# Duct Sizing

• posted on October 7, 2005, 2:01 pm
Hi,
We had a new furnace installed last year and we're planning to change the ducts. We have a few questions we're hoping you can answer.
Furnace Info:
50K in, 46K out.
From the manual, at a static pressure of .5", CFM is 1093. (Fan Speed Med-Hi - Heat Only)
Currently set to Low Fan Speed due to undersized ducts.
Existing Layout:
Plenum is approx. 22" x 20"
Main Heating Duct is 5" x 11" x 16' (Height and Width severely undersized)
7 runs off Main Heating Duct. 4" diameter. Each run is approx. 14 ft.
3 runs off Plenum. 4" diameter. Each run is approx. 14 ft.
3 runs into Cold Air Return. 8" diameter.
Questions:
What size should the Main Duct be? 18" x 8", 18" x 10", smaller, larger?
Should the Main Duct reduce in size farther from the furnace to maintain pressure? If yes, when should it reduce? Existing Main Duct is 16' long.
What diameter should the runs be? 5" was suggested?
Steve.
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• posted on October 7, 2005, 7:37 pm
Steve Groulx wrote:

Sorry not that easy. You need to do a manual D. I suggest you also should do a manual J first. These are mathematical calculations based on your home and HVAC equipment.

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Joseph Meehan

Dia duit
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• posted on October 7, 2005, 4:13 pm

For the inquiring reader, do the calculations (heating only) go in the following way?
Calculate the heat loss of the entire structure given the dimensions and construction details. Choose a furnace whose capacity equals the heat loss on the coldest day of the year. Break the heat loss down room by room.
The furnace will specify a given flow rate at a given pressure drop. Divide the total flow rate among the rooms proportional to the heat loss for each room. Lay out the return and supply ducting, and determine the flow rate through each duct segment. Choose acceptable pressure drops for each duct segment (representing fittings by their equivalent lengths), so that the end to end total pressure drop from any return grill to any supply outlet never exceeds the furnace specification pressure drop (often 0.5" wc?). Now the minimum cross sectional profile of each duct segment can be determined from the total pressure drop allowed for that segment, the length of the segment, the flow rate through that segment, and a guideline for the maximum linear velocity of the air in the segment to avoid excessive noise.
Is that basically how it goes? Are these reasonable maximum linear flow rates: register 250 FPM, supply branch and return trunk 500 FPM, supply trunk 750 FPM? For a gas furnace or a hydronic furnace, how does one determine the appropriate air temperature for the pressure drop calculations?
Thanks, Wayne
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• posted on October 7, 2005, 6:47 pm
Hi Wayne,
Although it would be very time consuming, I believe it would work. I found this online that I believe may. http://www.freecalc.com/ductloss.htm
Thanks, Steve.
wrote:

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• posted on October 7, 2005, 10:31 pm
Wayne Whitney wrote:

http://www.accaconference.com/Merchant2/merchant.mv?Screen=PROD&Store_Code ¬COA&Product_Code3-8&Category_Code=M--Joseph MeehanDia duit
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• posted on October 7, 2005, 6:21 pm

Yes, I could spend \$125 for Manual J and \$46 for Manual D. But I'd rather not. Since you seem to be familiar with these references, could you at least indicate whether I have the basic idea of duct sizing procedures correct? I'd check the books at my local library, but they don't seem to have them.
Yours, Wayne
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• posted on October 8, 2005, 2:33 am
Wayne Whitney wrote:

Sorry, I did not mean for you to buy the books, only to use the site to get a better idea about what we are talking about.
Steve, I believe, has posted an on-line version that may help you.
I do not personally use the calculations, which is why I did not try to explain it myself.
I do suggest that you may want to have a professional do the actual calculations, especially if your home is not "typical" as they can use their experience to make good judgment calls. If you just want to get ball park information, then I would go ahead and use the on line version. Come to think of it, maybe I will give it a try and see what happens, I will be facing a system replacement in the not too distant future.
--
Joseph Meehan

