abs ] kg 76 cm Hg|¿|¿

Wp = 1.137

KN −m kj = 1.137 kg kg

1 kn kPa−m2

a) Assuming pump work negligible: Rankine cycle efficiency, Ng=

h1−h 2 h 1−hf 3

Ng=

2919−2252 2919−252.8 = 0.25 ans.

b) Brake thermal efficiency of the cycle

3600 Ntb =

kj kw−hr

kg nb ( h 1−hf ) kj / kg kw−hr

x 100

And also: 3600 Ntb =

kj kw−hr

kg kj nb ( h 1−hf ) kw−hr kg

x 100

Ntb = 0.15 ans. e) Combined thermal and electrical efficiency; From eq. 8-6, p. 214, Ne =

3600 ; Ne = ? Wk(h 1−hf 3)

kg bhp−hr But Wk = kw 0.746 0.90 hp 6.7

kg = 9.979 kw−hr Thus,

3600 Nc =

9.979

kj kw−hr

kg kj ( 2919−252.8 ) kw−hr kg

x 100

Nc = 0.135 ans.

Problem 8-11 Find the overall heat rate and thermal efficiency of the plant shown in the diagram below. Boiler efficiency of the plant shown is 75%. Steam flow to pump when generating 350 kW is 181.4 kg per hr. working head on pump = 112.8 m. Neglect the mechanical losses of the pump.

Solution: Enthalpy of the steam at 863 kPa a, dry and saturated, h1 = 2775 kJ/kg Enthalpy of feedwater at 98.8°C hfg = 414 kJ/kg a) Bhp of engine at 350 kW output

350 kW =510 hp (0.746 kW /hp)(0.92) Steam flow through the engine, 510 bhp x 11.3

kg kg =5763 bhp−hr hr

Steam flow through the pump at same head 181.4 kg/hr (given)

Total steam generated, 5944.4 kg/hr Heat input to plant per kg of steam generated, ¿

h 1−hf 4 boiler eff .

( 2775−414 ) ¿

kJ kg

.75 ¿ 3148

Overall heat rate, ¿

Total heat input kW hrs ( generated)

(¿ 5944.4 kghr )(3148 kJkg ) 350 kW

¿ 53465

kJ ans . kW −hr

thermal efficiency,

kJ kg steam

¿ Plant heat output ¿ generator

kw kJ 3600 hr kw−hr kg kJ 5944.4 3148 hr kg

(

(510 hp ) .746 ¿

(

¿ plant ¿ Input ¿

)( )(

)

)

= 0.073. Ans.

Problem 8-16 Calculate the efficiency of an ideal generative cycle operating with steam at 3103 kPa a, 398.8°C, exhaust pressure, 25.4 mm Hg abs, tf = 187.8°C. Sketch this cycle on a T-S plane. Solution: Using eq. 8-7, p. 220. Power Plant Engineering by Morse, the regeneration cycle efficiency,

For steam at 3013 kPaa and 398.8°C, h1 = 3222 kPa After isentropic expansion to 25.4 mm Hg, x2 = 80% (quality): see Mollier chart hfg2 = 2441 kJ/kg (Latent heat of vaporation after expansion)

Tf = 187.8°C + 273 = 460.8°K (abs. feedwater temperature) Te = 26.1°C + 273 = 299.1°K (saturation temp. at exhaust pressure) Substituting values, we have: 460.8

Nreg ¿

[3222−.80 ( 2441 ) ]−4.187 ( 460.8 )+ 1143.1+4. 187 ( 299.1 ) ln 299.1 3222−4.187 ( 460.8 ) +1143.1

Nreg = .42 ans. Sketch of generative cycle on T-S plane

Problem 8-18 The flow heater of a one-heater regenerative is shown in Fig. 8-18P. Calculate the necessary quantities and draw and scaled heat steam of the plant based on 1-kg coal input. Scale 1cm = 2500kJ. All pumps are motor driven and combined, take 1% of the generator output.

Solution: at 2758 kPa a and 337.8°C; h1 = 3094 kJ/kg at 6.86 kPa a, 0.898 quality: h2 = 162.2 + .898(2409) = 2325 kJ/kg at 551.2 kPa a and 182.2°C; h3 = 2815 kJ/kg at 6.86 kPa a: hf4 = 162.2 kJ/kg at 148.9°C: hf5 = 627.2 kJ/kg at 6.86 kPa a: vf4 = 0.0010087 m3/kg

Total pump work wp = vf4 (P1 - P4) = 0010087m3/kg (2758 - 6.86)kN/m2 = 2.775 kJ/kg with pump work neglected hf4' = hf4 hf6 = hf5 Computation for heater

Heat balance

Amount of condensate is 1 - w1 = 1 – 0.175 ¿ 0.825

kg−condensate kg−boiler steam

Heat steam calculation: Heat absorbed in boiler per kg – steam, = h1 - hf6 = 3094 – 627.2 = 2466.8 kJ/kg Kg steam produces per kg coal, ¿

Hv of coal x boiler efficiency Heat absorbed boiler

kJ x 0.75 kg coal ¿ kJ 2466.8 steam kg 2791

¿ 8.486

kg−steam kg−coal

Total bled steam per kg – coal ¿ 8.48

kg steam kg bled steam x 0.175 kg coal kg boiler steam

¿ 1.48

kg bled steam kg coal

Total condensate per kg coal, ¿ 0.825

kg condensate kg steam x 8.48 kg boiler steam kg coal

¿ 6.996

kg−steam kg−coal

Indicated work of turbine per kg boiler steam, = 1 kg (h1 – h3) + (1 – w1) kg (h3 - h2) = 1 (3094 - 2815) + (1 - 0.175)(2815 - 2325) = 683.25 kJ/kg boiler steam Indicated work per kg of coal, ¿ 683.25

kJ boiler steam x 8.48 kg boiler steam kg coal

¿ 5793.96

kJ kg coal

Work delivered to generator per kg coal, ¿ 5793.96 ¿ 5446.3

kJ 0.94 kg coal

kJ kg coal

Plant output per kg coal with 1% going to pump driver,

= work delivered to generator x ( 1 – 1%) x generator efficiency ¿ 5446.3

kJ ( .99 ) ( .92 ) kg coal

¿ 4960.5

kJ kg coal

Heat to condensing water per kg coal, = (total condensate) (h2 - h4)

¿ 15.130

kJ kg coal

¿ 0.825 x 8.48

kg condensate kJ ( 2325−162.2 ) kg coal kg

Summary:

Problem 8-27 An ideal reheating cycle, based on Rankine cycle, has initial state 87.9 kg/cm2 ab, 398.9°C; reheat at 28.9 kg/cm2 ab to 398.9°C; exhaust at 25.4 mm Hg abs. Find the thermal efficiency of this cycle and compare with that of a Rankine cycle operating between the same terminal conditions .

P4 = 25.4 mm Hg abs x 0.001359 kg/cm2 abs/mm Hg ¿ 0.0345

kg ab . cm2

.: ℘=0.001004

kg−m 1 kJ x kg 101.95 kg−m

℘=882.17

℘=8.65

m3 ( kg || 87.9−0.0345 ) 2 ¿ ¿ kg cm mm Hg

kJ kg

Thus, thermal efficiency of reheat cycle, ¿

h 1−h 2+h 3−h 4−℘ h1−hf 5+h 3−h 2−℘

¿

3110.4−2843+233.7−2075−8.65 3110.4−109.34+3233.7−2843−8.65

¿ 0.419 ans . Computation for Rankine: Wp same as before, 8.65 kJ/kg h12 (with S = c from 1to 25.4 mm Hg ab) ¿ 1872.7

kJ kg

Thus, ¿

h 1−h 21−℘ h1−hf 5−℘

¿

3110.4−1872.7−8.65 3110.4−109.34−8.65

¿ 0.411ans .

.: The thermal efficiency of this cycle is greater by ( 0.419 - 0.411) or 0.008 than the Rankine cycle. Ans.

Problem 8-29 In a Central power station having reheat and regeneration, the steam generator delivers 195,045 kg steam per hr at 103 kg/cm 2 ab and 537.8°C to the turbine. After partial expansion of flow of 162,552 kg/hr is returned for reheating from 27.63 kg/cm 2 ab, 363.9 °C to 25.52 kg/cm2 ab 537.8°C, after which it is readmitted to the turbine for complete expansion. Feedwater is generatively heated to 230.6°C, load 64,511 kw. Boiler blowdown, 1950 kg per hr. Generator efficiency, 96%. Find the heat rate of the vapor cycle

Solution: h1 at 103

kg kJ ab, 537.8°C) = 3466.3 cm2 kg

h2 at 27.63

kg kJ ab, 363.9°C) = 3155.8 cm2 kg

h3 at 25.52

kg kJ ab, 537.8°C) = 3544.3 cm2 kg

hfw at 230.6°C = 993 hf1 at 103

kJ kg

kg kJ ab – 1412 cm2 kg

Steam rate:

=

=

kg−steam hr 64,511 kw

195,045

kg steam 3.023 kw−hr

Reheated steam in terms of steam flow: kg reheated steam hr kg boiler steam 195,045 hr

162,552 =

kg reheated steam = 0.833 kg boiler steam Blow Down: kg blowdown hr = kg 195,045 boiler steam hr 1950

kg blowdown = 0.00999 = 0.01 kg boiler steam Total heat input to cycle per kg steam,

= Heat to produce 1 kg steam in boiler + heat to replace blowdown + heat in reheater Assume that the blowdown is at 230.6°C: Then: Heat input to cycle (per kg steam), = 1 kg (h1 – hfw) + 0.01 kg (hf1 -hfw) + 0.833 kg (h3 – h2) Substituting values, we have: heat input: =1 kg (3466.3 – 993)

kJ kJ kJ + 0.01 kg (1412 – 993) + 0.833 kg (3544.3 – 3155.8) kg kg kg

kJ = 2801.1 kg steam Heat rate: = Heat input × steam rate = 2801.1

kJ kg steam × 3.023 kg steam kw−hr

kJ = 8367.73 kw−hr In kJ per bhp – hr Heat rate

= 8468

kJ kw × 0.746 × 96% kw−hr hp

= 6064

kJ ans. bhp−hr

Problem 8-32 The capacity of a power plant is to be increased by superposing a high pressure addition on the existing 17.58 kg/cm2 ga. plant existing turbines are rated to take 369,678 kg steam per hr at 287.8°C. Assume average stage efficiency of superposed turbines will be 76%; mechanical-electrical efficiency, 0.93. What initial steam conditions would suffice to add 20,000 kw to the plant capacity?

Let ɳA = enthalpy of steam at existing plant at 17.58

kg Ga., 287.8°C cm2

ɳA = 3000

kJ kg

With a steam flow of 369,678

kJ kg

The ideal additional output per kg steam to add 20,000 kw to the capacity. kJ kw−hr kg 369,678 hr

20,000 kw =

kJ = 194.76 kg Frow Ex. 1.p. 239 of the Text, actual average flow through superposed turbine is 97% to 98% of its throttle flow (take 97.5% ave.) Thus, with mechanical-electrical efficiency = 93%. Then:

∆hAB = hB – hA =

kJ kg 0.975 x 0.93 194.76

kJ ∆hAB = 214.8 kg kJ kJ ∴hB = 214.8 kg + 3000 kg kJ hB = 3214.8 kg Average stage efficiency: ɳs =

∆ h AB ∆ h s=c

hD = hB -

=

h B−h A hB −h D

hB −¿h ¿ ɳs A

214.8 kJ ∴hD = 3213.8 - 76 % = 2932 kg

with PD = 17.58

SD = 6.64

kg 2 ga. (exhaust pressure of superheated turbine) cm2 kJ kg−° K

Then: with hc = hB = 3214.8

kJ kg

kJ and SC and SD = 6.64 kg−° K From molier chart: PC = 54.13

kg ab cm2

tc = 4232°C ans.

Chapter 9 Energy Flow In The Steam Power Plant

Problem 9-2 Deuteron is an atomic particle composed of a neutron and a proton held together by nuclear binding energy. On the atomic mass scale, the deuteron mass is 2.0146. Find the binding energy in Mev per deuteron.

Solution: From Table 1-1, page 6, PPE by Morse, Rest mass of one neutron

=

1.00894

Rest mass of one proton

=

1.00758

Total Mass

=

2.01652

Mass of Deuteron

=

0.00192

From Table 3, pages 17-03, from Kent’s Mechanical Engineering’s handbook (power volume) 1 mass unit = 931 Mev therefore: Binding Energy = 0.00192 mass unit x 931 Mev per mass unit = 1.79 Mev ans

Problem 9-6 A flat partition is made in two layers, X, Y in close contact (Fig. 9.6 p). Thermocouple measurements are taken at station as shown. material Y is known to have K = 7.2 kcal per hr-m°C. Find: (a) kx; (b) contact resistance in hr – m 2 - °C per kcal.

Solution: a) for simple conduction of heat through flat wall: By Fourier’s Law, q = kA

kx = Thus,

ɵ d

(30.6−25.6) (48.9−41.5) = 7.2 6.4 6.4

kx =

7.2(48.9−41.5) (30.6−25.6)

kx = 10.656 kcal / hr – m° - C

ans.

b) For the two layered walls; over heat transferred coefficient, 1 u = ky 1 dx + + kY h kx Where: 1m dy 1000 mm = Resistance of material Y kcal ky 7.2 hr−m−° C 10.2mm

dy hr−m3−° C = 0.001417 ky kcal 1 hr−m 3−° C = Contract resistance h kcal 1m dx 100 mm = kcal kx 10.656 hr −m−° C 14 mm

dx hr−m2−° C = 0.00131 kx kcal Since: q = UA ∆t U=

q A ∆t

q Kɵ But: k = d =

kcal (48.9−41.5)° C (7.2 hr−m−° C) (6.4 mm)(

1m ) 1000 mm

kcal q = 8325 k hr−m 2 Thus: kcal U= hr−m2 ( 48.9−25.6 ) ° C 8325

And also, 1 kcal 357.3 hr−m−° C = 0.001417+ 1 + 0.00131 h 0.001417 +

1 1 + 0.00131 + 0.00131 h 357.3

1 hr−m 2−° C Therefore, h = 0.000067 kcal

ans.

Problem 9-8 Given Fourier’s Law as q dr = kAdɵ; derive Eq 9-7. Solution:

Shown is a curved cylinder: r1

= Internal radius; mm

r2

= External radius; mm

k

= Thermal conductivity of material in kcal/hr-Ft20/C/Ft. thickness or kcal/hr – Ft2 - °C

L

= Length of cylinder; mm

Q1; Q2 = Temp. on each surface of the cylinder q

= Heat flow through wall of cylinder; kcal/hr q dr = - kAdɵ q

dr = - kdɵ A

q dr = - kdɵ A But: A = 2π rL

q. dr = (- 2π L) k.dɵ q.

dr = 2π L k dɵ r

2

2

dr q. ∫ = (- 2πL).k∫ dɵ 1 r 1

q }21 = - kɵ }21 In r 2 πL

q 2 πL (In r2 – In r1) = - k (t2 – t1)

r q ( 2 ) = -k (t1 – t2) In 2 πL r1

2 πkl(t 1−t 2) r q= ¿ 2 r1 if ɵ = t1 – t2

;

2 πkɵL r2 ∴q= ans. ¿ r1

Problem 9-9 A thin, flat metal wall is bathed with water on one side, air on the other. It is found that 3300 kcal are transferred to the water in forced convection per m 2 per hour. Mean temp. difference, 28.8°C. The water film conductance is estimated at 4638 kcal per hrm2-°C. Make an estimate of the air film conductance.

Solution: q = UAΔt U=

q AΔt

kcal q = 3300 given A m2−hr Thus; kcal 3300 U = 27.8 = 118.7 hr−m2−° C We may neglect its resistance, Thus; 1 1 1 = + u ha hw 1 1 1 = – ha u hw 1 1 1 = – ha 118.7 4638

1 -3 ha = 8.20 x 10 ∴ ha = 121.8

kcal ans. hr−m2−° C

Problem 9-11 Air at an average steam temperature of 82°C is flowing at 762 m/min. in a rectangular duct 38 cm x 76 cm. Estimate the conductance of air film.

Solution: The equation to be used in Eq. 9-10, page 268, otherwise known as Nusselt’s equation is, Nu = CRexPry The important to consider is that Nu, Re, and Pr are dimensionless numbers. We shall evaluate each separately first. Nu =

hDe k

Where: h

= the conductance of the air film; kcal/hr-m2-°C

De

= equivalent diameter of a characteristics dimension of the fluid conduit; m

∴ De = 4 [

=4[

Free cross sectional Areaof flow ] Perimeter of this Area

( 38 x 76 ) c m2 ] 2 ( 38+76 ) (10000)

= 50.667 cm. – 0.507 m. k

= thermal conductivity of air at 82°C = 7.5 x 102 kcal/hr-m-°C from Fig. 9-6, page 270 (text) ∴ Nu =

hDe k

0.507 m kcal = 7.5 x 102 hr−m−° C Re

= Reynolds Number =

Where De V

DeV ρ u

= 0.507m as previously computed = velocity of flow = 762 m/min x

1 min 60 sec

= 12.7 m/sec ρ

= density of air at 82°C from the characteristic gas equation pv = WRT

W

=

pv RT

at, T = 82 +273 = 355 and 4m3

( 101.325 ) (1) ( 0.287 ) (355)

W

=

W

= 0.9945 kg/m3

From the table 9-2, for longitudinal flow in tubes and ducts, the value of the constants for Nusselt’s equation are

c

= 0.023

x

= 0.8

y

= 0.4

Substituting all known values in Nusselt’s equation, we have h (0.423) = (144,660.41)0.3 (2.44)0.4 (0.023) h = 1043.81 kcal/hr-m3-sec2 ρ =

W 0.9945 kg/m 3 g = 9.81 m/sec 3

= 0.1013 kg/m2-sec2 from Fig. 9-6 page 270 U = Viscosity = 22.5 x 107 kg-sec/m 0.9945 kg−sec 2 /m2 DeV ρ ( 0.507 m )( 12.7 m/sec ) 9.81 ∴ Re = = u 7 22.5 x 10 kg−sec /m

(

)

Re = 2.90 x 10-9 Prandtl Number; Pr =

CU K / 3600

Where: C

= Specific heat = 9.81 x 0.24 kcal/hr-m2-°C

U

= 22.5 x 107 kg-sec/m2

K

= thermal conductivity in kcal/hr-m2-°C

Pr

=

( 9.81 x 0.24 ) ( 22.5 x 107) 7.5 x 102 /3600

∴ Pr = 2,542,752,000

From Table 9-2 for longitudinal flow in tubes and ducts, the value of the constant for Nusselt’s equation are: Substituting all values found in Nusselt’s equation. Nu = cRexPry h(

0.507 ) = 0.023 (2.9 x 109)0.8 (2.54 x 109)0.4 7.5 x 102

Solving for Conductance h = 7.305 x 1012 kcal/hr-m2-°C ans.

Problem 9-13 Find the mean temp. difference in a steam condenser where the absolute pressure is 0.12 kg/cm2 and the condensing water terminal temp. are 18.3°C and 31.1°C.

Solution:

θm = ? Where: θm = log mean temp. difference =

At 0.12

θ max = 49.4 – 18.3 = 31.1°C θ min = 49.4 – 31.1 = 18.3°C ∴ Mean temp. difference = θ max

−θ m∈¿ ¿ θ m ax ln( )¿ θ m∈¿

31.1−18.3 31.1 θm = ln( ¿) ¿ 18.3 ∴ θm = 24. 13 °C ans.

Problem 9-15

−θ m∈¿ ¿ θ m ax ln( )¿ θ m∈¿

kg ab; saturation temperature or tsat. of steam is 49.4°C cm2

Thus:

θm =

θ max

Repeat Prob. 14, except make the case one of parallel flow. Solution: data from Prob. 14: one fluid rises from 82°C to 121°C. the other changes from 321°C to 218°C.

