A plane layer of coal of thickness L=1 m experiences uniform volumetric generation at a rate of q=20W/m3 due to slow oxidation of the coal particles. Averaged over a daily period, the top surface of the layer transfers heat by convection to ambient air for which h=5W/m2⋅K and T[infinity]=25∘C, while receiving solar irradiation in the amount GS=400W/m2. Irradiation from the atmosphere may be neglected. The solar absorptivity and emissivity of the surface are each αS=ε=0.95. (a) Write the steady-state form of the heat diffusion equation for the layer of coal. Verify that this equation is satisfied by a temperature distribution of the form T(x)=Ts+qL22k(1−x2L2). From this distribution, what can you say about conditions at the bottom surface (x=0)? Sketch the temperature distribution and label key features. (b) Obtain an expression for the rate of heat transfer by conduction per unit area at x=L. Applying an energy balance to a control surface about the top surface of the layer, obtain an expression for Ts. Evaluate Ts and T(0) for the prescribed conditions. (c) Daily average values of GS and h depend on a number of factors, such as time of year, cloud cover, and wind conditions. For h=5W/m2⋅K, compute and plot TS and T(0) as a function of GS for 50≤GS≤500W/m2. For GS=400W/m2, compute and plot TS and T(0) as a function of h for 5≤h≤50W/m2⋅K.

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.  

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