Dia duit
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• posted on October 11, 2005, 1:54 pm

Vee MUST do a manual D!!! :-)

Sounds good. The ASHRAE Handbook of Fundamentals lists coldest days for lots of cities in 99 and 97.5 percentiles, eg 6 and 9 F in Boston. The outdoor temp would be warmer than the 99% winter "design temp" 99% of the time in an average year heating season, eg all but 50 hours in a 5000 hour season. The 97.5 temp can reduce the required furnace capacity if you have a high-thermal-mass house or you are willing to wear a sweater once in a while.

That works for water, but not for air, according to Kreider and Rabl's 1994 "Heating and Cooling of Buildings":
Leq = Kf/fxD. This relationship shows the fundamental shortcoming with the equivalent length approach. Even though Kf and D are constant for a given pipe fitting under various flow conditions, the friction factor f is not, unless one is operating in the fully turbulent region where f is independent of Re... Since this is not always the case, the equivalent-length method must be used with caution. Later, when we discuss the flow of air in ducts where the Reynolds number is lower, the equivalent-length method is even a poorer approximation. The equivalent-length method is NOT RECOMMENDED for use in HVAC design.
Sounds like it might work for high-velocity ducts...

Is this where all the friendly professional HVAC net.folk jump in to help? :-)
In the most common "equal friction duct design" (vs static regain, velocity reduction, balanced-pressure or constant velocity), you might budget a design pressure drop of 0.1 "WG (enough to support a 0.1" water column) per 100' of duct and use round ducts in 1" diameter increments. (A W"xH" rectangular duct has an equivalent diameter D = 1.3(WH)^0.625/(W+H)^0.25, eg W = 14.5" and H = 3.5" makes D = 7.3".)
Example 11.2, closely-paraphrased from Kreider and Rabl:
Room C needs 1500 cfm and F needs 700 and H needs 1000, and it's 50' from the blower to T B which goes 30' to room C, and another 60' from T B to T D, which goes 40' to elbow E, then 40' more to room F, and it's 40' from T D to elbow G and 60' more to room H, like this, viewed in a fixed font:
H 1000 cfm | | | C 1500 cfm |60' | | |30' | | | 50' | 60' D 40' | blower --------------------------------- G B | | |40' | 40' | F 700 cfm---------- E
Say the pressure loss at each branch outlet is equivalent to 20' of duct, (a US grill manufacturer's traditional wacky equivalent-length spec that fits traditional wacky US duct design practice.) Let's ignore pressure losses due to duct size transitions for now.
The pressure loss in each fitting is of the form dP = CPv, where C is a fitting coefficient and velocity pressure Pv = (V/4005)^2 "Wg for "standard air," with V in fpm. We can look up these coefficients in Table A5.6(b) (oh, you don't have one of these? :-) For instance, C = 0.22 for an r/D = 1.0 elbow. So we make Table 11.3...
duct fitting length flow loss diam vel Pv Cf loss Ptotal (feet) (cfm) ("Wg) (in.) (fpm) ("WG) ("WG) ("WG)
