Hvac Pulley Math

HVAC Pulley Math Page 1 of 18 By William Greco [email protected] Warrington, Pa. 8/20/09 Executive Summary The basis of this essay is to algebraically combine Pulley and Fan Laws into single equations. (Note: Pulley pitch diameter and pulley diameter are used interchangeably, sheave and pulley will also be used interchangeably.) Main Except for direct drives, fans and blower sets have a driver pulley on the motor shaft and a driven pulley on the fan shaft with belt or belts to convey power to the fan. The mathematical relationships between motor and fan speeds, driver and driven pulley diameters, horsepower and fan static pressures will be investigated in this report. Also equations will be developed for fan operation when dealing with changing capacity requirements. Fan laws will be substituted and rearranged against pulley laws to predict fan performance in relation to modifying characteristic variables. Pulley Equation: DPM MPR  RFP1  equation-1 DPF where : RFP1  Existing RPM of fan pulley DPM  Diameter of motor pulley DPF  Diameter of fan pulley MPR  RPM of motor pulley A change in pulley sizes will affect the performance of the fan rpm which in turn will influence characteristics of all the other variables of the fans operation. Fan Laws The Fan Laws are the basic proportional analogy between fan RPM, CFM, Total Pressure developed by the fan, and required Brake Horsepower. They are most useful for calculating the impact of extrapolating from an existing fan performance to a new fan performance. Fan Horsepower with respect to RPM 3

HP1  RFP1   equation-2  HP2  RFP2  where: HP1  exisitng brake horsepower HP2  new brake horse power RFP1  RPM of fan with HP1exisitng brake horsepower  RFP2  RPM of fan with HP2 new brake horsepower 

HVAC Pulley Math

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By William Greco [email protected] Warrington, Pa. 8/20/09 Solving for RFP1 3 HP1  RFP1     HP2  RFP2   RFP13  3  RFP2

 HP1   HP2  RFP13 cross multiply  3  RFP2 DIVIDE BOTH SIDES BY HP2



HP2 RFP13  HP2

CUBE ROOT OF BOTH SIDES

  3

RFP1 

RFP13 HP2   HP1  HP2 RFP13   HP1RFP23    3 RFP2 HP1 HP2 

3

RFP1

3

HP1RFP23 





HP2 3

HP1RFP23  HP2

3

 RFP1 

HP1 RFP2  3

HP2

HP1 RFP2  3

HP2

1   RFP2   B  AB 3  A  A and A     RFP1  HP1 1   C C  HP2 3  1   RFP2  RFP1  HP13  (equation-3) 1    HP2 3  n

1 n

RFP1 from equation-1 = RFP1 from equation-3 Solving for HP2: 1

HP13 RFP2  DPM MPR   B  AB A     1 DPF C C 3 HP2 1

CROSS MULTIPLY

HP13 RFP2 DPF 

  HP2 1 HP2 3 DPM MPR 

1 3

1 3

DPM MPR   HP1 RFP2 DPF

1 divide by  DPMMPR 

HP2 3  DPM  MPR 

DPM  MPR 

1



HP13 RFP2 DPF 

DPM MPR 

1   3 HP1 RFP2 DPF    n n a  b   a  b  HP2   DPM MPR    

3

equation-4

HVAC Pulley Math

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By William Greco [email protected] Warrington, Pa. 8/20/09 1    HP13 RFP2 DPF  HP2     DPM MPR    

3

equation-4

Equation-4 must now be modified to conform with the physical relationships of existing pulley diameters versus modified new pulley diameters. The RFP2 variable which indicates the new RPM of the fan must have a substitution of DPF2, DPM2 and MPR2. DPF becomes DPF1, DPM becomes DPM1 and MPR becomes MPR1. 3

