# Homework Help: Heat flow rate with non-constant k through plane wall

1. Oct 9, 2014

### Maylis

1. The problem statement, all variables and given/known data
Consider a large plane wall with thickness L whose thermal conductivity varies in a specified temperature range as
$$k(T) = k_{0}(1 + \beta T)$$
where $k_{0}$ and $\beta$ are two specified constants. The wall surface at $x=0$ is maintained at a constant temperature of $T_{1}$, while the surface at $x = L$ is maintained at $T_{2}$. Assuming steady one dimensional heat transfer by conduction through the wall only, obtain a relation for the heat transfer rate through the wall.

Sketch temperature and heat flux profiles for zero, positive and negative coefficients. Explain briefly why gases usually have a positive coefficient a, but liquids and solids show a negative coefficient. Water is an exceptional liquid that shows a positive coefficient β, explain.

2. Relevant equations

3. The attempt at a solution
I tried this two different ways, and am stuck a little bit. The first approach I did a general energy balance
FIRST METHOD
$$\frac {dE}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x$$
$$\frac {d(\rho \hat c_{p} A \Delta x T)}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x$$
$$\rho \hat c_{p} A \Delta x \frac {dT}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x$$
divide through by $A \Delta x$
$$\lim_{\Delta x \to 0} \rho \hat c_{p} \frac {dT}{dt} = \frac {1}{A \Delta x} (\dot Q_{x} - \dot Q_{x + \Delta x}) + \dot e_{gen}$$

Assuming that no heat is generated and using Fourier's law, $\dot Q = -kA \frac {dT}{dx}$, this reduces to

$$\rho \hat c_{p} \frac {dT}{dt} = \frac {-1}{A} \frac {d \dot Q}{dx}$$
$$\rho \hat c_{p} \frac {dT}{dt} = \frac {-1}{A} \frac {d}{dx} (-kA \frac {dT}{dx})$$
The Area terms cancel out as well as the negative terms on the RHS,
$$\rho \hat c_{p} \frac {dT}{dt} = \frac {d}{dx} (k \frac {dT}{dx})$$
With the assuming of steady state and the problem statement stating that $k = k_{0}(1 + \beta T)$, we get

$\frac {d}{dx}[k_{0}(1 + \beta T) \frac {dT}{dx}] = 0$

Now I am hesitant to try to integrate this thing, with the $\frac {d}{dx}$ not combining to make $k_{0}(1+ \beta T) \frac {d^{2}T}{dx^2}$

SECOND METHOD
For this, I just went straight away with Fourier's Law
$$\dot Q = -kA \frac {dT}{dx}$$
$$\dot Q = -k_{0}(1 + \beta T)A \frac {dT}{dx}$$
$$\frac {\dot Q}{-k_{0}A} dx = (1 + \beta T) dT$$
$$\frac {\dot Q}{-k_{0}A} \int dx = \int (1 + \beta T) dT$$
This integrates to be
$$\frac {\dot Q}{-k_{0}A}x = T + \frac {\beta}{2}T^{2} + C$$
Using the boundary condition of $T = T_{1}$ at $x = 0$, this will be
$$0 = T_{1} + \frac {\beta}{2}T_{1}^{2} + C$$
$$C = -T_{1} - \frac {\beta}{2}T_{1}^{2}$$
Putting back into the original equation
$$\frac {\dot Q}{-k_{0}A}x = T + \frac {\beta}{2}T^{2} -T_{1} - \frac {\beta}{2}T_{1}^{2}$$
$$\dot Q = \frac {-k_{0}A}{x} [T + \frac {\beta}{2} T^{2} - T_{1} - \frac {\beta}{2}T_{1}^{2}]$$
So I do get $\dot Q = f(T)$, but it seems like I completely ignored the second boundary condition. It seems like this should have two constants of integration, which would be closer to what I saw in the first method. So I am both not sure how to integrate the first method, and here unsure of my boundary conditions for the second method.

Last edited: Oct 9, 2014
2. Oct 9, 2014

### Staff: Mentor

I think that in your second method you can make both integrals definite, incorporating your boundary conditions.
$$\frac {\dot Q}{-k_{0}A} \int_0^L dx = \int_{T_1}^{T_2} (1 + \beta T) dT$$

Last edited: Oct 9, 2014
3. Oct 9, 2014

### Staff: Mentor

Both methods are correct, and give the same answer.
This equation is correct. The term in brackets must be a constant if its derivative is zero. That constant is minus the heat flux.
This result is correct so far. Just substitute the other boundary condition at x = L, and you will get the value of Q dot (which is constant, and independent of both x and T).

You were soooo close to getting the right answer with both methods.