θ1 = 321°C – 82°C θ1 = 239°C θ2 = 218°C – 121°C θ2 = 97°C a)

Arithmetic mean temp. difference, θ =

θ1+ θ2 239+97 336 = = 2 2 2

= 168 ans.

b)

True logarithmic mean temp. difference, θm

θ1−θ2 239−97 142 θ1 = 239 = = ln ln 1.36 ln 97 θ2 = 455.5°C ans.

Problem 9-17 The rate of water discharge from a 5.4 mm copper alloy tube 3.05 m long surrounded by steam was 97 kg/min. Water terminal temperatures, 18.3°C and 28.9°C, Steam temperature, 104.4°C. What coefficient of conductance was developed?

Solution: Heat absorbed by water: q = WcΔt Where: W

= mas flow rate

c

= specific heat

Δt

= temperature increase

kg 60 min 1 kcal x x q = 97 min hr kg−° K x (28.9 – 18.3) °K kcal q = 61.692 hr

heat transfer from steam to water

q = UAθ U=

q Aθ

for mean temp. difference,

θ max

=

104.4 – 18.3 = 86.1°C

θ min

=

104.4 – 28.9 = 75.5°C

Thus; θ=

θ max

−θ m∈¿ ¿ θ m ax ln( )¿ θ m∈¿

86.1−75.5 86.1 θ= ln 75.5 θ = 80.7°C And so; kcal hr U= π ( 0.0254 m ) ( 3.05 m) ( 8.07 ℃) 61.692

kcal U = 3141 hr−m 2−° C

ans.

Problem 9-18 With the use of Fig. 9-10, determine the required surface for a condenser if 43,090 kg steam per hr are to be condensed at 50.8 mm Hg abs., using water at 21.1̊C, which will be heated to 4.4̊ less than steam temperature. Steam quality, 0.0875; water velocity, 2.44 m/sec., 19.1 mm; 1.24 mm Muntz tubes

Solution Saturation temperatures at 50.8 mm Hg abs. = 38.41̊C Temperature of water leaving condenser

= 38-1-4.4

Mean temp. difference

ϴ max = 38.1-21.1 = 17.31̊C ϴ min = 38.41-34.01 = 4.4̊C Thus, ϴ=

ϴmax−ϴ min ln ϴmax ¿ ¿ ¿

17.31−4.4 17.31 ϴ= ln 4.4 ϴ = 9.425̊C From Steam Tables: For saturated steam at 50.8 mm Hg abs., hf = 160.7 kj/kg hfg = 2411.5 kj/kg With 87.5 % quality: h = hf + x hfg h = 160.7 + 0.875 (2411.5) h = 2270.76 kJ/kg

Consider the condenser:

Heat rejected by steam = heat absorbed By cooling water By water cooling

ϴ = h - hf ϴ = 2270.76 – 160.7 ϴ = 2210.06 kJ/kg

Corresponding Heat Transfer Rate: q = 4,090 kg/Ar (2110.06 KJ/kg) (0.2388 kcal/KJ) q = 21.7 x 106 kcal/hr From Fig. 10, page 279; for 191.1 mm tubes and 2.44 mps water velocity, Uncorrected U = 3700

kcal hr−m2−℃

For 1.24 mm thick Muntz tubes, Multiplier = 0.96 Temp. correction factor = 1.00 ∴ corrected U = 3700 x 0.96 x 1.00 U = 3552

kcal hr−m2−℃

Therefore, Using the formula q = UAϴ The required surface area is A=

q Uϴ

6

A=

(

kcal hr

21.7 x 10 kcal 3552 (9.425 ̊ C) hr−m2−̊ C

)

A = 648 m 2 ans.

Problem 9-19 Determine the overall U of a condenser from individual predictions of Uv, Uw, Ut, and Us. Data 22.2 mm x 1.24 mm tubes, 2.13 m/sec water velocity, 7̊C mtd; 23.9 C ̊ average water temperature. Solution Vapor Film Coefficient as per eqn. 9-22, p. 276, PPE by Morse: 111910 kcal 2 q Uv = ( ) hr −m −℃ Av Conductance of Tube: Ut =

kt from table 9-1, page 267, text, for condenser tubes: dt kt = 93.7

∴Ut =

kcal hr−m2−℃

93.7 kcal per hr −m 2−℃ 0.00124 m

Ut = 75564

kcal hr−m2−℃

Scale Coefficient: Us = 14,650

kcal (Using Average Value) hr−m2−℃

Water Film Coefficient: Uw = br0.73 = b = 1379.7

J 0.24 Dw 0.27

J 0.24 b = 1379.7 Dw 0.27 From table 14.8, page 606, text: for water at 23.9 ̊C μ = 0.95 centipoise ∴γ =

1 1 100 = = centipoise -1 μ 0.95 95 Dw = OD – 2t Dw = 0.0222 m – 2 (0.00124 m) Dw = 0.01972 m

Thus, .100 .24 ) 95 b = 1379.7 (0.01972)0.27 (

b = 4032 And so, Uw = 4032 (2.13) 0.73 Uw = 7000

kcal hr−m 2−℃

Therefore: θ=

γ ℃=

1 2 Dv Dv Dv q [ + + + ] Av Uv Ut (Dw + Dv) Us Dw Uw Dw

q ¿ Av

7¿

q ¿] Av

7 ¿ ¿ ¿) 783,350 ¿ ¿ + 28.16 ( Let

q ) Av

q =x Av

Thus, x1.188 + 28.16x – 783,350 = 0 By Newton’s Method of Approximation; x 2=x 1 - (

( f ) ( x1 )

( f 1) ( x 1 )

)

For first trial value, note that 1.188 is close to 1.0 So that X1 + 28.16 X1 = 783,350 X1 =

783,350 1+ 28.16

X1 = 26, 860 Take: f (x) = x1.188 + 28.16x – γ 83,350 A f1 (x) = 1.188 x0.188 + 28.16 Then, f (x1) = (26,860) 1.188 + 28.16(26,860) – 783,350 f (x1) = 155,740 f1 (x1) = 1.188 (26.860) 0.188 + 28.16 f1 (x1) = 36.24

X2 = 26,860 -

155,740 36.24

X2 = 22,560 Continuing the Approximation Process: x 3=x 2 -

f ( x2 ) f 1 ( x2 )

f (x2) = (22,560) 1.188 + 28.16 (22560) – 783,350 f (x2) = 453.6 f (x2) = 1.188 (22,560)0.188 + 28.16 f (x2) = 35.98 Thus, X3 = 22,560 -

453.6 35.98

X3 = 22,550 f (x3) = (22,550) 1.188 + 28.16 (22,550) – 783,350 f (x3) = 94; this is close enough considering the great value of X. Thus, kcal qx = 22,550 Av hr−m2 From q = UAϴ We have, U= Thus,

q Aθ

U=

kcal hr −m2 7℃

22,550

Or U = 3220

kcal ans. hr−m 2−℃

Problem 9-20 Calculate the film coefficient of steam flowing through 38.1 mm x 2.67 mm tubes at 10.54 kg/cm2 ga 232 ℃,at the rate of 1220 m/min. From Fig, 9.6; page 270, PPE by Morse, at 10.54 kg/ cm 2 + 1.033 = 11.573 kg/cm 2 and 202 ̊ C U = 23.25 x 107 kg-sec/m. K = 7.75 x 102 kcal/hr-m ̊C The specific heat at Cp will be calculated from the steam table using the equation, Cp =

∆h ∆t

kg h at 11.573 x cm2 h

101.325 kPa 1.1 MPA kg 230 ℃ 1.033 2 = 1.1 MPa] cm

1.1 MPa kJ = 2904.9 235℃ kg ∆ h = 2904.9 – 2893.4

kJ kg

= 2893.4

kJ kg

∆ t = 235 – 230 = 5℃ Thus; Cp =

11.5 kJ = 2.3 -℃ 5 kg

Reynolds No.; Re DVp u

= Where:

De = Tube inside diameter in m =

38.1−2(2.67) 1000

= 0.03276 m V = Velocity of Flow; m/sec =

1220 m/min m = 20.33 60 sec /min sec

ρ = Density; kg∙sec2/m4 =

1 V 1 x 9.81

V1 at MPa & 232 ℃ = 202.03 x 10-3

ρ=

1

( 202.03 x 103 ) (98.1)

= 0.505 kg – sec2/m4

Solving For Re: Re =

( 0.03276 ) ( 20.33 ) (0.505) 23.25 x 10

Re = 144660.41

m3 kg

Cp U k Prandtl. Number; Pr = 3600 =

( 9.81 )( 2.3 )( 23.25 x 10+7) 7.75 x 10 2 /3600

= 2.44 x 1010 = 2.44

Nusselt’s Number; Nu = h

=h[

D K

0.03276 ] = h ( 0.423) 7.75 x 10−2

from Eq. 9-10, p. 268,Nusselt’s equation Nu = CRex Pry From Table 9-2, p.270, and for longitudinal flow in tubes,we have C = 0.023 x = 0.8 y = 0.4

Substituting all known values in Nusselt’s equation and solving for the film coefficient, h (0.423) = 0.023 (144.535)0.3 (2.44)0.4 h = 1047.81 kcal/hr – m2 – ℃ ans.

Problem

9-21 Determine, with equation 9-10, the film coefficient of conductance of water in a 50.8 mm x 3.18 mm steel tube. Average water temperature, 110℃; velocity, 1.52 m/sec. Compare result with a coefficient calculated from Eqn. 9-20.

Solution: By Eq.9-10, Nu = CRex Pry Where:

Nu =

hDe k

Re =

DeVρ μ

Cμ Pr = k 3600 De = OD-zt De = 0.0508 m – 2 (0.00138m) De = 0.0444 m From Table 9-1 page 269, PPE by Morse For water, k = 0.53

kcal hr−m 2−℃

(see Also Fig. 9-6,PPE by Morse,page 270) Water Velocity V = 1.52

m given sec

Density of Water at 110℃:

ρ = 958.3

kg ( see Steam tables) m3

Absolute Viscosity of Water at 110℃ from Fig. 9-6, page 270, PPE by Morse μ = 3 x 10-5 kg -

sec m2

Note: 1 Centipoise 1.02 x 10-4

kg−sec m2

From Kent’s Handbook, Power Vol. page 6-42; Specific heat of water, Cp = 1.0

kcal kg−℃

From Table 9-2, PPE by Morse, page 270, C = 0.023; x = 0.8 y = 0.4. For longitudinal flow in tube: Nu =

ADe K

kcal h (0.0444 m) ( hr−m −℃ ) Nu = 2

0.53

kcal hr −m2−℃

Nu = 0.0838 h Re =

DeVp μ m kg ) 953.3 3 sec m ( 3 x 10−5 ) (9.81 m 2 ) sec

( 0.0444 m ) (1.52 Re =

(

)

Re = 220,000 Pr =

3600Cμ K (3600

Pr =

sec kcal kg−sec )(1.0 ) 3 x 10−5 hr kg−℃ m2

(

( 0.53

kcal ) hr −m2−℃

)

1

(9.81 secm ) 2

Pr = 2.0 Nu, So By Eq.9-10, page 268, (PPE byMorse) Nu = CRex Pry 0.0838 h = (0.023) (220,000)0.8 (2.0)0.4 0.0838 h = 570.27 ∴h = 6800

kcal hr−m 2−℃

b) By Eq.9-20, page 276, PPE by Morse Uw = bV0.73 Where: J 0.24 b = 1379.7 Dw 0.27 J=

1 (μ in centipoise) μ

μ = 3 x 10

-5

1 Centipoise kg−sec ( 1.02 x 10−4 kg−sec ) m2 m2

μ = 0.294 centipoise ∴J =

(3.4)0.24 1 = 3.4 thus, b = 1379.7 μ (0.04444)0.27

and so; Uw = 4289.8 (1.52)0.73 kcal hr−m2−℃

Uw = 5283

ans.

Problem 9-24 Employ Eq. 9-19 to predict q/AV of a condenser heat transfer surface where the tubes are 22.2 mm x 1.24 mm copper alloy t 1, 32.2°C; t1, 16.7°C; to 27.8°C. Water velocity, 1.83 m/sec Us, 9765 kcal/m2-hr-°C. Solution: Eq. 9-19, p 276, PPE by Morse; θ=

D q 1 2Dv 1 1 + + v + Av U V U t ( D w +D v ) D w U s U w

[

(

)]

By: Eq. 9-22, page 276, PPE by Morse Uv=

111,910 q 0.188 Av

( )

By: Eq. 920: U w =b V 0.73

( J )0.24 b=1379.2 0.27 Dw Dv=0.0222 m Dw=Dv−2 t Dw = 0.0222 – 2 (0.00124) Dw = 0.1972 m J=

1 (μ in centipoise) μ

Ave. Water Temperature: 1 ¿ ( t1 + t0 ) 2 ¿

1 ( 16.7 + 27.8 ) = 22.5°C 2

From Table 14-8 p. 606 text. Interpolate:

1.15

t

μ

21.1

1.0 μ−1.0

{22.25} 5.6

μ

26.7

0.9

μ -10 1.15 1.15 = = -0.10 5.6 5.6 μ=1.0+ ( -.10 )

(1.15 5.6 )

μ=0.97946 Centipoise Thus,

-1.10

J=

1 =1.02097 μ

And

( 1.02097 )0.24 b = 1379.7 0.27 ( 0,01972 ) b = 4002.4 Therefore:

Uw = 4002.4 ( 1.83 ) Uw =6222

0.73

kcal m 2 - hr -°C

For condenser tubes: kt =93.7

kcal hr -m-°C

dt=0.00124 m given Thus; Dt=

93.7 0.00124

Ut=75,564.5

kcal hr - m 2 -°C

Mean temp. difference:

θ max = 32.2 - 16.7=15.5°C θ min = 32.3 - 27.8 = 4.4°C θ=

θ max - θmin θmax ln θmin

(

θ=

)

15.5-4.4 =8.815°C 15.6 ln 4.4

( )

Substituting values, we have q 0.188 Av q 2 (0.0222) 0.0222 8.815°C= + + Av 111,910 75,564.5 (0,04192) 0.01972

[

8.815 °C=

]

( )

q q 8.935 ×10−6 Av Av

[

Let x =

0.188

( )

986,600 =

q Av

0.188

( )

( 19765 + 16222 )

+ 34.72

+1.407 × 10−5 +2.962× 10−4

(qAv )

(qAv )

f ( x ) = x 1.188 +34.72 x −986,600=0 f 1 ( x ) = 1.888×❑0.188+ 34.72

]

By Newton’s Method x2=x 1−

( f )( x1) (f1 )( x 1 )

For First Trial Value: 35.72 x 1=986,600 ∴ x 1 = 27,600 Thus; f ( x1 ) =( 27,600 )1.188+ 34.72 ( 27,600 ) −986,600 f ( x 1) =160,384 f 1 ( x 1 )=1.188 ( 27,600 )1.188+ 34.72 f 1 ( x 1 )=42.84 Then: x2=27,600−

160,384 42.84

x2=24,000 x3 = x2−

( f ) ( x2 ) (f 1 )( x2 )

f ( x 2) =( 24,000 )1.188+ 34.72 ( 24,000 ) −986,600 f ( x 2) =6520 f 1 ( x 2 )=1.188 ( 24,000 )1.188+ 34.72 f 1 ( x 2 )=42.63 Thus:

x3 =24,000−

6520 4263

x3 =23,850 f ( x3 ) =( 23,850 )1.188 +34.72 ( 23,850 ) −986,600 f ( x3 ) =128 (close enough) Thus; q kcal =23,850 ans. Av hr - m 2

Problem 9-25 Find the ERS and EPRS of radiant surfaces of tubes on a furnace wall in arrangement (a) Fig 9-14. There are 20 tubes 88.9 mm dia. x 3.05 m long. Assume a slag factor of 0.92. Solution: Projects Area of Tubes, Ap = Dia. X Length x No. of tubes = 88.9 mm x Ap = 5.422 m 2

1m ( 3.05 m ) (20) 1000mm

From Eq. 9.27, p. 285, PPE by Morse, Equivalent Radiant Surface, ERS = A r = S δ Ap Where: S = Slag Factor=0.92 δ = Area Factor ¿

π For arrangement shown (see page 285 2

∴ ERS=0.92

(π2 ) (5.422 )

¿ 7.835 m 2 ans . EPRS = S E effectiveness x Refractory Wall area for tube spacing to diameter ratio of L, effectiveness = 100% from Fig. 9-15 page 286.

Problem 9-26

Find the ERS of a furnace envelope consisting of 3 walls arranged as in Fig. 9-15 (3), one wall refractory, floor refractory, and roof bank. BS area of walls, 10.7 m 2 each; of floor and roof, 8.4 m2 each; δ = 0.55π; for walls, 1.5π; for tube bank. S = 0.90 D = 76.2 mm; L-178 mm Solution: From Eq. 9-24, page 285, PPE by Morse, A r = S δ Ap Where: Av = Equivalent Radiant Surface (ERS) Ap = heating surface projected on the plane of the furnace envelope S = slagging factor 1 for clean tubes, 0.90 for normal operation δ = area factor For Roof: Ap = BS = 8.4 m2 ∴ A r = S δ Ap Ar = 0.9 x 1.5 π x 8.4 m2 For Each Wall: Ap = BS

D L

Ap = 10.7m 2

76.2 178

Ap = 4.58 m 2 ∴ ERS of Walls = S δ Ap ¿ ( 3 × 4.58 )( 0.90 )( 0.55 ) ¿ ¿ 13.74 m 2 ( 0.90 ) ( 0.55 ) ¿

¿ 21.37 m 2 ∴ Total ERS = 35.62 + 21.37 = 56.99 m 2 ans. Problem 9-27 Solve Ex. 9-26. Finding EPRS instead of ERS Solution: L 178 mm = =2.34 D 76.2 mm From Fig. 9-15, p. 286, for walls (Curve 3) Effectiveness = .55 EPRS of 3 walls ¿ .55 ( 3× 10.7 ) m 2=17.65 m 2 For roof (Arrangement assumed as curved (1), Fig. 9-15, p 286) Effectiveness = .97 for

L =¿2.34 D

EPRS of Roof = 0.97 x 8.4 m2 = 8.14 EPRS of 3 Walls

= 17.65 m2

EPRS of Roof

= 8.14 m2 25.79 m2 ans.

Note: To determine EPRS from eq. EPRS = BS x Effectiveness BS = Is only for surface containing surfaces Problem

9-28 Find the fraction cold of the furnace of problem 9-26 Solution: See problem 9-26 Projected Area of Roof= 8.36m2 Projected Area of 3 walls= 3 x 10.7 x

76.2 178

=13.48 m2 Total Ap

=8.36 m2 =13.48 m2 21.89 m2

BS of Roof X Floor = 2 x 8.36 = 16.72 m2 BS of 4 walls

= 4 x 10.7 = 42.8 m2 Total BS= 59.52 m2

FC=

AP 21.89 m2 = = 0.36 ans. BS 59.52m 2

(For evaluating F.C, BS is total refractory area of furnace envelope with or without tubes.)

