# Dela Cruz Assignment

• Uploaded by: Mikhail Roy Dela Cruz
• March 2021
• PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form.

### More details

• Words: 3,208
• Pages: 11
DIESEL CHIMNEY

POWERPLANT

AND

0.44 = Pi / 0.05(44,000) Pi = 968 kW

ASSIGNMENT: 1. Determine the indicated mean effective pressure of an engine in psi having a brake mean effective pressure of 750 kPa and 80% mechanical efficiency. a. 136 psi*

b. 137 psi

c. 138 psi

d. 140 psi

4. A 750 kW diesel electric plant has a brake thermal efficiency of 34%. If the heat generated by fuel is 9,000,000 kJ/hr, what is the generator efficiency? a. 85.33%

b. 65.88%

c. 75.55%

d. 88.23%*

Solution: ntb = Pb / mf Qh

Solution:

0.34 = Pb / (9,000,000 / 3600)

nm = Pmb / Pmi

Pb = 850 kW

0.80 = 750 kPa / Pmi Pmi = 937.5 kPa x (14.7 psi / 101.325 kPa) Pmi = 136 psi 2. Determine the friction power of an engine if the frictional torque developed is 0.3 kilonewton-meter running at 1200 rpm. a. 40.6 kW

b. 37.7 kW

c. 36.5 kW

d. 50.3 Kw

nG = 750/850 nG = 88.23% 5. A 16-cylinder V-type diesel engine is directly coupled to a 5000 kW AC generator. If generator efficiency is 90%, calculate the brake horsepower of the engine. a. 7447 hp*

b. 6468 hp

c. 8542 hp

d. 7665 hp

Solution:

Solution: Pf = 2πTn

nG = Gen Output / Pb

Pf = 2π(0.3)(1200/60)

0.9 = 5000 / Pb Pb = 5555.55 kW (1 hp / 0.746 kW)

Pf = 37.70 kW 3. What is the power developed in the cylinder if indicated thermal efficiency is 44%, the engine uses 0.05 kg/s fuel with heating value of 44,000 kJ/kg?

Pb = 7447.12 hp

a. 1000 kW*

b. 775

6. Determine the brake power of the engine having a brake thermal efficiency of 35% and uses 25 ͦAPI fuel consumption of 40 kg/hr.

c. 36.5 kW

d. 588 kW

a. 165.84 kW

b. 173.52 kW*

c. 186.52 kW

d. 160.67 Kw

Solution:

Solution:

nti = Pi / mf Qh

Qh = 41,130 + 139.6 (API)

Qh = 41,130 + 139.6 (25)

w2 = 1.0947 kg/m3

Qh = 44,620 kJ/kg

nm = Pb / Pi

ntb = Pb / mf Qh

0.85 = 500 / Pi

0.35 = Pb / (40/3600) (44,620)

Pi = 588.23 kW

Pb = 173.52 kW

Pf = P i - Pb

7. Determine the specific gravity of fuel oil having a heating value of 44,899.2 kJ/kg.

Pf = 588.23 - 500 = 88.23 kW

a. 0.90*

b. 0.80

c. 0.877

d. 0.893

(Pi2 / Pi1) = (w2 / w1) (Pi2 / 588.23) = (1.0947 / 1.1765) Pi2 = 547.336 kW Pb2 = 547.336 - 88.23

Solution: Qh = 41,130 + 139.6 (API)

Pb2 = 459.106 kW

API = 27

9. What is the displacement volume of 300mm x 4400mm, 4-stroke, 1200 rpm, 8-cylinder diesel engine?

SG = 141.5 / (131.5+API)

a. 0.243 m3/s

b.

c. 5.75 m3/s

d. 1.25 m3/s

44,899.2 = 41,130 + 139.6 (API)