1    DPM2 MPR2    HP13   DPF1   DPF2   HP2   equation-5  DPM1MPR1       where: HP2 = New brake horsepower HP1 = Existing brake horsepower DPM1 = Existing diameter of motor pulley DPF1 = Existing diameter of fan pulley MPR1 = RPM of motor pulley DPM2 = New or existing diameter of motor pulley DPF2 = New or existing diameter of fan pulley MPR2 = New RPM of motor pulley (in cases where a VFD is involved the existing motor rpm will not equal the new motor rpm) Example-1: (Pulley Change motor speed constant- Find New Horsepower) An existing fan has a motor pulley with an 8 inch diameter, a fan pulley of 6 inch diameter, and is operating at 3.7 brake horsepower. The motor is running at 1,740 RPM and is not on a variable frequency drive. The fan pulley is being replaced with a 5 inch diameter pulley, the new brake horsepower can be calculated using equation-5. The existing fan rpm with the 6" pulley is:

8 1740  new fan 8 1740   2,340 rpm   2,784 rpm rpm 6 5 3

 1  8 1740     3.7 3   6   5     6.4 Brake Horsepower HP2    8 1740        The new brake horsepower is now 6.4 due to the increase in fan rpm.

HVAC Pulley Math

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By William Greco [email protected] Warrington, Pa. 8/20/09 Example-1: (continued) Also note that if you plug the original 6 inch pulley back into the equation: 3

 1  8 1740     3.7 3   6   6     3.7 Brake Horsepower HP2    8 1740       The calculation can be worked out using the pulley equation-1 in the first step then applying the fan laws as step 2, however equation-5 allows a one step calculation. VFD calculations can also be worked out using equation-5. Example-2: (VFD speed change) A fan set has a fan pulley that is 6.75 inches in diameter and a 7.5 inch motor pulley, the motor is on a variable frequency drive and can be operated at 30% to 110% of the motors rated 1,750 rpm. The VFD is presently controlling the motor to run at 1,250 rpm (71.4% of rating) and the brake horsepower has been worked out to be 8.6 at the 71.4% rating. If the VFD speed is required to be increased to 90% of the 1750 rpm rating what will the new brake horsepower be ? (0.90 x 1750 = 1,575 rpm) 3

 1  7.5 1250     8.6 3   6.75    6.75   HP2     8.6 Brake Horsepower @ 71.4% 7.5 1250           3

 1  7.5 1575    8.6 3   6.75    6.75   HP2     17.2 Brake Horsepower @ 90% 7.5 1250          

The brake horsepower went from 8.6 BHP to 17.2 BHP (50% increase in brake horsepower) with only a 20% increase in motor speed.

HVAC Pulley Math

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By William Greco [email protected] Warrington, Pa. 8/20/09 Example-3: (VFD Change in motor speed and pulley modification-Find New Horsepower) A fan set has a fan pulley that is 6.5 inches in diameter and a 5.5 inch motor pulley, the motor is on a variable frequency drive and can be operated at 35% to 115% of the motors rated 1,750 rpm. The VFD is presently controlling the motor to run at 1,050 rpm (60% of rating) and the brake horsepower has been worked out to be 19.3 at the 60% rating. If the VFD speed is decreased to 55% of the 1750 rpm rating and the motor pulley is changed from 5.5 to 5.25 inch diameter what will the new brake horsepower be ? (0.55 x 1750 = 997.5 rpm) 3

1    5.5 1050   19.33   6.5    6.5   HP2     19.3 Brake Horsepower @ 60% of rating 5.5 1050        3

1    5.25 997.5  19.33  6.5      6.5   HP2     14.392 Brake Horsepower @ 55% of rating 5.5 1050       

The equation for fan brake horsepower can be substituted for the HP1 variable in equation-5. The fan brake horsepower equation: BHP 

CFM SP_inwg Gas_spec_grav 

6356 Fan_eff 

(equation-6)

where: BHP = Brake horsepower CFM = Cubic feet (of specified gas) per minute SP_inwg  Static Pressure developed by the fan (inches water gage) 6356 = a constant (see "Derivation of the Nine Major HVAC Constants") By Bill Greco - 2-16-07 Fan_eff=combined efficiency of fan and motor drive Gas_spec_grav  Specific Gravity of gas being transported by fan

HVAC Pulley Math

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By William Greco [email protected] Warrington, Pa. 8/20/09 Substituting the Fan Horsepower equation-6 for the HP1 variable into equation-5.

