Sample-problem

Sample Problem: 1. A fan draws 1.42 m3 per second of air at a static pressure of 2.54 cm of water through a duct 300 mm diameter and discharges it through a duct of 275 mm diameter. Determine the static fan efficiency if total fan mechanical is 70% and air is measured at 25 0C and 760 mmHg. Given: Q = 1.42

𝑚3 𝑠

hw = 2.54 cm d = 300 mm d = 275 mm ηm = 70% T = 25 0C P = 760 mmHg Required: ηs = ? Solution: 𝜌𝑎 = 𝜌𝑎 =

𝑃 𝑅𝑇

101.325 𝑘𝑃𝑎 𝑘𝐽

(0.287 𝑘𝑔 𝐾)(25 + 273)𝐾 𝑚3 𝑘𝑔

𝜌𝑎 = 1.184727451 

Solving for Static Head: ℎ𝑠 =

ℎ𝑠 =



(2.54 𝑐𝑚 𝑥

ℎ𝑤 𝜌𝑓 𝜌𝑎 1𝑚

𝑘𝑔

)(1000 𝑚3 ) 100𝑐𝑚 𝑘𝑔

1.184727451 𝑚3

ℎ𝑠 = 21.43953022 𝑚 Solving for Vs and Vd: 𝑄 𝑉𝑠 = = 𝐴𝑠

1.42 𝜋 4

(300𝑚𝑚 𝑥

𝑚3 𝑠 1𝑚 1000𝑚𝑚

𝑉𝑠 = 20.08889059

𝑚 𝑠

)2

𝑄 𝑉𝑑 = = 𝐴𝑑

1.42

𝑚3 𝑠

𝜋

(275𝑚𝑚 𝑥 4

1𝑚 1000𝑚𝑚

𝑉𝑑 = 23.90744005 

)2

𝑚 𝑠

Solving for h: ℎ𝑣 =

𝑉𝑑 2 − 𝑉𝑠 2 2(𝑔) 𝑚 2

ℎ𝑣 =

𝑚 2

(23.90744005 𝑠 ) − (20.08889059 𝑠 ) 𝑚

2 (9.81 𝑠2 ) ℎ𝑣 = 8.56280144 𝑚 ℎ = ℎ𝑠 + ℎ𝑣 ℎ = 21.43953022 + 8.56280144𝑚 ℎ = 30.00233166𝑚



Solving for ηT :

𝑘𝑔

0.70 =

𝜂𝑇 =

Ɣ𝑎 𝑄ℎ 𝐵𝑃

𝑚

1𝑘𝑁

((1.184727451 𝑚3 )(9.81 𝑠2 )(1000𝑁)(1.42

𝑠

)(30.00233166𝑚)

𝐵𝑃 𝐵𝑃 = 0.7073474155 𝑘𝑊 𝜂𝑠 = 𝑘𝑔

𝜂𝑠 =

𝑚3

𝑚

Ɣ𝑎 𝑄ℎ𝑠 𝐵𝑃 1𝑘𝑁

((1.184727451 𝑚3 )(9.81 𝑠2 )(1000𝑁)(1.42

𝑚3 𝑠

)(21.43953027𝑚)

0.7073474155 𝑘𝑊 𝜂𝑠 = 0.5002168273 𝑥 100 𝜂𝑠 = 50.0217%

Sample problem: A fan delivers 4.7 m3/s at a static pressure of 5.08 cm of water when operating at a speed of 400 rpm. The power input required is 2.963 kW. If 7.05 m3/s are desired in the same fan and installation, find the pressure in cm of water. Given:

Q1 = 4.7 m3/s h1 = 5.08 cm of water N1 = 400 rpm Pi = 2.963 kW Q2 = 7.05 m3/s

Required: h2= ?

Solution: 

Solving for N2: 𝑄1 𝑁1 = 𝑄2 𝑁2

4.7

𝑚3

7.05

𝑠 𝑚3 𝑠

=

400 𝑟𝑝𝑚 𝑁2

𝑁2 = 600 𝑟𝑝𝑚 

Solving for h2: ℎ1 𝑁1 = ( )2 ℎ2 𝑁2 5.08 𝑐𝑚 400𝑟𝑝𝑚 2 =( ) ℎ2 600𝑟𝑝𝑚 ℎ2 = 11.43 𝑐𝑚 of water

