HVAC Duct Friction Equations

HVAC Duct Friction Equations William A. Greco [email protected] June 7, 2009

Page 1 of 6 Trident3

Executive Summary As air flows through any conduit there is an associated pressure drop. The Carrier System Design Manual, Air Disribution Part 2 describes an equation for calculating the pressure drop. The only variable that is solved for is the pressure drop. This report works out all the other variables that are involved in the equation. All ductalator’s and other hand held ductwork calculators are based on this equation. The basic Equation: Found in The Carrier System Design Manual, Air Disribution Part 2, 10th Printing 1974, page 2-31 is the following equation: 1.82

 L  V  P = 0.03 f  1.22    d  1000  where: P = friction loss (inches water gage) f = interior surface roughness (0.9 for galvanized duct) L = length of duct (feet) d = duct diameter (inches), equivalent diameter for rectangular ductwork V = air velocity (feet per minute) Solving for L (length) : If we solve strictly for galvanized duct then we can eliminate the first variable f 0.9 x 0.03 = 0.027. The equation now looks like this: 1.82

 L  V  P = 0.027  1.22    d  1000  Next: 10001.82 = 288403 And : 0.027 V1.82 L P = 288403 d1.22

(equation-A)



Solving for L (length) : (Continued) Multiply both sides by : 288403 d1.22  0.027 V1.82 L 288403 d 1.22   1.22  288403       d 1.22 P 288403 d  = 0.027 V1.82 L P 288403 d1.22  =

Divide both sides by 0.027 V1.82 P 288403 d1.22 

=

0.027 V 1.82 L 0.027 V 1.82

0.027 V1.82 P 288403 d1.22  L= 0.027 V1.82

or L =

P 288403 d1.22  0.03 f  V1.82

(equation-1)

Solving for d (diameter): The same equation-A is used as solving for L :

0.027 V1.82 L (equation-A) 288403 d1.22 Multiply both sides by 288403 d1.22  P =

0.027 V1.82 L 288403 d 1.22 

 =  288403 d P 288403 d  = 0.027 V L P 288403 d

1.22

1.22

1.22

1.82

Divide both sides by 288403 P 



 = 0.027 V

P 288403 d1.22  P  288403 

1.82

L P 288403 1 1

d1.22

0.027 V1.82 L = and A N = B then A = B N  d = P 288403

 0.027 V1.82 L 1.22    P 288403 

(equation-2)



Solving for V (velocity): 0.027 V1.82 L (equation-A) 288403 d1.22 Multiply both sides by 288403 d1.22  P =



P 288403 d

1.22

0.027 V1.82 L 288403 d 1.22 

 =

288403 d 1.22

Divide both sides by 0.027L:





P 288403 d1.22  0.027 V1.82 L = 0.027 L 0.027L

1

V1.82 =



P 288403 d

1.22



0.027L

and A N = B then A = B

1 N





 P 288403 d1.22   V=   0.027L 

1.82   

(equation-3)

It must be remembered that this equation for Velocity represents a velocity dependant on the pressure drop and length of duct and is not an instantaneous velocity calculation where CFM is simply divided by the duct cross section. Example-1: A galvanized duct with an equvilant length of 79 feet is required by design have a pressure drop of 0.125 inches water gage static friction, the duct velocity is 980 feet per minute. Find the exact required equivalent diameter of the duct. 1

 0.027 V1.82 L 1.22 equation-2 will be used to solve this problem  d =   see equation-2  P 288403  1

 0.027 980 1.82 79 1.22  d=  = 9.927"  (inches diameter)  0.125 288403    2

2

 r  144

to check the cfm =

 9.927    feet  2   = square feet of area =  = 0.537 feet 2  980  = 526.3 CFM 144 min  



A circular duct does not have 90 degree square corners, which cause vortices. A recatngular duct of the same cross section area as a round duct will have a higher friction loss and will be able to handle less capacity in cfm (cubic feet per minute) than a round duct. That is the reason all ductwork is sized by equivalent diameter from the following equation found in The “Carrier System Design Manual”, Air Disribution Part 2, 10th Printing 1974, page 2-34: 0.625

de

ab  = 1.3 a + b 0.25

(equation-4)

where: a = side a of a rectangular duct b = side b of a rectangular duct d e = equivalent diameter of the rectangular duct

Example-2: Using the 9.927 inch diameter round duct from example-1, find a square duct which is equivalent. (Note: it is mathematically impossible to solve for a or b from equation-4, therefore either side a or b must be known or assumed after which the problem becomes an interative process.) The square duct equivalent for a 9.927 inch diameter duct is: 0.625

de

9.075 9.075 = 1.3

9.075 + 9.0750.25

= 9.92 inch diameter

a 9.075 inch square duct is equivalent to a 9.92 inch diameter duct A different way of finding an equivalent square duct size corresponding to an equivalent round duct is to produce a table and develope a constant from equation-4 as found on page-5.



Circular Equivalent Table Of Square Ducts Based on equation-4 (in inches) Side a Side b Circular Equivalent de 10 10 10.932 Dia. 11 11 12.025 Dia. 12 12 13.118 Dia. 13 13 14.211 Dia. 14 14 15.304 Dia. 15 15 16.397 Dia. 16 16 17.491 Dia. 17 17 18.584 Dia.

Note that with each additional inch of square duct the circular equivalent is an additional 1.093 inches of diameter. 1.093 is a constant. To get a circular equivalent of 10 x 10 duct simply multiply 1.093 x 10 = 10.93 inches in equivalent diameter. A 14 x 14 square duct equivalent diameter would be = 14 x 1.093 =15.304 inches in equivalent diameter. For rectangular ducts where a  b, tables, duct calculation devices or equation-4 must be used.

Example-3: A mechanic has a duct passing through a firewall and finds that his air flow is lower than normal, he suspects that a fire damper is partially closed but cannot prove it. He measures off a straight section of the 22 x 10 duct of 20 feet and then takes a velocity measurement of 400 fpm. The mechanic now takes a static pressure reading of both ends of his 20 foot duct. The recorded readings are +2.17 inwg on the upstream side and +2.16 inwg on the downstream side yielding o (2.17 -2.16)- .01 inwg pressure drop in 20 ft. Find the length of duct from the readings that have been taken. (see page 6 of 6 for the solution)



Example –3 (continued): Solution: First find the equivalent diameter by using equation-4: 0.625

de

22 10  = 1.3

0.25

22 + 10

= 15.91 inch diameter

Using equation-1 L =

0.01288403 15.911.22  0.03 0.9  4001.82

= 57.4 feet

(equation-1)

From the calculation, 57.4 ft. > 20ft., it seems that a fire damper is probably partially closed or something is blocking the duct. Example-4: An air to air heat exchanger manufacturer requires a velocity above 1,200 fpm to calibrate a new piece of equipment. The engineering department has determined that a velocity below 1,200 fpm will result in laminar flow causing the test to be of no value. The test equipment has a 12 x 12 inch 55 foot long duct that has a measured pressure drop of 0.1 inwg, using equation-3 what is the calculated velocity. 12 1.093 = 13.116 inches diameter 1





 0.1 288403 13.1161.22  1.82  V=  = 1,275 fpm   0.027 55    The velocity is above the 1,200 fpm.

William A. Greco 2404 Greensward N. Warrington, Pa. 18976 Reference: Carrier System Design Manual, Air Disribution Part 2, 1974

HVAC Duct Friction Equations

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