Chet

4. Oct 9, 2014

### Maylis

So then, that means

$$\dot Q = \frac {-k_{0}A}{L} [T_{2} + \frac {\beta}{2} T_{2}^{2} - T_{1} + \frac {\beta}{2} T_{1}^{2}]$$

I didn't know you can just substitute two boundary conditions into one equation in sequence, that seems strange...unless there were two constants of integration

Last edited: Oct 9, 2014
5. Oct 9, 2014

### Staff: Mentor

In method 1 there actually are two constants of integration. In method 2, you already recognized that the heat flux is constant, and, to find out what that constant is, you needed to apply the condition on the other boundary.

Chet

6. Oct 10, 2014

### Maylis

I'm now trying to sketch the temperature and heat flux profiles for $\beta = 0$, $\beta < 0$, and $\beta > 0$. I think for $\beta = 0$, the temperature profile will be a straight negative slope, and heat flux is just a straight horizontal line.

However, I am not sure about the other two

If the thermal conductivity of the wall is changing, how can the heat flux be constant conceptually? I think the conductivity will tend to go down as temperature goes up. At the beginning the temperature T1 > T2, the thermal conductivity will be at its lowest value, but will start decreasing as it goes along. If the material isn't uniformly conducting heat, how can the heat flow rate be a constant?

I was doing another problem, and I came across something peculiar. When thermal conductivity is a small number, it seems like the temperature profile has a very steep slope. How can this be? When you have a small thermal conductivity, you shouldn't conduct heat very well, so the temperature should not change very much.

Also, then next part of this question
Consider a 1.5-m-high and 0.6-m-wide plate whose thickness is 0.15 m. One side of the plate is
maintained at a temperature of 500 K while the other side is maintained at 350 K. The thermal
conductivity of the plate can be assumed to vary linearly in that temperature range as
$k(T) = k_{0}(1 + \beta T)$, where $k_{0}$ =18 W/m.K and $\beta$ = 8.7 x 10^-4 K^-1. Disregarding the edge effects and assuming steady one-dimensional heat transfer, determine the rate of heat conduction through the plate.

I just used the equation for $\dot Q$ that I derived earlier, and got -1300 J/s. How can this be physically interpreted? If I used a convention that positive $\dot Q$ is from $T_{1}$ to $T_{2}$, where $T_{1} > T_{2}$, doesn't this mean that the heat is flowing backwards, from the cold side of the wall to the hot side of the wall??

Last edited: Oct 11, 2014
7. Oct 11, 2014

### Staff: Mentor

With a constant heat flux (or heat current), the temperature change per unit distance through the wall will vary with the local conductivity. So long as all the temperature changes add up to the total temperature difference between one side and the other of the wall nature is satisfied.

It’s very much like a series electrical circuit where there’s a potential difference across a string of resistors. The current only depends upon the total resistance and total potential difference. While the resistors may have different values along the way, the current is the same through each resistor. Each resistor will have its own potential change depending upon the current and the resistor’s value.
When you wrote Fourier’s Law the $Q$ represented the heat on the “source” side of the wall and $\dot Q$ the change in heat on the source side. So a negative value represents heat leaving the source. By the way, when I run your numbers through the formula I don’t get your –1300 J/s. I get something quite a bit larger. So it might be worthwhile checking your calculation there.

8. Oct 11, 2014

### Maylis

So then the heat flux profile for all 3 situations will just be a straight horizontal line. Only the temperature profile will change with $\beta$

Also,
$\dot Q = - \frac {(18)(0.9)}{0.15}[350 + \frac {8.7 \times 10^{-4}}{2} (350)^2 - 500 + \frac {8.7 \times 10^{-4}}{2} (500)^2] = -1300 W$

How do you calculate it different??

9. Oct 11, 2014

### Staff: Mentor

Shouldn’t the second T2 term be negative?

10. Oct 11, 2014

### Maylis

Woops, I was reading my post #4, I changed that sign accidentally. I get about 22,190 J/s. However, from what you said, that means heat is coming into the source, which means that heat is flowing into the hot part of the wall??

Last edited: Oct 11, 2014
11. Oct 11, 2014

### dotmatrix

Can't even been to comment on your equations, but its my understanding that as heat goes up resistance goes down.

12. Oct 11, 2014

### Staff: Mentor

I retract my interpretation of sign for the direction of Q flow. Sorry about that bit of fuzzy thinking Dunno where my brain went to for that one. Your original interpretation is correct.

13. Oct 11, 2014

### Staff: Mentor

This presents what Gneill said in a slightly different way.
The temperature gradient decreases in such a way, that, when multiplied by the thermal conductivity, their product (the heat flux) is constant. The heat has nowhere else to go, so the local heat flux through the wall must be constant (as a function of position x).
If the temperature difference is fixed, then, for smaller thermal conductivity, the heat flow will decrease. If the heat flow is fixed, then for smaller thermal conductivity, the temperature gradient will have to be steeper. Again, the product of the temperature gradient and the thermal conductivity is equal to the heat flux.

