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___ 5 ____ּא ½” Diameter
1600 rpm = 22” Diameter
½ x 22 = 1600 x 5 ּא
½ x 22 1600 x 5 ּא
22 =
22
1600 = ּא x 5 ½
22
400 = ּא
__ 8 ____ּא ” Diameter
400 rpm =
10” Diameter
x 10 = 400 x 8 ּא
x 10 = 400 x 8 ּא
10 10
400 = ּא x 8
10
320 = ּא
Pulley/RPM Adjustment # 1
A 1725 rpm motor drives a fan pulley fly wheel that is 8” diameter. If the desired fan rpm is 620, what
is the motor pulley size? ____ּא____ ____620_
1725 rpm = 8” Diameter
1725 x 620 = ּא x 8
1725 x 620 ּא x 8
1725 = 1725
620 = ּא x 8
1725
2.875 = ּא Ø
Pulley/RPM Adjustment # 2
A 1800 rpm motor has a 3.25” diameter pulley installed on it. If you require a fan rpm of 700rpm, what is the fan pulley size?
700 rpm 3 1/4” Diameter
1800 rpm =ּא
700 x 1800 = ּא x 3 ¼”
700 x 1800 ּא x 3 ¼
700 = 700
1800 = ּא x 3 ¼
700
8.357 = ּא Ø
Pulley/RPM Adjustment # 3
If a motor having an rpm of 1750 has a pulley installed on it that is 3 ½ “ dia. What is the flywheel size if the desired fan rpm must be 860?
860 rpm 3 ½ ” Diameter
1750 rpm =ּא
860 x 1750 = ּא x 3 ½ ”
860 x 1750 ּא x 3 ½”
860 = 860
1750 = ּא x 3 ½”
860
7.122 = ּא Ø
FAN PULLEYMOTOR PULLEY
DIA. R.P.M. DIA. R.P.M.
a. ? 300 6" 400
b. ? 250 9" 750
c. 12" 150 ? 450
d. 8" 600 ? 200
e. 4" ? 3" 1600
f 12" ? 8" 800
g. 6" 900 9" ?
h. 12" 450 9" ?
DRIVER GEAR DRIVEN GEAR
No. of Teeth
R.P.M.No. of Teeth
R.P.M.
a. 16 100 ? 50
b. 48 200 ? 100
c. ? 300 96 200
d. ? 420 28 360
e. 56 250 70 ?
f 28 390 32 ?
g. 24 ? 20 135
h. 14 ? 21 1200
FAN PULLEY MOTOR PULLEY
DIA. R.P.M. DIA. R.P.M.
a. 8 300 6" 400
b. 27 250 9" 750
c. 12" 150 4 450
d. 8" 600 24 200
e. 4" 1200 3" 1600
f 12" 533 8" 800
g. 6" 900 9" 600
h. 12" 450 9" 600
DRIVER GEAR DRIVEN GEAR
No. of Teeth
R.P.M. No. of Teeth R.P.M.
a. 16 100 32 50
b. 48 200 96 100
c. 64 300 96 200
d. 24 420 28 360
e. 56" 250 70 200
f. 28" 390 32 341.25
g. 24 112.5 20 135
h. 14 1800 21 1200
Fan Law 1: CFM varies directly with RPMFan Law 1 says that air volume (CFM) varies directly with rotational speed of the fan in RPM(revolutions per minute):
CFM2 RPM2
CFM1 = RPM1
CFM1 = original CFM RPM1 = original RPM
RPM2 = new RPM CFM2 = new CFM
Fan Law 1 means that if the RPM of a fan is increased, the CFM increases by the same percentage. Figure 14 shows the relationships in Fan Law 1. If the CFM is increased 50%(from 10,000 to 15,000) the RPM increases 50% (from 388 to 582).
Sheet metal workers often use Fan Law 1 because it predicts RPM or CFM when fanchanges are made. For example, the CFM produced by a fan needs to be raised from 25,000to 30,000. The current RPM is 970. What RPM should he used to produce the 30,000 CFM needed? CFM2 RPM2
CFM1 = RPM1
RPM2 30,000 970
= 25,000 Therefore to obtain 30,000 CFM, the fan speed must be
increased to 1165 RPM. RPM2 x 25,000 = 970 x 30,000
RPM2 = 970 x 30,000 25,000RPM2 = 1164 (Round to 1165)
R.P.M. C.F.M.
388 10,000
582 15,000
776 20,000
970 25,000
1164 30,000
1358 35,000
Fig. 14: CFM varies directly with RPM
In the same way, the new CFM can be predicted if the RPM is changed to a specific amount. For example, a fan is producing 21,420 CFM at 801 RPM. If the RPM is changed to 700 RPM, what CFM will be produced by the Fan?
CFM2 RPM2
CFM1 = RPM1
CFM2 700
21,420 = 801
CFM2 x 801 = 21,420 x 700
CFM2 = 21,420 x 700
801
CFM2 = 18,719 (Round to 18,720 CFM)
FAN LAW #1 QUESTIONS
1. A fan is delivering 3,750 CFM @ 945 RPM. If the speed is increased to 1,175 RPM, how many CFM will be delivered?
2. A fan is delivering 12,225 CFM @ 1,150 RPM. If the speed is decreased to 775 RPM, how many CFM will be delivered?
3. A fan is delivering 3,600 CFM at 1,000 RPM. If the fan speed is increased to 1,250 RPM, how many CFM will be delivered?