AB 50 3200 0.05 22 1210 0.091 0.050 BC 30 1500 0.03 16 1070 0.071 0.48 0.034 0.064 C 20 1500 0.02 1070 0.071 0.020 BD 60 1700 0.06 17 1080 0.073 0.011 0.001 0.061 DG 40 1000 0.04 14 935 0.055 0.013 0.001 0.041 GH 60 1000 0.06 935 0.055 0.22 0.012 0.072 H 20 1000 0.04 14 935 0.055 0.020 DE 40 700 0.04 890 0.049 0.51 0.025 0.065 EF 40 700 0.04 12 890 0.049 0.22 0.011 0.051 F 20 700 0.02 12 890 0.049 0.020
Since we've decided on 0.1 "WG/100', the duct loss column is 0.001 times the length of the duct section. Given the flow rate, we can find the duct diameter from Figure 11.9. If you don't have one of these, you can use a Darcy/Altshul/Tsai approximation. For instance, for AB, guessing D = 22" makes the duct loss DP = 0.0412 "WG.
10 PI=4*ATN(1) 20 D"'est. duct diameter (inches) 30 CFM200'airflow 40 VÏM/(PI*(D/24)^2)'air velocity (fpm) 50 PV=(V/4005)^2'velocity pressure ("WG) 60 RE=8.560001*D*V'Reynolds number 70 FP=.11*(12*.0003/D+68/RE)^.25'friction factor 80 IF FP>.018 THEN F=FP:GOTO 100 90 F=.85*FP+.0028 95 LP'duct length (feet) 100 DP=F*12*L/D*PV'pressure drop ("WG) 110 PRINT DP
4.123591E-02
With less guessing, if 0.05 = f*12*50/D(V/4005)^2 for segment AB, f = 1337D/V^2 = 3.88x10^-9D^5 = 0.11(12x0.0003+68/58609^2)^0.25, so D = 15.4(d^2+268.8)^(1/21). Plugging in D = 20 on the right makes D = 20.99 on the left. Repeating makes D = 21.05, then 21.055... Casio's fx-260 calculator (\$8.76 at Wal-Mart) does 21st roots.
Given the duct diameters and airflow volumes, we can find velocities. For instance, duct AB has a Pi(22/2/12)^2 = 2.64 ft^2 cross section, so V = 3200ft^3/m/2.64 ft^2 = 1210 ft/m, and so on.
Given the air velocities, we can find velocity pressures. For instance, duct AB has 1210 fpm, so Pv = (1210/4005)^2 = 0.091 "WG, and so on.
Now we need the velocity ratios to find the straight-through loss at Ts B and D... 1080/1210 = 0.89 at B and 935/1080 = 0.87 at D, which makes their coefficients 0.011 and 0.013, from the table labeled "main" in the "wye, diverging" entries in Table A5.6(i). (Oh, you don't have one of those?), with fitting pressure losses CfPv. The tables also supply branch loss coefficients 0.48 and 0.51 at B and D...
The final column of table 11.3 is the sum of straight duct and fitting pressure losses. The 3 branch total pressure drops are Pabdgh = 0.244", Pabc = 0.134", and Pabdef = 0.247", the maximum loss, so that's used for the blower requirement, along with other pressure drops, eg fans and filters.
Kreider and Rabl continue:
The shorter duct with only half the pressure drop of the other branches will require a balancing damper to provide approximately another 0.113 "WG pressure drop. This will result in a system in which the pressure drop in each branch is balanced...
This seems odd. Larger ducts cost more. Why not use a smaller duct for the extra pressure drop? That isn't part of "equal friction duct design," nor "modified equal friction." Manual D designers might stop here... but
