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 –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
in order to obtain results to graph the temperature distribution in the Trombe wall after 2, 4, 6, and 8 hours of exposure.
GIVEN
• Trombe wall suddenly exposed to sunlight
FIND
(a) Graph the temperature distribution in Trombe wall after 2, 4, 6 and 8 hours of exposure.
See the figure to the right for the arrangement of control volumes and nodes and symbol definitions.
The nodes are located at
and the time steps are given by
For the half control volume at i = 1, the explicit form of the energy balance is
where h is the heat transfer coefficient on the room-side of the wall. For the half control volume at
i = N, the explicit form of the energy balance is
where Uo is the combined heat transfer coefficient to outside ambient.
Solving for TN, m + 1
For all the interior nodes, i = 2, 3, 4, ... N – 1, the energy balance is
The solution of the above problem is obtained by discretization in matlab.
Following is the matlab code for problem.
L=0.02; % in m
N=11;
t=2; % in hour
delt=0.5 ; % in seconds
delx=L/(N-1); % in m
qabs=500; % in W/m^2
h=10;% W/(m^2 K)
crho= 2.6*10^5;
alpha=5*10^-7; % in m^2/s
k=0.13; % W/(m K)
Tinf=15; % Celsius
T(1:N,20)=10; % Celsius
Tout=-15; % Celsius
for m=1:1:(3600*t/delt)
for i=N-1:-1:2
Tf=T;
T(i,m+1)=T(i,m)+(alpha*delt/delx^2*(T(i-1,m)-2*T(i,m)+T(i+1,m)));
end;
T(1,m+1)=T(1,m)+(2*delt/(crho*delx)*(k/delx*(T(2,m)-T(1,m)+h*(Tinf-
T(1,m)))));
T(N,m+1)=T(N,m)+(2*delt/(crho*delx)*(k/delx*(T(N-1,m)-T(N,m)+qabs-
h*(T(N,m)-Tout))));
% count=0;
% for i=1:N
% if abs(Tf(i)-T(i))<10^-4
% count=count+1;
% end
%
% end
% if count==N
% break
% end
end
for i=1:1:N
x(i)=(i-1)*delx;
end
plot(x,T(:,14401))
The temperature distribution for 2 hrs is obtained is shown below.
Since the heat flow becomes steady state in less than 2 hours, the temperature distribution for
4, 6 and 8 hrs are same as above.
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