Problem 9-29

With the Orrok equation, find the temperature of gases at furnaces exit for a case of combustion where bituminous coal of QL= 6255 kcal/ kg is burned on 15:1 A:F Ratio. Air temperature, 26.7°C; Cr, 9.8 kg/m 2 ERS. Allow for combustion heat losses of 280kcal/kg coal. Cp. Of gases= 0.26 kcal/kg-°C. 1 Wa √ Cr 1+ 60 Where: Ω = fraction of the available heat in the furnace which is transmitted by radiation Wa= kg air used/ kg coal = 15 (As give) Cr= kg coa / sq. m Equivalent Radiant Surface = 9.8 kg m2 ERS(As Given) Substituting this values in Eq. 9-26 above

¿

Ω

1 15 √ 9.8 1+ 60

=

1 1 = = 0.3666 1+ 1.73 2.73

Lower Heat Value of Coal used

= 6255 kcal/kg

Combustion Losses

= 280 kcal/kg coal

Available Sensible Heat Heat Absorbed in Furnace through radiation: = Ω x Heat available Heat carried away by the flue gas = (1-Ω)(Heat Available) = (1- 0.366)(6535) =4143.19 kcal/kg coal

= 6535 kcal/kg coal

Flue gas = 1+15 or 16 kJ/kg coal

Heat Absorbed by Flue Gas = 16 x 0.26 x (tg – 26.7°c) = 4143.19 (tg = 26.7) =

4143.19 16 x 26

=

4143.19+ ( 26.7 )( 4.16) 4.16

tg= 1022.659

Temperature of flue gas at furnace exit; tg= 1022.659 +26.7 tg= 1049.359°C ans. c= 0.023 x= 0.8 y= 0.4 Substituting all known values in Nusselt’s equation h(0.423) = (144,660.41)0.8 (0.023)(2.44)0.4 h= 1043.81 kcal/m – m2 - °C W ρ= = g

kg m3 kg = 0.1013 m m2−sec 2 9.81 sec 2

0.9945

from fig. 9-6 U- Viscosity = 22.5 x107 kg−sec /m

Re=

DeVp = u

(

m 0.9945 kg−sec2 ❑ sec 9.81 m2 7 22.5 x 10 kg−sec/m

( 0.507 ) 12.7

)(

)

Re= 2.90 x 10-09

Prandtl Number; Pr=

CP U K 3600

Where: C= Specific heat = 9.81 x 0.24 kcal/hr-m2-°c U= 22.5x107 kg-sec/m2 K= thermal conductivity in (BTU/sec- FT - °F) kcal/hr –m 2-°C

( 9.81 x 0.24 ) ( 22.5 x 107) PR= 102 7.5 x 3600 PR= 2,542,752,000 ans.

Chapter 10 Steam Generators

Problem 10-5 A fuel of Qh = 6888 kcal per kg was burned in the furnace of 223 m 2 water tube boiler which was generating 5443 kg steam per hr at 14.06 kg per cm 2 abs, d & s, from 109°C feed, at a time when fuel consumption was 612 kg per hr. a. Find kcal rating and percent rating; b. Calculate overall thermal efficiency Solution: For dry and saturated steam of 14.06 kg per cm2 ab, from steam tables hs= 2787 kJ per kg For feed water at 109 °C; hf= 456.44kJ per kg Therefore: a) Kcal Rating = ws (hs=hf)

[

=5443

kg steam kJ kcal ( 2787−456.44 ) [0.2388 ] hr kg kJ

= 3.03 x 106

][

kcal ans. hr

]

Rared Bo. Hp: Heating Surface RBo Hp= 0.91 m2 of heating surface enginer Bo. Hp 223 m 2 = m 2 = 245 Bo Hp ans. 0.91 BoHp b) Developed Bo Hp Ws( hs−hf ) = DBoHp= 35,322

kg kJ ( 2787−465.44 ) hr kg kJ 35322 BoHp−Hr

5443

DBoHp= 359 BoHp ans. Percent Rating: %R = %R=

D Bo Hp x 100 % R Bo Hp 359 x 100 % 245

R= 146% ans. Overall Thermal Efficiency kg kJ kcal 5443 )( 2787−456.44 ) (o .2388 ) ( hr kg kJ Ws(hs−hf ) μBO= = Wf Qh ) ( 612 kghr )(6888 kcal kg μBO=0.72 ans

Problem

10-8 A simple water tube boiler of 1858 m2 heating surface is offered by salesman for installation in a steam plant where 36287 kg of d & s steam are to be produced at 13.36 kg/cm2 ga. from 99°C feed water. Using the Bobcock formula, verify whether this area is adequate by assuming mass flow 24412 kg per hr per m 2, per hr per sq. ft., furnace and chimney temperatures of 982 °C and 232 °C. Solution: From steam tables For d & s steam at 13.36 kg/cm2 ga.; hs= 2789 kJ per kg For feed water @ 99°C; hf= 414kJ per kg. Heat required to produce 36, 287 kg/hr & steam =[36,287 kg/hr] [(2789 - 414) = 86.2 X 106

kJ ] kg

Kj hr

By Bobcock Formula: (Eqn. 9-14, PPE by Morse) U= A + BG Where: U = Coefficient of heat transfer A & B = constant as follow for boiler; A = 9.76

kcal hr−m 2−° C

B= 0.0014

kcal kg−° C

And G = mass flow, kg per hr per m2 Thus,

U = 9.76 + 0.0014 x 24412 U = 43.937

kcal hr−m−° C

Θmax−Θ min Θmax Θ= ln( ) Θmin 787−37 737 = 245°C Θ= ln( ) 37 Heat transfer surface required: Kj ) ( 86.2 X 10 kcal )( 0.230 kcal kJ 6

A=

Q = UΘ

(

43.937

kcal (245.3 ° C) hr −m2−° C

)

A = 1911.5 m2 Therefore, the boiler being offered is not adequate to produce the specified steam rate, its surface are being less than the required heating area based on a mass flow of 24,412 kg per hr per m 2. However, if the mass flow can be increased, the overall

heat transfer coefficient will increase, thus, the required heating area will be less, the boiler can, thus, become adequate. For the new mass flow rate: G=

U− A B

Θ −A G= sa SA=Surface Area B ) ( 86.2 x 10 kJhr )(0.239 kcal kJ 6

G=

( 1858 )( 245.3 ° C ) (0.0014

kcal ) kg−° C

kcal hr−m2−° C = kcal 0.0014 kg−° C 9.76

G = 25,316 kg per hr per m2 Problem 10-9 A boiler like Fig. 10-6 (except no wall cooling) has 101.6 mm x 6m long tubes in a bank 22 wide x 16 deep with 152.4 mm horizontal spacing. Transverse baffles divide the tube surface into three equal gas passes. Steam, 12.3 kg/cm 3 ga., d & s; 93.3°C feed water. Fuel: coal at Qs = 2500 kcal. Using 19 kg air per kg coal. a) Use the Bobcock formula to find the overall heat transfer coefficient b) Assume that when burning 3402 kg coal per hr the furnace temperature will be 1010°C. Then, find the probable temperature of gases leaving the setting, also the rate of steam generation, kg per hr. Solution:

a) Bobcock formula for overall heat transfer coefficient: U = A + BG where: for boiler: A = B=

9.76 kcal per m2 – hr - °C 0.0014 kcal per kg - °C

And G = Gas mass flow rate,

kg hr−m 2

For G: At 19 kg air per kg coal; Total flue gas = (19 + 1) (3402 kg/hr) = 68,040 kg/hr Flow area per element; Ae = (

L ¿ (s−d) np

Where: L = length of tube Np = number of passes s = spacing between tubes D = outside diameter of tubes Thus; Ae =(

6m )(0.1524 – 0.1016)m 3

Ae = 0.1016m2 For a tube bank 22 tubes wide;

Total flow area, At = 23 (0.1016m2) At= 2.3368 m2 Thus; kg hr G= 2.3368 m2 68,040

G = 29116.7

kg hr−m 2

And so, U = 9.76

kcal kcal kg +0.0014 29116.7 2 kg ° C hr−m −° C hr−m 2

U = 50.52

kcal ans. hr−m2−° C

Total heat transferred to the tubes consists of: 1. Radiation to 1/3 of the tubes in first pass 2. Conduction and convection to all the tubes. Heat transferred by radiation: Effective Radiant Surface. ERS = SδAp Where: S = Slagging factor, 1 for clean tubes, 0.9 for normal operation δ = Area Factor Ap = Projected Area For S, use 0.9

For δ, use 2 ( see eq. 9.27, PPE by Morse and Fig. 9-14 d) For Ap: Ap =

δ D nt nρ

np= no of tubes, np = no of passes

And so, Ap = (

6m )(0.1016 m )(22 tubes) 3

Ap = 4.4704 m2 Thus, ERS = 0.9 x 2 x 4.4704 m2 ERS = 8.0467 m2 Fraction of available heat in surface which is transmitted by radiation Ω=

1 W a √ Cr 1+ 60

Where: Wa = kg air used per kg coal Cr = kg coal per hr per m2 or ERS

Cr=

kg coal hr 8.0467 m2

3402

Thus; Ω=

And so:

1 =0.133 19 √ 422.78 1+ 60

Heat transmitted by radiation QR = Ωwf Qwf Qs QR = 1,131,200

kcal hr

Heat transmitted by convection and conduction: QC= W.Cp (t1-t2) Where: W = kg flue gas per hr W = 68,040 kg per hr as computed previously Cp = 0.24

kcal (actually for air but assumed ¿be applicable for flue gas) kg−℃

t1 = furnace temp = 1010 ℃ given t2 = temp. of flue gas leaving the boiler tubes, ℃ and so: Qc = (68,040

kg kcal )(0.24 )(1010 – t2) hr kg−℃

∴ Qc=16,329 ( 1010−t 2 ) Solve for t2:

kcal hr

Tsat for steam at 12.3 kg/cm2 G = 192℃ θ max ¿ 1010−192=918 θ min ¿ t 2 −192 Thus: θ=

918−(t 2−192) ln ¿ ¿¿

Total Heat Transfer Area = π (0.1016m) (6m) (22 x 16) = 674m2 Total Heat Transferred Q = QR + Qc Q = 1,131,200 + 16,329.6 (1070 – t2)

kcal hr

From the formula: Q = UAθ [1,131,200 + 16, 329.6 (1010-t2)]

= 50.52

kcal hr

kcal (674 m 2 )¿ 2 hr−m −℃

Let (t 2−192 ¿=x Thus:

1,131,200 + 16,329.6 (818 – x ) = 34,050.0

(918−x) 918 ln x

( 918−x ) 69.3 + (818 – x) = 2.085 918 ln x 1914−2.085 x 918 887.3 – x = ln x (887.3 – x)(ln 918 – lnx) = 1914 – 2.085 x (887.3 – x)(6.822 – lnx) = 1914 - 2.085 x 6053.16 – 6.822 x – 887.3 lnx + x lnx = 1914 – 2.085 x x lnx – 887.3 lnx – 4737x + 4139.16= 0 By Newton’s Method: x 2=x 1−

f ( x 1) f 1 ( x 1)

Take: f(x) = x lnx – 887.3 lnx – 4.737 x + 4139.16 f 1 ( x )=x

( 1x )+ln x−887.3 ( 1x )−4.737

f 1 ( x )=ln x−887.3

( 1x )−3.737

For first trial value, x; Take t 2=300 ℃ since t 2 is close ¿192 ℃ ∴ x1=300−192=108 Thus: f( x 1 )=108 ln 108+ ln 108 (−887.3 )−4.737 ( 108 ) + 4139.16=−21.22 f 1 ( x 1 )=ln 108− Thus:

887.3 −3.737=−727 108

x 2=108−

(−21.22) (−7.27)

x 2=105 f( x 2 ¿=105 ln 105+ln 105 (−887.3 )−4.737 ( 105 ) +4139.16=09.8 close ¿ zero ¿ take x = 105

thus, t 2=192+105

t 2=297 ℃ ans Thus: Q = UAθ = 1,131,200 + 16,329.6 (1010- 297) Q = 12,774,205

kcal hr

Rate of Steam Generation: Hs = at 12.3 kg/cm2 GA= 2786 kJ/kg Hf = at 93.3℃ = 390

kJ kg

Steam Rate: w s=

Q h s−h f 12, 774 , 205

W s=

] [ ( 2786−390 ) kJ ] [0.2388 kcal kJ

w s=22,326

Problem 10-13

kcal hr

kg ans. hr

Testing of a stoker-fired boiler revealed that, during a period when 2064 kg of 3000 kcal coal were burned, refuse accumulated in the ashpit to the extent of 225 kg, although the coal as contained only 10% ash by analysis. Determine the gate efficiency of this stocker (no fly ash). Solution kgs refuse in ash pit per kg of coal = =

225 kgs refuse 2064 kgs refuse

=

0.109 kgs refuse kgs coal

total kgs refuse total kgs refuse

Kgs combustible in refuse per kg coal = =0.109 – 0.10 = .009

kg refuse−kg ash kg coal

kg combustible kg coal

Therefore, grate efficiency from Eq. 10 – 1 p. 296 PPE by ref

Morse is grate efficiency = 1 – c

( Qc Qh )

where: Cref =

kg combustible ∈refuse kg coal burned

Qc = 33,820 kj/kg (assuming combustible in refuse is fixed carbon) Qh = Heat value of coal Thus, kj kj x 0.2388 kg kg kcal 3100 kg

33.820 grate efficiency = 1-.009

33,820 kj/kg 0.2388 kcal/kj 3100 kcal/kg

grate efficiency = 0.98 ans. Problem 10-15

The boiler outlet steam of large unit is 99% dry at 105.55 °C. superheater outlet state is 98.4

kg abs, feedwater 216 cm²

kg abs, 538°C. The unit also includes a steam cm²

reheater, the inlet and outlet states of which are 9.1

kg kg abs, 249°C and 7.0m abs, cm² cm²

371°C, respectively. Compute the percentage of the total heat transfer contributed by each element of the surface. Solution X1 = 99%, p1 = 105.5

kg ab cm²

P3 = 9.1

kg abs cm²

t3 = 249°C P4 = 7.0

kg abs cm²

t4 = 371°C P2 = 98.4

kg abs cm²

tg =538°C at 105.5

kg abs, & x = 99% cm²

h1= 2690 at 98.4

kj kg

kg ab & 538°C cm²

h2 =3478.3 at 9.1

kg ab & 244°C cm² h3 = 2945

at 7.0

kj kg

kj kg

kg ab & 371°C cm² h4 = 3207.9

for feedwater at 216°C, hf= 923

kj kg

kj kg

Thus: Heat absorbed in boiler = h1 – hf = (2690-923) = 1767

kj kg

kj kg

Heat absorbed in superheater = h3 – h1 = 3473.3-2690 = 783.3

kj kg

Heat absorbed in reheater = h4 – h3 = 3207.9-2945 = 262.9 Total Heat Absorbed = 1767 + 783.3 + 262.9 = 2873.2

kj kg

Therefore, Percentage of Total Heat Contributed by Boiler =

1767

2813.2 Superheater = 783

x 100% =62.8 % ans x 100% = 27.8% ans

2813.2 Reheater =

262 2813.2

x 100% - 9.4 % ans

Problem 10-19 The furnace of a certain large boiler is diagrammed in Fig. 10-19, p.357, PPE by Morse. Ap of tube bank = 7.6 m x 3m. Side walls each shielded by 76.2 mm x 3m tubes on 152 mm centers. Front wall refractory, Bridge wall, refractory covered tubes, having Ar = 3.7 m^2. Fuel consumption, 2668 kg per hour good Bituminous Coal; A; F = 12. Preheated air at 93.3°C, QL= 2625 kcal, assume Cp = 0.24 kcal per kg-°C mean

between 15.6°C and 93.3°C, 0.26 between 93.3°C and t°C. Find the temperature of the gases at the furnace, aperture, temperature, t°C, using Eq. 9-26. Solution: 8 = 1.2

Front Wall Refractory Bridge Wall Refractory Cover Wall Calculated for total ERS;

1. Tube Bank: Ar = S. 8 . Ap = 0.85 x 1.2 x 7.6 m x 3m Ar = 23.256 m² 2. Bridge Wall Ar = 3.7 m² given 3. Side Water Walls

1m

Ap = 2 x 20 x 76.2 mm

1000mm Ap= 9.144 m² Therefore Ar= S8Ap Ar = 0.75 π x 0.95 x 9.144 m² Ar = 20.468 m² Total Equivalent Radial Surface A = 23. 256 + 3.7 + 20.468 A = 47.424 m²

x 3m

Assuming 10% ash per kg coal, then gas per kg coal =12 + 1 – 0.1 = 12.09 kg per kg coal Total flue gas for a fuel consumption of 2668 kg per hr = 2668

kg coal kg gas x 12.9 hr kg coal

= 34,417.2

kg gas hr

Assuming 15% of heat value of fuel lost by radiation and combustion: Qs = Ql 9 1-1.5%) Qs = 2625kcal/kg (1-1.5%) Qs = 2585.6 kcal/kg (Available sensible heat) kcal kg coal Therefore, Available sensible heat per kg gas = kg gas kg 12.9 kg coal 2585.6

= 200.4 kcal/kg gas Fraction of sensible heat transmitted by radiation: 1 Ω= Wa √ Cr 1+ 60

kcal kg coal hr = 56.26 Cr = hr−m ² 47.424 m ² 2668

kg coal hr-m^2

1 Therefore, Ω=

1+

12 √ 56.26 60

= 0.4

Sensible heat left in gas at furnace exit = (1-Ω) x available heat/kg gas.

(

= (1-0.4) 200.4

kcal kg gas

)

kcal kg gas

= 120.24

Temperature of gasses at furnace aperture. Qs¹ = Cp (t-93.3) °C Therefore, t =

Qs + 93.3 Cp

kcal kg gas t= = 93.3 kcal 0.26 kg−° C 120.24

t = 555.76°C ans.

Problem 10-23 Estimate the required furnace volume for pulverized coal-fired stem generator where 1375 x 10⁴ kcal per hour heat transfer at 80% thermal efficiency is the expected thermal performance. Boiler has water walls. Ash fusion temperature is 1204°C. Solution Heat Liberated by fuel per hr at 80% thermal efficiency =

1375 x 104 kcal per hr 0.8

= 1719 x 10⁴

kcal hr

From the table 10-1, .314 for pulverized coal, ash fusion temperature of 1204°C and with water walls, Design value of furnace heat release rate = 195,800 kcal per m² per hr.Thus, total volume of furnace required, kcal hr = kcal 195,800 3 m −hr 1719 x 104

=87.8 m³ ans.

Problem 10-25

Find the kcal per hr-m2 heat loss through a furnace wall from 982.2 oC hot face temperature to 18.3 oC ambient temperature. The wall is composed of 228.6 mm fire clay brick, 114.3 mm diatomaceous silica brick, and 203.2 mm red brick.

