# Heat Equation in cylindrical coordinates

1. Jan 29, 2013

### 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.

2. Jan 29, 2013

### Staff: Mentor

First integrate once to get:

$$r\frac{dT}{dr}-(r\frac{dT}{dr})_{r=0}=-\frac{q r^2}{2k}$$

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.

3. Jan 29, 2013

### 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.

Last edited: Jan 29, 2013
4. Jan 29, 2013

### Staff: Mentor

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 rB equal the outer radius of your bale, and let Tb equal the temperature at the outer radius. Then, you need to satisfy the boundary condition:
$$-k(\frac{dT}{dr})_{r = r_b}=hT_b$$

So substitute in and solve for T0.

5. Jan 29, 2013

### 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.

6. Jan 29, 2013

### Staff: Mentor

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.

$$T_0 -\frac{qr_b^2}{4k}=T_b$$

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

7. Jan 30, 2013

### 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.

8. Jan 30, 2013

### Staff: Mentor

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