2.262

m3/s

SG = 141.5 / (131.5+27) SG = 0.8927 8. A 500-kW diesel engine operates at 101.3 kPa and 27 ͦC in Manila. If the engine will operate in Baguio having 93 kPa and 23 ͦC, what new brake power will develop if mechanical efficiency is 85%? a. 600 kW

b. 754 kW

c. 459 kW*

d. 971 Kw

Solution: w1 = P/RT w1 = 101.325 / (0.287) (27+273) w1 = 1.1765 kg/m3 w2 = P/RT w2 = 93 / (0.287) (23+273)

Solution: VD = (π/4) D2 L N c VD = (π/4) (0.3)2 (0.4) (1200/ 2x60) (8) VD = 2.262 m3/s 10. What is the friction horsepower of a 300-kW diesel engine having a mechanical efficiency of 86%?

a. 86.5 hp

b. 87.5 hp

c. 90.5 hp

d. 65.5 hp*

Solution: nm = Pb / Pi 0.86 = 300 / Pi Pi = 348.84 kW Pf = P i - Pb

Pf = 348.84 - 300

Solution:

Pf = 48.84 kW x (1 hp/ 0.746 kW)

mg = ma + mf - mash

Pf = 65.46 hp

A/F = ma / mf

11. Determine the output power of a diesel power plant if the engine and generator efficiency is 83% and 95%, respectively. The engine uses 25 ͦAPI fuel and has a fuel consumption of 0.08 kg/s.

16 = ma / mf

a. 2795 kW

b. 8642 kW

c. 9753 kW

d. 2815 Kw*

Solution: Qg = m f Qh Qg = 0.08 [41,130+139.6(25)] Qg = 3569.6 kW Generator Output = 3569.6 (0.83) (0.95)

ma = 16 mf mg = 16 mf + mf - 0.1mf mg = 16.9 mf mg = 16.9 (2500) mg = 42,250 kg/hr 14. The gas density of chimney is 0.75 kg/m3 and air density of 1.15 kg/m3. If the driving pressure is 0.25 kPa, determine the height of chimney. a. 54.6 m

b. 63.7 m*

c. 74.6 m

d. 68.5 m

Solution:

Generator Output = 2814.63 Kw

hw = H (da - dg)

12. Determine the piston speed of a 250 mm x 300 mm diesel engine running at 1200 rpm.

0.25 = H (1.15 - 0.75)

a. 6 m/s

b. 12 m/s*

c. 18 m/s

d. 5 m/s

Solution:

Piston Speed = 2 L N Piston Speed = 2 (0.30) (1200/60) Piston Speed = 12 m/s 13. A boiler uses 2500 kg of coal per hour and air required for the combustion in 16 kg per kg coal. If ash loss is 10%, determine the mass of the gas entering the chimney. a. 42,250 kg/hr*

b. 78,300 kg/hr

c. 85,452 kg/hr

d. 33,800 kg/hr

H = 63.71 m

15. The actual velocity of gas entering in a chimney is 8 m/s. The gas temperature is 25 ͦC and pressure of 98 kPa with a gas constant of 0.287 kJ/kg - K. Determine the chimney diameter if mass of gas is 50,000 kg/hr. a. 1.57 m

b. 1.81 m

c. 3.56 m

d. 1.39 m*

Solution: PgVg = mgRgTg 98(Vg) = (25+273)

(50,000/3600)

Vg = 12.12 m3/sec

(0.287)

Vg = A x v 12.12 = (π/4) D2 (8) D = 1.39 m

HYDRO-ELECTRIC & GAS TURBINE POWERPLANT. 1. A hydro-electric power plant consumes 52,650,000 KW-hrs. per annum. Expected flow is 1665 m³/min and overall efficiency is 65%. What is the net head? a. 30 m

b. 31 m

c. 32 m

d. 34 m*

SOLUTION: 𝐺𝑒𝑛.𝑂𝑢𝑡𝑝𝑢𝑡

Eff.net = 𝑊𝑎𝑡𝑒𝑟𝑃𝑜𝑤𝑒𝑟 0.65 =

52650000/8760 𝑊𝑎𝑡𝑒𝑟𝑃𝑜𝑤𝑒𝑟

Water Power = 9246.575 KW Water Power = wQh 9246.575 = 9.81(1665/60) h = 33.966 m / 34 m