PROBLEM: A blower operating at 1500 rpm compresses air from 68°F and 14.7 psia to 10 psig. The desired flow is 1350 cfm and at this point, brake horsepower is 80 hp. Determine the over-all blower efficiency at design point if k=1.3395. GIVEN: N= 1500 rpm T1= 68°F P1= 14.7 psia P2= 10 psig v= 1350cfm BP= 80 hp k= 1.3395 REQUIRED: Overall blower efficiency, ŋ𝑜 SOLUTION: Solving for work using isentropic compression

𝑊=

𝑘 𝑃2 𝑘−1 (𝑃1 𝑉1 ) [( ) 𝑘 − 1] 𝑘−1 𝑃1

1.3395 𝑙𝑏 12𝑖𝑛 2 𝑓𝑡 3 1𝑚𝑖𝑛 𝑊= 𝑥 [(14.7 2 𝑥 ( ) ] (1350 )𝑥 1.3395 − 1 𝑖𝑛 1𝑓𝑡 𝑚𝑖𝑛 60𝑠 1.3395−1 1.3395

10 𝑝𝑠𝑖𝑔 + 14.7 𝑝𝑠𝑖 [( ) 14.7 𝑝𝑠𝑖𝑎 𝑊 = 26416.01081

− 1]

𝑙𝑏 − 𝑓𝑡 1ℎ𝑝 𝑥 𝑙𝑏−𝑓𝑡 𝑠 550 𝑠

𝑊 = 48.0291057 ℎ𝑝

Solving for overall blower efficiency

𝐵𝐻𝑃 = ŋ𝑜 = ŋ𝑜 =

𝑊 ŋ𝑜

𝑊 𝑥100% 𝐵𝐻𝑃

48.0291057ℎ𝑝 × 100% 80ℎ𝑝

ŋ𝑜 = 60.03638821%

PROBLEM: A 50 kW motor is used to drive a fan that has a total head of 110 m. If fan efficiency is 70%, what is the maximum capacity of the fan?

GIVEN: BP= 50 kW HT= 110m Ŋf= 70% REQUIRED: Maximum capacity of the fan, Q SOLUTION: For the total air power, TAP

ŋ𝑓 =

𝑇𝐴𝑃 𝐵𝑃

𝑇𝐴𝑃 = (ŋ𝑓 )(𝐵𝑃) 𝑇𝐴𝑃 = (0.70)(50𝑘𝑊) 𝑇𝐴𝑃 = 35𝑘𝑊

Using the density of air at 21.2°C, 𝜌𝑎 = 1.20029845 kg/m3

𝑇𝐴𝑃 = 𝛾𝑄𝐻𝑇 ; 𝛾 = 𝜌𝑔 𝑄= 𝑄=

𝑇𝐴𝑃 𝜌𝑔𝐻𝑇

35000𝑊 𝑘𝑔

𝑚

1.200029845 𝑚3 (9.81 𝑠2) (110𝑚) 𝑚3 𝑄 = 27.02802454 𝑠

A fan described in a manufacturer’s table is rated to deliver 500 m 3/min at a static pressure (gage) of 254 cm when running at 250 rpm and requiring 3.6 kW. If the fan speed is changed to 305 rpm and air handled were at 65℃ instead of standard 21℃, find the power in kW. Given: N1 = 250 rpm N2 = 305 rpm T1 = 21℃ T2 = 65℃ Required: P2 Solution : At 305 rpm and 21℃; P1 N1 = ( )3 P2 N2 3.6kW 250 rpm 3 =( ) P2 305 rpm P2 = 6.5370528kW At 305 rpm and 65℃; ρ=

P RT

ρ1 𝑃⁄𝑅𝑇1 = ρ2 𝑃⁄𝑅𝑇2 BP1 ρ1 𝑇2 = = BP2 ρ2 𝑇1

6.5370528kW 65 + 273𝐾 = P2 21 + 273 𝐾 P2 = 5.686075512kW

The volume of flow of air delivered by fan is 20 m3/sec and 180 mm water gage. The density of air is 1.185 kg/m3 and the motor power needed to drive the fan is 44 kW. What is the efficiency?

Given: h = 180 mm 𝑄 = 20 m3/sec ρ = 1.185 kg/m3 MP = 44 kW Required: Fan efficiency (𝜂) Solution: 1000 kg⁄m3

ℎ𝑠 = 0.18m(1.185 kg⁄m3) ℎ𝑠 = 151.8987342m

Solving for Air Power; Air Power = γQh Air Power = (1.185 kg⁄m3 × 0.00981 × 200 m3 ⁄s × 151.8987342m) Air Power = 35.316kW

𝜂=

Air Power Motor Power

𝜂=

35.316kW 44kW

𝜂 = 0.8026363636 × 100% 𝜂 = 80.26363636%

A fan whose static efficiency is 40% has a capacity of 60,000 ft 3/hr at 60°F and barometer of 30 in Hg and gives a static pressure of 2 in of water column on full delivery. What size of electric motor should be used to drive the fan?

