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
MedHi  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?
Thanks for your help.
Steve.
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.
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
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:
Here is a reference that should answer your questions.
http://www.accaconference.com/Merchant2/merchant.mv?Screen=PROD&Store_Code ¬COA&Product_Code38&Category_Code=MJoseph
MeehanDia duit
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
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 online 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.
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 highthermalmass 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
equivalentlength method must be used with caution. Later, when we
discuss the flow of air in ducts where the Reynolds number is lower,
the equivalentlength method is even a poorer approximation. The
equivalentlength method is NOT RECOMMENDED for use in HVAC design.
Sounds like it might work for highvelocity 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, balancedpressure 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, closelyparaphrased 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 equivalentlength 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
4.123591E02
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 fx260 calculator ($8.76 at WalMart) 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 straightthrough 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 QBC00
140 VBC=QBC/(PI*(DBC/24)^2)
150 PVBC=(VBC/4005)^2
160 CBL=.52+(VBC/VAB.8)/(.81)*(.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 QBD00
260 VBD=QBD/(PI*(DBD/24)^2)
270 PVBD=(VBD/4005)^2
280 CB=.02+(VBD/VAB.8)/(.81)*.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)/(.81)*(.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 QDH00
500 VDH=QDH/(PI*(DDH/24)^2)
510 PVDH=(VDH/4005)^2
520 CD=.02+(VDH/VBD.8)/(.81)*.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 LDH0
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 IMBALANCE0*(DPMAXDPMIN)/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 Tpivot 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?
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.
On 12 Oct 2005 17:53:56 0400, snippedforprivacy@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 /
It's the official ASHRAE 552004 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/hF would need (7030)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)(70T)^1.5 = (T30)9.6 makes
T = 70((T30)/38.1)^(2/3). Plugging T = 69 in on the right
makes T = 68.984 on the left. Repeating makes T = 68.985.
Nick
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 /
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