1    DPM2 MPR2    HP13   DPF1      DPF2   HP2    DPM1MPR1       becomes:

3

equation-5

1     CFM SP_inwg Gas_spec_grav  3  DPM2 MPR2       DPF1  DPF2 6356Fan_eff      HP2      DPM1 MPR1         

3

equation-7 

where: HP2 = New brake horsepower DPM1 = Existing diameter of motor pulley DPF1 = Existing diameter of fan pulley MPR1 = RPM of motor pulley DPM2 = New or existing diameter of motor pulley DPF2 = New or existing diameter of fan pulley MPR2 = New RPM of motor pulley (in cases where a VFD is involved the existing CFM = Existing Cubic Feet (of specified gas) per minute Gas_spec_grav = Specific Gravity of gas being transported by fan Fan_eff = Existing Efficiency of the fan (1 <) SP_inwg = Existing Static pressure developed by the fan 6356 = A constant

HVAC Pulley Math

Page 7 of 18

By William Greco [email protected] Warrington, Pa. 8/20/09 Example-4: (Existing Horsepower Variables are all known- Find New Horsepower) An existing belt driven fan is moving 9,980 cfm and producing 4.1 inches water gage (inwg) of pressure. The fan is on a variable frequency drive (VFD) and is running at 1,150 rpm which is 65.7 % rating of the 1,750 rpm motor. Specific gravity of the air being moved by the fan is taken at unity (1). The fan sets efficiency is 78% (0.78). The existing fan pulley diameter is 8.1” and the existing motor pulley diameter is 7.5”. The VFD is to be adjusted to 73% of the motor 1,750 rpm (0.73 x 1,750 = 1,278 rpm) the motor pulley is to be changed to a 8” diameter and the existing fan pulley is to be changed out for a 7.625” diameter pulley. Applying equation-7 what will the new brake horsepower be ? DPM1 = 7.5 DPF1 = 8.1 MPR1 = 1,150 MPR2 = 1,278 DPF2 = 7.625 DPM2 = 8 CFM = 9,980 SP_inwg = 4.1 Gas_spec_grav = 1 for air Fan_eff = 0.78 (78%) 9,980 4.11

6356 0.78

 8.25 existing BHP by equation-6 3

1     9,980 4.11 3  7.5 1,150       8.1  63560.78   8.1    8.25 existing BHP by equation-7 HP2      7.5 1,150       3

1      9,980 4.11 3  8 1, 278       8.1  63560.78   7.625    16.48 new BHP by equation-7 HP2      7.5 1,150        The modified operating parameters of the fan set call for the motor to produce 16.48 brake horsepower.

HVAC Pulley Math By William Greco [email protected] Warrington, Pa. 8/20/09

Page 8 of 18

Equation-7 can be modified by substituting the brake horsepower variables for HP2, then solving for CFM or inwg. To simplify the calculation assume that air is the only gas being transported and Gas_spec_grav = 1. The number 2 will be used after the CFM, SP_inwg and Fan_eff to identify the HP2 variables. 1

 CFM2 SP_inwg2 1  3 HP 2     6356 Fan_eff2  

Thus : 3

1     3  DPM2 MPR2   CFM SP_inwg 1     1  DPF1    DPF2  CFM2 SP_inwg2 1  3   6356 Fan_eff          DPM1MPR1  6356 Fan_eff2         Solving for CFM2: For algebraic manipulation simplicity the following substitutions were made : C1 for CFM (existing condition) C2 for CFM2 (new condition) D1 for DPM1 (existing condition) D2 for DPM2 (new condition) E1 for Fan_eff (existing condition) E2 for Fan_eff2 (new condition) F1 for DPF1 (existing condition) F2 for DPF2 (new condition) M1 for MPR1 (existing condition) M2 for MPR2 (new condition) S1 for SP_inwg (existing condition) S2 for SP_inwg2 (new condition)