Solution

From Eq., 10-2; p. 318, heat radiated from the outside wall

qr Kcal =C . Fc ( T o 4−T a4 ) A hr −m 2 Where: c= coefficient of radiation; dependent on outside surface (see values on p.318, text) For plain brick wall, c = 464.1x10-10 Fc = convection factor, average value about 1.5 To = cold face temp Ta = Ambient air temperature, Eq, 10-3; Heat transferred by conduction:

t i−t o t i−t o qc Kcal 4 4 = qr qc C . F c ( T o −T a )= 2 A d 1 d 2 d 3 hr−m with = d1 d2 d3 A A + + + + k 1 k 2 k3 k1 k2 k3 Since ti – to=Ti –To

(

Ti−¿=

d 1 d 2 d3 4 4 + + CFc ( T o −T a ) k1 k2 k3

)

Thus:

[(

Ti−¿=

d1 d 2 d 3 4 4 + + CFc ( T o −T a ) ( 982+ 273) k1 k2 k3

)

]

From Fig. 10-16; p.319, text for fire clay brick: K1= 0.25 at 300oC to 0.4 at 1427oC

For red brick K1= 0.8 at 30oC to 1.2 at 871oC

In the absence of mean temp. value for each material, assume k value as follows:

K1 = 0.3 K2 = 1.3 K3 = 0.8 All in

Kcal hr−m−° C

Thus: d1 = k1

0. 2286 m hr−m 2−° C =0.762 Kcal Kcal 0.3 hr−m−° C

d2 = k2

0.1143 m hr−m2−° C =0.0879 Kcal Kcal 1.3 hr−m−° C

d3 = k3

0.2032 m hr −m2−° C =0.254 Kcal Kcal 0.8 hr−m−° C

Therefore: Ta= 18.3+273 = 291.3oK

1255.2 - To= (0.762+0.0879+0.254) x 464.1x10-10 x 1.5 x (To – 7.2x109)

hr−m2−° C Kcal

hr−m2−° C Kcal

solving for To we have

1255.2-To = 7.685x10-8 (To4-7.2x109) 1255.2-To = 7.685x10-8 To4 – 553.3 7.685x10-8 To4 + To – 1808.5 = 0

Let: f(To) = 7.685x10-8 To4 + To – 1808.5 f1 (To) = 4 x 7.685x10-8 To3+1

By Newron’s method

¿ 2=¿ 1 –

f ( ¿1 ) f 1(¿ 1)

Since To>Ta, Let To1 = 300oK Then: f(To1) = 7.685x10-8 (300)4+ 300 -1808.5 f(To1) = -886 f1(To1)= 4 x 7.685x10-8(300)3 + 1 f1(To1) = 9.3 Thus:

To2 = 300 –

(−886 ) 9.3

To2 = 395oK f(To2) = 7.685x10-8 (395)4+ 395-1808.5 f(To2)= 457.3 f1(To2)= 4 x 7.685x10-8(395)3 + 1 f1(To2)=19.9

therefore To3 = 395

−457.3 19.9

To3 = 372 f(To3) = 7.685x10-8 (372)4+ 372-1808.5 f(to3) = 35.2 – close enough

Take To = 370ok ot to= 97oC

qr

Thus A =464.1 x 1 0

−10

x 1.5 x ¿

qr Kcal =803.5 A hr−m 2 Solve for TA δ TB

q t i−t a = A d1 k1 d1 k1

q A

( )( )

Therefore tA = ti −

[(

Thus tA = (982.2oC) − 803.5

Kcal hr−m2−° C 0.762 Kcal hr−m2

)(

)]

tA = 370 oC for tB: q t A −t B = A d2 k2 d2

( qA )( k )

Therefore tB = tA −

2

[(

tB = 370oC − 803.5 tB = 300oC t B −t 0 q = d3 check for t0: A k3

( )

Kcal hr−m2−° C 0.0879 Kcal hr−m2

)(

)]

d3

( qA )( k )

therefore t0 = tB −

3

t0 = 300 - (803.5) (0.254) t0 = 90oC – close enough

Check for assumed conductivities:

Mean temp. for fre clay brick =

t i +t A 982.2+370 = =676.1 ° C 2 2

At this temp., k1= 0.27 (close) Mean temp for silica brick =

t A + t B 370+300 = =335° C 2 2

Actual K2 = 1.0 Mean temp. for red brick =

t B +t 0 300+96 = =198° C 2 2

We may still use 0.8 for k3 or a lower value! Since the assumed values for the k’s are close enough to the proper values, the computed value of Heat transfer rate is already sufficient, however, if further accuracy is desired, a recumputation will be necessary.

Problem 10-29

A convection superheater with tube elements 38.1 mm x 2.7mm x 6 m long is to be placed in a boile setting where it will be able to superheat steam of 24.6 kg/cm 2 ga,

98% dry to 260oC when Gs= 195kg steam per sec-m2. Coefficient of conductance expected to be realized is 73 Kcal/m2-hr-oC. Wg/Ws= 1.8: Cg= 0.25. Find the required ti. Given 22680 kg per hr rate of steam flow, how many elements should a superheater have?

Solution

a) Temp of steam is at 24.6 kg/cm2 ga, 98% dry= 224.2oC Entalphy of steam is entering a superheater at 24.6 kg/cm 2ga, 98% dryness, h1 = 965=0.98x1.837=2665 kj/kg h2 = 2908 kj/kg let ti1 = temp of gas entering superheater to1 = temp of flue gas leaving superheater heat absorbed by steam = heat given off by flue gas Ws (h2 – h1) = Wg (0.25) (ti1- to1) h2 – h 1 = 0.239

Wg (0.25)(ti 1−¿ 1) Ws

Kcal kj kcal ( 2908−2665 ) =1.8 0.25 ( t i 1−t o 1) ° C kj kg kg ° C

(

ti1- to1 = 12906oC Mean temp difference Θ=

Q UA

Q = Ws (h2 – h1) Q = Ws ( 2908−2665 )

(

Q = 243 Ws U = 79

kj hr

kj kg

) (0.239 kcal kj )

kcal m −hr−° C 2

A = Nt π Do L: Nt = number of tubes

)

But Nt

( π4 D ) Gs ( 3600)=Ws i2

Then : kcal hr ( π D 0 L ) m2 Ws Gs (3600 ) π Di 4

243 Ws0.239 Θ=

kcal (73 m −h−° C) ( 2

kcal hr Θ= ¿ ( 73 ) ( 4 ) (6) ( 0.0381 ) (243)(0. 239)

Θ= 653oC

2

)

(ti¿¿ 1−260)−( ¿1−224.2) ti1−260 ln 1 ¿ −224.2

therefore θ=

(

¿

)

Thus 653 ln

(

ti 1−260 = ti1- to1- 35.8 1 ¿ −224.2

)

But to1= ti1 – 129.06 Then 653 ln

(

ti 1−260 = 129.06 – 35.8 = 93.26 ti 1−353.26

)

93.26 ti1−260 =e 1 653 ti −353.26 ti1 – 260 = 1.1535(ti1 – 353.26) ti1 (1.1535-1) = (1.1535 x 353.26 – 260) ti1= 960oC temperature of flue gas entering the super heater. mass fow per second π b) Number of tubes or elements = Gs x x Di 4 23600

Gs= 195

kg steam given sec−m2

kg hr kg Therefore Nt = 195 ¿ 2 π sec−m ¿ ¿ 4 22680

Nt = 38 tubes ans.

Problem 10-30 In a proposal counter flow steel tube economizer the Wg/Ww ratio is to be 1.5., ti1= 316oC , ti= 32.23oC. It is desired to proportion the economizer so that

t is 54.4 oC.

Find the necessary sqm. Heating surface per 454kg per hr feed water. Gas mass flow, 58,563kg per hr m2 Solution For economizer U = A+ BG From p.330 text, for smooth tube economizers: A = 1.22

kcal m −h−° C 2

kcal B = 0.00075 hr−° C Thus; U = 1.22

kcal m −h−° C 2

kcal

kg

+ (0.00075 hr−° C ¿(58563 m2−hr ¿

U = 45.142

kcal m −h−° C 2

From p.330 Eq. 10-4. Z= 4.45 Ww/Wg - 0.10

( 1.51 )−0.10

where: t1 = gas temp oC

Z= 4.45

t= water temp oC

Z= 2.867

i= in o= out

Then ti1−¿1 =2.867 ¿−ti Then 316oC – t1o = 2.867 (54.4oC) t1o = 316 – 2.867 (54.4) t1o = 160.05

Thus

179.4−77.85 179.4 Θ= ln ⁡ 77.85

(

)

Θ = 121.64 oC Heat absorbed by 454 kg/hr feed water = (454 kg/hr)(1

kcal 54.5oC) kg−° C

kcal hr

= 24,698 Then A=

Q Uθ 24698

A=

(

kcal hr

45.142 kcal (121.64 ° C ) m2−hr −° C

)

A = 4.5m2 ans. Dry flue gas per coal , Wdg

[

11CO 2+ 8CO 2+ 7(N 2+CO) 3 5 C1 + S + S 3( CO 2+CO) 8 8

[

( 11 X 12 ) + ( 8 X 7 ) +7 ( 80+ 1 ) 3 ( 12−1 )

=

=

][

]

3 5 = 0.5543+ 8 0.0132 + 8 (0.132)

[

]

=12.88 kg/kg coal burned

]

(2) Heat loss de to dry flue gas = Wdg x cp x

t

= 12.88 x 0.24 (260- 21.1) = 738.497 kcal/kg coal % heat loss due to dry gas 738.497 x 100 %=11.74 % 6293 (3) Heat loss due to moisture in the fuel = (H2o n fuel) x (1089 – ta + 0.46 tg) = 0.1285 x (1089 – 21.1 + 0.46(260) = 121. 86

kcal lb coal

% heat loss due to moisture =

121.86 x 100 =1.936 % 6293

(4) Heat loss due to H2O from free H2 =

9H ( 1089−ta +0.46 tg ) for tg <302 ° ∁ 100

= 9 (0.04) x [948] = 341

% heat loss de to H2 =

kcal lb coal

341 x 100% = 5.418% 6293

(5) Heat loss due to CO (ncomplete combustion) 10160(CO x C 1) = C O 2+ CO

10160(1 x 0.5543) kcal =433.21 12+ 1 kg coal

% Heat loss due to incomplete combustion, 433.21 = 6293 x 100 %=6.88 % (6) Heat loss due to combustion in fly ash = 0.053067 x 14600 = 774.78kcal per kg % heat loss due to combustion in fly ash =

774.78 x 100 %=12.31 % 6293

(7) Heat loss due to radiation and convection Referring to fig 10-41 p. 353. For 22680kg per hr actual evaporation. 29487 kg/hr capacity and two walls water cooled corresponding heat loss =1.61% (8) Unaccounted for losses by difference = 1.61%

SUMMARY OF HEAT BALANCE Steams 1. Heat absorbed by unit

kcal/kg coal

%

4620

73.41

2. Heat loss due to dry gas

738.494

11.74

3. Heat loss due to H2O in fuel

121.86

1.936

341

5.418

4. Heat loss due to H2 in fuel

5. Heat loss due to CO

433.21

6.88

6. Heat loss due to combustion m fly ash

774.78

12.3

7. Heat loss due to radiation.

--

8. Unaccounted for loses

--

Heal from fuel

7029

100

Problem 10-31 The temperature of a certain convection superheaterrises fom 371 oC at 27216 to 454 oC at 54432 kg per hr steam flow through it. 42kg 1/cm2 abs. Spray water, 93.3oC what constant superheat can be maintained by desuperheat control for a delivery range of 27216 to 54432 kg per hr? What part of the 54432 kg flow will be evaporated spray water? By what percent does the regulated steam volume at 54432 kg per hr flow differ from the super heater discharge volume Solution (a)

The constant superheat can be maintained by desuperheat control is the lower end and is equal to 371oC

(b)

The superheater temp equation may be derived from the characteristics given, and is as follows: Superheater temp, t = 371 +

=371 +

[

454 ° ∁−371° ∁ x ( 1−27216 ) 27216

( L−27216 ) °∁ 204.44

Where L = kg/hr steam flow through superheater

]

454−371 27216

= temp rise above 371oC in superheater,per kg steam flow above 27216kg/hr

Assuming a preliminary value of t = 445oC (this value will be re checked later) Enthalpy of steam leaving superheater at 42 kg/cm 2 abs and 445oC = 790.02 kcal/kg Enthalpy of spray water at 93.33 oC = 93.14 kcal/kg Enthalpy of steam through desuperheater of 42 kg/cm 2 abs and 371oC = 748.7

Heat entering desuperheater = heat leaving L = (790.02) +( 54432 – L ) 93.14 = 54432 x 748.7 L= (790.02 – 93.14) = 54432 (748.7 – 93.14) L=

54432(655.58) 696.86

= 51,207.60 kg/hr

Substituting this value of L in our derived eq. for supereheater temperature. t

=

371 + (1 – 27216 )

=

371 + (51207.60 – 27216 )

=

371 + 117.35 = 430.97OC which is close enough to the assumed value 445oC

Spray water required

= 54432 – L = 54432 – 51207.60 = 3224.4 kg/hr

ans.

(c) Specific volume at 42 kg/cm2 abs and 445oC =0.0765 cu. meter/kg Specific volume at 42 kg/ cm2 abs 371oC = 0.0666 cu. m/kg Volume of flow through superheater = 51207.60 x 0.0765 =3917.38 m 3/hr Volume of regulated steam from desuperheater = 54432 x 0.0765 = 3625.17 m 3/hr Difference = 292.21 m3/hr

% difference in volume of flow =

292.21 x 100 % 3917.38

= 7.46% Problem

ans.

10-38 Calculate a boiler heat balance from these data: Coal Analysis: C, 6074; H2= 4.00; S, 1.32, O2, 8.24; N2, 1.15; H2O, 1285; ash, 11.70; Qh 6293 kcal/per kg. Flue Gas Analysis: CO2, 12; CO, 1; O2, 7; N2, 80. Atmospheric temperature, 21.1ºC; flue gas temperature 260ºC Heat transferred to steam per kg coal, 4620. Refused in pit per ton coal burned 42.9 kg, free of combustible, Fly-ash sampling showed 25% combustible in it. The generating unit has capacity of 29484 kg per hr, was tested at a load of 26680 kg per hr. Two of the furnace side walls are water colored; Solution: 1. % heat absorbed by units:

4620 × 100 %=73.41% 6283

2. Heat loss due to dry flue gas: Ash per metric ton coal = 1000 ×.117=117 kg ¿ 42.2 kg

Refuse in pit/coal Ash in fly ash per ton coal by diff

¿ 159.2

kg m toncoal burned

Total fly ash per ton coal

¿

159.2 636.80 kg = 1−0.25 3 metric ton coal

Combustible on fly ash ¿

636.80 ×.25 3

¿ 53.067 kg per mton coal∨0.053067 kg per kg coal C1 = carbon per kg coal ¿ 0.6074−0.053067 ¿ 0.554333 kg per kg coal

Chapter 11 Steam Prime Movers

Problem 11-1

Find the required bore x stroke of a high speed simple steam engine for direct connection to a 60-cycle, 25-kw generator. Governor can be adjusted to operate in 300-350 rpm range. P1= 7.03 kg/cm2ga, saturated steam; atmospheric exhaust. Clearance, 5%; cutoff, 25%; nme= 0.77. Solution:

VC= 0.05 VD r=

v2 , ratio of expansion v1

r=

V C +V D V C +0.25 V D

Thus,

¿ ¿

1.05 0.30

r =3.5 P1=7.03

kg kg ga+ 1.034 2 2 cm cm

P1=7.0634

kg |.| cm2

P3=1.0334

kg |.| cm2

Diagram Factor, DF: From table 11-1, p. 368, PPE by Morse

( 0.05+1 ) V D ( 0.05+0.25 ) V D

Take average value for High speed, simple automatic 1 DF = ( 0.70+0.85 )=0.775 2 Then, Pmep =DF∗Pmi 1+ lnr −P 3 r

[ ( ) ] [ ( )

¿ DF P1

Pmep =0.775 7.0634 ¿ 2.7725

1+ln 3.5 −1.0334 3.5

]

kg |.| cm2

Therefore the engine could be: 30 cm× 35 cm× 300 rpm ANS .

Problem 11-7 Given a 30.5 x 45.7 cm controlled compression horizontal unaflow having ŋ e based on 1 hp = 0.70 when operated between 8.8

kg gm . and atmosphere; steam cm2

initially dry saturated. Piston rod 44.5 mm; tail rod, 31.8 mm; ŋ m = 0.90. When this is connected to a standard 240 rpm NEMA a-c generator, what is the proper KW rating? Find Wk. Consider rated load to be such that r = 4.0. Solution: a.

P1=8.8

kg kg ga+1.0334 2 2 cm cm P1=9.8334

kg |.| cm2

Exhaust Pressure:

P2=1.0334

kg |.| cm 2

Ratio of expansion, r = 4 (given) DIAGRAM FACTOR: From table 11-1, p. 368, text. For controlled compression, non-condensing, DF= 0.80 to 0.85 Take: DF= 0.825 (average)

[

Pmep =DF P1

( 1+ ln r ) −P2 r

[

¿ 0.825 9.8334

Pmep =3.9872

]

( 1+4ln 4 )−1.0334 ]

kg cm2

But from equation 11-6, pg. 367, PPE by Morse, we have IHP=

Pmep LεA N n 449702 39872

¿

kg rev ( 45.7 cm ) 240 2 min cm π 2 2 × [ (30.5 cm ) −( 9.45 ) ] kg−cm 4 449702 hp−min

(

)

+π ( 30.5 cm)2− (3.18 )2 4 IHP=139.8 hp BHP=ŋm IHP ∴ BHP=0.9 ×139.8 hp ¿ 125.8 hp

]

To determine the kw rating of the generator, the efficiency of the generator must have first to be known. Assume generator efficiency of 85%∴The corresponding generator output

(

¿ 125.8 hp 0.746

kw 0.85 hp

)

¿ 79.8 kw=80 kw From Fig. A-15, appendix, text, for 240 RPM, generator efficiency: 50 kw- 86.13% 100 kw- 88% By interpolation: 50 86.3 ❑50 30 80 m ŋ=86.3 100 88

[[

] ]

1.7

∴

ŋ−86.3 30 = 1.7 50

ŋ=86.3+1.7

( 3050 )

ŋ=87.32 % Less standard deductions, table 6-3, p. 185, text ∴ ngen =( 87.32−4 )=83.32% Then, kw rating of generator ¿ 125.8 ×0.746 × 0.8332 ¿ 78.2 kw ANS . b. Engine Efficiency

ŋe =

632.4 W s ( h 1−h2 )

Where: Ws= steam rate, based on Ihp, kg per Ihp-hr h1 = enthalpy at intake = 2775.4 kJ/kg h2 = enthalpy at exhaust, after isentropic expansion =2384.6 kJ/kg Thus: W s=

632.4 ŋe ( h 1−h2 )

kcal hp−hr W S= kJ kcal 0.8332 (2775.4−23846 ) ×0.239 kg kJ 632.4

¿ 8.125

kg hp−hr

Combined Steam Rate: W

k=¿

8.125 kg =14.52 ANS .¿ 0.9×0.8332 ×.746 kw−hr

Problem 11-18 How much steam will 76-hp mechanical drive turbine need per hr? Steam supplied ay 11.6 kg/cm2 ga saturated. Exhaust, 0.35 kg/cm 2 ga. Single stage, 1800 rpm type. Solution: From fig. 11-16, p.382, PPE by Morse, Brake Engine Efficiency, ŋeb for 1800 rpm single stage turbine of 76-hp rating is 23%.

For steam at ( 11.16+ 1.0334 ) h1 =2787

kg |.|saturated cm 2

kJ kg

For Exhaust steam after isentropic expansion: h2 =2408 ŋeb =

kJ kg

632.4 wb ( h1−h 2 )

Thus, wb=

632.4 ŋeb ( h1−h 2 )

kcal bhp−hr ¿ kJ kcal 0.23 (2787−2408 ) × ( 0.239 ) kg kJ 632.4

¿ 30.35

kg bhp−hr

Total Steam Required: ¿ 30.35

kg (76 hp ) bhp−hr

¿ 2307

kg hr

Problem 11-1 A large, 1800 rpm, non condensing turbine uses 158, 760 kg steam per hr at 3 20,280 shaft HP 1, P1 = 28.1 kg per cm ab, 315.6 ° c

, 2.1 kg /cmgas exhaust. Is

the developed engine efficiency in agreement with Fig. 11-17? Assume nm= 0.97 Solution: P1 28.1 = P2 2.1+ 1.0334 = 8.97 Specific volume at 28.1

kg | | ¿ cm₃

m₂ = 0.0023 kg

TOTAL THROTTLE FLOW kg m₃ = 158,760 hr x 0.0923 kg m₃ = 14,652 hr From Fig 11-77, p. 383 text Engine efficiency = 79.5% Correction for pressure ratio other than 6.0 With

P1 = 8.97, add 0.7% P2

(From Fig 11-77, p. 383 text) Thus, n s=79.5 + 0.7 =80.2% ans At 28.1

kg abs, 315.6 °C, cm2

kJ h1 = 3040.4 kg

At 2.1

kg ga, after isentropic expansion from exhaust cm2 kJ h2 = 2593.9 kg

BREAK STEAM RATE: kg hr = kg 7,828 hp−hr 158,760

Indicated engine efficiency from actual test 2658 = 7,828

kJ hp−hr x 100 %

kg kJ x 0.97(3040.4−2593.9) hp−hr kg

= 79.2%

ans.