Making BC 15 vs 16" makes Vbc = 1217 fpm, so Vab/Vc = 1 and the branch loss becomes 0... f = 0.11(0.0036/15+68/(8.56x1217x15)^0.25 = 0.0177, with duct loss Pbc = 0.0177x12x30/15(1217/4005)^2 = 0.039 "WG and loss Pc = 20x0.039/30 = 0.026 "WG at the C grill, which makes Pabc = 0.0115 vs 0.134 "WG.
Now we might try reducing Pabdgh and Pabdef with a little BASIC program:
20 PI=4*ATN(1) 30 DAB":DBC:DBD:DDF:DDH'duct diameters (inches) 40 QAB200'AB airflow (cfm) 50 VAB=QAB/(PI*(DAB/24)^2)'AB velocity (fpm) 60 PVAB=(VAB/4005)^2'AB velocity pressure ("WG) 70 RE=8.560001*DAB*VAB'AB Reynolds number 80 FPAB=.11*(12*.0003/DAB+68/RE)^.25'AB friction factor 90 IF FPAB>.018 THEN FAB=FPAB:GOTO 110 100 FAB=.85*FPAB+.0028 110 LABP'AB duct length (feet) 120 DPABúB*12*LAB/DAB*PVAB'AB friction loss ("WG) 130 QBC00 140 VBC=QBC/(PI*(DBC/24)^2) 150 PVBC=(VBC/4005)^2 160 CBL=.52+(VBC/VAB-.8)/(.8-1)*(.52-.42)'B branch line loss coeff. 170 DPBLËL*PVBC'B branch line loss ("WG) 180 RE=8.560001*DBC*VBC 190 FPBC=.11*(12*.0003/DBC+68/RE)^.25 200 IF FPBC>.018 THEN FBC=FPBC:GOTO 220 210 FBC=.85*FPBC+.0028 220 LBC0 230 DPBCûC*12*LBC/DBC*PVBC 240 DPC *DPBC/LBC'C grill loss ("WG) 250 QBD00 260 VBD=QBD/(PI*(DBD/24)^2) 270 PVBD=(VBD/4005)^2 280 CB=.02+(VBD/VAB-.8)/(.8-1)*.02'B through coeff. 290 DPBË*PVBD'B through loss ("WG) 300 RE=8.560001*DBD*VBD 310 FPBD=.11*(12*.0003/DBD+68/RE)^.25 320 IF FPBD>.018 THEN FBD=FPBD:GOTO 340 330 FBD=.85*FPBD+.0028 340 LBD` 350 DPBDûD*12*LBD/DBD*PVBD 360 QDFp0 370 VDF=QDF/(PI*(DDF/24)^2) 380 PVDF=(VDF/4005)^2 390 CDL=.52+(VDF/VBD-.8)/(.8-1)*(.52-.42) 400 DPDLÍL*PVDF 410 RE=8.560001*DDF*VDF 420 FPDF=.11*(12*.0003/DDF+68/RE)^.25 430 IF FPDF>.018 THEN FDF=FPDF:GOTO 450 440 FDF=.85*FPDF+.0028 450 LDF€ 460 DPDFýF*12*LDF/DDF*PVDF 470 DPE=.22*PVDF'E loss ("WG) 480 DPF *DPDF/LDF 490 QDH00 500 VDH=QDH/(PI*(DDH/24)^2) 510 PVDH=(VDH/4005)^2 520 CD=.02+(VDH/VBD-.8)/(.8-1)*.02 530 DPDÍ*PVDH 540 RE=8.560001*DDH*VDH 550 FPDH=.11*(12*.0003/DDH+68/RE)^.25 560 IF FPDH>.018 THEN FDH=FPDH:GOTO 580 570 FDH=.85*FPDH+.0028 580 LDH0 590 DPDHýH*12*LDH/DDH*PVDH 600 DPG=.22*PVDH 610 DPH *DPDH/LDH 620 DPABDGH=DPAB+DPB+DPBD+DPD+DPDH+DPG+DPH 630 DPABC=DPAB+DPBL+DPBC+DPC 640 DPABDEF=DPAB+DPB+DPBD+DPDL+DPDF+DPE+DPF 650 PRINT VAB,VBC,VBD,VDF,VDH 660 IF DPABDGH<DPABC THEN DPMIN=DPABDGH ELSE DPMIN=DPABC 670 IF DPABDEF<DPMIN THEN DPMIN=DPABDEF 680 IF DPABDGH>DPABC THEN DPMAX=DPABDGH ELSE DPMAX=DPABC 690 IF DPABDEF>DPMAX THEN DPMAX=DPABDEF 700 IMBALANCE0*(DPMAX-DPMIN)/DPMIN 710 FANPOWER=QAB*DPMAX 720 PRINT DPABDEF,DPABC,DPABDGH,IMBALANCE,FANPOWER
DAB":DBC:DBD:DDF:DDH ("equal friction) makes
Vab Vbc Vbd Vdf Vdh 1212.208 1074.296 1078.509 891.2676 935.4413 fpm
DPabdgh Dpabc Dpabdef Imbalance % Fanpower ..230287 .1239497 .2149416 "WG 85.79069 736.9182
DAB!:DBC:DBD :DDF:DDH (eyeballing) makes
Vab Vbc Vbd Vdf Vdh 1330.405 951.6254 779.2226 501.338 565.8841 fpm
DPabdgh Dpabc Dpabdef Imbalance % Fanpower ..1148923 .1195638 .1139547 "WG 4.922268 382.6043
Eyeballing works a lot better here, with 5% vs 86% imbalance and half the blower power. "Static regain with T-pivot optimization" might work even better, but that requires more serious software.