Given: 𝑒𝑠 = 40% 𝑄 = 60,000

𝑓𝑡 3 ℎ𝑟

𝑇 = 60°𝐹 𝑃𝑎𝑡𝑚 = 30 𝑖𝑛 𝐻𝑔 ℎ𝑤 = 2 𝑖𝑛 Required: 𝐻𝑝 Solution: ℎ𝑠 =

ℎ𝑤 Ɣ𝑤 Ɣ𝑎 (2 𝑖𝑛 𝑥

ℎ𝑠 =

1 𝑓𝑡 12 𝑖𝑛

) (62.4

Ɣ𝑎

𝑙𝑏 𝑓𝑡 3

)

10.4 ℎ𝑠 =

𝑙𝑏 𝑓𝑡 2

Ɣ𝑎

𝐴𝑖𝑟 𝑃𝑜𝑤𝑒𝑟 = Ɣ𝑎 𝑄ℎ Ɣ𝑎 (60,000 𝐴𝑖𝑟 𝑃𝑜𝑤𝑒𝑟 =

𝑓𝑡 3 ℎ𝑟

1 ℎ𝑟

10.4

) (60 𝑚𝑖𝑛) (

𝑙𝑏 𝑓𝑡2

Ɣ𝑎

)

𝑓𝑡−𝑙𝑏 𝑚𝑖𝑛

33,000

1 ℎ𝑝

𝐴𝑖𝑟 𝑃𝑜𝑤𝑒𝑟 = 0.3151515152 ℎ𝑝 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑢𝑠𝑒 1 ℎ𝑝 𝑚𝑜𝑡𝑜𝑟 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

Air is flowing in a duct with velocity of 7.62 m/s and static pressure of 2.16 cm of water gauge. The duct diameter is 1.22 m, the barometric pressure 99.4 kPa and the gage fluid temperature and air temperature are 30°C. What is the total pressure against which the fan will operate in cm of water? Given: 𝑣 = 7.62

𝑚 𝑠

ℎ𝑤 = 2.16 𝑐𝑚 𝑑 = 1.22 𝑚 𝑃𝑎𝑡𝑚 = 99.4 𝑘𝑃𝑎 𝑇 = 30°𝐶 Required: ℎ Solution: ℎ𝑣 =

ℎ𝑣 =

𝑣2 2𝑔 (7.62

𝑚 2 𝑠

2 (9.81

)

𝑚 𝑠2

)

ℎ𝑣 = 2.959449541 𝑚

Ɣ= Ɣ=

𝑃 𝑅𝑇 99.4 𝑘𝑃𝑎 (0.287

𝑘𝐽 𝑘𝑔𝐾

) (30 + 273𝐾)

Ɣ = 1.143041133

𝑘𝑔 𝑚3

𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑐𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 1.143041133 ℎ𝑣 = 2.959449541 𝑚 ( 𝑘𝑔 1000 𝑚3 ℎ𝑣 = 3.382772558𝑥10−3 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 ℎ𝑣 = 0.3382772558 𝑐𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 ℎ = ℎ𝑠 + ℎ𝑣 ℎ = 2.16 𝑐𝑚 + 0.3382772558 𝑐𝑚 ℎ = 2.498277256 𝑐𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

𝑘𝑔 𝑚3

)

1. Air enters a fan through a duct at a velocity of 6.3 m/s and an inlet static pressure of 2.5 cm of water less than atmospheric pressure. The air leaves the fan through duct at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. If the specific weight of the air is 1.20 kg/m 3 and the fan delivers 9.45 m3/sec, what is the fan efficiency when the power input to the fan is 13.75 kW at the coupling? GIVEN: 𝑣𝑠 = 6.3

𝑚 𝑠

𝑣𝑑 = 11.25

𝑚 𝑠

ℎ𝑠 = 2.5 𝑐𝑚 ℎ𝑑 = 11.25 𝑐𝑚

𝑤 = 1.20

𝑘𝑔 𝑚3

𝑚3 𝑄 = 9.45 𝑠

REQUIRED: Fan Efficiency SOLUTION: ℎ = ℎ𝑠 + ℎ𝑣

ℎ=(

ℎ=[

𝑃𝑑 −𝑃𝑠 𝑣𝑑 2 −𝑣𝑠 2 )+( ) 𝑤 2𝑔

0.0762 𝑚 − (−0.025 𝑚) 1.20

𝑘𝑔

] (1000) + [

(11.25

𝑚3

𝑚 2

) − (6.3 𝑠

2 (9.81

𝑚

) 𝑠2

𝑚 2 𝑠

)