First simplify the right hand side of the equation:

1     C1S11  3 D2 M2   F1    F2   6356 E1     D1M1        

3

1    C1S1 3   a D2 F1 M2     c  6356 E1   ac    b  numerator is a fraction, compound fraction can be eliminated  =    d bd D1F2 M1      

3

HVAC Pulley Math

Page 9 of 18

By William Greco [email protected] Warrington, Pa. 8/20/09 3

1      C1S1 3    D2 F1M2  1     n 6356 E1 3   a   n n move exponent into the numerator and denominator   =a b    D1F2 M1 b         3 a 1   c D2 F1M2 C1 S13  ac numerator is a fraction, compound fraction can be eliminated  b =  1   d bd 3  D1 F2 M1 6356 E1       n

power of a product, move exponent out  ab  1  3 D2 F1 M2 C 1 S1   reorder factors   1 1   6356 3 D1 E13 F2 M1 

    

3

3

n

power of a product, move exponent out  ab  1  13  3 C1 D2 F1 M2 S1   reorder terms   1 1   6356 3 D1 E13 F2 M1   

1   3 D2 F1 M2 C 1 S1       =a n b n   1 1   D1F2 M16356 3 E13   

1 1   3  D2 F1 M2 C1 S13  n n =a b   1 1   6356 3 D1 E13 F2 M1   

3

3

3

n

numerator and denominator terms moved out  ab 

1  13  3 C1 D2 F1 M2 S1    an   = n  3 1 1 b   3 3  6356 D1 E1 F2 M1   

HVAC Pulley Math

Page 10 of 18

By William Greco [email protected] Warrington, Pa. 8/20/09 3

n

terms moved out of brackets  ab 

 13  3 3 3  C1  D2 F1 M2  =a n b n   3 1 3    13  3 3  6356  D1  E1      1 3

C 2 S2

remove braket from left hand side of equation 

1 3

1

1

6356 3 E2 3 3

3

F23 M13

C1 D23  

F13 M23 S1

6356 D13 E1 F23 M13

 13  3 3 3  C1  D2 F1 M2 1  multiply and cancel exponents 3x  1   3 1 3 3    13  3 3  6356  D1  E1        C1 D23 F13 M23 S1 right hand side of equation simplified  6356 D13 E1 F23 M13 3 3 3  C 2 S2  C1 D2 F1 M2 S1    3 3 3  6356 E2  6356 D1 E1 F2 M1 solving for C2 (New CFM) from equation-8 :

 13   S1   

 13  S1 

  

3

F23 M13

(equation-8)

C1 D23 F13 M23 S1  C 2 S2   cross multiply    6356 D13 E1 F23 M13  6356 E2  C 2 S26356 D13