Problem 11-20 Estimate the stage efficiency in the dry steam stages of a pressure compounded impulse turbine of average economy. Do the same for a reaction turbine of high commercial efficiency. Solution: a.) From table 11-4. P.356, PPE by Morse, for pressure compounded turbine ( curve A) in Fig. 11-19 Blading velocity ratios used in practice is 0.40 to 0.45 but take the average value. =

1 ( 0.40 + 0.45 ) 2

= 0.425

From Fig. 11-19, P. 366, text: Blading efficiency = 0.89 thus, stage efficiency , =0.94 x Blading efficiency = 2% to 3% (rotation and leakage allowance) Take rotation and leakage allowance 3.5% Then, n s = 0.94 x 0.89 = 0.033 n s = 0.8016 or

n s = 80.16%

ans.

b.) For reaction turbine Blading Velocity ratios used in practice = 0.60 to 0.80 table 11-4 P. 386 text; at high efficiency, take 0.80 from Fig. 11-19, P. 386; blading efficiency = 0.95 assuming 2% rotation losses: n s= 0.94 x 0.95 -,0.02 n s = 0.873 Or n s = 87.3%

Problem 11-29 Find the speed charger settings in terms of no-load rpm, that will enable machines A & B to divide a 10,000 kw load at 60 cps as follows: A. of 5,000-kw capacity; 1.5% regulation; takes 3000kw; B.) 8000kw capacity, 1.0% regulations; takes; 7000 kw. Both have 4-pole alternators. Solution: For no load speeds:

rpm =

120 f 120(60) = = 1800 rpm n 4

PERCENT REGULATION % reg. = Thus. SNL = 1800 rpm. (1 + %reg.) For A: SNLₐ = 1800 rpm (1 + 0.015) = 1827 rpm For B: SNLb = 1800 rpm (1 + 0.10) = 1818 rpm

For machine A. Y A 1827−1800 = 3 5

SNL−1800 1800

3 [( 1827 rpm )−( 1800 rpm ) ] 5

YA

=

YA

= 16.2

Thus, new speed charger setting for machine A at no load to carry 3000kw at 1800 rpm, = 1800 + 16.2 = 1816.2 rpm

ans.

For Machine B. Y B 1818−1800 = 7 8 7 (18) 8

YB

=

YB

= 15.75

Speed charger setting for B at no load to carry 7000kw at 1800 rpm = 1800 + 15.75 = 1815.75 rpm

ans.

Problem 11-30 Three turbo-alternators are sharing a 33,000kw, 60 cycles station load. All are normal 1800-rpm units. Were the load then to decrease to 22,000 kw with no change in governor setting. How would the load be divided between the three machines and what would the frequency be? Unit

Speed Regulation

Capacity

Load at 60 CPS

A

0.9%

15,000kw

12,000kw

B

1.2%

15,000kw

14,000kw

C

1.9%

10,000kw

7,000kw

Solution: For no load speeds: SNL = 1800 rpm (1 = reg.) Thus, SNLₐ = 1800 rpm (1 + 0.9%) = 1816.2 rpm SNLₐ = 1800 rpm (1 + 1.2%) = 1821.6 rpm SNLₐ = 1800 rpm (1 + 1.9) = 1834.2 rpm NO LOAD SETTINGS @ 60 CYCLES FOR GIVEN LOADS

For A: SNL A−1800 15 = YA−1800 12 Y A −1800=

12 (SNL A −1800) 15

Y A −1800=

12 ¿ 15 )

Y A = 1812.96 rpm For B: Y B−1800 14 = SNL B −1800 15 Y B =1800+

14 ( SNLB −1800) 15

¿ 1800+

14 ¿ 15 )

Y B = 1820.2 rpm For C: Y c −1800 7 = SNLC −1800 10 Y C =1800+

¿ 1800+

7 ( SNLC −1800) 10

7 ¿ 10 )

Y C = 1823.94 rpm

When the load is decreased to 22,000 kw at the same governor setting, the speed will have to increased to reduce the loads in each machine. The new speed is represented by the horizontal dashed line N 1 (above figure and the new loads by X a, X b, X c. Note that X a + X b+ X c = 22,000kw for N 1. By Similar Triangles: Y A−N 1 Y A −1800 − XA 12 Y B −N 1 Y B−1800 − XB 14 Y C −N 1 Y C −1800 − XC 7

Thus, 12(Y A −N 1) X A= Y A −1800 14 ( Y B−N 1) X B= Y B−1800 XC =

7 (Y C −N 1 ) Y C −1800

Substituting Values of Y A , Y B ,∧Y C we have. 12(1812.96−N 1) = 1678.67 – 0.926 N 1 X A= 1812.96−1800 X B=

14(1820.2−N 1) = 1261.52 - 0.693 N 1 1820.2−1800

7 (1823.94−N 1) = 533.32 - 0.292 N 1 XC = 1823.94−1800

Since X a + X b+ X c = 22,000kw ∕ 1000 (1678 – 0.926 N 1) + 1261.52 – 0.693 N 1 ¿+ ( 533.32−0.292 N 1 ) =

22,000 1000

3473.51 – 1.911 N 1=22 N 1=¿ 3473.51

−22 1.911

N 1=¿ 1806.13 rpm Thus,

X A =1678.67−0.926 ( 1806.13 )=6.196 MW X B=1261.52−0.693 (1806.13 )=9.873 MW X C =533.32−0.292 ( 1806.13 )=5.932 MW

For New Loads: A: 6,196 kw B: 9,873 kw C: 5,932 kw Total: 22,000 kw New Frequency: f 1=

1806.13 (60 cps ) 1800

f 1=60.2 cps

Chapter 12 The Gas Loop

ans.

Problem 12-2 All open coal yard has dimensions 36.6 m x 15.2 m. How many tons of coal can be stockfield on it to a depth of 3.05 m side slopes 35ºC?

Value of X: S sin 35º = 3.05 m S=

3.05 m sin 35º

Thus: x = 36.6 m – 2 (5 cos 35º) y = 36.6 m – 2 (3.05 m/sin 35º) cos 35º x = 27.89 m

b = Area of Top = xy b = (27.89)(6.49) m2

B = Area of Base B = (36.6 m) (15 – 2 m)

B = 556.32 sq. m.

Volume of Frustum; V

V=

1 (B + b + √ Bb ) 3h

1 = (3.05 m)(556.32m2 + 181 m2)√ ( 556.32 ) m2 (181 m2 ) 3 V = 1072.2 m3, average value.

Total Weight = ȣy ȣ = Specific Weight → 800 to 880 kg/m3 Use 840 kg/m3 average value. Total Weight = 840 kg/m3 (1072.2 m3) = 900 650 kg = 900.65tons

Problem 12-4

Find the length of a suspension bunker to contain 181 tons of coal without supercharge: width, 4.6m, depth, 4.3m. What width of conveyor belt would have the capacity to fill it in 2hours, running at 45.7 m/min? Solution:

Volume =

( 181tons ) (1000

kg ) ton

800 kg/m 3 V = 226.25 m3 Using a Berquist Section (See Figure12-5) Level Capacity =

5 BD; m3 /m Length 8

Where: B = Width D = Depth ⸫ Level capacity =

5 (4.6 m)(4.3 m) 8

a. Length of Bunker Required 226.25 m3 L= 5 ( 4.6 m)(4.3 m) 8 L = 18.30 m b. From Table 12.2, p.452, Capacity of Conveyor Belt: = 0.000404 Sb2 tons/hr. S = Lineal speed of conveyor b = Belt width

⸫ 181tons/2hrs. = (0.000404) 45.7 m/min. (b)2

1

181 ¿ ¿2 b =( ( 2 ) (45.7)(0.000404) b = 70 cm

ans.

Problem 12-6 Calculate the principal data for a belt conveyor to transport 18 tons/hr at 20º. Length to be sufficient to give 33.5 m rise. 1750 rpm motor, V-Belt drive to helical gear speed reducer. Belt speed, 61 m/min. Discharge over tail pulley.

From Table 12-2, p. 452, PPE by Morse, Capacity of Belt Conveyor = 0.000404 tons/hr. S = Lineal speed of belt = 61 m/mm b = Belt width in inches 18 Tons = (0.000404 tons/hr) (61 m/min) (b2) b2 =

18 = 730.4 cm2 (0.000404)( 61) √

b = 27 cm From Table12-3, p. 453, PPE by Morse, take standard belt width of 35.6 cm; k = 22.3 Tripper Hp = 0.61 + 0.0039 T Belt type: 5 ply, 28 duck

Conveyor Length L = 33.5m/sin 20º = 98 m From Table 12-2; p.452, text; Horse Power Required to Drive Shaft (HPDS) = HpDS =

L+ 45.72 [0.06 KS + T] 9000 98+45.72 [ 0.06 x 22.3 x 61+18 ] 9000

= 1.6 Hp Power of Lifting =

Th sine of angle of inclination =(T . S)( ) 273.7 273.7

= (18) ¿¿ = 1.37 Tripper Power = 0.61 + 0.0039 x 18 = 0.68 Actual Power required to drive conveyor shaft = 1.6 + 1.37 + 0.68 = 3.65 Hp As given motor drives helical gear reducer through V-belts, assume further that the gear reducer drives the conveyor through roller chains. Then there will be 1 pair of V-belt sheaves 1 pair of sprocket 1 set of reducer For which we should add 5% of each = 15% total

Then Hp required = 3.65 (1.15) = 4.2 Hp From Table 12-1, p. 450, PPE by Morse, Min. pulley for 28cm. 6 ply is 76.2 Carrier spacing = 1.22 m Idler spacing = 3.05 m Wt. of material/ply/cm Width/m = 0.0119 kg For 1.6mm Rubber cover = 0.0102 kg Length of belt = 2 x 98 m = 196 m Total Wt of belt with 2 Rubber Belts = (6 x 0.0119 + 2 x 0.0102) kg/cm-m

(35.6 cm x 196 m)

= 464.6 kg Pulley Speed = 61 m/min = ℼ · Δ · N RPM, N =

61 m/min = 28.3 RPM ℼ ( 0.686 ) m

Overall speed reduction from 1750 RPM =

1750 = 61 : 8 : 1 28.3

Summary of Specifications: Belt: 35.6 cm wide x 6 ply x 28 ducks with 21.6 mm Rubber covers carrier spring = 1.22 m on centers Idler spacing = 3.05 m on centers drive : 15HP 62 : 1 reduction. Length of conveyor = 196 m.

Problem 12-7 Calculate the principal data for a 10.7 m horizontal screw conveyor for 2.7 tons coal per hr. about 60 RPM. Worm gear speed reducer. 1750-RPM Motor. Solution Convert 2.7 tons/hr to m/min (2.7)(909) =40.905 kg/min 60 At 60 RPM, kg/min =

40.905 =0.68175 kg /min 60

Since the 60 RPM given is approximately, the RPM will be made to conform to a standard screw. Referring to table 2-2 p. 452, for 15-24 cm screw. Capacity = 0.5454 kg/REV RPM = Corresponding Speed Reduction =

40.905 =75 RPM 0.5454

1750 RPM =23.33:1 75 RPM

For 20.32 Screw, Capacity = 1.5 kg/Rev. RPM = Speed Reduction =

40.905 = 27.27 RPM 1.5

1750 = 64:1 27.5

64:1 is too high. Ratio of 23 : 33 : 1 is better speed reduction for worm gear reducer = 21

From Table 12-2 p. 452, Hp to drive shaft = 0.001L (1.331 N + 7.333T) L = Conveyor Length = 10.67 M N = Drive sgaft speed = 83.3 RPM T = Capacity = 3 tons/hr ⸫HP = (0.001)(10.67) [1.331 (83.3) + 7.333 (3)] = 1.38 Hp Since reducer is worm gear type,add 15%. For losses (refer footnote to table 122, p. 452) Hp of Motor = (1.38)(1.15) = 1.6 say

2Hp next standard size

Specification: 15.24 cm screw 10.67 m long 2 Hp motor worm gear reducer with speed ratio of 23 : 33 : 1

Problem 12-9 The individual buckets of a vertical elevator carry 3.64 kg coal and are spaced 30.5 cm apart on the chain. Sprocket wheels 61 cm diameter. Chain speed 79.2

m . Height min

between sprocket 9.1 m drive 1750 RPM motor through sprockets and steel roller chain. Specify the drive and diagram the conveyor. From Table 12-2, P – 452, Maximum capacity T = 0.07

Wbs tons ; b hr

Wb

= Individual bucket load

S

= Lineal speed

B

= Bucket Pitch

T = 0.07 (3.6) (79.2) (0.305) T = 65.44

tons hr

kg 1 hr 1000 ( 9,1+ 0.61 ) m (65.44 tons )( )( hr ton 60 min ) Theoretical Hoisting Power

=

(

kg−m 6116.3 kw−min

)

kw HP kw 0.746 HP 0.746

( )

=

1.73 kw From Table 12-2, p. 452, Hp to drive conveyor drive shaft = 1.8 (Theoretical Hoisting Power) = 1.8 (2.32) = 4.18 Hp Speed of Conveyor Sprocket

N=

s 2 πr

m min N= rad 0.61 m 2π rev 2 79.2

(

)

N = 41.3 RPM Speed Reduction =

1750 RPM =42.4 :1 41.3 RPM

This drive using all chains will require at least 3 pairs of sprockets using steel roller chains, add 5% per pair for losses.

∴ Required Motor Hp = (1.15)(4.18 Hp) = 4.81 Hp, Use 5 Hp Final specification of drive; 5 Hp motor with total speed reduction of 42.5 Hp

Problem 12-11 The power plant with the load given in Prob. 3, Chapter 2, is a steam plant burning no. 6 Fuel oil, 15 ° Be, of SSF 250 at 15 ° C. Rotary-cup type burner. Boiler efficiency, 80%. Plant steam rate 10

kg . Factor of evaporation, 1.15. Find maximum kw−hr

rate of fuel flow, daily consumption preheat temperature.

Solution: For Problem 3, Chapter 2 Energy produced/day = 4850 kw – hr Maximum Demand = 365 kw From EQN 5 – 8, p. 124, PPE by Morse, Qh = 42450 – 93 (Be’ + 10); 5 per g at 15 ° C. = 42450 – 93 (Be’ + 10) = 40125

J cal kCal or 4649400 = 46494 g g g

Using Eqn 5 – 2, p.119,

∴

° Be ’

=

140 −130 S . G at 15.6 /15.6

S.G

=

140 +15=0.965Factors of Evaporation 130

=

Heat Absorbed Kg. steam kCal 745.6 kg Heat absorbed per kg steam =

(1.15) (745.6)

=

857.44

kCal kg

Heat absorbed/ kw – hr at a steam rate of 10

kg kw−hr

kCal kw−hr

= (857.44)(10)(100) = 857440

Assuming boiler efficiency at all loads constant at the given efficiency of 80% Then

kg Fule = kW −hr

857440 kg =0.023 kw−hr 4494 x 8(100)

For maximum demand of 365 kw, kg of fuel at maximum demand min

(

= (365) 0.23

= 0.1399

kw kw−hr

kg min

hr )( 601 min )

gallon Maximum = min

kg min .1399 = (8.39)( S−G) ( 8.33 ) (.965)

= 1.26GPM For total output of 4850

kw−hrs day

Kg fuel used = (4850)(0.023) = 115.55

Gas Fuel Since

=

kg day

115.55 Gal =14.37 day ( 0.965 ) ( 8.33 )

SSF = 600 SSU 250 SSF =

250 x 600 SSU 62

= 2420 SSU From Figure 12-21, p. 460, PPE by Morse, for a viscosity rating of 2420 SSU at 50 ° C , Preheat Temperature = 67.78°

ans.

Problem 12-12 A No. 5 residual oil rated 250 SSU@ 37.8° is to be burned in a centrifugal type atomizer. Will it require to be burned in a pressure atomizing burner, how many watts electrical input would be required for the preheater when flow is 3.79 S.G of 0.93 Sp. Heat, 0.55 Solution:

kCal kg per ° C

liters ? Assume min

a)

From Figure 12 – 21, p. 460 At 250 SSU and 37.8 ° C, no preheating will be necessary for a centrifugal type atomizer

b)

For a pressure atomizing burner and again using Fig. 12 – 21, the oil would have to be preheated to 44.4 ° C 1GPM at 0.93 SG = 8.39 x 0.422 = 3.52

To heat 3.52

kg min

kg min

kg oil from a tank temperature of 65.55° C to 37.8° C min

will require (3.52) (0.55) (37.8 – 15.55) = 929

Kj min

This is equivalent to (929)(60) =3.91 kw 3913 For heater 90% efficient, required electrical output is 3.90 + 0.90 = 4.34 kw

Problem 12- 13 Find the steam consumed/hr to preheat heavy oil for burning in a steam atomize/ 1 GPM flow. Assume specific gravity – 0.97, steam at 586.36 kg/cm 2 saturated, oil in tank heated to minimum required temperature for pumping. Viscosimeter test on oil SSU 1150 at 43.3 ̊C.

Solution:

Using Fig. 12-21, for SSU at 43.3 ̊C minimum temperature for easy pumping = 22.22 ̊C. From the same figure, the required preheat temperature for atomization is 60 ̊C. At 1 GPM flow, wt of oil = 1 x 8.33 Sp. G. = 3.66 kg/ min In 1 hour; the fuel flow = 3.66 kg. Area of Samplling Nozzle = πD2/4 =

Area of the Orifice =

π (3.5) 2 = 9.68 cm2 4

π ( D) 2 π x L = = 5.06 cm2 4 4

Rate of flow into nozzle supply = Velocity of Gas x Area of Nozzle

=

(1353.66

m )(9.68 cm2 ) min = 84538 m3 / min 0.1550

Assuming the volume of the gas is inversely proportional to the absolute temperature, then at 143.33 C ̊ .

Rate of flow through orifice =

(84538) ( 143.33+273 287.78+273 )

= 62762.056 m3/min

Also Orifice Flow = Orifice Area x Coefficient x Ideal Velocity Orifice Area = 0.059 m2 as previously computed Coefficient = 0.65 as given Ideal Velocity = √ 29 H H = Head of gas through orifice

kg ) = m3 y 0.877 (3.28) (1000

= 969

Y = Differential reading on the manometer in meter of water 0.877 = ∫ of Flue gas at 143.33 Total Heat Absorbed by the Fuel = WCpDt Assuming Cp. = 0.3861 Heat Absorbed = 3.66 kg x 0.386 (60 ̊C – 22.22 ̊C) = 75781.08 kJ/hr Assuming that the steam condenses @ saturation temperature. Then heat given up/kg = Latent Heat = 7716 kJ/kg @ 14.1 kg/cm 2 Total steam consumed for heating =

75781.08 7716

= 9.82 kg/hr

Problem 12- 14 Given following Bagtest sampler data, find required differential manometer pressure for a test spot where pitot-static in the duct gives a reading of 1.65 cm water. Sampling nozzle, 3.5 cm, orifice dia. 2.5 cm; coefficient, 0.65. Flue gas 287. 78 ̊C, sample cooled to 143.33 ̊C. Solution:

Referring to Fig. 12-14 p. 495 and for coal fire density of flue gas at 287.78 C ̊

= 0.665 kg per

Density of Flue Gas at 143.33 ̊C = 0.877 kg/m3 Heat of flue gas at point of pitot static location corresponding to 1.65 cm water H=

1.65 28.4 x = 25.15 m of gas 3.66 0.665

Assuming pitot static coefficient of 1, Velocity of Gas = √ 29 H = √ 19.62(25.15) = 1353.66 m/ min.

