
re: insulated trombe wall
12 may 1999
sylvan butler wrote:
nicksanspam@ece.vill.edu wrote:
>>you might make them each about 2% of the area, if using natural convection.
>i don't know how to interpret that "2% of the area" phrase. do you mean
>2% of the interior volume? or 2% of the glass frontal area?
i was suggesting making each vent hole about 2% of the glass frontal area.
>is there any advantage to assymetric inlet vs outlet sizes? e.g. should
>there be a larger outlet to accomodate less dense higher temp air?
the smaller hole tends to limit performance. i'd make 'em equal. i'd figure
it this way, for starters: one empirical chimney formula predicts airflow
cfm = 16.6 av square_root(h dt), where av is the area of each vent hole in
square feet, h is the chimney height in feet, and dt (f) is the temperature
difference between the holes. the heatflow in btu/h is about equal to the
fahrenheit temperature difference times the airflow in cfm.
suppose the average outdoor temperature is 30 f, and 1,000 btu/ft^2 of sun
falls on a south wall on an average january day (nrel's longterm phila
average) and the sunspace is 8' tall, and it has adequate depth for free
vertical airflow (at least 8'/15 = 6", according to steve baer), and its
single r1 glazing layer transmits 90% of the sun, which all arrives over a
6 hour solar collection period, ie a 1' width of 8' tall sunspace transmits
1'x8'x1000btu/ft^2x90%/6h = 1200 btu/h, which equals the useful heat flowing
through the vent holes into the house plus the heat loss from the glazing
to the outdoors, ie 1200 = (t70) 16.6 av sqrt(h(t70))+(t30)8ft^2/r1, or
av = (180t)/(5.875(t70)^1.5.) (1)
over 24 hours, the net heat delivered to the house (subtracting the loss
through the r1 dampers at night) is q = 6hx47av(t70)^1.518h(70f30f)2av/r1
= av(282(t70)^1.51440) = (180t)(48245(t70)^1.5) (substituting (1.))
using satanic calculus:
dq/dt = 0 = (1)(48245(t70)^1.5)+(180t)(245)(1.5)(t70)^2.5),
= 48+245(t70)^1.5+(66150368t)(t70)^2.5, so
48(t70)^2.5 = 245(t70)+66150368t = 49000  123t, and
t = 70 + (10212.56t)^0.4
which makes t = 84.42 with an initial guess of 90 f and 84.53 with another
iteration. from (1), av = 0.293 ft^2 per linear foot of sunspace, which
corresponds to 3.7% of the glazing area and results in a net solar collection
efficiency of 52%, ie q = 4160 btu/day per linear foot of 8' tall sunspace.
the efficiency doesn't depend much on the vent area. making t = 90 f makes
av = 0.171 ft^2 (2.1% of the glazing) and q = 4073 btu/day.
as another example, a 16' tall x 12' deep sunspace with r2 double glazing
with 80% solar transmission and r10 dampers (eg 2" of styrofoam with
thermofor passive vent openers) and 620 btu/ft^2 of sun that falls on
a horizontal surface in january transmits 0.8(620x12+1000x16)/6h = 3125
btu/h per linear foot of width. with an average sunspace temperature t,
66av(t70)^1.5 btu/h enters the house. if the curved sunspace has 20ft^2
of glazing per linear foot of width, (t30)20ft^2/r2 btu/h is lost through
the glazing, so 3125 btu/h = 66av(t70)^1.5 + 10(t30), and
av = (343t)/(6.6(t70)^1.5) (2).
over an average night, the dampers lose 18h(70f30f)2av/r10 = 144av btu.
q = av(6hx66(t70)^1.5144) = (343t)(6021.8(t70)^1.5) of net heat enters
the house (substituting (2.)) dq/dt = 0 means t = 70+(1620.182t)^0.4, which
makes t = 77.4 f for an optimum net heat delivery of q = 15647 btu/day when
av = 2 ft^2 (10% of the glazing), and a net collection efficiency of 67%.
putting some dark mesh screen inside the sunspace might improve efficiency
by reducing reradiation and keeping the cooler air near the glazing...
nick

