Brick Test


How do you think the temperature of a brick will respond to sunlight?
Well this of course depends on sunlight intensity, ambient temperature and a few other factors. To increase the bricks sensitivity to sunlight let's paint the top surface of the brick black and encase the bottom and sides of the brick in insulation. Let's also cut the brick in half and place glazing over one half-brick and leave the other half-brick exposed. Let's also drill a 2" deep hole in the brick sections where we can place temperature probes to obtain accurate average temperature readings of each section. Our test fixture will look like this:

The test fixture box is made from cedar boards lined with 1/2" isocyanurate foam. Both bricks are painted black and installed inside the insulated framework. One brick has a Kalwall glazed cover and the other brick is exposed to the atmosphere. Since the sides and back of the bricks are well insulated (R=4) compared to the tops of the bricks we can assume that most of the heat gains and heat losses of the bricks  take place through the top surface area.

To calculate the solar heat gain we'll first calculate the heat loss when no sunlight is available AND THEN we'll calculate the heat gain when sunlight strikes the bricks. This will be done for both the glazed brick and the un-glazed brick.


Qg = HEAT GAIN    10AM to 11AM

SOLAR HEAT GAIN = Q gain + Q lost


The above graph contains all the data we need to calculate the solar heat gain and efficiency of heat gain from the glazed and unglazed bricks, but we'll need to analyze data in bright sunlight and data in the absences of sunlight to understand the nature of solar heat gain. The orange bars indicate the areas where data will be taken and analyzed. The first orange bar HEAT GAIN (10AM-11AM) is the bright sunlight time period with a solar flux density of about 250 BTU/ft^2. The second orange bar HEAT LOSS (6AM-7AM) is the time period without sunlight.

To calculate solar heat gain Qs we'll need to know the heat loss Ql and the heat gain Qg. We will also need to know the mass (m) of the brick 2.5 lbs. and the heat capacity of the cement material.  Cp =.2 BTU/lb/*F

HEAT LOSS  glazed brick             Ql = (T1-T2) x Cp x m  = (44*-36*) x.2 x 2.5 lb  = 5 BTU/hr
HEAT LOSS  glazed brick/*F        Ql/*F = 5 BTU* /  (t1-t2) =  5 / (40*-22*)          = .28 BTU
HEAT LOSS  un-glazed brick        Ql = (T1-T2) x Cp x m  = (32*-24*) x.2 x 2.5 lb  = 4 BTU/hr
HEAT LOSS  un-glazed brick/*F    Ql/*F = 4 BTU* /  (t1-t2) =  4 / (26*-22*)         = 1 BTU


HEAT GAIN  glazed brick             Qg = (T1-T2) x Cp x m  = (92*-70*) x.2 x 2.5 lb  = 11 BTU/hr
SOLAR HEAT GAIN glazed brick    Qs = Qg + (t1-t2) x Ql/*F = 11 BTU + (81*-32*) x .28  = 25 BTU
HEAT GAIN  un-glazed brick             Qg = (T1-T2) x Cp x m  = (82*-64*) x.2 x 2.5 lb  = 9 BTU/hr
SOLAR HEAT GAIN un-glazed brick    Qs = Qg + (t1-t2) x Ql/*F = 9 BTU + (73*-32*) x 1  = 50 BTU


OK so according to these calculations the solar heat gain of the unglazed brick is twice the solar heat gain of the glazed brick even though the glazed brick has a net heat gain of 11 BTUs and the unglazed brick has a net heat gain 9 BTU...  Now if we assume that the solar flux density between 10AM and 11AM is 250 BTU/ft2 and the surface area for heat collection is .08 ft^2 the solar energy available to the brick is .08 x 250 = 20 BTU

so     the efficiency of the unglazed brick between 10AM and 11AM is 9/20 = 45%
AND  the efficiency of the glazed brick between 10AM and 11AM is 11/20   = 55%

This all makes sense and is about what I expected, but can you explain a solar heat gain of 50 BTUs/hr when only 20 BTUs of solar energy are available to the brick? Is it possible that the Cp of the brick is .60 BTU/*F/lb. ????

Jug Test