[Greenbuilding] forced air heating vent placement

[Nick Pine]

<home-nrg at dnaco.net> wrote:

>You might check out the research done by Build America, some years back. As 
>I recall,they were able to design a single family house with all of the 
>ducts in a central wall cavity and, by using a more a focused grill, place 
>the supply grill on the inside (duct) wall and through conditioned air to 
>the far wall.

Some of Harry Thomason's solar houses had no blowers or fans. A large hot 
water tank in the basement was surrounded by stones with vertical ducts 
inside an inside wall that helped warm air rise and exit at the first floor 
ceiling level. Cooler air returned to the basement via floor registers near 
outside walls. Why fight the natural wintertime convective loop in which 
cool air slides down the inside of windows and warmer air rises up next to 
inside walls?

Thermal chimneys inside inside walls could also increase the output of 
hydronic baseboard radiators 
http://www.supplyhouse.com/Slant-Fin-104002040-4-ft-Baseline-2000-Baseboard-3-4-5591000-p
If 1' of 180 F radiator moves 600 Btu/h into 65 F air with a 1.5 
Btu/h-F-ft^2 fin airfilm resistance, ie 600 = 1.5Af(180-65), the equivalent 
total 2-sided fin area Af = 3.5 ft^2.

A T (F) chimney exit air temp makes the average chimney air temp Tb = 
(65+T)/2 and the average chimney-room air temp diff dT = Tb-65 with a C = 
16.6Asqrt(HdT) CFM flow. With 2" fins and an A = 2"x12"/144 ft^2 minimum 
chimney cross sectional area and height H = 8' between vents, C = 
7.8sqrt(dT) CFM and heatflow Q = 7.8dT^1.5 Btu/h, approximately.

Figure 1 of chapter 23 of the 1981 ASHRAE handbook (in pdf nrcc23738 on the 
web) shows an approximate surface airfilm conductance U = 1.5+V/4 Btu/h-ft^2 
with air velocity V in mph, with Q = (1.5+V/4)3.5ft^2(180-65) = 604+101V 
Btu/h. and V = 0.0682C mph, ie 0.532sqrt(dT) mph, so 604+101x0.532sqrt(dT) = 
7.8dT^1.5, and dT = (77.4+6.89sqrt(dT))^(2/3). Plugging in dT = 10  on the 
right makes dT = 21.4 F on the left. Repeating makes dT = 22.8, 22.9, and 
22.95 F.

So Q = 7.8dT^1.5 = 857 Btu/h, 43% more than the 600 Btu/h without the 
chimney. V = 2.5 mph and Tb = 87.95 F and T = 110.9 F.

It's hard to imagine a problem with condensation on the inside of windows in 
this scheme. Andersen's Guide-to-Understanding-Condensation pdf says an 
average family of 4 generates a total of 18 gallons of water vapor per week 
(0.8925 lb/h) by perspiration and breathing (12 gallons per day), showers 
(1/2 lb each), and so on. This could be lowered if people breathed less or 
took fewer showers or mopped floors less or cooked with lids on pots or 
fixed rain leaks or plumbing leaks or damp basements.

With 30 cfm of outdoor air ventilation and wi pounds of water per pound of 
dry indoor air, 60x30x0.075wi = 0.8925 lb/h makes humidity ratio wi = 
0.0066, with water vapor pressure Pa = 29.921/(1+0.62198/wi) =  0.315 "Hg = 
Re^(17.863-9621/(70+460), ie relative humidity R = 42% at 70 F, with a Tdp = 
1/(1/(70+460)-ln(0.42)/9621))-460 = 45.8 F dewpoint.

The inside glazing of a double glazed R2 window with an R1/3 indoor air film 
would be 45.8 F if the room air were 70 F and (70-45.8)3 = 72.5 Btu/h-ft^2 
flowed through the window, with an outdoor air temp T and (70-T)/2 = 72.5 
Btu/h, ie T = minus 75 F, if I did that right.

Nick