Heat Equation in cylindrical coordinates

In summary: The maximum temperature occurs at r = 0. That's the point at which the temperature is 6.27 K. But this doesn't seem like a reasonable result. If the hay is dry, then there should be no heat generation. Therefore, the temperature should be the ambient temperature, or 0 degrees Celsius. I think you've made a mistake somewhere in your calculations. In summary, the problem involves finding the maximum steady-state temperature of cylindrical bales of hay that are used to feed livestock in the winter months. The hay is stored end-to-end in long rows and microbial energy generation occurs within the hay. The farmer must be careful not to bale the hay in a too-wet condition,
  • #1
eurekameh
210
0
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.
 
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  • #2
eurekameh said:
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.
 
  • #3
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:
  • #4
eurekameh said:
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 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:
[tex]-k(\frac{dT}{dr})_{r = r_b}=hT_b[/tex]

So substitute in and solve for T0.
 
  • #5
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
eurekameh said:
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]
 
  • #7
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
eurekameh said:
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.
 

FAQ: Heat Equation in cylindrical coordinates

1. What is the heat equation in cylindrical coordinates?

The heat equation in cylindrical coordinates is a partial differential equation that describes the flow of heat through a cylindrical object or system. It takes into account the variables of temperature, time, and spatial coordinates in the cylindrical system.

2. What are the assumptions made in the heat equation in cylindrical coordinates?

The heat equation in cylindrical coordinates assumes that the material being studied is homogeneous, isotropic, and has a constant thermal conductivity. It also assumes that the thermal properties of the material do not change with time.

3. How is the heat equation in cylindrical coordinates derived?

The heat equation in cylindrical coordinates is derived using the principles of conservation of energy and Fourier's law of heat conduction. It involves taking the dot product of the gradient of temperature with the thermal conductivity and setting it equal to the time derivative of temperature.

4. What are the boundary conditions for the heat equation in cylindrical coordinates?

The boundary conditions for the heat equation in cylindrical coordinates depend on the specific problem being studied. They can include prescribed temperatures at the boundaries, prescribed heat fluxes, and insulated boundaries where no heat can enter or leave the system.

5. What are the applications of the heat equation in cylindrical coordinates?

The heat equation in cylindrical coordinates has many practical applications, including in the study of heat transfer in pipes, heat exchangers, and other cylindrical systems. It is also used in modeling the temperature distribution in cylindrical objects during processes such as welding or cooling. Additionally, it is an important tool in understanding the behavior of fluids and gases in cylindrical systems.

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