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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
π‘š3
𝑠
hw = 2.54 cm
d = 300 mm
d = 275 mm
ηm = 70%
T = 25 0C
P = 760 mmHg
Required:
ηs = ?
Solution:
πœŒπ‘Ž =
πœŒπ‘Ž =
𝑃
𝑅𝑇
101.325 π‘˜π‘ƒπ‘Ž
π‘˜π½
(0.287 π‘˜π‘” 𝐾)(25 + 273)𝐾
π‘š3
π‘˜π‘”
πœŒπ‘Ž = 1.184727451
ο‚·
Solving for Static Head:
β„Žπ‘  =
β„Žπ‘  =
ο‚·
(2.54 π‘π‘š π‘₯
β„Žπ‘€ πœŒπ‘“
πœŒπ‘Ž
1π‘š
π‘˜π‘”
)(1000 π‘š3 )
100π‘π‘š
π‘˜π‘”
1.184727451 π‘š3
β„Žπ‘  = 21.43953022 π‘š
Solving for Vs and Vd:
𝑄
𝑉𝑠 =
=
𝐴𝑠
1.42
πœ‹
4
(300π‘šπ‘š π‘₯
π‘š3
𝑠
1π‘š
1000π‘šπ‘š
𝑉𝑠 = 20.08889059
π‘š
𝑠
)2
𝑄
𝑉𝑑 =
=
𝐴𝑑
1.42
π‘š3
𝑠
πœ‹
(275π‘šπ‘š π‘₯
4
1π‘š
1000π‘šπ‘š
𝑉𝑑 = 23.90744005
ο‚·
)2
π‘š
𝑠
Solving for h:
β„Žπ‘£ =
𝑉𝑑 2 − 𝑉𝑠 2
2(𝑔)
π‘š 2
β„Žπ‘£ =
π‘š 2
(23.90744005 𝑠 ) − (20.08889059 𝑠 )
π‘š
2 (9.81 𝑠2 )
β„Žπ‘£ = 8.56280144 π‘š
β„Ž = β„Žπ‘  + β„Žπ‘£
β„Ž = 21.43953022 + 8.56280144π‘š
β„Ž = 30.00233166π‘š
ο‚·
Solving for ηT :
π‘˜π‘”
0.70 =
πœ‚π‘‡ =
Ζ”π‘Ž π‘„β„Ž
𝐡𝑃
π‘š
1π‘˜π‘
((1.184727451 π‘š3 )(9.81 𝑠2 )(1000𝑁)(1.42
𝑠
)(30.00233166π‘š)
𝐡𝑃
𝐡𝑃 = 0.7073474155 π‘˜π‘Š
πœ‚π‘  =
π‘˜π‘”
πœ‚π‘  =
π‘š3
π‘š
Ζ”π‘Ž π‘„β„Žπ‘ 
𝐡𝑃
1π‘˜π‘
((1.184727451 π‘š3 )(9.81 𝑠2 )(1000𝑁)(1.42
π‘š3
𝑠
)(21.43953027π‘š)
0.7073474155 π‘˜π‘Š
πœ‚π‘  = 0.5002168273 π‘₯ 100
πœ‚π‘  = 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:
ο‚·
Solving for N2:
𝑄1 𝑁1
=
𝑄2 𝑁2
4.7
π‘š3
7.05
𝑠
π‘š3
𝑠
=
400 π‘Ÿπ‘π‘š
𝑁2
𝑁2 = 600 π‘Ÿπ‘π‘š
ο‚·
Solving for h2:
β„Ž1
𝑁1
= ( )2
β„Ž2
𝑁2
5.08 π‘π‘š
400π‘Ÿπ‘π‘š 2
=(
)
β„Ž2
600π‘Ÿπ‘π‘š
β„Ž2 = 11.43 π‘π‘š 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, Ε‹π‘œ
SOLUTION:
Solving for work using isentropic compression
π‘Š=
π‘˜
𝑃2 π‘˜−1
(𝑃1 𝑉1 ) [( ) π‘˜ − 1]
π‘˜−1
𝑃1
1.3395
𝑙𝑏
12𝑖𝑛 2
𝑓𝑑 3 1π‘šπ‘–π‘›
π‘Š=
π‘₯
[(14.7 2 π‘₯ (
) ] (1350
)π‘₯
1.3395 − 1
𝑖𝑛
1𝑓𝑑
π‘šπ‘–π‘› 60𝑠
1.3395−1
1.3395
10 𝑝𝑠𝑖𝑔 + 14.7 𝑝𝑠𝑖
[(
)
14.7 π‘π‘ π‘–π‘Ž
π‘Š = 26416.01081
− 1]
𝑙𝑏 − 𝑓𝑑
1β„Žπ‘
π‘₯
𝑙𝑏−𝑓𝑑
𝑠
550
𝑠
π‘Š = 48.0291057 β„Žπ‘
Solving for overall blower efficiency
𝐡𝐻𝑃 =
Ε‹π‘œ =
Ε‹π‘œ =
π‘Š
Ε‹π‘œ
π‘Š
π‘₯100%
𝐡𝐻𝑃
48.0291057β„Žπ‘
× 100%
80β„Žπ‘
Ε‹π‘œ = 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
𝐡𝑃
𝑇𝐴𝑃 = (ŋ𝑓 )(𝐡𝑃)
𝑇𝐴𝑃 = (0.70)(50π‘˜π‘Š)
𝑇𝐴𝑃 = 35π‘˜π‘Š