E1 F23 M13   6356 E2 C1 D23 F13 M23 S1

SUBTRACT 6356  C 2 S2  6356  6356  D13 E1 F23 M13    6356  6356  E2C1 D23 F13 M23 S1 FROM BOTH SIDES

C 2 S2 D13 E1 F23 M13  E2 C1 D23 F13 M23 S1

3

3

3

divide S2 D1 E1 F2 M1  from both sides

C 2 S2 D13 E1 F23 M13 S2 D13 E1 F23 M13



E2 C1 D23 F13 M23 S1 S2 D13 E1 F23 M13

HVAC Pulley Math

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By William Greco [email protected] Warrington, Pa. 8/20/09 re-applying the original variable names = Fan_eff2 CFM DPM23 DPF13 MPR23 SP_inwg  CFM2  (equation-9) SP_inwg2 DPM13 Fan_eff DPF23 MPR13 where: CFM = Existing CFM CFM2 = New CFM DPF1 = Existing diameter of fan pulley DPF2 = New or existing diameter of fan pulley DPM1 = Existing diameter of motor pulley DPM2 = New or existing diameter of motor pulley Fan_eff = Existing Fan Set Efficiency (1 <) Fan_eff2 = New Fan Set Efficiency (1 <) MPR1 = Existing RPM of motor pulley MPR2 = New RPM of motor pulley (in cases where a VFD is involved the existing SP_inwg = Existing Static pressure developed by the fan SP_inwg2 = New Static pressure developed by the fan

Example-5 (Existing and New Variables are Known to find New CFM) An existing fan set is producing 10,500 cfm @ 4.3 inches water gage, the motor is on a variable frequency drive (VFD) and is running at 70% of it’s 1,750 rpm rating or 1,225 rpm. The motor pulley’s have a pitch diameter of 10 inches and the fan pulley’s diameter’s are 9.3 inches. Fan set efficiency is 79%. The VFD speed is to be increased to 75% of it’s capacity or 1,750 x 0.75 = 1,312 rpm, the fan pulley diameter’s are to be increased from 9.3 inches to 10.25 inches and the motor pulley’s (sheaves) are to be increased from 10 inch diameter to 10.5 inches in diameter. The new static pressure produced by the fan is assumed to be 4 inches water gage (inwg). Fan set efficiency is assumed to remain at 79%. Find the new CFM.

CFM2 

0.79 10,500 10.53 9.33 1,3123 4.3 4 103 0.79 10.253 12253

 11,990 CFM

HVAC Pulley Math Page 12 of 18 By William Greco [email protected] Warrington, Pa. 8/20/09 An equation to determine the static pressure developed by the fan can be worked out using Equation-8 and solving for S2 (SP_inwg2): 3 3 3  C 2 S2  C1 D2 F1 M2 S1   3 3 3  6356 E2  6356 D1 E1 F2 M1 C1 D23 F13 M23 S1   C 2 S2   6356 E2   cross multiply  6356 D13 E1 F23 M13  

C 2 S26356 D13

(equation-8)

E1 F23 M13   6356 E2 C1 D23 F13 M23 S1

SUBTRACT 6356  C 2 S2  6356  6356  D13 E1 F23 M13    6356  6356  E2C1 D23 F13 M23 S1 FROM BOTH SIDES

C 2 S2 D13 E1 F23 M13  E2 C1 D23 F13 M23 S1

3

3

3

divide C2 D1 E1 F2 M1  from both sides

S2 

S2 C2 D13 E1 F23 M13 C2 D13 E1 F23 M13

E2 C1 D23 F13 M23 S1 C2 D13 E1 F23 M13

SP_inwg2 



E2 C1 D23 F13 M23 S1 C2 D13 E1 F23 M13

(equation-10)

Fan_eff2 CFM2 DPM23 DPF13 MPR23 SP_inwg  CFM DPM13 Fan_eff DPF23 MPR13

(equation-10)

Example-6: An existing fan set is producing 11,500 cfm and producing 3.5 inches water gage, the motor is not on a variable frequency drive (VFD) and is running at 1,750 rpm. The motor pulley’s have a pitch diameter of 11.5 inches and the fan pulley’s diameter’s are 10 inches. Fan set efficiency is 80%. The fan pulley diameter’s are to be decreased from 10 inches to 9.25 inches and the motor pulley’s (sheaves) are to remain at the existing 11.5 inch diameter. The CFM is being reduced to 9,500 cfm. Fan set efficiency is assumed to remain at 80%. Find the new static pressure developed by the fan.