Substitute all the known values 0.573 =

(

=

0.785 x 0.65 x √ 64.4 x 96 y 144 0. 573 0.278

2

)

= 0.11 m of water

Problem 12- 15 The gas velocity through a dust sampler having been properly adjusted it was operated 20 min @ 0.03 m water pressure across the 0.02 m θ orifice and collected 0.06 kg. Dust. The sampler nozzle is 0.027 m θ. Sample temperature, 115.55 C ̊ ; Flue gas temperature, 232.22 ̊C. Assume orifice coefficient = 0.61. Find gas loading in grains per cu. m. Solution: From Fig. 12-44, p. 495, Text, Flue gas ρ and @ 232.22 C ̊ = 0.73 kg/m3 Flue gas and @ 115.55 ̊C = 0.94 kg/m 3 Orifice area = 0.0375 sq. m. Sampler nozzle area = 0.065 sq. m. @ .03 m pressure across the orifice, the flu gas ρ of 0.94 kg/m 3 @ 115.55 C ̊ . Head in m of gas @ orifice

H=

1.221 62.4 x 3.66 0.94

= 33.384 meters Ideal velocity of gas through orifice = √ 2GH = √ 33.384 ( 62.4 ) = 1531 m/ min Orifice Flow = Area x Coefficient x Ideal Velocity = (0.61) (1531) (11.06 x w-3) = (10.68 CFM) (0.028) = 0.299 m3/ min Rate of flow through nozzle (Assuming volume proportional to absolute temperature) =

(10.68) (0.028) ( 232.22+273.00 115.55+ 273.00 )

= (13.884 CFM) (0.028) = 0.389 m3/ min Problem 12 – 18 An air preheater consist of 6.3 mm flat plates 91.4 cm wide by h. m high plates spaced 31.8 m ° C both edge passages are for air. Mass flow, 26860 45359

kg gas . Flow: gas, m 2−hr

kg kg ; air 38555 . Gas in at 371.1° C , out at to. Air in at 15.6° C; out at 148.9° C hr hr

a) Find t0’ if Cp = is 24 for air and 0.26 for gas b) Find the necessary heating (value) surface using Eq. 12 – 1.

Solution: a)

Heat rejected by flue gas = Heat absorbed by air CgMg (ti’ – t0’) = CaMa (t0 – t1) kg (45359 ¿(371.1 - t ’)° C = (0.24)(38555)(148.9 – 15.6) ( 0.26kg °kCal C ) hr 0

∴ t0’ = 266.5° C

b)

ans.

Over all heat transfer coefficient U = A + BG For Air preheater, A = 1.95 B = 0.00045, Text p.472 ∴ U = 1.95 + (0.00045)(26860) U = 14.037

kCal hr−m2−° C

Mean Temperature Difference

371.1 ℃ 148.9 ℃

θ

Gas 2

Air

θ

1

15.6 ℃

θ 1 = 266.5 – 15.6

= 250 g℃

θ 2 = 371.1 – 148.g

= 222.2℃

θ1−θ 2 250.9−222.2 = θm = θ1 250.9 ln ln θ2 222.2

( )

(

)

∴ θm = 236.26℃

Total Heat Transferred, Q = MaCa ∆ ta Q = 38555

kg kCal 0.24 ( 148.9−15.6 ) ℃ hr kg ℃

(

Q = 1233450 From Q = UAθ m

)

kCal hr ,

A=

1233480 A=

kCal hr

kCal (236.26 ℃ ) (14.037 hr−m ℃) 2

Problem

Q Uθm

12 – 21 Test data on a 3047.12 sq.m plate type reheater are: Air flow, 157045.45 37.78 ℃ ; gas, in 336.67 ℃ ; out, 182.22℃ , all ℃ . Solution: θ 1 = 182.22 – 37.78

= 126.66 ℃

θ 2 = 336.67 – 226.11 = 92.79 ℃

Mean Temperature Difference T

336.67

Gas

O2 226.11

182.22

θ1 Air

Flow

θ1−θ 2 θ1 ln θ2

θm =

( )

126.66−92.78 126.66 = ln 92.78

(

)

= 107.78 ℃ Total Heat Transferred = Q

kg air in; hr

Q = 117,649,080 Also

kJ hr

= U A θm

Q 117649080

= U x 3047.12 x 107.78 U=

117649080 ( 3047.12 ) 107.78

= 358.23

kJ hr−sq .−℃

Over all Coefficient of Condunctance

ans.

Problem 12 – 24 What pressure, cm water, will give air at 93.33 ℃ and 10.16 cm. static pressure, a velocity increase from 0 to 731.7

m ? min

Solution: Absolute static pressure on the air = 101.325 = 30.39

kn ( 13.378 ) + ( 0.10 )( 19.02 ) m2

kg m2

Using the characteristics gas equation PV = MRT Density =

w P = at 93.33 ℃ v M

w 30.39 = v ( 287.08 ) ( 93.33+273 ) −4 ∴ Density of Air = 2.9 x 10

kg m3

Assuming Velocity Coefficient c = 1 and with initial velocity V 1 = 0, then final velocity V2 = √ 2 gh=43902.4

m hr

Required head for this velocity, 2 meter V 2 ( 439024 ) hr H= = 2g ( 2.54 x 1010 )

= 7.56 meter air Friction Factor f = 1.15 x 0.014 for rectangular section Flue gas against steel (see p. 477)

Then, Substituting values, we have D4 =

2 0.769 1.15 x 0.94 ( 20.67 ) ( 30.48) 10 2 ( 9.8 ) (0.758)

[

D4 = 1.085 cm H20

Problem 12-27

ans.

]

Flue gas (R = 51.5) enters a tapering breeching at 15.24 m/sec 5.588 cm water vacuum, 304.44°C. Breeching is short /and well insulated. It tapers from 1.718 m at inlet to 3.02 m* at exit. Estimate the exit vacuum, assuming incompressible flow. Solution: Static Pressure of Gas at 5.588 cm vacuum

¿ 101.325

kg 5.588 0.192 −( x kg/m 2 ) 2 1000 1000 m ¿ 10316.8

kg m2

Density of this pressure and temperature 304.44°C p2 =10316.8 / (282.18 )( 304.44+ 273 ) rt ¿ 0.06

kg m3

Since breeching is short and well insulated, heat loss and pressure loss will be negligible. Density of the gas will remain constant and any change in head will be due to the change in velocity. Velocity is inversely proportional to cross sectional area for incompressible fluids. Velocity Exit = (5.588) ¿ 15.24 m/sec

1.718 3.02

= 8.69 m/sec Head EQN for inlet and exit of duct (Pressure Head + Velocity Head) at inlet = (Pressumed Head + Velocity Head) at Exit

P1 V 21 P2 V 22 + = + W1 2g W 2 2g P1 P2 V 21−V 22 − = m ; Difference of Pressure Between Exit and Inlet Or W1 W 2 2g

(

(

)

P1 P 2 15.24 m/sec 2−8.69 m/ sec 2 − = W1 W 2 2( 9.81)

)

= 7.96 meters of gas LetY w = cm water column equivalent to 26.1 ft or 7.96 m Yw 0,192cm = =7.96 m 12 ( 1000 ) 0.6396 kg /m3 Y w=

( 7.96 ) (12)(0.6396)(1000) 0,192 cm

Y w =6.06 x 10− 4 cm Since pressure @ exit is 6.06 x 10−14 cm higher that at inlet Vacuum at Exit = Vacuum at Inlet = 6.06 x 10−4 cm ¿ 5.588 cm−6.06 x 10−4 cm = 29.828 cm water

Problem 12-30 The products of combustion of 635 kg Pocahontas coal/hr pass through breeching at 260°C and 7.62 m/sec. Excess air, 50%. Assume square section breeching and find its dimensions. Sea level. Solution

From table 5-4, p. 127, Text, Ultimate analysis of Pocahontas coal is as follows: (Coal No. 5) Sulfur ………………………………………… 0.55% Hydrogen ………………………………….. 4.5% Carbon ………………………………………. 84.02% Nitrogen ……………………………………. 1 17% Oxygen ……………………………………… 6.03% Ash …………………………………………... 3.73%

Theoretical air required for combustion: ¿ 11.5 C+ 34.5¿) + 4.3 S ¿ 11.5 ( 0.8402 ) +34.5(0.045−

0.0603 )+ 4.3 (0.0055) 8

=11.0 kg/kg coal

With 50% Excess Air, Total Air = 11.0 x 1.50−16.5 kg

kg air kg coal

Fraction of fuel in flue gas assuming no combustible in refuse = 1 – Ash = 1 – 0.0373 = 0.9627 kg/kg coal Flue gas formed, = 16.50 + 0.9627

= 17.4627 kg per kg coal From Figure 12-44, p. 495, density of flue gas – coal fired at 260°C and sea level ¿ 0.68

kg m3

∴ Total volume of combustion products gas ) (¿ 635 kghrcoal )(17.4627 kgkgcoal 0.68 kg/m

3

¿ 16310 m 3 /hr Cross sectional of Breeching, =

Volume of Gas Velocity of Gas

m3 hr cm 7.62 ¿¿ sec 16310

(

)

Dimensions if square = 77cm x 77cm Recommended speed then at 107.22°C = (13.64 m/sec) (0.776) = 10.58 m/sec At this velocity the required duct section is

( 3528000CFM )( 0.028) = m 10.58 (3600) sec

(

)

¿ 2.60 m2 For a square duct, dimensions of one side ¿ √ 2.60 m2

= 1.6 m or 162.56 cm Required section is therefore 162.56 cm x 162.56 cm b.) Using Eqn. 12-3, p.477, Frictional loss in duct D4 =

d f .V2.H ( ) 10 2. g . R

d=0.94

kg m3

= Density

F = Friction Factor V = Gas Velocity 2

2

V =( 10.58 )

= (0.005) (0.127cm) = 10.58 m/sec = 111.94

R = Hydraulic Radius ¿

(1.6)(1.6) Area = Perimeter 4 (1.6) = 0.405 m

H = Length of duct = 27.13 m ∴ D 4=

( 0.94 ) ¿ 10

= 0.204 cm ans.

Problem 12-33 All air ducts conveys 33181.82 kg air/hr between an air heater and a pulverized coal burner. Air temperature 107.22°C delivering to a burner at 5.08 cm water. Length 23.42 m with two 90° bends, 0.038 m magnesia heat insulation. a.) Find dimensions of duct, if square

b.) Find total air head at heater c.) Estimate heat loss per hour. Solution: a) Static pressure of air at burner

¿

(

5.08 cm cm 1000 m

¿ 10423.7

()

0

1000 .192cm + ( 101.325 ) ❑

)

kg sq . m

Density of Air at 107°C w P = v M ¿

10423.7 k /m2 287.08(107+273)

¿ 0.94 kg/m 3

Total volume of air delivered to burner ¿

kg air /hr kg 0,94 3 m

( 35280000 CFM ) ( 0.028 )=98784 m3 /min From table 12-5, p. 480, the recommended velocities in air ducts for a static pressure of 5.08 cm is 81.82 m/min or 13.64 meter/sec at 21°C. However, the velocity maybe increased 1% for each -12.2°C above 21°C.

∴ at 107.22°C, the velocity maybe increased 107.22−21 x 1 %=16.7 % 12.2 From footnote, p. 477, add 0.127 cm water loss for each 90° bend or 0.254cm for 2 90° bends, hence Total draft loss in duct due to friction = 0.204 cm + 0.254 cm = 0.458 cm water

Velocity Head at Exit =

V2 2g

m 2 sec ¿ =5.7 m air 19.62m/sec 2 ❑

(

10.58

)

Let Y w = cm water equivalent of 5.7 m air when at the computed density of 0.94 kg/m 3 Yw x

1012.98 kg/m3 =5.7 m 0.94 kg /m 3

Y w =0.072 cm H 2 O , velocity head at burner Total Head at Burner = Pressure Head + Velocity Head = 0.072 + 5.537 ¿ 5.609 cm H 2 O

Total Head at Heater = Heat at Burner + Draft Loss = 5.609 + 0.458 ¿ 6.067 cm H 2 O

ans.

c.) Total surface area of air duct = (1.6 m) (27.18) (4) = 176.51m 2

Assuming the air duct to be at same temperature as the air temperature difference between hot surface. = 77°C – 59.44 = 17.56°C (Air consumed = 59.44°C)

From Table 12-6, p. 483, for 1 ½ magnesia heat insulation and interpolating for 65.5°C temperature difference between hot surface and air corresponding heat loss in sq. ft per in = 33.35, we have Total heat loss for 176.51 m2 = (33.35)(176.51 m2) = 5886.6 kcal/hr

ans.

Problem 12-37 A forced draft fan when driven by a constant speed motor produces a pressure of 20.32 cm. When Q= 1415.8 m3/min. What pressure would it produce on same air if Q were made 2265.4 m3/min. What would be the relative speed and power? Solution: Let

D1= 20.32 cm H2O = Initial Pressure

Q1= 1415.8 m3/min = Initial Flow N1= Initial Speed, RPM P1= Initial Power d1= Initial Density

A2 B2 a) Then by: = B1 A1

x

d2 d1

( )( )

y

Take A=D ; B=Q and from Table 12-10 Since d2= d1 x=2 D2 Q2 Then: = Q1 D1

2

( )

D2= D1

Q2 Q1

2

( )

D2= 20.32

( 2265.4 1415.8 )

D2= 52.0 cm H2O

b) Takes A= N From Table 12-10, x=1 Q2 D2 = Q1 D1

( )

N2 = N1

Q2 2265.4 = N1 1415.8 Q1

(

)

N2= 1.6 N1 This indicates a 60 % increased in speed

c) Takes A= P From Table 12-10, x=3 P2 Q2 = Q1 P1

3

( )

P2 = P1 (1.6)3 P1 P2 = 4.1 P1

Problem 12-41 Find the power required and delivery of a fan at 1200 RPM when its 1080 RPM characteristics are (53,000 CFM) 1484 m3/min, 25.4 cm static pressure, kw= 82.06. Temperature constant at 260°C. Efficiency constant. Solution: a) Use fan characteristic Eqn 12-10. p. 488, and the accompanying table A2 B2 = B1 A1

x

d2 d1

( )( )

y

A2 B1 When d1= d2 this reduces to: = B2 A1

x

( )

Let B1= Original RPM = 1080

B2= New RPM = 1200 A1= Original m3/min = 1484 A1= New m3/min = Q2 For these conditions, x=1 1484 1080 = Q2 1200

(

Q2=

1

)

(1484) ( 1200 1080 )

= 1648.89 m3/min delivery at 1200 RPM b) Let B1 and B2 same as above A1 =Original

kw= (110)(0.746)

A2 = New

kw= P2

Then x=3 A1 B1 And = B2 A2

x

( )

;

P2= (82.06 kw)

1200 1080

(

P1 N1 P2 N2

3

( )

3

)

= 112 kw, power required at 1200 RPM

Problem 12-44

ans

The products of combustion of a designated coal from Table 5-4 are discharged through a guyed steel stack 91.44 cm inside dia. X 22.86 m high (above breeching entrance). Gas temperature in breeching 315.55°C, atm 15.55°C, sea level. Rate of firing 818.18 kg. per hr. Find the available draft when (a) 60 % (b) 100 % excess air is used. Solution: Arbitrarily choosing coal No. 3 from Table 5-4,ultimate analysis is as follows: Sulfur...................................................................................... 1.35 % Hydrogen................................................................................ 4.74 % Carbon.................................................................................... 78.51 % Nitrogen................................................................................... 1.19 % Oxygen..................................................................................... 5.37% Air............................................................................................. 8.84 % Theoretical air per kg of coal for combustion

(

= 11.5 C + 34.5 H −

O + 4.3 S 8

)

(

= 11.5 (0.7851)+ 34.5 0.0474−

0.0537 + 4.3 (0.0135) 8

= 9.08 + 1.4 + 0.058 = 10.538 kg/kgcoal a) At 60% excess air Actual air used

= 10.538 x 1.6

Available Draft

= 16.85 kg/kgcoal

)

Assuming no combustible in ash pit, Fraction of coal in flue gas

= 1-Ash = 1-0.0884 = 0.9116 kg/kgcoal

Flue gas/ kgcoal

= 16.85 + 0.9116 = 17. 7616 kg

Using Fig. 12-44, p. 495, Density of flue gas at 315.55 °C =0.633 kg/m3 Volume of flue gas formed/ minute, kg kgcoal kg 0.633 3 m

17.76

= (818.18 kg/hr)

( )

=6.356 m3/s Area of chimney =

(0.91)2 π = 0.71 m2 4

Velocity of gas in chimney ;

V=

Volume Area

V=

6.365 = 9.695 m3/sec 0.71

V5= (9.695)5= 85652.3 V5 Q

1/ 2

( )

= (13456.76)1/2 = 116

From Figure 12-44, p.495, Density of air at 15.55 sea level = 9.076 x 10 -4 kg/m3 From p. 494 available draft factor k= 2.4 for steel stacks using Eqn. 12-12, p-494 and substituting the known values, available draft 30.48 m

D30 = k(da-dg) – .007578 dg

V5 Q

1/ 2

( )

= 2.4 (9.67 x 10 -4 – 0.633) – 0.007578 dg (0.633)(116) = 0.896 cm of water / 30 m

For 22.87 m,

( 22.87 30 )

=(0.896)

= 0.081 cm water

ans.

b) At 100 % Excess Air Actual air used = 10.538 x 2 = 21.076 kg/kgfuel Kg flue gas formed/kgcoal = 21.076 + 0.9116 = 21.9876 kg/kgcoal Density of flue gas (formed), same as in (a)= 0.633 kg/m 3 Volume of flue gas formed/second,

( 21.987 0.633 )

Q=(818.18)

= 7.868 m3/sec Velocity in chimney, V=

7.868 = 12 m/sec 0.71

V5 = (12)5 = 248832 V 5 248832 = = 31625.83 7.868 Q V5 Q

1/ 2

( )

= (31625.83)1/2 = 117.84

Using again Eqn. 12-12, p. 494. D30 = k(da-dg) – .007578 dg

V5 Q

1/ 2

( )

= 2.4 (9.67 x 10 -4 – 0.633) – 0.007578 dg (0.633)(117.84) = 0.592 cm / 30 meter Total available draft for 22.87 high chimney = 0.592

( 22.87 30.48 )

= 44.323 cm water

ans.

Problem 12-47 The products of combustion of 636.36 kg/hr Pocahontas coal to pass through a breeching at 260° C and 7.62 m/sec, then through a masonry chimney 36.58 m high and 1.5 cm inside cm inside diameter at the top sea level location, 30% excess air. 21.1° C atmosphere. a) Find the required dimensions of a square breeching b) Find the available chimney draft Solution: a) This same problem as No. 30 p. 507. Please refer to that no. for the solution and answer b) From the figure 12-44 p. 495 dg = Density of the flue gas at 260° C = 0.69 kg/ m3 da= Density of Air at 21.1° C = 1.2 kg/ m3 Chimney Area =

(1.5)2 =1.76 m2 4

Volume of Gas = 16016 m3/ hr. (Refer to problem30’s solution) Velocity of gas in chimney 3

( 5.72 ) 160.16 m hr V= =2.52 m/sec 3 ( 1.76 m ) ( 3600 )

(

)

V5 = (2.52)2 = 101.62

101.62 −6 V 5 ( 5.72 ) (16016) =1.76 x 10 = Q 3600 1

¿ = (1.76 x 10−6) 2 = 1.33 x 10−3

Using equation. 12-12 p.494; we have

D30 = k (da –dg) – 0.007578 ¿ Where; k = 2.7 Buck Chimney D30 = 2.7(1.2 – 0.69) – 0.007578 (0.69) (1.33 x 10 3) = 1.37 cm/ 30m Total available draft for 36.58 m = (1.37 cm)

(

36.58 m 30 m )

= 1.67 cm water

ans.

Chapter 13 The Feed Water Loop

Problem 13-1 The feed water to a boiler is 92 % condensate and 8 % make-up containing 270 ppm solids. What weight of solids enters the boiler/hr. at 22680 kg/hr. steam evaporation? Solution: Total make-up per hour = 8 % of 22680 kg/hr. = 1814.4 kg/hr. make up H2O Solids contain in make-up = (270 x 10-6) (1814.4 kg/hr.) Solids = 0.49 kg solids/hr.

