# Unsteady state heat transfer differential equation

1. Feb 22, 2015

### Mangoes

I'm going through the solution to a problem that was assigned to my class and there's a step I don't really understand which I think is a concept I'm misunderstanding.

1. The problem statement, all variables and given/known data

The curved surface of a cylinder of radius R and length L is insulated. The face at x = L is maintained at To and a constant heat flux qo enters the rod at x = 0. The temperature profile in the cylinder is at steady state.

At t > 0, the current flow through the wire generates heat at a volumetric rate S. Determine the transient unsteady temperature profiles in the rod after this sudden change. The density, specific heat, and thermal conductivity of the rod are ρ, C, and k respectively.

2. Relevant equations

Fourier's Law:

$$q = -k\frac{dT}{dx}$$

3. The attempt at a solution

For unsteady state problems, I've been taught to first solve the steady state component first and then tackle the unsteady state.

I apply conservation of energy with respect to the rate of accumulation of energy in the cylinder. Since the system is in steady state, accumulation is 0. For the steady state, there is no generation of energy. I also use the definition of flux (flux = (flow rate)/(area)) to express rate. I use a differential shell to write the balance.

Rate Accumulation = Rate In - Rate Out + Rate Gen

$$0 = q[x]*A - q[x+Δx]*A$$

Note that q[x] means flux evaluated at x and q[x+Δx] means flux evaluated at x + Δx. A is area. I divide out A and divide by Δx and take the limit as Δx→0 to get:

$$0 = \frac{dq}{dx}$$

Applying Fourier's law,

$$0 = \frac{d^2T}{dx^2}$$

This is a simple ordinary differential equation which can easily be solved for. In solving the two constants, I use the initial conditions q[x=0] = qo and T[x=L] = To. The solution isn't important to where I'm getting stuck at, but I use a similar approach for the unsteady state which is why I bothered even including this.

Now, for the unsteady state, there's a heat generating term and accumulation isn't 0 anymore.

Rate Accumulation = Rate In - Rate Out + Rate Gen

$$\frac{dE}{dt} = q[x]*A - q[x+Δx]*A + S*A*Δx$$

S is given as a rate per unit volume, so multiplying it by volume will give the rate of heat generation.

Now I note that

$$\frac{dE}{dt} = (A*Δx)(ρ)(C)\frac{dT}{dt}$$

I can replace my accumulation term with something in terms of temperature now. Once again, dividing out A and dividing by Δx and taking the limit as Δx→0,

$$ρC \frac{dT}{dt} = -\frac{dq}{dx} + S$$

Applying Fourier's law once again:

$$ρC \frac{∂T}{∂t} = k\frac{∂^2T}{∂x^2} + S$$

Now, basically the main insight I'm supposed to make at this point is that eventually my system will get to steady state. This means I should be able to split my temperature profile equation into two components: a component dependent on space only which will be the only term as t becomes very large, and a component dependent on both space and time which will be important during the time in which the system is in unsteady state.

So,

$$T(x,t) = T_1(x) + T_2(x,t)$$

Applying the postulate to the equation above it with the partials gives me.

$$ρC \frac{∂T_2}{∂t} = k\frac{∂^2}{∂x^2}(T_1 + T_2) + S$$

Note that I killed off the space-dependent term on the LHS because it's being differentiated with respect to time.

Here's where I'm getting confused.

I can write two equations from the above. One for when steady state is reached and one for unsteady state.

In steady state, nothing is going to be changing with respect to time, so the LHS is 0 and the following must be true:

$$0 = k\frac{∂^2T_1}{∂x^2} + S$$

Meanwhile, in unsteady state, I would think that the following equation is true:

$$ρC \frac{∂T_2}{∂t} = k\frac{∂^2}{∂x^2}(T_1 + T_2) + S$$

But according to the solution I'm being given, the equation is supposed to be:

$$ρC \frac{∂T_2}{∂t} = k\frac{∂^2T_2}{∂x^2}$$

I'm not understanding why this is. Why would we be able to neglect both the change in T1 and the volumetric rate S?

Apologies for the long post but I couldn't really think of a way to cut this down much.

Last edited: Feb 22, 2015
2. Feb 23, 2015

### Orodruin

Staff Emeritus
You are not neglecting them. You are using the fact that these terms cancel due to the steady state differential equation.

3. Feb 23, 2015

### Mangoes

I'm not really understanding why they cancel. The only thing I can see being canceled in steady state considerations are time-dependent factors, which neither S or T1 is.

4. Feb 23, 2015

### Orodruin

Staff Emeritus
You have written down the differential equation the stationary state should satisfy. Use it!

Note that I am not saying they are zero, I am saying that they cancel because of the conditions you put on the steady state.

Edit: To put it in another way: Solve for S in the steady state equation and insert it into your new equation...

5. Feb 23, 2015

### Mangoes

Ohhh, I was looking at it completely wrong. I was misinterpreting where the last equation was coming from. What you're saying makes much more sense. Thanks a lot, much appreciated!

6. Feb 23, 2015

### Orodruin

Staff Emeritus
On an additional note, you could have let the stationary solution have any inhomogeneity, which would have resulted in a different PDE for the dynamic solution. However, the one chosen removes the inhomogeneity from the dynamic PDE, which makes it easier to solve.