Using the density of air at 21.2°C, πœŒπ‘Ž = 1.20029845 kg/m3
𝑇𝐴𝑃 = 𝛾𝑄𝐻𝑇 ; 𝛾 = πœŒπ‘”
𝑄=
𝑄=
𝑇𝐴𝑃
πœŒπ‘”π»π‘‡
35000π‘Š
π‘˜π‘”
π‘š
1.200029845 π‘š3 (9.81 𝑠2) (110π‘š)
π‘š3
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 (πœ‚)
Solution:
1000 kg⁄m3
β„Žπ‘  = 0.18m(1.185 kg⁄m3)
β„Žπ‘  = 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
πœ‚=
Air Power
Motor Power
πœ‚=
35.316kW
44kW
πœ‚ = 0.8026363636 × 100%
πœ‚ = 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
𝑓𝑑 3
β„Žπ‘Ÿ
𝑇 = 60°πΉ
π‘ƒπ‘Žπ‘‘π‘š = 30 𝑖𝑛 𝐻𝑔
β„Žπ‘€ = 2 𝑖𝑛
Required:
𝐻𝑝
Solution:
β„Žπ‘  =
β„Žπ‘€ Ɣ𝑀
Ζ”π‘Ž
(2 𝑖𝑛 π‘₯
β„Žπ‘  =
1 𝑓𝑑
12 𝑖𝑛
) (62.4
Ζ”π‘Ž
𝑙𝑏
𝑓𝑑 3
)
10.4
β„Žπ‘  =
𝑙𝑏
𝑓𝑑 2
Ζ”π‘Ž
π΄π‘–π‘Ÿ π‘ƒπ‘œπ‘€π‘’π‘Ÿ = Ζ”π‘Ž π‘„β„Ž
Ζ”π‘Ž (60,000
π΄π‘–π‘Ÿ π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
𝑓𝑑 3
β„Žπ‘Ÿ
1 β„Žπ‘Ÿ
10.4
) (60 π‘šπ‘–π‘›) (
𝑙𝑏
𝑓𝑑2
Ζ”π‘Ž
)
𝑓𝑑−𝑙𝑏
π‘šπ‘–π‘›
33,000
1 β„Žπ‘
π΄π‘–π‘Ÿ π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 0.3151515152 β„Žπ‘
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑒𝑠𝑒 1 β„Žπ‘ π‘šπ‘œπ‘‘π‘œπ‘Ÿ (π‘ π‘‘π‘Žπ‘›π‘‘π‘Žπ‘Ÿπ‘‘)
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
π‘š
𝑠
β„Žπ‘€ = 2.16 π‘π‘š
𝑑 = 1.22 π‘š
π‘ƒπ‘Žπ‘‘π‘š = 99.4 π‘˜π‘ƒπ‘Ž
𝑇 = 30°πΆ
Required:
β„Ž
Solution:
β„Žπ‘£ =
β„Žπ‘£ =
𝑣2
2𝑔
(7.62
π‘š 2
𝑠
2 (9.81
)
π‘š
𝑠2
)
β„Žπ‘£ = 2.959449541 π‘š
Ζ”=
Ζ”=
𝑃
𝑅𝑇
99.4 π‘˜π‘ƒπ‘Ž
(0.287
π‘˜π½
π‘˜π‘”πΎ
) (30 + 273𝐾)
Ζ” = 1.143041133
π‘˜π‘”
π‘š3
π‘†π‘œπ‘™π‘£π‘–π‘›π‘” π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘’ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ β„Žπ‘’π‘Žπ‘‘ 𝑖𝑛 π‘‘π‘’π‘Ÿπ‘šπ‘  π‘œπ‘“ π‘π‘š π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ
1.143041133
β„Žπ‘£ = 2.959449541 π‘š (
π‘˜π‘”
1000 π‘š3
β„Žπ‘£ = 3.382772558π‘₯10−3 π‘š π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ
β„Žπ‘£ = 0.3382772558 π‘π‘š π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ
β„Ž = β„Žπ‘  + β„Žπ‘£
β„Ž = 2.16 π‘π‘š + 0.3382772558 π‘π‘š
β„Ž = 2.498277256 π‘π‘š π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ
π‘˜π‘”
π‘š3
)
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
π‘š
𝑠
𝑣𝑑 = 11.25
π‘š
𝑠
β„Žπ‘  = 2.5 π‘π‘š
β„Žπ‘‘ = 11.25 π‘π‘š
𝑀 = 1.20
π‘˜π‘”
π‘š3
π‘š3
𝑄 = 9.45
𝑠
REQUIRED:
Fan Efficiency
