The rate of heat flow (conduction) between two points

The rate of heat flow (conduction) between two points on a cylinder heated at one end is given by dQ/dt=λA*(dT/dx) where λ = a constant, A = the cylinder’s cross-sectional area, Q = heat flow, T = temperature, t = time, and x = distance from the heated end. Because the equation involves two derivatives, we will simplify this equation by letting dT/dx=(100(l-x)(20-t))/(100-xt) where L is the length of the rod. Combine the two equations and compute the heat flow for t = 0 to 25 s. The initial condition is Q(0) = 0 and the parameters are λ = 0.5 cal cm/s, A = 12 cm2, L = 20 cm, and x = 2.5 cm. Plot your result.

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[aren rajan]

yes i did yet i don't get the answers

[aren rajan]

Answer 1

[Ameer Hamza]

You can solve it as follow:

1) Finding an analytical solution

lambda = 0.5;
A = 12;
L = 20;
x = 2.5;
syms Q(t)
eq = diff(Q,1) == lambda*A*100*(L-x)*(20-t)/(100-x*t);
heatFlowEquation = dsolve(eq, Q(0)==0);
heatFlow = double(subs(heatFlowEquation, 25) - subs(heatFlowEquation, 0))
heatFlow =
 2.2610e+04

2) Numerical Solution:

lambda = 0.5;
A = 12;
L = 20;
x = 2.5;
[t, Q] = ode45(@(t,Q) lambda*A*100*(L-x)*(20-t)/(100-x*t), [0 25], 0);
heatFlow = Q(end) - Q(1);
heatFlow =
 2.2610e+04

Answer 2

[KAMVELIHLE JARA]

A furnace wall is built up with 200 mm thick refractory bricks and 150 mm insulating bricks. The temperature of the surrounding is 40°C, whereas that inside the furnace is 1000°C. The thermal conductivities of the refractory bricks and insulating bricks are 5W/m K and 0.5W/mK respectively. If the coefficients of heat transfer for the furnace gas and air is 80 and 40W/K, determine the rate of heat flow per square metre.