Ans.

Problem 13-10 The 1370 mm diameter steam drum on aa boiler is 2440 mm long and has a 250 mm gauge glass at mid run level. Find thee maximum steam generation that could be cared for by a blowdown of half a water gauge each 8-hr shift. Pressure, 17.5 kg/cm 2ga. Sf = 150 ppm, Sb = 2000 ppm Solution:

Volume of water blowdown = 1.370 m x 2.440 m x .5(0.250m) = 0.41785 m3 Specific Volume of Liquid at 17.5 kg/cm 2ga. = 0.00117 m3/kg Mass of blowdown 0.41785 m 3 Wb = m3 = 357 kg 0.00117 kg Let Ws= kg of steam flow blow down

Total feed water between blow down = Ws+ Wb= (Ws+ 357) kg With boiler contractions, Sb = 2000 PPM and feed water concentration, S4 = 150 ppm By Mass Balance: St (Ws + Wb) = SbWb

150 (Ws+ 357) = 2000(357) Ws=

( 200−150 ) (357) 150

Ws= 4403 kg steam in 8 hrs Average steam generation:

=

4403 kg 8 hrs = 550.37 kg

Ans.

This is the maximum rate of steam generation that can be cared of by the given blowdown

Problem 13-14 Problem 13-16 Problem 13-20

Problem 13-21 Evaporation of 2272.73 kg of raw water is the desired performance of singlestage evaporator. How much live steam will it use? Live steam at 7.33kg/cm 2 ab, 16.37°C superheat. Raw water at 15.6 °C, Ɵ = 15.6 °C on the heating surface. 5% blowdown.

= 0.05 W

Let Ww = kg raw water per kg steam tb = sat. temp at 7.33 kg/cm2 = Ɵ = 165.92 – 15.6 = 153.32°C h1 at 7.33 kg/cm2 and 18.33°C; superheat = 2843.325 kj/kg hf2 at 733 kg/cm2 = 6°C = 99.48 kj/kg hb at 132.54°C = 2717.89 kj/kg hfb at 132.54°C = 554.719 kj/kg hfa at 15.6°C = 64.99 kj/kg

Heat given off by steam = heat absorbed by vaporized water + heat lost in blowdown. Substituting values, we have 1(h1 – hf2) = 0.95 Ww (hb – hfa) + 0.05 Ww (hfb – hfa) 2843.225 kj/kg – 699.48 = 0.95 Ww (2717.89 – 64.99) + 0.5 Ww (554.719 – 64.99)

2,143.745 = 0.95 Ww (2652.9) + 0.05 Ww (489.729) Ww =

2,143.745 = 0.808 kg raw water/ live steam 2652.9

Total steam required for 2272.73 kg raw water entering the evaporator =

¿ 2272.73 0.808

= 2812.78 kg/hr

Ans.

Problem 13.23 Calculate the maximum quantity of distilled water that can be obtained from the operation of a multiple effect evaporator operating between 7.045 kg/cm 2 ab, dry and saturated live steam state and 1.03 kg/cm 2 ab. condenser vapor pressure. Neglect blowdown Ɵ about 21.1 °C. Consider that the input is 0.45 kg. of steam 13.23m. Solution: Saturation temperature at 7.045 kg/cm2 = 164.33°C Saturation temperature at 1.03 kg/cm2 = 100°C Total temperature difference between vapor leaving last evaporator and sat. temp. of live steam entering first evaporator = 164.33 – 100 =64.33°C If 0 = 3.33°C (Terminal for evaporator's difference = mean temp. diff.) Approximate no. of evaporators =

64.33 = 19.3 say 20 3.33

actual Ɵ =

¿ 64.33 = 3.32°C terminal difference per stage 19.3

Make a schematic diagram of the 3 stage evaporator showing all temperatures and states of the flow. W1, W2, W3 will represent the lbs of raw water entering each stage referred 1 kg live steam entering the first evaporator.

hfg at 7.045 kg/cm2

=

2062.9 kJ/kg

hg at 142.89°C

=

2731.3 kJ/kg

hfg at 142.89°C

=

2130.678 kJ/kg

hg at 121.44°C

=

2701.644 kJ/kg

hfg at 121.44°C

=

2193.345 kj/kg

hg at 1.03 kg/cm2

=

2669.15 kj/kg

hf at 15.55°C

=

64.989 kJ/kg

Make a heat balance for each evaporator. In general, heat given off by vapor in condensing = heat absorbed by raw water in evaporating.

Evaporator (A): 2062.9 = W, (2731.3 - 64.989) W1 =

¿ 2062.9 2666.311

= 0.7736 kJ/kg live steam Evaporator (B): W1 (hfg) = W2 (hg at 121.44 at 15.55°C) W2 =

¿ 1647 2636.655

= 0.627 kg/kg live steam Evaporator (C): W2 (hfg at 121.44) = W3 (hg at 1.03 kg/cm2 - hf at 15.55°C) W3 =

¿ 1375.23 4604.161

= 0.527 kg/kg live steam

Total distilled water that can be obtained per kg live steam (from raw water) = W1+ W2 + W3 = 0.723 + 0.627 + 0.527 = 1.927 kg

Since the steam condensate is also distilled water, the total distilled water produced / kg steam is 1.927 + 1 = 2.927 kg

Ans.

Problem 13-24 An evaporator is inserted into the heat balance of an industrial plant as shown in Fig. 13-24P. A make up of 10% is required and this is supplied by the evaporator. Calculate the boiler feedwater temperature, and all flows (kg per hr) for a generator load of 100 kw. hexh = 2529.89 J/g.

(A) Evaporator h2 at 6.0 kg/cm² = 1213.25 kJ/kg hf2, at 157.78°C = 622.39 kJ/kg hf8 at 183.33°C = 76.593 kJ/kg hg9 at 131.78°C = 2724.39 kJ/kg Make a heat balanced for evaporator: heat given off = heat absorbed

W2 (hs - hf4) = 0.1 (hg9 - hf8) W2 (1213.797) = 590.92 W2 =

¿ 590.92 = 0.132 kg steam/kg throttle steam 2647.797

(B) Industrial process hg5 at 65.55°C = 273.48 kJ/kg Make mass balance for process: W3 = 1 - W 2 = 1 - 0.132 = 0.868 kg/kg throttle steam W5 = W3 - 0.08 = 0.868 - 0.08 = 0.788 kg/kg throttle steam

(c) Storage W4 = W2 = 0.132 kg/kg Make mass balance for storage: W7 = W4+ W5 + W6 = 0.132 + 0.788 + 0.1 = 1.02 kg/kg throttle Make heat balance for storage: W7hf7 = w4 hf4 + W6hf6 + W5hf5

Since hf4, at 157.78°C = 622.33 kJ/kg hf6 at 137.78°C = 577.929 kJ/kg hf5 at 65.55°C = 273.62 kJ/kg 1.02 hf7 = 0.132 (622.33) + 0.1 (571.929) + 0.788 (273.62) hf7 =

¿ 355.59 = 348.62 kJ/kg 1.02

(D) Feed water heater: hf7 = 348.62 kJ/kg from above hg9 = 2724.39 kJ/kg at 137.78°C hf10 = unknown hf6 at 137.78°C = 577.929 kJ/kg Make heat balance for heater: heat in = heat out W7hf7 + W9hg9 = W10hf10 + W6hf6 1.02(348.62) + 0.1(2724.39) = 1.02 hf10 + 0.1 (577.929) 1.02 hf10 = 955.59 + 272.439 - 57.79 = 570.26 hf10 =

¿ 570.26 = 559 kj/kg 1.02

(a) Boiler feed water temp. = hf10 - 425 = 559 - 425 = 134°C

(b) Work done per kg throttle steam = 1(h1 - h2) + (1 - W2) (h2 - h3) h1 = 5780.558 kJ/kg h2 = (1213.25 kJ/kg or 2678/434 kJ/kg h3 = 2529.85 kJ/kg W2 = 0.132 kg as previously computed W = 1(2780.558 - 2678.434) + (1- 0.132)(2678.434 2529.85) = 102 + 128.97 230.97 kJ/kg At 0.79 Mechanical-electrical efficiency, total work done by steam for 100 kw output, =

¿ 100 x 3607 kJ/ hr 0.70

Total throttle flow =

( 100 ) (3607) (0.79)(99.5)

= 4408.77 kg/hr Total feedwater = W10 x 4408.77 = 1.02 x 4408.77 = 4680.54 kg Total extraction = W2 x 4408.77 = 0.132 x 4408.77 = 581 kg.hr Total make up = 0.1 x 4408.77 = 440.079 kg/hr Total process return = W5 x 4408.77 = 0.788 x 4408.77 = 3474 kg/hr

Problem 13-27 Find the cold water TDH for a pump through which 473 li/m of 33.33 °C water are passing discharge 5.08 cm dia., suction, 75mm. Discharge pressure is 3.5 kg/cm 2 ga. Measured 0.6 m above pump centerline. Suction, 1.42 kg/cm 2 ga. measured 0.45cm below center line.

Solution: Vf at 33.33 °C = 0.001005 m3/kg Vf at 15.55 °C = 0.001001 m3/kg

Specific gravity at 33.33 °C Water referred to 15.55 °C =

0.001001 =1.0 0.001005

Convert 135 gpm to m3/sec

gal/min ( 125 60 sec/ min )

kg gal 28.38 kj m3 3.7

( )

( 0.028m3m3 )=7.6 x 10

-3

m3/sec

Velocity at discharge; Vs2 =

Q A at Discharge

= π /4(0.0508)2 = 3.74 m/sec Velocity at suction, Vs1 = π /4(0.0762)2 =

1.64 m/sec

Let: TDH = total dynamic head Pressure head = p/w Velocity head =

V s2 2g

Static head = Z

Then using the pump center line as reference and assuming functional losses between gage location negligible 2

(V s 2) P1 + Hp = P2/w + Z2 + (Vs2)2/2g +Z 1+ w 2g 3.4−1.42 (3.74)2−(1.64)2 Hp = + (0.6 + 0.45) + 1 19.62 = 23.23 m of 93.33° C water Cold water TDH = Hp (Sp. Gr) = 23.23 x 1 = 23.23 m

Problem 13-28 Calculate drive hp for pumping 1703 l/min cold water to a tank. Suction 127 mm Hg vacuum, deliver at 5.3 kg/cm2 ga, both measured close to pump, ɳp = 0.65

Let hd = total head at discharge hs = total head at suction hp = pump head By Bernoulli’s Energy Eqn.: Hd =

Pd Vd 2 + + Zd w 2g

Pd Vd 2 Hs = + + Zd w 2g Hp = hd – hs Hp =

Pd + Ps Vd 2 +Vs2 + +(Zd −Zs) w 2g

Neglect velocity head and elevation head changes (This is check if Ɵ of suction and discharge are equal and is Zs = Zd)

Then hp = pd – ps/w Pd = 5.3 kg/cm2 ga Ps = -127 mmHg x

1.0334 kg /cm2 760 mm/kg

= 0.1727 kg/cm2 So that, 5.3+0.1727 kg /cm2 2 ¿) Hp = 1000 kg /m 3 = 54.727 m lm3 kg )( 1000 3 )(54.724 m) 1000 lit m kg−m 0.746 kw (6116.3 )( ) kw−min hp

1703l/min( BHP =

= 20.43 hp Drive hp for pump Bhp hp

= =

20.43 0.65

= 31.4 hp Ans.

Problem 13 -29

Estimate discharge rate (li/min) and maximum suction lift of a 95mm x 127mm x 152 mm duplex, direct acting steam pump. ɳv = 0.90 Sea level. Could this be a boiler feed pump? Solution: By Eqn 13-11, p.549, PPE by Morse, commercial pump speed, Vp = 1.38√ 1 ; m/min of piston speed Where; L = stroke in mm. Vp = 1.38√ 1.52 = 17.0 m/min Piston displacement of pump, PD = no. of cylinders (area) (Vp) = 2 (π/4)(0.127 mm Hg)2(17.0 m/min) = 0.431 m3/min By Equation 13-12, p.549, discharge capacity, Q=

ɳ v PD l/min 1000

Q=

0.90(0.431)(100)3 100

Then:

Q = 387.9 l/min b)

Steam Cylinder diameter 95 = = 0.748 < 1.6 Water cylinder diameter 127 For boiler feed pumps, Ds/Dw is about 1.6, see p. 547. This pump is therefore, not boiler feed pump.

Problem

13-30 Select a suitable dimension for duplex direct-acting steam pump for the ff boiler service. 265 lit/min against 8.8 kg/cm2ab. boiler pressure. H2O at 93.33°C, boiler water level, 5.49 m above pump. Installation of suction in accordance with Figure 13-27. Altitude, 457 m. Water cylinder D = L, Nv = 0.90. Neglect velocity head and pipe function. Find whp. Solution: From Figure 13-27, p. 549, submergence for 457 m altitude and 93.33°C = 3.8m

Barometric pressure will be decreased by approximately 25.4 mmHg per 304.89 m. above sea level. Approximate barometric pressure at 457 altitude ¿ 39−

457 x 25.4 mmHg=0.928mmHg 304.89

Or in kg/cm2 , barometric pressure 0928 x 1.03 kg/cm 2 ¿ 39 ¿ 0.0245 kg /cm2 Boiler gage pressure = 8 kg/cm2 – 0.00245 = 8.7755 Vf at 15.55°C = 1.001 x 10-3 m3/kg Vf at 33.33°C = 1.038 x 10-3 m3/kg Vf ratio ¿

1.001 x 10−3 =0.964 1.038 x 10−3

Writing the head eqn of the pump, we have Z1 +hp=Z 2+

P2 w

kg cm2 cm 2 3.80+ Hp=5.49+ (100 ) kg m 1000 3 m 7.814

Hp=5.49−3.80+78.14 ¿ 79.83 m (a) WHP=

( gpm x 7.786 x 0.964 ) (79.83) ; but ( 6116.9 ) (0.746)

But gpm=70∨

3.79 lit / min 1 gpm

= 265.3 lit/min ¿

(70 x 7.786 x 0.964 )(79.83) =9.192 Hp ( 6116.3 ) (0.746)

(b) From 13-12, p 549 Q=

´ Nv PD ; li/min 1000

´ 1000 Q ∴ Piston Displacemen , PD= Nv ¿

1000 ( 265 ) 0.9

¿ 2944.44 cm3 Also

´ PD=no . of cylinders x area of bore x piston speed ¿2 x

πD w 2 x(Vp x 12) 4

For commercial pumps, Vp=23 √ L x factor for water temp (From Kent’s Hdbk, factor = 0.06 for 33.33°C Vp=23 √ L x 0.66 ¿ 23 √ Dw x 0.66 since Dw=L ´ Substitute this value of Vp and PD=294777.78 cm , so 294777.78=2 x

Dw 2.5=

π ( D¿¿ w)2 x 23 √ D x 0.66 x 12¿ 4

( 294777.78 ) (4) 2 π (23) ( 12 ) (0.66)

= 1030.7 =13.33 cm For boiler feed service steam cylinder here Ds ≥1.6 Dw Ds ≥1.6 ( 13.33 cm ) ≥ 25.66 cm

Use Ds = 22.86 cm nearest standard size L = Dw = 13.33 cm ; use 16 cm to get nearest standard size 22.86 cm x 13.33 cm x 15.25 pump nearest standard size

Problem 13-31 Results of a test on 25.4 cm x 15.24 cm x 38.48 cm duplex direct-acting pump are given. Find η,

whp, duty. Test data: time 60 min., steam used, 568 kg of 14

kg/cm2ab. saturated steam. Exhaust, 1.03 kg/cm2ab.; Speed, 38 strokes/min per cylinder. Water pump = 22727.27 kg. TDH, 113.64m of 32.2°C H2O. Solution: h1at 14 kg/cm2

=

2782.18 kJ/kg

hf at 1.03 kg/cm2

=

12.69 kJ/kg

Vf at 15.55°C

=

1.0 x 10-3 m3/kg

Vf at 32.22°C

=

1.03 x 10-3 m3/kg

Specific gravity at 32.22°C referred to (60°F) 15.22°C ¿

1.0 x 10−3 =0.971 1.03 x 10−3

(a) Piston displacement of water cylinder per 60 min = No. of cylinder x area of bore x stroke / (60 min – cylinders) x stroke π x ( 15.24 )2 ¿2 x x ( 38 x 60 ) x 30.48 cm 4 ¿ 25353599.25 cm 3

Actual volume of water pumped out ¿

22727.27(103.87) =108843 m3 ( 28.3 ) (.971)

Volumetric efficiency, N 2=

¿

water pumped out piston displacement

28572374.5cm3 25353599.25cm3

¿ 1.13

(b)

WHP=

¿

kg . water pumped/min x head kg−m 6116.3 kw−min

( 22727.27 kg ) (113.64 ) ( 60 ) ( 6116.3 ) ( 0.746 )

¿ 9.43 WHP (c) Total kg. work done ¿ ( 22727.27 ) (113.64 m) ¿ 3150926.96 kg−m per 60 min Total heat supplied ¿ kg steam used x(h1−hf ) ¿ 568.18 ( 2782.18−12.69 ) ¿ 1573568.828 kJ per 60 min Duty=

¿

3.15 x 106 kg−m 1.57 x 106 kJ

= 2.0 x 106 kg.m per 106 kJ

output ∈kg−m input ∈million kJ

Problem 13-32 What size of 1750 rpm motor should be used to drive a 100 mm x 200 mm single acting triplex pump? Np = 0.88; Nm = 0.90; Nv = 0.95. What V-belt pulley diameters should be used? Water temperature, 110°C; TDH = 24.6 kg/cm2? Solution: Referring to Equation 13-13, p 551, usual pump speed for triplex power pumps N

=

907 L-1/2 rpm

L

=

Stroke in mm

Since L = 20.3 cm L-1/2 =

N=

1 1 = −1 /2 1.51 203

907 =201.55 rpm 4.51

This is also the same rpm given in table 2, p573 of Kent’s handbook, power volume. However, this speed is for cold water and for 110°C water, the speed should be reduce by 36.4% according to table 3 p. 573 Kent’s handbook, so that for 110°C water, pump should be run only at N1 = 261 x (1 – 0.364) = 204.55(0.636) = 128.18 rpm At 201.55 rpm speed reduction from 1750 rpm ¿

1750 =8.683 201.55

At 128.78 rpm, speed reduction from 1750 rpm ¿

1750 =13.65 128.78

Ratio of cold water to hot water; RPM =

201.55 =1.575 128.18

It will be seen from the above that it would not be practical to drive the pumps directly through V- belts due to the large speed ratios. It would be more practical to use gear reducer with the given ratios. To enable the pump to handle its full rated capacity when handling cold water, a good system would be to a select a motor sufficient to handle full capacity at 21.11°C. The drive being a 1.1 V-belt drive from motor 27.4:1 speed ratio for 110°C to be handled by the pump, the reducer pulley maybe changed with another approximately 1.575 times as diameter as the original. Of course, longer V-belts will have to be used and provisions made of readjusting center distance The motor pulley ∅ will be dependent on the section of belts which in turn will depend on the motor Hp. This must be determined first therefore. Based on our assumption of designing for maximum capacity that is at 201.55 rpm displacement of the pump of this speed.

