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re: duct sizing
11 oct 2005
wayne whitney   wrote:

>joseph meehan  wrote:
>> sorry not that easy.  you need to do a manual d.

vee must do a manual d!!! :-)

>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.

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.
>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)...

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...

>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?

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'    |
            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=22'est. duct diameter (inches)
30 cfm=3200'airflow
40 v=cfm/(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 l=50'duct length (feet)
100 dp=f*12*l/d*pv'pressure drop ("wg)
110 print dp


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 dpabdghdpabc then dpmax=dpabdgh else dpmax=dpabc
690 if dpabdef>dpmax 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

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=21:dbc=17:dbd=20:ddf=16:ddh=18 (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.

>are these reasonable maximum linear flow rates: register 250 fpm, 
>supply branch and return trunk 500 fpm, supply trunk 750 fpm?

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?

>for a gas furnace or a hydronic furnace, how does one determine
>the appropriate air temperature for the pressure drop calculations?

rtfm? :-)


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