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The following article was published in ASHRAE Journal, May 2009. ©Copyright 2009 American Society of Heating, Refrigerating and AirConditioning Engineers, Inc. It is presented for educational purposes only. This article may not be copied and/or distributed electronically or in paper form without permission of ASHRAE.

Equal Friction Solution for Duct Sizing By Fred W. Dougherty, P.E., Life Member ASHRAE

T

 his article presents an automated quick and simple equal friction solution to calculating duct diameter and pressure loss, using an iterative solution to

the Colebrook equation. An initial duct diameter is calculated for duct segment airflows at a Dp/100 ft (Dp/30 m) to be input by the designer. Typical values for supply and return ducts would be 0.09, 0.095, or 0.100 in. w.g. per 100 ft (22.41, 23.66, or 24.9 Pa per 30 m) of duct. In the calculations that follow, this parameter is referred to as dp100. The complete solution requires running two VBasic macros in Excel. A copy of the Excel workbook with the macros can be downloaded at www.ashrae.org/ ashraejournal/duct_size_by_dp100.xls. The solutions and macros presented here will primarily be of interest to 50

ASHRAE Journal

the independent design engineer who is designing HVAC systems for small commercial projects under 20,000 ft2 (1858 m2), generally using DX unitary equipment. Commercial duct design software normally is used for larger, more complex projects. Reviewing the

algorithms in the spreadsheet will help readers understand how the commercial software works, so that it isn’t simply a “black box.” The method outlined here can be applied to the static regain design method, but that exercise is left to the reader. The Designer’s Problem

After laying out the ductwork for an air-conditioning project and computing the airflow to each terminal, the designer knows the airflow and length for each supply and return duct segment, and can select a duct height that is compatible with the geometric constraints in the area where the ducts will be routed. (A segment can be defined as a run of duct About the Author Fred W. Dougherty, P.E., is a building systems consulting engineer in Panacea, Fla.

a s h r a e . o r g

May 2009

The Excel workbook discussed here is available online at www. ashrae.org/ashraejournal/duct_ size_by_dp100.xls. from one branch to the next. A branch may be a major junction or a take-off for a terminal diffuser or return grille.) The next challenge is selecting a diameter for the duct segment, finding a width if the duct is to be rectangular, and estimating the duct friction and fitting dynamic losses for each segment of the critical duct routing path. The sum of losses of all of the supply and return segments will make up the external static pressure requirement for the air-handling unit. If the filter is external to the unit, the filter and bracket may be included as a segment, or the filter pressure loss may be added if it is internal to a packaged air handler. For small projects, this procedure is often done manually using charts and fitting loss coefficients provided in the 2005 ASHRAE Handbook—Fundamentals.1 Even if a duct design computer program is available, the input/output is often too cumbersome to be practical for small projects. On the other hand, a thorough analysis using manual methods will also consume resources in a tight design budget, and can lead to potentially serious errors or unnecessarily oversized ducts. A Simple Spreadsheet Solution

Calculating “newductsize” V

= Duct Velocity, fps

Q

= Airflow, cfm

D

= Hydraulic Diameter, in.

Re

= Reynolds number

f

= Moody Friction Factor

e

= Roughness Coefficient

L

= Duct Length, ft

h

= Duct Inside Height, in.

w

= Duct Inside Width, in.

V

= Q/((PiD2/4)/144)

Re

= 8.56DV

Start Macro newductsize

Clear Contents of Calculated Cells

Read cfm, h,e,r, dp100

Go to Next Segment

Guess dia (D)

Calc V, Re

Guess f

factor = 1/f  0.5

Change dia

Calc Factor, factor1

factor1 = 2 log(12e/3.7/D +2 .51/Re/f  0.5)