Problem 13-35 A 2-stage centrifugal pump delivers 190 lit/min of 107.22°C water against 76m head at 3500 rpm. What is specific speed? What head could a geometrically similar pump produce if operated at 3000 rpm at 151 lit/min. flow? Summary Vf at 29.44°C = 1.004 x 10-3 m3/kg Vf at 107.22°C = 1.049 x 10-3 m3/kg Specific gravity of 107.22°C water referred to 29.44

1.004 x 10−3 m 3 /kg = = 0.957 1.049 x 10−3 m 3 /kg N

=

rpm of pump shaft = 3500

Q

=

50 GPM or 190 lit/min

H

=

head del stage at 29.44°C

=

76.22m x sp . gr . 2

=

76.22m x 0.957 = 36.47 m 2

Specific speed, Ns =

=

N √Q H 3/ 4

3500 √ 190 3500 ( 13.76 ) ¿ 14.84 (36.47 m)0.75

= 3250,96 rpm (29.44°C) (b)

At

N = 3000 Q = 151.6 lit/min

H0.75

=

N √Q Ns

=

3000 √151.6 3000 ( 12.288 ) ¿ = 11.339 3250.96 3250,96

H= 11.3394/3 = 25.38 m per stage (29.44°C) for 107.22°C water, and for 2 stages Total head ¿

( 25.39 )( 2 ) 0.957

= 53.06 m of 107.22°C water

Problem 11-36 Determine the maximum practicable static head of hot water on the suction of a double suction, centrifugal heater condensate pump. Water at 85°C, 92530 kg/hr flow. Heater pressure, 0.68 kg/cm2 ; discharge at 2.818 kg/cm2ab. Neglect pipe friction Solution: Vf at 29.44°C = 1.004 x 10-3 m3/kg Vg at 85°C = 1.032 x 10-3m3/kg Specific gravity at 85°C referred to 29.44°C ¿

1.004 x 10−3 =¿ 0.973 1.032 x 10−3

Water flow in lit/min ¿

92530 ( 3.79 ) ( 60 )( 3.786 ) (0.973) ¿ 1586.63 li/min

Vapor pressure at 85°C = 0.59 kg/cm2 Available suction head = heater pressure = 0.68 kg/cm 2 Net positive suction head ¿

Dynamic pressure at suction−sat . pressure Sp .Gr .

¿

( 0.68+0.59 ) (100)2 ¿ 0.9 m 1000

Using figure 13-34 and fro 1591.8 lit/min double suction pumps, the minimum positive suction head for various rpm’s are given For example

a) For 3500 rpm, minimum positive suction head¿ 2.13 m Required minimum static head therefore for this rpm ¿ 2.13−0.9 ¿ 1. 23 m b) For 1750 rpm, minimum positive suction head = 0.73 m Since NPSH is already 0.9 m, no static head is required. Therefore minimum practicable static head is 0 c) The above will indicate a requirement for static heads when running at 2900 rpm and 3500 rpm and move at 1750 rpm and below

Chapter 14 Piping System

Problem 14-2 Would schedule 120, 203.2 mm pipe made to A53-SA specification be acceptable on a line operating at 17.6 kg/cm2ga, 400°C? Solution: Referring to Figure A-6, p.665, PPE by Morse, and for schedule 120, 203.2 mm pipe, outside diameter,

Do = 219.08 mm

pipe wall thickness

dt = 18.24

working pressure,

P = 17.6 kg/cm2ga.

Allowance for corrosion

C= 165

also,

Substituting these values in Eqn. 14-3, p.575, dt=

pDo +C ; mm. 2 S+ 0.8 p

Where: dt

=

mnimum pipe wall thichness; mm.

p

=

working pressure ; kg/cm2ga.

Do

=

outside diameter; mm.

S

=

working stress (Table 14-1, p 574)

C

=

1.27 for pipe 25.4 mm and smaller; 1.65 for 31.8 mm. and larger.

∴ 18.24=

2 S +0.08 ( 17.6 )=

2 S +14.08=

17.6(219.08) + 0.0655 2 S +0.8(17.6)

17.6(219.08) 18.24−1.65

17.6(219.08) 16.59

∴ S=109.15 kg /cm2 working stress in pipe

From Table 14-1, p 574, A-53 SA specifications has maximum allowable stress of 843.7 kg/cm2, therefore it is safe. But pipe seems uneconomical ; Try schedule 80 Do

=

219.08mm

dt

=

12.7mm

C

=

1.65

Using equation 14-3 again, we have dt=

pDo +C 2 S+ 0.8 p

12.7=

17.6 (219.08) +1.65 2 S+0.8(17.6)

2 S +14.08=

¿

17.6(219.08) 12.7−1.65

17.6(219.08) 11.05

= 334.86 2S = 334.86 – 14.08 = 320.78 S = 160.39 kg/cm2 ; which is less than the working stress 843.7 kg/cm2 ∴Schedule 80 is also safe, and schedule 120 is show to be too heavy and uneconomical

Problem 14-5 A certain pipe line covered with heat insulation has exposed flanges. These are found to be steel and to measure 279.4 mm dia. x 33.3 mm thick, and to have 8 bolts. Without removing any of the insulation, it is desired to determine whether it would be proper to put 24.6 kg/cm2 ga steam through this line. Assume line is carbon steel flanges properly selected for the pipe weight. Solution: Refer to Fig. A-7, p. 665, and find pipe size for which the given flanges correspond.

Corresponding pipe size is 127 mm. for data given. Pressure class is 21.08 kg/cm2 based on 400℃ (See also footnote, p.669)

From table 14-1, p 574, this could be either A-53, S-A or A-53, S-B. Assume it is A-53, S-A.

The weaker material whose maximum allowance stress given as 843.7kg/cm2 Also, to be on the safe ride, assume schedule 40 Then, from Fig. A-6, p. 665 Do

=

Outside Diameter

=

141.3 mm

dt

=

wall thickness

=

6.55 mm

Using Eqn. 14-3, p. 575, with C = 1.65, P = 24.60 kg/cm dt

=

pDo +C 2 S+.8 p

6.55

=

24.60(141.3) + 1.65 2 S+ 0.8(24.6)

=

24.6(141.3) 6.55−1.65

=

24.6(141.3) = 709.38 4.90

2S + 19.68

2S = 709.38 – 19.68 = 689.7 S =

689.7 = 344.85 kg/cm2 < 843.7 kg/cm2 2

∴ Pipe is safe for 24.6 kg/cm2

Problem 14-14 A portion of a manufacturer’s safety valve rating table is reproduced herewith. From this line, specify 3 safety valves and settings suitable for use on a stoker-fired, water-tube boiler generating at 17.6 kg/cm 2 ga. Boiler heating surface, 427.4 sq m2

including 55.7 sq m2 water wall area. Steam rating, 10886 kg per hour; design pressure, 21.1 kg/cm2 ga.

Valve size: mm

38.1 x 50.8

50.8 x 76.2

Rated Ws.v.

16.9 kg/cm2ga

4,196

6,895

kg per hour

16.9 kg/cm2ga

4,364

7,149

at pressures

16.9 kg/cm2ga

4,527

7,439

of

16.9 kg/cm2ga

4,695

7,711

63.5 x 101.6

76.2 x 101.6

101.6 x 152.4

9,843

15,241

23,201

10,206

15,830

24,113

10,614

16,447

25,029

10,986

17,055

25,936

Solution: Let Wmax

=

Max. rated steam generating capacity; kg per hour

=

10,886 kg per hour (given)

Wt

=

total required relieving capacity of safety valves; kg per hour

A1

=

Total heating surface of boiler (excluding water walls); sq. ft

=

427.4 – 55.7 = 371.67 sq. m

=

Total heating surface of water walls; sq. ft

=

55.7 sq. m (given)

A2

Then encoding to ASME Boiler Code, 1) Wt ≥

Wmax is one criterion for relieving capacity,

another is 2) Wr =

C1A1 + C2A2 (eqn. 14-4, p. 589, PPE by Morse)

The larger value of Wt found, determines the choice. 1) From the above one value of Wt =Wmax = 10,886 kg/hr or using Eqn. 14-4 above, and with C1= 8, C2= 12 from table on page 590 corresponding to stoker-fired water tube boilers; we have Wt

= 39.6 (371.7) + 53.59 (55.7) = 14,518.602 + 2984.96 = 17,503.56

Use Wt

= 17,503 kg/hr, since it is the larger value.

Item 3 Art. 14-7, p. 589 states that “if mounted singly on the drum, the smaller safety valve must have at least half the relieving capacity of the larger.” Item 6, states “setting of the highest-set safety valve on the boiler shall not exceed 103% of the design pressure of the steam generator. The range of pressure (pop-off) between lowest and highest set value, shall not exceed 10% of the set value of the highest pressure valve. From Item 6, the max. setting of pressure valve allowed is 211.08 x 1.03 = 21.71 kg/cm2g and the corresponding low pressure setting for this is 21.71 – 2.171 = 19.54

kg/cm2ga. However, these settings are too high for the operating conditions given. If the setting are made at 17.92 kg/cm2g, 18.03 kg/cm2g, and 18.62 kg/cm2ga Item 6 and the conditions of operation will both be satisfied. The relieving capacity should be based on 17.57 kg/cm 2ga for the three valves. Based on item 3, if the smallest valve used is 38.1 x 50.8; capacity, 4372.7 kg per hr.; the next should have a capacity less than double this, say 50.8 x 762 capacity 7,163.6 kg per hr. If the third valve is also 50.8 x 76.2 this will satisfy all the requirements.

Summarizing, valves to be used are: one 38.1 x 50.8 set at 17.92 kg/cm 2ga capacity at 17.57 kg/cm2ga

=

4,3727 kg/hr

one 50.8 x 76.2 set at 18.62 kg/cm 2g capacity at 17.57 kg/cm2ga

=

7,163.3 kg/hr

one 50.8 x 76.2 set at 18.62 kg/cm 2ga capacity at 17.57 kg/cm2ga total Ws v.

=

7,163.6

=

18,700 kg/hr

ans.

Note: It will be found that the above selection is the most economical choice for singly mounted valves since it represents the smallest size that will satisfy the code.

Problem 14-17

A 254 mm schedule 80 A-53 steel pipe is clamped between two rigid supports 6.1 m apart when its temperature is 15.6℃. A liquid at 93.3℃ is then pumped through the line. What thrust force is developed at the reactions? Solutions: For the steel pipe material, modulus of elasticity E = 2.11 x 106 kg/cm2 From Fig. A-6, p. 665, we get the following outside diameter and wall thickness of 254 mm, schedule 80 pipe; Do

=

273.05 mm

Thickness

=

15.06 mm

∴ Outside cross-sectional area of pipe =

π Do2 4

=

3.1416 (273.05)2 4

= 58,556.52 sq. mm Inside diameter of pipe, Di

= 273.05 – 2 x 15.06 = 273.05 – 10.12 = 242.93 mm

Inside cross-sectional area of pipe: =

π Di2 4

= π∨3.1416 x ¿¿

=

74,556.3(π ) 4

= 46,350.37 sq. mm. Area of material under stress, A = =

58,556.52 – 46,350.37 12,206.15 sq. mm2

Length of material between supports, L

=

6.1 m (

1000 mm ¿ = 6,100 mm 1m

Coefficient of linear expansion for steel =

6.5 x 10-6 in./(in-℉) as used in most of strength of material textbooks and machine design books.

∴ Total elongation of pipe if allowed to expand freely due to temperature change, assuming that the pipe will also be at 93.3℃ is ∂

=

6.5 x 10-6 x 6,100 mm

Due to the supports, the pipe will be kept from expanding and the supports will exert a force P, such that

Or

∂

=

∂ AE = L

6.5 x 10−6 ¿ (6,100)(12,206.15)(30 x 10 6)

¿ 6,100 mm .

P

=

∂ AE = L

6.5 x 10−6 ¿ (6,100)(12,206.15)(30 x 10 6)

¿ 6,100 mm .

=

2,380,199.25 kg/sq. mm. or 23,801.99 kg/cm2

Ans.

Note: The answer found will be good only for the value of coefficient of expansion = 6.5 x 10-6 Depending on the references used, there is bound to be some variations on these valves. However, the answer gives an indication of the large thrust produced by thermal stress.

Problem 14-22 How much condensation is produced per hour by heat leakage from a Schedule 40 steel pipe 355.6 mm OD x 25.9 m long. Saturated steam at 17.6 kg/cm 2ab; ambient temperature, 15.60C. Cover is Double Standard thick sectional magnesia.

Solution: Saturation temperature at 17.6 kg/cm 2 = 2050C = ti Total heat loss for 25.9m pipe is 0.144 x 25.9 = 3.72 KCAL/HR Latent heat of vaporation, hfg at 17.6 kg/cm2ab, = 207.997 kcal/kg Steam condensed due to heat loss = =

Problem 14.27

3.72 = 6.02 kg/hr I ans. 207.997

total heat loss h fg

An 203.2 mm main steam header is to receive an “economic” thickness of insulation No. 4, Fig. 14-21, p 599, PPE by Morae, Steam temperature, 371.1 0C; header in continuous use. Insulation cost 79c per bd. M.; steam cost 200c per 10 6 kcal. Fixed charges, 12%. Find the thickness and calculate the heat loss per m of pipe.

Solution: Insulation No. 4 is diatomaceous earth bounded with asbestos fiber temp. at internal surface of insulation, tf = 371.10C. Assume for a preliminary estimate, temp. of external surface, to = 55.60C Mean temp. of insulation =

55.6+371.1 2

= 213.330C For which from Fig. 14-21, p. 599, k – 0.063 kcal/hr-m-0C per m, thickness

Assume ambient temp. of 320C = ta Temp. difference, ti – ta = 371.1 -32 = 339.110C 339.1 = 1 +0.53 0.043 339.1 1 m2 2 = = 0.163 kcal/hr- m x 2073.43 3.281 ft 2 To verify, the heat loss from the outer surface = h (t0 – ta)

= 1.893 (55.56 – 32) = 44.60 kcal/hr-ft2 – 0F Which is very close. No recalculation will be necessary Heat loss per ft. of pipe =

422.75 x 0.16 1000

= 0.212 kcal/hr-m

Note: 1.) The solution and answer will depend a great deal on the ambient temperature assumed, although it does not seem to affect the economic thickness very much. 2.) The lower the ambient temperature, the greater will be the heat loss; and the higher the ambient temp, the lower the heat loss, Or for ta < 3220 C loss > o.212 kcal/hr-m And for ta > 3220 C loss < 0.212 kcal/hr-m

Problem 14-31 Find the least size of Schedule 80 pipe that will convey 1325 li/min water a short distance, including four 900 elbows and one gate valve, without ΔP exceeding 0.35 kg/cm2. = 0.0035 + 0.0007562 (

1 ¿ -0.424 (75.6)( 4.92)

= 0.003561 Using Equation 14-10, p. 604, per friction head,

H=

=

2 fLV 2 in m. gD 2 x 0.003561 x 10.25 x 24.206 9.81 x 0.0756

= 2.38 m Friction loss in psi = (2.38)(1000)(

1 ¿ 10,000

= 0.0238 kg/cm2 Therefore, the best size of schedule pipe that can be used is 76.2 mm. ans.

Note: 1.) Larger size would of course do 2.) Elbows have been assumed to be standard. Much less friction loss would result using meium sweep or long sweep elbows.

Problem 14-35 Find the kg/cm2 friction loss in a 30.5 m copper pipe, 19.1 mm ID, in which there is a flow of 544 kg per hr of 24 °. Be fuel oil at 21.1 ℃. Solution: Viscosity of fuel oil at 24° Be and 21.1 ℃ = 40 centipoises Specific gravity of 24 °Be, S.G. = Volume of flow per sec

=

140 = 0.91 130+24

wt. per. hr 3600 x 62.4 x S.G.

Velocity of flow, v =

= z DVS

=

544 3600 x 62.4 x S.G.

=

1.66 X 10-4 m3/sec

1.66 x 10 -4 π x (19.1) 2 4 x (1000)2 0.58 m/s

40 0.0191 x 0.58 x .91

=

Therefore, flow is viscous. Using Equation 14-14, p. 605, for viscous flow, friction factor,

f

[

z DVS

=

0.000016026

=

0.000016026 x

=

0.0636

]

40 0.0191 x 0.58 x .91

Using Equation 14-10, p. 604, PPE by Morse, friction head, H

=

2fL V 2 gD

=

2 x 0.0636 x 9.81 x 0.0191

=

30.5 x

(0.58)2

6.96 m

Friction loss in kg/cm2 = =

H x

density of water x

(6.96 m)(1000

kg m3

)( 0.91)

sp. Gravity x

(

1m 100cm

2

)

=

0.633 kg/cm2

Ans.

Problem 14-38 Find the maximum support spacing safe to use with a long horizontal 152.4 mm Schedule 40 bare cold water pipe of A-72-BW material. What is the maximum deflection of this pipe when so supported? Solution: From Table 14-1, p. 574, PPE by Morse, the maximum allowable stress for this material is S = 421.8 kg/cm2

From Fig. 9-6, same reference; for a 152.4 mm, Schedule 40 pipe OD

=

168.275 mm

Wall thickness

=

7.11 mm

Wf.

=

28.28 mm

Internal diameter of pipe

=

168.275 mm - 2(7.11 mm)

=

154.06 mm

Water pipe when completely full, per m of pipe = =

π 1 cm 154.06 x 4 10 mm

(

18.64 kg/m

x

1m 100 cm

2

)

x 1 x 1000 kg/m3

Total weight of pipe within full = =

28.28 + 18.64 46.92 kg/m

Use Equation 14-20, p. 609, PPE by Morse, to find maximum support spacing: S1 =

W L2 Do 24I 46.92 = 0.4692 kg/cm 100

W

=

wt. per m of pipe

L

=

distance between supports, hence unknown

Do

=

outside diameter of pipe = 16.8275 cm

Di

=

inside diameter of pipe = 15.406 cm

I

=

moment of inertia of pipe

=

π (Do4 – Di4) 64

=

0.491 (16.8275 4 - 15.4064)

=

1170.49 cm 4 (same as in Machinery’s handbook)

Substituting all known values in Eq. 14-20, we have, =

0. 4692 x L2 24 x 1170.49

L2

=

421. 8 x 24 x 1170.49 0.4692 x 16.83

L

=

1224.76 cm

421.8 kg/cm2

Where S

=

x 16.8275

S1

Assuming operating temp. at 15.56 ℃ (cold water) From p. 609, PPE by Morse, E

=

(20.5 - 0.0076t) x 5 kg/cm 2

=

(20.5 - 0.0076 x 15.56) x 10 5 kg/cm2

=

20.38 x 105 kg/cm2

Using Eq. 14-19, p.609 same reference, deflection y =

W L4 384 EI

= =

0. 4692 x 1224.764 384 x 20.38 x 10 5 x 1170.49 1.15 cm

Problem 14- 40 A certain 18.3 m section of insulated 234 mm Schedule 80 steel pipe when warmed to operating temp from a cold state of 4.4° F produces initial condensation which a trap should drain out as the line heats up. Assume this takes 5 minutes. Steam saturated at 28.1 kg/cm2 ga. Estimate the discharge capacity of the trap, in kg/hr. Solution: From Fig. A-6, p. 665, PPE by Morse, 254 mm schedule 80 steel pipe will weigh 96.04 kg/m ∴ Total weight of 18.3 pipe = 96.04 x 18.3 = 1757.6 kg Saturation temp. at 28.1 kg/cm2 ga. = 229.21°C Specific heat of steel = 0.484 kt/kg °C (this wil vary with some textbooks) Total heat absorbed by steel pipe in raising its temp. from 4.4°C (40°F) to 229.21 °C (444.58°F).

=

W Cp ∆T

=

1757.6 x 0.484

=

191.166 kg in 5 min

=

191. 166 kJ 5 min

=

38, 233.2 kJ/min

x

(229.21-4.44)

If we assume that for this period the heat loss through the insulation is neglible, then the heat lost by the steam is also equal to the heat absorbed by the pipe.

Latent heat of vaporization at 28.1 kg/cm 2 ga is hfg

=

1917.9 kJ/kg

This means that for every heat lost in the amount of 1917.9 kJ/kg, will result in the condensation of 1 kg or for

Total condensation = =

191.166 5

kJ/min heat lost,

191.166 5 x 1917.9 19.93 kg/min

Discharge capacity of steam trap, kg per hr is 19.93(60)

=

1195.8 kg per hr

Ans.