SOLUTION:
β„Ž = β„Žπ‘  + β„Žπ‘£
β„Ž=(
β„Ž=[
𝑃𝑑 −𝑃𝑠
𝑣𝑑 2 −𝑣𝑠 2
)+(
)
𝑀
2𝑔
0.0762 π‘š − (−0.025 π‘š)
1.20
π‘˜π‘”
] (1000) + [
(11.25
π‘š3
π‘š 2
) − (6.3
𝑠
2 (9.81
π‘š
)
𝑠2
π‘š 2
𝑠
)
]
β„Ž = 88.76108563 π‘š
π΄π‘–π‘Ÿ π‘π‘œπ‘€π‘’π‘Ÿ = π‘€π‘„β„Ž
π‘˜π‘”
π‘š3
π΄π‘–π‘Ÿ π‘π‘œπ‘€π‘’π‘Ÿ = (1.20 3 π‘₯ 0.00981) (9.45
) (88.76108563 π‘š)
π‘š
𝑠
π΄π‘–π‘Ÿ π‘π‘œπ‘€π‘’π‘Ÿ = 9.874262475 π‘˜π‘Š
πΉπ‘Žπ‘› 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
πΉπ‘Žπ‘› 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
π΄π‘–π‘Ÿ π‘π‘œπ‘€π‘’π‘Ÿ
𝑃𝐼𝑁
9.874262475 π‘˜π‘Š
13.75 π‘˜π‘Š
πΉπ‘Žπ‘› 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, 𝑀 = 1.2
π‘˜π‘”
π‘š3
Solving for the density at 93oC and 735 mm Hg
𝑀=
𝑃
𝑅𝑇
101.325 π‘˜π‘ƒπ‘Ž
𝑀=
735 π‘šπ‘š 𝐻𝑔 ( 760 π‘šπ‘š 𝐻𝑔 )
0.287
π‘˜π½
π‘˜π‘”−𝐾
(93 + 273)
𝑀 = 0.9328834256
π‘˜π‘”
π‘š3
Using Fan Laws:
𝑀
π΅π‘Ÿπ‘Žπ‘˜π‘’ π‘π‘œπ‘€π‘’π‘Ÿ 𝑖𝑛𝑝𝑒𝑑 = (
) (π΅π‘Ÿπ‘Žπ‘˜π‘’ π‘π‘œπ‘€π‘’π‘Ÿ π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘)
π‘€π‘Žπ‘–π‘Ÿ
0.9328834256
π΅π‘Ÿπ‘Žπ‘˜π‘’ π‘π‘œπ‘€π‘’π‘Ÿ 𝑖𝑛𝑝𝑒𝑑 = (
π‘˜π‘”
1.20 π‘š3
π‘˜π‘”
π‘š3
) (620 π‘˜π‘Š)
π΅π‘Ÿπ‘Žπ‘˜π‘’ π‘π‘œπ‘€π‘’π‘Ÿ 𝑖𝑛𝑝𝑒𝑑 = 481.9897699 π‘˜π‘Š
π‘†π‘‘π‘Žπ‘‘π‘–π‘ π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ = (
𝑀
) (π‘†π‘‘π‘Žπ‘‘π‘–π‘ π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’)
π‘€π‘Žπ‘–π‘Ÿ
0.9328834256
π‘†π‘‘π‘Žπ‘‘π‘–π‘ π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ = (
π‘˜π‘”
1.20 π‘š3
π‘˜π‘”
π‘š3
) (310 π‘šπ‘š)
π‘†π‘‘π‘Žπ‘‘π‘–π‘ π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ = 240.9948849 π‘šπ‘š
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 ο€½
P
98Kpa
ο€½
RT (0.287 kJ / kg - K) (32 + 273)K
w2 ο€½ 1.1195kg / m 3
Solving for the new brake power of the fan,
BP2 w2
ο€½
BP1 w1
BP2
1.1195kg / m 3
ο€½
100kW
1.2kg / m 3
BP2 ο€½ 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 ο€½
(hw2 ο€­ hw1 )d w
da
(0.15m ο€­ (ο€­0.03m))1000kg / m 3
hs ο€½
1.2kg / m 3
hs ο€½ 150m
Solving for velocity head,
v ο€­ vs
hv ο€½ d
2g
2
2
(7 m / sec) 2 ο€­ (3m / sec) 2
2(9.81m / sec 2 )
hv ο€½ 2.038m
hv ο€½
Solving for the total head of fan,
h ο€½ hs  hv
h ο€½ 150m  2.038m
h ο€½ 152.038m
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