SP_inwg2 

0.80 11, 500 11.53 103 17503 3.5 9,500 11.53 0.80 9.253 1, 7503

 5.35 inches water gage

HVAC Pulley Math

Page 13 of 18

By William Greco [email protected] Warrington, Pa. 8/20/2009 Pulley Size Changes Utilizing the following fan law which relates CFM to Fan RPM the driving and driven Pulley diameters can be determined. Capacity varies as the speed:  DPM1MPR1    DPF1 CFM1 RPM1 CFM1      CFM2 RPM2 CFM2  DPM2 MPR2     DPF2  

(equation-11)

where: CFM1 = Existing CFM, CFM2 = New CFM, DPF1 = Existing diameter of fan pulley DPF2 = New or existing diameter of fan pulley DPM1 = Existing diameter of motor pulley DPM2 = New or existing diameter of motor pulley, MPR1 = Existing RPM of motor pulley MPR2 = New RPM of motor pulley (in cases where a VFD is involved) Simplifying the right hand side of the equation:  DPM1MPR1        DPF1 DPM1 MPR1   a 1 1   multiply 1 over   a  the factor b DPF1  DPM2 MPR2   b  DPM2 MPR2        DPF2  DPF2  

multiply

DPM1MPR1 DPM1DPF2 MPR1 multiply by the reciprocal of the fraction  DPM2 MPR2  DPF1MPR2 DPM2 DPF1 DPF2 DPM1DPF2 MPR1 CFM1  (equation-11a) CFM2 DPF1MPR2 DPM2

solving for DPM2: cross multiply  CFM1 DPF1 MPR2 DPM2  CFM2 DPM1 DPF2 MPR1

divide both sides by CFM1 DPF1 MPR2 

DPM2 

CFM2 DPM1 DPF2 MPR1 CFM1 DPF1 MPR2

(equation-11b)

CFM1 DPF1 MPR2 DPM2 CFM2 DPM1 DPF2 MPR1  CFM1 DPF1 MPR2 CFM1 DPF1 MPR2

(equation-12)

HVAC Pulley Math

Page 14 of 18

By William Greco [email protected] Warrington, Pa. 8/20/2009 Example-7: Find Motor Pulley Size based on CFM changes The CFM being moved by a fan set with a motor not controlled by a VFD must be increased from 11,000 cfm to 12,100 cfm, the existing fan pulley pitch diameter is 6 inches and the diameter of the motor pulley is 10 inches, the motor rpm is 1,750. Find the size of a new motor pulley to achieve the 12,100 cfm. 12,100 10  6 1,750  11,000 6  1,750

 11

change to an 11 inch diameter motor pulley

Example-8: Find Motor Pulley Size based on CFM, Fan Pulley, and VFD changes The CFM being moved by a fan set with a motor controlled by a VFD must be increased from 10,600 cfm to 12,144 cfm, the existing fan pulley pitch diameter is 6 inches and will be changed to 7.125 inches, the existing diameter of the motor pulley is 10 inches, the motor rpm will be increased from 1,312 rpm to 1,400 rpm. Find the size of a new motor pulley to achieve the 14,100 cfm. 12,144 10 7.125 1,312  10, 600 6 1, 400

 12.75 inch diameter motor pulley is required

With the use of equation 11-b, equations for CFM2 = New CFM, DPF2 = New or existing diameter of fan pulley and MPR2 = New RPM of motor pulley (in cases where a VFD is involved) can be easily found. To find CFM2 : CFM1 DPF1 MPR2 DPM2  CFM2 DPM1 DPF2 MPR1 (equation-11b) divide both sides by  DPM1 DPF2 MPR1 CFM2 

CFM1 DPF1 MPR2 DPM2 DPM1 DPF2 MPR1

CFM1 DPF1 MPR2 DPM2 CFM2 DPM1 DPF2 MPR1  DPM1 DPF2 MPR1 DPM1 DPF2 MPR1

(equation-13)