Sounds good to me... 400 fpm might use less blower power. But why use ducts at all, vs making air flow through floor grates and closets and rooms and hallways?

RTFM? :-)
Nick
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• posted on October 12, 2005, 7:29 pm
wrote:

Nick, your post is very informative as always. I'll have to try to get ahold of a copy of that book. Thanks for the informative example.
Cheers, Wayne
P.S. One suggestion, equations written as above are hard to parse. It would be alot easier to read as "L_eq = K_f/f*D", assuming I parsed it correctly.
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• posted on October 12, 2005, 9:49 pm
White space would definitely improve the readability also
wrote:

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• posted on October 12, 2005, 9:53 pm

It's over \$100. You might start with a recent edition of the ASHRAE Handbook of Fundamentals, about \$35 on ebay.
Nick
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• posted on October 13, 2005, 2:11 am
On 12 Oct 2005 17:53:56 -0400, snipped-for-privacy@ece.villanova.edu wrote:

Or just use Nick's programs, which are free, and damned near worth what they cost !
Click here every day to feed an animal that needs you today !!!
http://www.theanimalrescuesite.com /
Paul ( pjm @ pobox . com ) - remove spaces to email me 'Some days, it's just not worth chewing through the restraints.' 'With sufficient thrust, pigs fly just fine.' HVAC/R program for Palm PDA's Free demo now available online http://pmilligan.net/palm /
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• posted on October 12, 2005, 9:10 am

PE Drew Gillett (deanegathotmail.com) writes:

In the US, vs 2.5 cfm for a tight 2400 ft^2 Canadian house.

This makes it hard to estimate room-by-room losses.

Hmmm.
People don't say this often enough.

Doors also allow zoning...
Nick
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• posted on October 12, 2005, 7:26 pm
Thanks Nick. I appreciate the details and your help with this matter. Love the BASIC program.

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• posted on October 12, 2005, 9:15 pm

You are welcome.

It's the official ASHRAE 55-2004 standard language :-)

Maybe less. A 10x10' room with 2 R20 exterior walls and an 8 ft^2 R4 window and a thermal conductance of 9.6 Btu/h-F would need (70-30)9.6 = 384 Btu/h to stay 70 F. If 70 F air thermosyphons through a 3'x6' door with 384 = 16.6x9ft^2sqrt(6')dT^1.5, dT = 1 F.
More accurately, 16.6x9xsqrt(6)(70-T)^1.5 = (T-30)9.6 makes T = 70-((T-30)/38.1)^(2/3). Plugging T = 69 in on the right makes T = 68.984 on the left. Repeating makes T = 68.985.
Nick
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• posted on October 13, 2005, 12:25 am
Too bad he's a fucking lunatic, and his numbers are all wrong as usual.
Have fun trying to build his nightmare.
On Wed, 12 Oct 2005 15:26:39 -0400, "Steve Groulx"

Click here every day to feed an animal that needs you today !!!
http://www.theanimalrescuesite.com /
Paul ( pjm @ pobox . com ) - remove spaces to email me 'Some days, it's just not worth chewing through the restraints.' 'With sufficient thrust, pigs fly just fine.' HVAC/R program for Palm PDA's Free demo now available online http://pmilligan.net/palm /