2. In a hydro-electric power the tail water level fixes at 480 m. The net head is 27 m and head loss is 4% of the gross head. What is the head water elevation? a. 500.34 m

b. 508.12 m*

c. 456.34 m

d. 567.34 m

SOLUTION:

SOLUTION: hg = H.W. Elev – T.W. Elev. hg = 195 – 160 hg = 35 m h = hg + hL h =35 – (0.05)(35) h = 33.25 m

h = hg + hL

Water Power = wQh

27 = hg + 0.04hg

Water Power = 9.81(10)(33.25)

hg = H.W. Elev – T.W. Elev.

Water Power = 3261.825 KW

28.125 = H.W. Elev – 480 H.W. Elev = 508.125 m 3. The available flow of water is 25 m³/sec at 30 m elevation. If a hydroelectric plant is to be installed with turbine efficiency of 85% and generator efficiency of 90%, what maximum power that the plant could generate?

5. The flow of the river is 20 m³/sec and produced a total brake of 6,000 KW. If it is porposed to install two turbines each has 85% efficiency, what is the available head? a. 35 m

b. 37 m

c. 39 m

d. 36 m*

a. 4658.5 KW

b. 3478.5 KW

SOLUTION:

c. 5628.5 KW*

d. 4758.5 KW

Water Power = 6,000/0.85

SOLUTION:

Water Power = 7058.82 KW

Water Power = wQh

7058.82 = 9.81(20)h

Water Power = 9.81(25)(30)

h = 35.98 m / 36 m

Water Power = 7357.5 KW

6. Two turbines generates a total brake power of 5000 KW. If one unit is thrice the capacity of the other, find the capacity of smaller unit.

Gen. Output = 7357.5 KW (0.85)(0.9) Gen. Output = 5628.5 KW 4. For a proposed hydro-electric plant, the tail water and head water elevation is 160 m and 195 m, respectively. If available flow is 10 m³/sec and head loss of 5% of water available head, what is the water power?

a. 1250 KW*

b. 3450 KW

c. 2456 KW

d. 5763 KW

a. 3261.8 KW*

b. 4254.6 KW

W2 = 3W 1

c. 5874.5 KW

d. 2456.5 KW

5000 = W 1 + 3 W 2

SOLUTION: WT = W1 +W 2

W1 = 1250 KW

SOLUTION:

7. In a hydro-electric plant the brake power is 1800 KW running at 450 rpm and net head of 30 m. Determine the specific speed of the turbine.

N=

a. 60.29 rpm

b. 65.29 rpm

c. 75.29 rpm

d. 71.29 rpm*

120𝑓 𝑃

300 =

120(60) 𝑃

P = 24 poles

SOLUTION: Hp = 1800 x 1Hp/0.746 KW Hp = 2414.87 Hp Ns =

𝑁√𝐻𝑃 ℎ^5⁄4

Ns =

450√2414.87 (30𝑥3.281)^5⁄4

10. The penstock of hydro-electric plant is 0.5x0.5 m with the velocity of 5.5 m/sec has a head of 20 m. What is the output of the turbine if turbine efficiency os 87%? a. 845.32 KW

b. 789.34 KW

c. 654.56 KW

d. 234.56 KW*

Ns = 71.29 rpm

SOLUTION:

8. The specific speed of turbine is 75 rpm and running at 450 rpm. If the head is 20 m and generator efficiency is 90%, what is the maximum power delivered by the generator.