Example-9: Find new CFM based on Fan Pulley, and VFD speed changes The CFM being moved by a fan set with a motor controlled by a VFD will change from 9,100 cfm. The existing fan pulley pitch diameter is 7.5 inches and will be changed to 8 inches, the existing diameter of the motor pulley is 10 inches, the motor rpm will be increased from 1,260 rpm to 1,390 rpm. Find the new CFM. 9,100 10 7.5 1,390  10 81, 260

 9, 411 cfm

HVAC Pulley Math

Page 15 of 18

By William Greco [email protected] Warrington, Pa. 8/20/2009 To find MPR2 = New RPM of motor pulley (in cases where a VFD is involved): CFM1 DPF1 MPR2 DPM2  CFM2 DPM1 DPF2 MPR1 divide both sides by  CFM1 DPF1 DPM2 MPR2 

CFM2 DPM1 DPF2 MPR1 CFM1 DPF1 DPM2

(equation-11b)

CFM1 DPF1 MPR2 DPM2 CFM2 DPM1 DPF2 MPR1  CFM1 DPF1 DPM2 CFM1 DPF1 DPM2 (equation-14)

Example-10: Find new VFD motor controlled rpm based on cfm and motor pulley change The CFM being moved by a fan set with a motor controlled by a VFD needs to change from 8,750 cfm to 10,000 cfm. The existing fan pulley pitch diameter is 7.5 inches, the existing diameter of the motor pulley is 10 inches and will be revised to 10.5 inches the existing motor rpm is 1,290 rpm. Find the new required (MPR2) motor rpm. by equation 14 

10, 000 10 7.5 1, 290  8750 7.5 10.5

 1, 404 rpm

Example-11: Find new VFD motor controlled rpm based on cfm change The CFM being moved by a fan set with a motor controlled by a VFD needs to change from 12,700 cfm to 16,100 cfm. The existing fan pulley pitch diameter is 6 inches and the existing diameter of the motor pulley is 10 inches. The existing motor rpm is 1,195 rpm. Find the new required (MPR2) motor rpm by equation-14.. (new) required rpm

16,100 10 7.5 1,195 

12, 700 7.5 10 To find DPF2 = New or existing diameter of fan pulley: FM1 DPF1 MPR2 DPM2  CFM2 DPM1 DPF2 MPR1

divide both sides by  CFM2 DPM1 MPR1

DPF2 

 1,514 rpm

(equation-11b)

CFM2 DPM1 DPF2 MPR1 CFM1 DPF1 MPR2 DPM2  CFM2 DPM1 MPR1 CFM2 DPM1 MPR1

CFM1 DPF1 MPR2 DPM2 CFM2 DPM1 MPR1

(equation-15)

HVAC Pulley Math

Page 16 of 18

By William Greco [email protected] Warrington, Pa. 8/20/2009

Example-12: Find new fan pulley pitch diameter based on CFM and motor pulley change The CFM being moved by a fan set without a VFD needs to change from 8,300 cfm to 9,965 cfm. The existing fan pulley pitch diameter is 5.625 inches, the existing diameter of the motor pulley is 8 inches and will be revised to 10.25 inches, the existing motor rpm is 1,750 rpm. Find the new required DPF2 = New diameter of fan pulley. By equation-15: 8,300 10.25 5.625 1, 750  9,965 81, 750

 6 inch pitch diameter for new fan pulley

Utilizing the following fan law which relates Amps to Fan RPM the driving and driven sheave (pulley) diameters can be determined along with Amperage. Substituting equation-1 for RPM into the speed varies as the amperage yields equation-16:   DPM1MPR1    3  DPF1 Amps1  RPM 1  Amps1        Amps2  RPM 2  Amps2  DPM2 MPR2        DPF2  

3

(equation-16)

Simplifying the right hand side of the equation:  DPM1MPR1  invert and multiply   DPF1  

