Heat Equation in cylindrical coordinates

[eurekameh]

Large, cylindrical bales of hay used to feed livestock in
the winter months are D = 2 m in diameter and are
stored end-to-end in long rows. Microbial energy generation
occurs in the hay and can be excessive if the
farmer bales the hay in a too-wet condition. Assuming
the thermal conductivity of baled hay to be
k = 0.04 W/m*K, determine the maximum steady-state
hay temperature for dry hay (q,dot = 1W/m3), moist hay
(q,dot = 10 W/m3), and wet hay (q,dot = 100 W/m3). Ambient
conditions are T,infinity = 0 degrees C and h = 25 W/m2*K.

I believe I have to use the heat equation in cylindrical coordinates for this problem. Simplifying most terms, I have: (1/r)(d/dr)(k*r*dT/dr) + q,dot = 0, where q,dot is the heat generation term. I integrated this equation and got: T(r) = (-q,dot*r^2)/(4*k) + C1*r + C2, where C1 and C2 are constants of integration. I'm having trouble figuring out the boundary conditions and so, can't find C1, C2. Once I get the constants of integration, I'd find where dT/dr = 0 to get the radial distance r in which there is a maximum temperature. If anyone has any idea, thanks in advance.

 

[Chestermiller]

Large, cylindrical bales of hay used to feed livestock in
the winter months are D = 2 m in diameter and are
stored end-to-end in long rows. Microbial energy generation
occurs in the hay and can be excessive if the
farmer bales the hay in a too-wet condition. Assuming
the thermal conductivity of baled hay to be
k = 0.04 W/m*K, determine the maximum steady-state
hay temperature for dry hay (q,dot = 1W/m3), moist hay
(q,dot = 10 W/m3), and wet hay (q,dot = 100 W/m3). Ambient
conditions are T,infinity = 0 degrees C and h = 25 W/m2*K.

I believe I have to use the heat equation in cylindrical coordinates for this problem. Simplifying most terms, I have: (1/r)(d/dr)(k*r*dT/dr) + q,dot = 0, where q,dot is the heat generation term. I integrated this equation and got: T(r) = (-q,dot*r^2)/(4*k) + C1*r + C2, where C1 and C2 are constants of integration. I'm having trouble figuring out the boundary conditions and so, can't find C1, C2. Once I get the constants of integration, I'd find where dT/dr = 0 to get the radial distance r in which there is a maximum temperature. If anyone has any idea, thanks in advance.

First integrate once to get:

[tex]r\frac{dT}{dr}-(r\frac{dT}{dr})_{r=0}=-\frac{q r^2}{2k}[/tex]

The second term on the left hand side is equal to zero. Then integrate again using the boundary condition T = T₀ at r = 0. You can find the value of T₀ by making use of the boundary condition at the outer radius of the bale.

 

[eurekameh]

I'm not really understanding the second term on the left hand side. Is it your first constant of integration?

I have T(r) = (-q*r^2)/(4*k) + T,0 after integrating again and using that boundary condition that T(0) = T,0.
To find the maximum, I should integrate this to get dT/dr = (-q*r)/(2k) and set it equal to 0. This only gives me r = 0. So the maximum temperature occurs at r = 0, where the maximum temperature is T(0) = T,0. How do I find T,0? Something seems to me.

 

[Chestermiller]

I'm not really understanding the second term on the left hand side. Is it your first constant of integration?

I have T(r) = (-q*r^2)/(4*k) + T,0 after integrating again and using that boundary condition that T(0) = T,0.
To find the maximum, I should integrate this to get dT/dr = (-q*r)/(2k) and set it equal to 0. This only gives me r = 0. So the maximum temperature occurs at r = 0, where the maximum temperature is T(0) = T,0. How do I find T,0? Something seems to me.

Yes. That's right. All I did was integrate once between definite limits r = 0 and r = arbitrary r. This shows that your constant C1 is equal to zero. Now, how do you calculate T₀ from the information given. Let r_B equal the outer radius of your bale, and let T_b equal the temperature at the outer radius. Then, you need to satisfy the boundary condition:
[tex]-k(\frac{dT}{dr})_{r = r_b}=hT_b[/tex]

So substitute in and solve for T₀.

 

[eurekameh]

How are you getting that boundary condition? Why is the heat flux due to conduction equal to the heat flux due to convection? Are you doing an energy balance?

I don't think the temperature at the outer radius Tb is given, unless Tb = T,infinity = 0 celsius.

 

[Chestermiller]

How are you getting that boundary condition? Why is the heat flux due to conduction equal to the heat flux due to convection? Are you doing an energy balance?

I don't think the temperature at the outer radius Tb is given, unless Tb = T,infinity = 0 celsius.

There's convection outside the cylinder, and conduction inside the cylinder. The heat flux due to conduction has to match the heat flux due to conduction at the boundary of the cylinder.

[tex]T_0 -\frac{qr_b^2}{4k}=T_b[/tex]

[tex]-k(\frac{dT}{dr})_{r=r_b}=\frac{qr_b}{2}=h(T_0 -\frac{qr_b^2}{4k})[/tex]

 

[eurekameh]

I solved for T,0 with the last equation and when plugging in values for q,dot = 1 W/m^3, I'm getting T,0 = 6.27 K. And so, T(r) = (-q,dot*r^2)/(4*k) + 6.27. Thus, T,max = 6,27 K, but this answer is wrong.

 

[Chestermiller]

I solved for T,0 with the last equation and when plugging in values for q,dot = 1 W/m^3, I'm getting T,0 = 6.27 K. And so, T(r) = (-q,dot*r^2)/(4*k) + 6.27. Thus, T,max = 6,27 K, but this answer is wrong.

For the data you gave, that's the correct answer.