___ 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?