|
|
re: frugal ac tips
3 aug 2002
wrote:
>change the filter in your ac to increase airflow across the coils and
>improve the efficiency of the unit...
maybe it's better to let the filter clog up or restrict the airflow
in order to make the ac into a more efficient dehumidifier that puts
the latent heat outdoors, vs a more efficient air cooler.
it's my understanding that ac standards have changed in the direction
of more airflow and warmer output air temps in order to make them more
efficient air coolers vs dehumidifiers. maybe that's the wrong direction
for energy conservation. ("consult your hvac criminal." :-)
ashrae's comfort standard, as i understand it, says people are equally
comfortable from about 67 to 81 f, at a temperature t = 87.2-1600w,
where w is the humidity ratio, ie pounds of water per pound of dry air.
for instance, at 74 f, w = (87.2-74)/1600 = 0.00825, so the partial
pressure of water vapor pa = 29.921/(1+0.62198/w) = 0.392 "hg. the
vapor pressure of 74 f air at 100% rh is ps = exp(17.863-9621/(460+74))
= 0.748 "hg, roughly, so rh = 100pa/ps = 52% for comfort.
say harry and nick live in identical houses with a thermal conductance
of 400 btu/h-f plus 100 cfm of air leaks, with no internal heat gain,
for simplicity. it's 100 f at 50% rh outdoors. harry's ac makes 60 f air
at w = 0.01117 with a thermostat set to 74. nick's moves less air, with
a partially clogged filter or several filters and a freezestat to avoid
icing and 40 f output air with w = 0.005279 and some kind of weirdass
controller that dehumidifies until t = 87.2-1600w, if possible.
who has the lower ac bill?
harry spends (100-74)400 = 10,400 btu/h on sensible cooling from heat
conduction. at about 0.075 lb/ft^3, the air that leaks into his house in
an hour weighs about 100cfmx60m/hx0.075 = 450 pounds. cooling it takes
another 450x0.24(100-74) = 2808 btu, so the total sensible cooling is
13,208 btu/h, and his ac recirculates 13,208/(0.24(74-60)) = 3931 pounds
per hour of air which might remove 3931(w-0.01117) pounds of water vapor
from the house. with wout = 0.0213, the incoming air contains 450wout
= 9.6 lb/h of water vapor, vs 450w in the house air that leaks out. if
the ac removes the difference, w = 0.0122, so rh = .0122/0.0825 = 77%.
in the house. harry finds that uncomfortable and cranks the thermostat
down to 68.4 f, and his total ac power becomes 20,356 btu/h (see below.)
nick's 40 f ac with output air at w = 0.00528 is mostly a dehumidifier.
he's happy at 73.3 f indoors, with a total ac power of 19,253 btu/h,
6% less than harry's (see below.)
moral: sloth pays. leave the filter alone.
nick
10 tout=100'outdoor temp (f)
20 rout=50'outdoor rh (%)
30 g=400'house conductance (btu/h-f)
40 cout=100'outdoor air leakage (cfm)
50 mout=60*cout*.075'air leakage (lb/h)
60 pout=rout/100*exp(17.863-9621/(460+tout))'outdoor vp ("hg)
70 wout=.62198/(29.921/pout-1)'outdoor hum rat
80 tac=60'ac air temp (f)
90 pac=exp(17.863-9621/(460+tac))'ac vp ("hg)
100 wac=.62198/(29.921/pac-1)'ac hum rat
110 for t = 68.2 to 68.6 step .1'room temp (f)
120 wc=(87.2-t)/1600'comfy w at temp t
130 q=(tout-t)*(g+.24*mout)'sensible cooling (btu/h)
140 p=q/(.24*(t-tac))'air moved (lb/h)
150 w=(mout*wout+wac*p)/(450+p)'actual w
160 l=mout*(wout-w)*1000'latent cooling (btu/h)
170 pcool=q+l'total cooling power (btu/h)
180 print t,1000*wc,1000*w,pcool
190 next
ac air room air comfy w actual w ac power
60 f 68.2 11.875 11.69504 20464.83
68.3 11.8125 11.7027 20410.59
68.4 11.75 > 11.71039 20356.33
68.5 11.6875 11.71811 20302.06
68.6 11.625 11.72588 20247.76
40 f 73.2 8.75 8.613618 19311.48
73.3 8.687501 > 8.631456 19252.65
73.4 8.625002 8.649376 19193.79
73.5 8.562502 8.667384 19134.88
73.6 8.500004 8.685474 19075.95
we might make very "cold air" with a desiccant solution
with solar regeneration, and keep the room close to 87 f.
|
|