Q=Axv

a. 450.5 KW

b. 650.5 KW

Water Power = wQh

c. 650.5 KW*

d. 780.5 KW

Water Power = 9.81(1.375)(20)

75 =

Q = 1.375 m³/sec

Water Power = 269.775 KW

SOLUTION: Ns =

Q = (0.5 x 0.5)(5.5)

Turbine Output = 269.775(0.87)

𝑁√𝐻𝑃 ℎ^5⁄4

Turbine Output = 234.70 KW

450√𝐻𝑝 (20𝑥3.281)^5⁄4

Hp = 968.92

11. An air standard Brayton cycle has a pressure ratio of 12. Find the thermal efficiency of the engine.

Gen. Output = (968.92 x 0.746)(0.9)

a. 34.23%

b. 50.83%*

Gen. Output = 650.53 KW

c. 56.32%

d. 65.23%

9. For a generator running at 300 rpm and 60 Hz, find the number of generator poles.

SOLUTION:

a. 24 poles* c. 18 poles

b. 8 poles d. 20 poles

e=1e=1-

1 (𝑅𝑝)

𝑘−1 𝑘

1 (12)

1.4−1 1.4

e = 50.83%

12. An air standard Brayton cycle has an air leaving the high-temperature heat exchanger at 850℃ and leaving the turbine at 310℃. What is the thermal efficiency?

T2

= (8)^ 298

1.4−1 1.4

T2 = 539.81°𝐾 T3 T4

P3

=( P4)^

1.4−1 1.4

T3 = 1100 + 273 a. 42.21%

b. 23.34%

T3 = 1373 °𝐾

c. 48.08%

d. 56.34%

1373

SOLUTION:

Wc = m(1)( 539.81 – 298)

T4 = 310 + 237

Wc/m = 241.81 KJ/kg

T4 = 583 °𝐾 = (rp) ^

WT = mCp(T3 – T4)

1.4−1

1123

= (rp) ^ 583

1.4

WT = m(1)( 1373 – 757.95)

1.4−1

WT/m = 615.05 KJ/kg

1.4

WNET = 615.05 - 241.81

rp = 9.919 e=1-

1.4

Wc = mCp(T2 - T1)

T3 = 1123 °𝐾

T4

1.4−1

T4 = 757.95°𝐾

T3 = 850 + 273

T3

T4

= (8)^

WNET = 373.24 KJ/kg

1 1.4−1 (9.919) 1.4

e = 48.08% 13. An air standard Brayton cycle has a pressure ratio of 8. The air properties at the start of compression are 100 kPa and 25℃. The maximum allowance temperature is 1100℃. Determine the net work.

14. The air standard Brayton cycle has a net power output of 100 kw. The working substance is air, entering the compressor at 30℃, leaving the high temperature heat exchanger at 750℃ and leaving the turbine at 300℃. Determine the mass flow rate of air. a. 1698 kg/hr*

b. 1543 kg/hr d. 2344 kg/hr

a. 373.24 KJ/kg*

b. 283.45 KJ/kg

c. 1543 kg/hr

c. 321.34 KJ/kg

d. 398.23 KJ/kg

SOLUTION:

SOLUTION: P2 P1

T1 = 303°𝐾

=8

T1 = 25 + 273

P2

=( P1)^ T1

T3 = 750 + 273 T3 = 1023 °𝐾

T1 = 298°𝐾 T2

T1 = 30 + 273

1.4−1 1.4

T4 = 300 + 273 T4 = 573°𝐾

T2

T3

= T4 T1 T2 303

=

1023 573

T2 = 540.96°𝐾 Wc = mCp(T2 - T1) Wc = m(1)( 540.96 – 303)

1. Mass flow rate of ground water in a geothermal powerplant is 1,500,000 kg/hr and the quality after the throttling is 30%. Determine the brake power of turbine if the change in enthalpy of steam at inlet and outlet is 700 KJ/kg. A. 68.5 MW B. 87.5 MW* C. 64.5 MW D. 89.5 MW

Wc = 237.96 m

SOLUTION:

WT = mCp(T3 – T4)

ms = x(mg) = 0.3 (1,500,000)

