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?
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.
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.
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.
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)
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.
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=22:DBC=16:DBD=17:DDF=12:DDH=14'duct diameters (inches)
40 QAB=3200'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 LAB=50'AB duct length (feet)
120 DPAB=FAB*12*LAB/DAB*PVAB'AB friction loss ("WG)
130 QBC=1500
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 DPBL=CBL*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 LBC=30
230 DPBC=FBC*12*LBC/DBC*PVBC
240 DPC=20*DPBC/LBC'C grill loss ("WG)
250 QBD=1700
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 DPB=CB*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=60
350 DPBD=FBD*12*LBD/DBD*PVBD
360 QDF=700
370 VDF=QDF/(PI*(DDF/24)^2)
380 PVDF=(VDF/4005)^2
390 CDL=.52+(VDF/VBD-.8)/(.8-1)*(.52-.42)
400 DPDL=CDL*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=80
460 DPDF=FDF*12*LDF/DDF*PVDF
470 DPE=.22*PVDF'E loss ("WG)
480 DPF=20*DPDF/LDF
490 QDH=1000
500 VDH=QDH/(PI*(DDH/24)^2)
510 PVDH=(VDH/4005)^2
520 CD=.02+(VDH/VBD-.8)/(.8-1)*.02
530 DPD=CD*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=100
590 DPDH=FDH*12*LDH/DDH*PVDH
600 DPG=.22*PVDH
610 DPH=20*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 DPABDGHDPMAX THEN DPMAX=DPABDEF
700 IMBALANCE=100*(DPMAX-DPMIN)/DPMIN
710 FANPOWER=QAB*DPMAX
720 PRINT DPABDEF,DPABC,DPABDGH,IMBALANCE,FANPOWER
DAB=22:DBC=16:DBD=17:DDF=12:DDH=14 ("equal friction) makes
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?
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.
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.
Too bad he's a f****ng lunatic, and his numbers are all wrong as usual.
Have fun trying to build his nightmare.
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Or just use Nick's programs, which are free, and damned near worth what they cost !
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