3

3

3

   DPM1MPR1DPF2  DPF2       DPM2 MPR2    DPF1DPM2 MPR2  

n

an DPM13 MPR13 DPF23 a n n n  and ab  a b      b bn DPF13 DPM23 MPR23

(equation-16a)

Amps1 DPM13 MPR13 DPF23  Amps2 DPF13 DPM23 MPR23 cross multiply  Amps1 DPF13 DPM23 MPR23  Amps2 DPM13 MPR13 DPF23

(equation-16b)

HVAC Pulley Math

Page 17 of 18

By William Greco [email protected] Warrington, Pa. 8/20/2009 Amps1 DPF13 DPM23 MPR23  Amps2 DPM13 MPR13 DPF23

(equation-16b)

Solving for Amps2: divide both sides by DPM1 MPR1 DPF2

 3 3 3

Amps2 

Amps1 DPF13 DPM23 MPR23 DPM13 MPR13 DPF23

Amps1 DPF13 DPM23 MPR23 DPM13 MPR13 DPF23



Amps2 DPM13 MPR13 DPF23 DPM13 MPR13 DPF23

(equation-17)

where: Amps1 = Existing Amperage, Amps2 = Revised Amperage, DPF1 = Existing diameter of fan pulley DPF2 = New or existing diameter of fan pulley DPM1 = Existing diameter of motor pulley DPM2 = New or existing diameter of motor pulley, MPR1 = Existing RPM of motor pulley MPR2 = New RPM of motor pulley (in cases where a VFD is involved)

Example-13: Find new amperage based on VFD speed change A fan set with a motor controlled by a VFD is ramped up from 1,150 rpm to 1,360 rpm. The existing fan sheave pitch diameter is 6 inches and the existing diameter of the motor sheave is 8 inches. The existing motor is pulling 13 amps Using equation-17 find the new amperage (Amps2).

    21.5 Amps 83 11503 63

13 63 83 13603

Example-14: Find new amperage based on VFD speed and motor pulley change A fan set with a motor controlled by a VFD is ramped down from 1,450 rpm to 1,325 rpm. The existing fan sheave pitch diameter is 6 inches and the diameter of the motor sheave is to be decreased from 10 inches to 9.5 inches. The existing motor is pulling 19.6 amps Using equation-17 find the new amperage (Amps2).

    12.8 Amps 103 14503 63

19.6 63 9.53 13253

HVAC Pulley Math

Page 18 of 18

By William Greco [email protected] Warrington, Pa. 8/20/2009 Solving for MPR2: Amps1 DPF13 DPM23 MPR23  Amps2 DPM13 MPR13 DPF23

divide both sides by  3 3

Amps1 DPF13 DPM23 MPR2

Amps1 DPF1 DPM2

Amps1 DPF13 DPM23

3



(equation-16b)

Amps2 DPM13 MPR13 DPF23 Amps1 DPF13 DPM23 1

3

a ba

1 b3

 Amps2 DPM13 MPR13 DPF23   MPR2     Amps1 DPF13 DPM23  

1 3  1  MPR2  3

1 Amps2 3

DPM1 MPR1 DPF2

1 Amps13

3

(equation-18)

DPF1 DPM2

Example-15: VFD controlled Motor RPM to achieve required amperage A fan set with a motor controlled by a VFD is running at 1,375 rpm. The existing fan sheave pitch diameter is 8 inches and the diameter of the motor sheave is 9.875 inches. The existing motor is pulling 19.5 amps Using equation-18 determine the VFD RPM setting to achieve 24.7 amps. 1 24.7 3

9.8751,375 8

1 19.5 3

By

 1,487.7 Rpm  1,488 RPM

88.875

William Greco 2404 Greensward N. Warrington, Pa. 18976

Reference: Air Conditioning Testing/Adjusting/Balancing –A Field Practice Manual John Gladstone 1981 pg. 11 Carrier System Design Manual Chapter 6, Air Handling Equipment 1968 5th Printing pg. 6-8