To more accurately model the energy input from the sun, suppose the absorbed flux in given by qabs (t) = t (375 – 46.875 t) where t is in hours and qabs is in W/m2. (This time variation of qabs gives the same total heat input to the wall i.e., 2000 W hr/m2). for qabs in place of the constant value of 500 W/m2. Explain your results. A Trombe wall is a masonry wall often used in passive solar homes to store solar energy. Suppose that such a wall, fabricated from 20-cm-thick solid concrete blocks (k = 0.13 W/(mK), ? = 0.05 × 10–5 m2/s) is initially at 15°C in equilibrium with the room in which it is located. It is suddenly exposed to sunlight and absorbs 500 W/m2 on the exposed face. The exposed face loses heat by radiation and convection to the outside ambient temperature of

Question

–15°C through a combined heat transfer coefficient of 10 W/(m2 K). The other face of the wall is exposed to the room air through a heat transfer coefficient of 10 W/(m2 K). Assuming that the room air temperature does not change, determine the maximum temperature in the wall after 4 hours of exposure and the net heat transferred to the room.

GIVEN
• Trombe wall with specified absorbed solar flux as a function of time
FIND
(a) Maximum temperature in the wall after 4 hours
(b) Heat input to the room

Answer 1

See the accompanying figure for the arrangement of control volumes and nodes and symbol

definitions. The nodes are located as



and the time steps are given by



For the half control volume at i = 1, the explicit form of the energy balance is



For the half control volume at i = N, the explicit from of the energy balance is



For all the interior nodes, i = 2, 3, 4, ... N – 1, the energy balance is



The maximum temperature in the wall at any time step m must be TN, m and the heat input to the room

after any time step m is given by



Since we have chosen an explicit method, we can use the marching procedure as described Also, the time step ?t is restricted. After setting up the computer program to step through the time steps, the energy balance on nodes i = 1, 2, and N were checked by hand to insure that the code was correct. A run was then made with N = 41, ?t = 5 seconds. The results indicate that the heat input to the room is –1834 joules per m2 of wall area and the maximum wall temperature is 54.16°C. In comparison with the results where 30630 J/m2 was delivered to the room, here the room has lost 1834 J/m2 to the wall. The reason is that for early times, before the absorbed solar flux becomes significant, the wall is losing heat to the outside and is rapidly cooling. The room-side face of the wall dips below the air temperature of 15°C and begins to remove heat from the room. Only at later times does the wall heat up sufficiently to begin transferring heat back to the room. For the short 4 hour run, the net effect is a loss of heat from the room to the wall.