4. A fan is delivering 7,500 CFM at 886 RPM. If 10,000 CFM is required, what will be the new RPM?
1. A fan is delivering 3,750 CFM @ 945 RPM. If the speed is increased to 1,175 RPM, how many CFM will be delivered?
CFM2 RPM2
CFM1 = RPM1
CFM2 1175
3750= 945
CFM2 x 945 = 3750 x 1175
CFM2 = 3750 x 1175
945
CFM2 = 4662.6984
There fore the new CFM will be 4665 CFM.
2. A fan is delivering 12,225 CFM @ 1,150If the speed is decreased to 775 RPM, how many CFM will be delivered?
CFM2 RPM2
CFM1 = RPM1
CFM2 775
12225 = 1150
CFM2 x 1150 = 12225 x 775
CFM2 = 12225 x 775
1150
CFM2 = 8238.587
There fore the new CFM will be 8240 CFM.
3. A fan is delivering 3,600 CFM at 1,000 RPM. If the fan speed is increased to 1,250 RPM, how many CFM will be delivered?
CFM2 RPM2
CFM1 = RPM1
CFM2 1250
3600 = 1000
CFM2 x 1000 = 3600 x 1250
CFM2 = 3600 x 1250
1000
CFM2 = 4500
There fore the new CFM will be 4500 CFM.
4. A fan is delivering 7,500 CFM at 886 RPM. If 10,000 CFM is required, what will be the new RPM?
CFM2 RPM2
CFM1 = RPM1
10000 RPM2
7500 = 886
7500 x RPM2 = 10000 x 886
CFM2 = 10000 x 886
7500
CFM2 = 1181.333
There fore the new RPM will be 1180 RPM.
Fan Law 2: SP varies as a square of RPM
Fan Law 2 is used to determine the static pressure a fan will produce after the fan RPM is changed. Fan Law 2 says that static pressure (SP) varies directly as a square of the rotational speed of the fan (RPM):
2 SP2 RPM2
SP1 RPM1
For example, a fan is producing 0.5” wg static pressure. The RPM is
changed from 500 RPM to 550 RPM. What will the new static pressure be?
2 2
SP2 550 SP2 = 0.5 x 550 SP2 = 0.605” 0.5 500 500
Fan law 2 says that the SP ratio is the square root of the RPM ratio. You can demonstrate this with the answer to the problem above. The SP ratio is 1.21:
0.605 1.21
0.5
The RPM ratio
550 1.1
500
The SP ratio is the square of the RPM ratio 1.1: 2
1.1 = 1.21
Because a change in RPM is directly related to a change in CFM (Fan Law 1), the following variation of fan law 2 can be used:
2 SP2 CFM2
SP1 CFM1
A fan is delivering 7,534 CFM @ 886 RPM @ 1.2 SP. If 10,000 CFM is required, what will be the new RPM? And what will be the new SP?
CFM2 RPM2
CFM1 = RPM1
10000 RPM2
7534= 886
10000 x 886 = 7534 x RPM2
10000 x 886 = RPM2
7534
RPM2 = 1176.002124 2
SP2 CFM2
SP1 CFM1
2
SP2 10000 2.1141218”
1.2 7534
Fan Law # 3 Bhp varies as the cube of the RPM• Fan law 3 is used to find the new Bhp (brake horespower) needed if the RPM is
changed, or it can find the RPM if the Bhp is changed. Fan Law 3 says that brake horsepower (Bhp) varies as the cube of the rotaitional speed of the fan (RPM):
3
Bhp2 RPM2
Bhp1 RPM1
For example, a fan motor is drawing 0.84 Bhp and is producing 7650 CFM at 360 RPM. The speed is increased to 396 RPM. What is the new Bhp?
3
Bhp2 RPM2
Bhp1 RPM1
3
Bhp2 = 0.84 x 396 360
Bhp2 = 1.118
• Fan Law # 3 says that the Bhp ratio is the cube root of the RPM ratio. You can demonstrate this with the answer to the problem above. The Bhp ratio is 1.331
1.118 = 1.331
0.84
The RPM ratio is 1.1
396 = 1.1
360
The Bhp ratio 1.331 is the cube root of the RPM ratio 1.1: 3
1.1 = 1.331
1. A system is operating under the following conditions: 7,255 CFM, 575 RPM, 1.36” SP, 7.75 Bhp. If the system required 9,250 CFM, what would be the new values for RPM, SP, Bhp?
2. A system is operating under the following conditions: 12,735 CFM, 976 RPM, 1.55” SP, 12.35 Bhp. If a 15 hp motor is installed, what are the maximum CFM, RPM,SP that can be obtained?
A system is operating under the following conditions: 7,255 CFM, 575 RPM, 1.36” SP, 7.75 Bhp. If the system required 9,250 CFM, what would be the new values for RPM, SP, Bhp?