WT = m(1)( 1023 – 573)

ms = 450,000 kg/hr = 125 kg/sec

WT = 450 m

Wt = ms (h3-h4) = 125(700) = 87,500 KW

WNET = W T + W c 100 = 450 m + 237.96 m

Wt = 87.5 MW

m = 0.4716 kg/s (3600) m = 1697.79 kg/hr 15. The compressor for an actual gas turbine requires 300 KJ/kg of work to quadruple the inlet pressure. The inlet air temperature is 100℃. Determine the compressor air exit temperature. a. 234°𝐾

b. 542°𝐾

c. 653°𝐾

d. 673°𝑲*

SOLUTION: T1 = 100 + 273 T1 = 373°𝐾 Wc = mCp(T2 - T1)

A. Ground water of geothermal powerplant has an enthalpy of 700 KJ/kg and turbine inlet is 2750 KJ/kg and enthalpy of hot water in flash tank is 500 KJ/kg . What is the mass of steam flow entering the turbine if mass flow of ground water is 45 kg/sec? A. 3.27 kg/s B. 2.27 kg/s C. 4.27 kg/s D. 9.27 kg/s SOLUTION: H2 = Hf + X(Hfg-hf) 700 = 500 + X(2750-500) X = 0.0888 ms = x * mg = 0.0888 * 45 = 4 kg/sec

Wc/m = Cp(T2 - T1) 300 = 1(T2 – 373) T2 = 673°𝑲

GEOTHERMAL

2. The enthalpy entering the turbine of a geothermal powerplant is 2750 KJ/kg and mass rate of 1 kg/sec. The turbine brake power is 1000 KW condenser outlet has an enthalpy of 210 KJ/kg. If the temperature rise of cooling water in condenser is 8◦C, what is mass of cooling water requirement?

A. 44 kg/sec C.46 kg/sec

B. 45 kg/sec D. 47 kg/sec

SOLUTION:

SOLUTION: WT = ms(h3-h4) ; (16,000/0.9*0.8) = ms (500) Ms = 44.44 kg/sec = 160,000 kg/hr

Wt = ms ( h3-h4); 1000 = 1(2750-h4); h4 = 1750 KJ/kg

160,000 = 0.20(Mg)

Qr = Qw

Mg = 800,000 kg/hr

1(1750-210) = mw(4.187)(8)

No. of wells = 800,000/200,000 = 4 wells

Mw = 45.97 kg/sec 3. In a 12 MW geothermal powerplant, the mass flow of steam entering the turbine is 26 kg/s. The quality after throttling is 25% and enthalpy of ground water is 750 KJ/kg. Determine the overall efficiency of the plant? A. 7.4 % B.9.6 % C.5.4 % D. 15.4 %* SOLUTION: Ms = x Mg 26 = 0.25 Mg Mg = 104 kg/sec

5. A geothermal powerplant draws a pressurized water from a well at 20 MPa and 300◦C. To produce a steam water mixture in the separator, where the unflashed water is removed, this water is throttled to a pressure of 1.5 MPa. The flashed steam which is dry and saturated passes through the steam collector and enters the turbine at 1.5 MPa and expands to 1 atm. The turbine efficiency is 85% at a rated power output of 10 MW. Calculate overall plant efficiency. A. 7.29 % B. 12.34 % C. 9.34 % D. 19.45 % SOLUTION:

%overall = 12,000/(104*750) = 15.38%

@ 1.5 MPa h3 = 2792.2 KJ/kg, s3 = 6.4448 @ 1 atm (100 C) sf = 1.3069, sfg = 6.048, hf = 419.04, hfg = 2257 H4 = 2336.4 KJ/kg X4 = 0.8495

4. A 16,000 KW geothermal plant has a generator efficiency and turbine efficiency of 90% and 80%, respectively. If the quality after throttling is 20% and each well discharges 200,000 kg/hr , determine the number of wells are required to produce if the change of enthalpy at entrance and exit of turbine is 500 KG/kg. A. 4 wells* B. 5 wells C. 6 wells D. 8 wells

