Answer:
Reduce general form of heat diffusion equation by applying given conditions and then substitute given temperature distribution equation to verify it. Use x = 0 condition in to the heat distribution equation to find conditions at the bottom. Set up energy equation to find surface temperature, and then using it find T.. Use the equations from b) to draw graphs.
Explanation:
Known:
plane layer of coal thickness L = 1 m
uniform energy generation q = 20 W/m^3
convection factor h = 5 W/m^2K
ambient air temperature T∞= 25°C
solar irradiation G= 400W/m^2
solar absorptivity and emissivity of the surface a_s = ∈ = 0.95
Temperature distribution equation:
T(x)=T_s+(qL^2/2k)(1-x^2/L^2)
a) General form of heat diffusion equation is:
d/dx(k*d/dt)+d/dy(k*dT/dy)+d/dz(k*dT/dz)+q=pc_p(dT/dt)
But since it is one dimensional steady state conduction, the general heat diffusion equation reduces to:
d/dx(dT/dx)+q/k=0
Substitute given temperature distribution in to the heat diffusion equation to verify that this equation is satisfied by the given temperature distribution:
d/dx{(d/dx)T(x)=T_s+qL^2/2k(1-x^2/L^2)+q/k=0
d/dx(qL^2/2k)*(-2x/L^2)+q/k=0
q/k=q/k
Use x = 0 in the heat distribution equation:
d/dx(qL^2/2k)*(-2*0/L^2)=q/k
q=0
That means there is no slope at x = 0 so the temperature gradient is 0, therefore there is no change in temperature at the bottom.
b) Find heat flux at the surface by equalize energy generation and heat flux by conduction:
q_x(x)=E_gen=q*x q_x(L)=qL
Set up energy balance equation considering every type of heat flux:
qL+αG_s=h(T_s-T∞)+∈σT_s^4
T_s=295.67 K==>22.67°
Using calculated T_s, find T(0):
T(0) =T_s + (qL^2/2k), thermal conductivity found using thermodynamic tables, k = 0.26W/mK
T(0) = 22.67+ 20.1 /2*0*0.26
T(0) = 61.13°C
c) Use the equations from case b) to draw graphs:
Red line, To
Blue line, T.