Change f Factor = factor1

No

Yes return f dp100A = 12L fe(V/1097)2/D L = 100

calc dp100A = f(f,r,V,dia)

dp100A = dp100

No

Yes Return dia h = 0

Yes

No Guess w

dia1 = (1.3(hw)0.625)/[h + w]0.25

Calc dia1 = f(h,w)

dia1 = dia

Change dia1 No

Yes Return w

As noted previously, this spreadsheet is useful for projects that are generally Figure 1: Flow diagram for macro “newductsize.” under 20,000 ft2 (1858 m2) and have more than one zone. Design airflow typically will be less than The spreadsheet runs two macros. The first is “newductsize,” 15,000 cfm (7079 L/s). Referring to Reference 1, Figure 9, which calculates duct diameter as a function of segment airflow the recommended range for dp100 is 0.08 to 0.6 in. w.g. per and desired dp100, and duct width as a function of input duct 100 ft (19.92 to 149.4 Pa per 30 m) of duct. Experience shows height. (That is, if it is rectangular. If duct height is input as that selecting a target dp100 near the low end of this range: zero or left blank, duct width will be set to zero.) Duct seg0.09 to 0.1 in. w.g./100 ft (22.41 to 24.9 Pa/30 m) results in ment length is also input into “newductsize” but is only used low energy use at a reasonable material cost. as input to the second macro. Figure 1 is a flow diagram showAt airflows greater than 15,000 cfm (7079 L/s), air ve- ing the architecture of the macro, and the formulas used for locity in the duct may become excessive and create noise the calculations. An example of the output of “newductsize” problems. This program does not address this contingency, is shown as Figure 2. (Shaded cells are calculated; clear cells so at these higher airflows, the user is advised to manually are user input.) calculate duct diameter based on an acceptable air velocAll of the “dp100” values are identical. This is the value ity, rather than relying on the diameter calculated based on selected by the designer for the calculation. Second, the values equal friction. for “dia” and “width” are not whole numbers. This is a problem May 2009

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51

Iteration Techniques

Figure 2: Input/output of macro “newductsize.” 1.000

dp100, in. w.g. per 100 ft

because in practice, duct dimensions are always presented in design documents as integer inches. This problem is corrected when the “friction factor” macro is run. Twelve segments are shown. The first six are supply duct segments, and the second six are return duct segments with cfm reduced by the prorated amount of outdoor air delivered in the supply. In the case shown, each room in the zone has a return corresponding to its supply, so the program automatically calculates the reduced return airflow, and assumes that the return duct height will be the same as the supply. This feature can be overridden if not applicable, allowing the user to input return segments with airflows and duct heights. Because this method is so simple, it is practical to run it for all of the duct routes in a branching system, to ensure finding the “critical” route, which will result in the maximum external static pressure on the air handler.

D versus dp100

0.100

The heart of the method presented here is an iterative solution to the Colebrook equation, which draws on that proposed in a previous ASHRAE Journal article.2 Although friction factor, denoted as f, 0.010 25.0 24.0 23.0 22.0 21.0 20.0 19.0 18.0 17.0 16.0 15.0 14.0 may have a wide range of values in pipDiameter (in.) ing problems, the range in the areas of interest in air-conditioning duct design Figure 3: dp100 variation with diameter—constant airflow. is fairly narrow, and an iterative solution is easier and more compatible as a simple IF loop embedded An example of the f iteration loop in VBasic follows: in a computer program. The Colebrook equation is presented ‘guess initial f in the ASHRAE Handbook3 and is reproduced here as follows: f = .02 ifloopf: 1/f   0.5 = 2 log10 (12e/3.7D + 2.51/Re/f  0.5) (1) ‘calculate fact and fact1 fact = 1/(f ^ 0.5) Lester suggests solving this equation using the built-in Excel Sheet2.Cells(15 + sbrcount, 13) = fact iteration algorithm by evaluating each side separately starting fact1 = -2 * ((Log(12 * e / 3.7 / dia + 2.51 / Re / f ^ 0.5)) / with an initial guess of f. This can be expressed as follows: Log(10)) Sheet2.Cells(15 + sbrcount, 14) = fact1 factor = 1/f  0.5 diff = (fact – fact1) factor1 = 2 log10 × (12e/3.7D + 2.51/Re/f  0.5) (2) Sheet2.Cells(15 + sbrcount, 15) = diff countf = countf + 1 However, it is not clear that the built-in iteration algorithm Sheet2.Cells(15 + sbrcount, 11) = countf can be called in a VBasic macro, so the method presented here If countf < 99 Then performs the iteration within the macro, as follows: If Abs(diff) > 0.0001 Then f = f * (1 + (fact – fact1) / fact) diff = factor – factor1 GoTo ifloopf if abs(diff) > 0.0001 then f = f  × (1 + (factor – factor1)/factor) End If repeat until diff