What Is the Governing Equation for Heat Transfer in a Composting Pile?

[edge333]

Homework Statement

One dimensional system in vertical direction
2 meter high composting pile @ 65 $$\circ$$C
Top of pile has wind @ 40 $$\circ$$C and h = 50 W/m²*K
Bottom of pile at ground temperature of 20 $$\circ$$C
Only conductive heat transfer WITHIN pile; no convection
Volumetric biochemical heat generation, Q = 7 W/m³
Compost has k = 0.1 W/m*K

a.) Setup the governing equation and boundary conditions
b.) Determine the temperature as a function of height from the ground
c.) Calculate the maximum temperature in the pile
d.) Calculate the top surface temperature of the pile

Homework Equations

General equation:​

$$\rho c_{p}\frac{\partial T}{\partial t} + \rho c_{p}\frac{\partial}{\partial x} ( uT ) = k ( \frac{\partial^{2}T}{\partial x^{2}} ) + Q$$​

Fourier's Law​

$$q_{x} = -k \frac{dT}{dx}$$​

Newton's Law of Cooling​

$$q_{x} = h ( T_{s} - T_{\infty} )$$​

The Attempt at a Solution

a.)
There is no storage and no convection within the pile, so the general equation reduces to:

$$\frac{d^{2}T}{dx^{2}} = -\frac{Q}{k}$$​

Boundary conditions:​

At x = 0, T = 20 $$\circ$$C​

$$At \ x = L, -k \frac{dT}{dx} = h( T_{s} - T_{\infty} )$$​

Heat flux at the top surface due to conduction is equal to heat flux at the top surface due to wind convection​

b.)
Integrating once and using the second boundary condition gives:

$$\frac{dT}{dx} = - \frac{Q}{k} ( x + C_{1} )$$
​

$$C_{1}=\frac{h( T_{s} - T_{\infty})}{Q}-L$$​

$$C_{1}=\frac{(50 \ W/m^{2} \cdot K)(40^{\circ} C - 65^{\circ} C)}{(7 \ W/m^{3})}-(2 \ m)$$​

$$C_{1}=-177 \ m$$​

From what my professor said, this value for C(1) is incorrect (even the units). Haven't gotten to parts c.) and d.) yet. So I just need some using the boundary condition to find the first constant of integration.

 

[Mapes]

Is it possible your professor has defined $C_1$ as

$$\frac{dT}{dx}=-\frac{Q}{k}x+C_1$$

I would that's the more typical way to do the integration, though your way is fine too.

 

[edge333]

Yeah, that is the way some of my peers have done it (edit: tried to do it). However, the answer he gave us was 80.06 W/K which I can't figure out. Our my temperatures correct?

 

[Mapes]

Yeah, that is the way some of my peers have done it. However, the answer he gave us was 80.06 W/K which I can't figure out. Our my temperatures correct?

Sorry, the answer for what? $C_1$? $C_1$ can be defined multiple ways, so a number alone doesn't have any meaning without the definition.

 

[edge333]

I realize that C₁ can be defined multiple ways but the units are in W/k. I have a feeling he made a typo on the units though because the value for C₁ works numerically for the following equation in order to determine the maximum temperature in part c.), however, the units don't make sense:

$$T=-\frac{Q}{k}\frac{x^{2}}{2}+C_{1}x+C_{2}$$

where C₂ = 20 degrees C found by using the first boundary condition.

Max Temperature:​

$$T_{max}, \ \frac{dT}{dx}=0$$​

$$0=-\frac{Q}{k}x+80.06$$​

$$x=1.14 \ m$$​

This answer, x, for T_maxis another answer he gave us as being correct.

It seems to me that C₁ should be in Kelvin per meter not watts per Kelvin[/INDENT][/INDENT] Then again that could just be a typo although I still cannot figure out how to solve for C₁ regardless of method to find the correct solution.