Wt = ms (h3-h4) %T 10,000 = ms (2792.2 – 2336.4)(0.85) ms = 25.81 kg/sec @ 20 MPa and 300 C h1= 1333.3 KJ/kg @ 1 MPa hf = 844.89, hfg = 1947.3, x2 = 0.25 ms = x2 mg

(25.81 x 3600) = 0.25 (mg) Mg = 371,664 kg/hr %t = (10,000)/(371,644/3600)(1333.3) = 7.26 %

6. A flashed steam geothermal powerplant is located where underground hot water is available as saturated liquid at 700 KPa. The well head pressure is 600 KPa. The flashed steam enters a turbine at 500 KPa and expands to 15 Kpa, when it is condensed. The flow rate from the well is 29.6 kg/s. Determine the power produced in KW. A. 430.13 KW B. 540.23 KW C. 370.93 KW D, 210.34 KW SOLUTION: H1 = 697.22 KJ/kg, h3 = 2748.7 KJ/kg, hf = 640.23, hfg = 2108.5 697.22 = 640.23 + (X) (2108.5) X2 = 0.027 ms = x mg ms = 0.027 (29.6) = 0.80 kg/sec Power produced = 0.80(2748.7 – 2211) = 430.16 KW

7. A 18,000 KW geothermal plant has a generator efficiency of 90% and 80% respectively. If the quality after throttling is 20% and each well discharges 400,000 kg/hr , determine the number of wells are required to produce if the change of enthalpy at

entrance and exit of turbine is 500 KG/kg. A. 4 wells wells D. 8 wells

B. 6 wells

C. 2

SOLUTION: WT = ms(h3-h4) ; (18,000/0.9*0.8) = ms (500) Ms = 50 kg/sec = 180,000 kg/hr 180,000 = 0.20(Mg) Mg = 900,000 kg/hr No. of wells = 900,000/400,000 = 2.25 wells 8. In an ideal Rankine cycle, the steam throttle condition is 4.10 MPa and 440◦C. If the turbine exhaust is 0.105 MPa, determine the thermal efficiency of the cyle. A. 20.34 % B. 27.55 % C. 34.44 % D. 43.12 % SOLUTION: h1 = 3305.7, s1 = 6.8911, x1 = 0.925, h2 = 2508.54, h3 = 423.24, h4 = 427.412 Qa = h1-h4 = 3305.7-427.412 = 2878.29 KJ/kg Wt = h1-h2 = 3305.7-2508.54 = 797.16 KJ/kg Wp = h4-h3 = 427.412- 423.24 = 4.172 KJ/kg Wnet = Wt-Wp = 797.16 -4.172 = 792.99 KJ/kg %t = Wnet/ Qa = 792.99/2878.29 = 27.55 % 9. In a Rankine cycle, saturated liquid water @ 1 bar is compressed isentropically to 150 bar. First by heating in a boiler, and then by superheating at constant pressure of 150 bar, the water substance is brought to 750 K. After adiabatic reversible expansion in a turbine to 1 bar, it is then cooled in a condenser to saturated liquid. What is the thermal efficiency? A. 23.45 % B. 16.23 % C. 34.24 % D. 18.23 %

SOLUTION: h1=3240.5, s1=6.2549, x=0.8176, h2= 2263.6, h4 = 433, h3= 417.46 Wp = h4=h3 = 433 - 417.46 = 15.54 KJ/kg Wt = h1-h2 = 3240.5 – 2263.6 = 976.9 KJ/kg Efficiency = (976.9-15.54)/(3240.5433) = 34.24 %

March 2021 0
January 2021 1
January 2021 0
March 2021 0
January 2021 0
February 2021 0

March 2021 0
February 2021 0
February 2021 0
February 2021 1
February 2021 1
March 2021 0