]

ℎ = 88.76108563 𝑚 𝐴𝑖𝑟 𝑝𝑜𝑤𝑒𝑟 = 𝑤𝑄ℎ 𝑘𝑔 𝑚3 𝐴𝑖𝑟 𝑝𝑜𝑤𝑒𝑟 = (1.20 3 𝑥 0.00981) (9.45 ) (88.76108563 𝑚) 𝑚 𝑠 𝐴𝑖𝑟 𝑝𝑜𝑤𝑒𝑟 = 9.874262475 𝑘𝑊

𝐹𝑎𝑛 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =

𝐹𝑎𝑛 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =

𝐴𝑖𝑟 𝑝𝑜𝑤𝑒𝑟 𝑃𝐼𝑁

9.874262475 𝑘𝑊 13.75 𝑘𝑊

𝐹𝑎𝑛 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 71.812818%

2. A fan is listed as having the following performance with standard air: Volume Discharge - 120 m3/s Speed – 7 rps Static Pressure – 310 mm water gage Brake Power Required – 620 kW The system duct will remain the same and the fan will discharge the same volume of 120 m3/s of air at 93oC and a barometric pressure of 735 mm Hg when its speed is 7 rps. Find the brake power input and the static pressure required. REQUIRED: Brake power input Static pressure SOLUTION: For standard air, 𝑤 = 1.2

𝑘𝑔 𝑚3

Solving for the density at 93oC and 735 mm Hg

𝑤=

𝑃 𝑅𝑇 101.325 𝑘𝑃𝑎

𝑤=

735 𝑚𝑚 𝐻𝑔 ( 760 𝑚𝑚 𝐻𝑔 ) 0.287

𝑘𝐽 𝑘𝑔−𝐾

(93 + 273)

𝑤 = 0.9328834256

𝑘𝑔 𝑚3

Using Fan Laws: 𝑤 𝐵𝑟𝑎𝑘𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 = ( ) (𝐵𝑟𝑎𝑘𝑒 𝑝𝑜𝑤𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑) 𝑤𝑎𝑖𝑟 0.9328834256 𝐵𝑟𝑎𝑘𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 = ( 𝑘𝑔 1.20 𝑚3

𝑘𝑔 𝑚3

) (620 𝑘𝑊)

𝐵𝑟𝑎𝑘𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 = 481.9897699 𝑘𝑊 𝑆𝑡𝑎𝑡𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = (

𝑤 ) (𝑆𝑡𝑎𝑡𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒) 𝑤𝑎𝑖𝑟

0.9328834256 𝑆𝑡𝑎𝑡𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = ( 𝑘𝑔 1.20 𝑚3

𝑘𝑔 𝑚3

) (310 𝑚𝑚)

𝑆𝑡𝑎𝑡𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 240.9948849 𝑚𝑚

At 1.2 kg/m3 air density, a fan develops a brake power of 100kW. If it operates at 98Kpa and 32oC with the same speed, what is the new brake power of the fan? Given:

w1=1.2 kg/m3 BP1=100kW P=98Kpa T=32oC

Required: BP2 Solution:

Solving for the new air density,

w2 

P 98Kpa  RT (0.287 kJ / kg - K) (32 + 273)K

w2  1.1195kg / m 3 Solving for the new brake power of the fan,

BP2 w2  BP1 w1 BP2 1.1195kg / m 3  100kW 1.2kg / m 3 BP2  93.29kW

A fan has a suction pressure of 30mm water vacuum with air velocity of 3m/sec. The discharge has 150mm of water gage and discharge velocity of 7m/sec. Determine the total head of fan if air density is 1.2 kg/m3. Given: hw1= 30mm = .03m vs= 3m/sec hw2= 150mm = .15m vd= 7m/sec da= 1.2 kg/m3 Required: Total head of fan Solution:

Solving for static head,

hs 

(hw2  hw1 )d w da

(0.15m  (0.03m))1000kg / m 3 hs  1.2kg / m 3 hs  150m

Solving for velocity head,

v  vs hv  d 2g 2

2

(7 m / sec) 2  (3m / sec) 2 2(9.81m / sec 2 ) hv  2.038m

hv 

Solving for the total head of fan,

h  hs  hv h  150m  2.038m h  152.038m