14. Oct 11, 2014

### Maylis

Ok, well both can't be constant then, right? But the temperature difference is fixed in this problem as well, but I think the heat flux is as well. For $\beta = 0$, then if the heat flux is a constant (q vs. x is a straight horizontal line), then

$\dot q = \frac {-k_{0}}{L} [T_{2} + \frac {\beta}{2} T_{2}^{2} - T_{1} - \frac {\beta}{2} T_{1}^{2}]$, which for a fixed $T_{1}$ and $T_{2}$ is a constant.

going back to $\dot q = -k_{0} \frac {dT}{dx}$, if $\dot q$ is known to be constant, then integration yields
$$\frac {\dot q}{-k_{0}} \int dx = \int dT$$
$$\frac {\dot q}{-k_{0}} x = T + C$$
from the boundary condition at $x = 0$, $T = T_{1}$
$$0 = T_{1} + C$$
$$C = -T_{1}$$
so then
$$T = \frac {\dot q}{-k_{0}} x + T_{1}$$
of course I just ignored the second boundary condition, so something still isn't write here, but it suggests the temperature gradient will be linear with a slope of $\frac {\dot q}{-k_{0}}$

I suppose if I wrote it the other way and ignored the boundary condition at $x=0$, I would get
$$\frac {\dot q}{-k_{0}} \int dx = \int dT$$
$$\frac {\dot q}{-k_{0}} x = T + C$$
$$\frac {\dot q}{-k_{0}}L = T_{2} + C$$
$$C = \frac {\dot q}{-k_{0}}L - T_{2}$$
hence,
$$T = \frac {\dot q}{-k_{0}}(x - L) + T_{2}$$

I just have a feeling these equations are wrong. At $x = L$, then $T = T_{2}$, which is fine. However, at $x = 0$, $T = \frac {\dot q L}{k_{0}} + T_{2}$, and I calculate $T = 555.45 K$, which is not correct, it should be 500 K. The other boundary conditions yields the same, giving the correct $T_{1}$ at $x = 0$, but seems to fall apart at $x = L$. By the way this is for when $\beta$ is that number given in the problem statement, 8.7*10^-4

Last edited: Oct 11, 2014
15. Oct 11, 2014

### Staff: Mentor

At steady state the heat flux and temperature difference are both constant.

The heat flux is the result of the fixed temperature difference over the net conductivity once the system settles out and the temperature profile (and thus conductivity profile) has settled into its steady state.

16. Oct 11, 2014

### Staff: Mentor

What you're missing is that $\dot{q}=-k\frac{(T_2-T_1)}{L}$ (for the case of constant conductivity). This satisfies both your equations.

Chet

17. Oct 11, 2014

### Maylis

How did I not get that when I did the integration? Where did I go wrong...plus I am trying to find equation for T, not $\dot q$. I want the temperature profile. By the way, is the heat flux rate constant no matter the sign of beta, or will the heat flux rate not always be constant.

18. Oct 11, 2014

### Staff: Mentor

The heat flux will be constant, irrespective of beta.
Go back to your final equation using Method 1:
$$\frac{d}{dx}\left(k_0(1+\beta T)\frac{dT}{dx}\right)=0$$
Integrating once gives:
$$(k_0(1+\beta T)\frac{dT}{dx}=C_1$$
Integrating again gives:
$$k_0(T+\beta \frac{T^2}{2})=C_1x+C_2$$
Use the boundary conditions to determine C1 and C2

Chet

19. Oct 11, 2014

### Maylis

Darn. Why the heck does method 2 not produce the same results? Or do I have to do some handwavy math to make it that way..

20. Oct 11, 2014

### Staff: Mentor

Method 2 does produce the same results. When you yourself did the integration for method 2, you implicitly assumed that Q dot is constant(check the math). And this assumption was correct. If you evaluate Q dot using both boundary conditions, and then eliminate Q dot from the equation for the temperature, you get the exact same result as with method 1.

Chet

21. Oct 11, 2014

### Maylis

Okay, so starting from
$$\frac {d}{dx}[k_{0}(1 + \beta T) \frac {dT}{dx}] = 0$$
$$\int d[k_{0}(1 + \beta T) \frac {dT}{dx}] = \int 0 \hspace{0.05 in} dx$$
$$k_{0}(1 + \beta T) \frac {dT}{dx} = C_{1}$$
Integrating again,
$$\int k_{0}(1 + \beta T) dT = \int C_{1} dx$$
$$k_{0}(T + \frac {\beta}{2} T^{2}) = C_{1}x + C_{2}$$
Using the first boundary condition that $T = T_{1}$ at $x = 0$,
$$k_{0}(T_{1} + \frac {\beta}{2}T_{1}^{2}) = C_{2}$$
Now using the second boundary condition that $T = T_{2}$ at $x = L$,
$$k_{0}(T_{2} + \frac {\beta}{2} T_{2}^{2}) = C_{1}L + k_{0}(T_{1} + \frac {\beta}{2}T_{1}^{2})$$
$$C_{1} = \frac {k_{0}}{L}(T_{2} + \frac {\beta}{2}T_{2}^{2} - T_{1} - \frac {\beta}{2}T_{1}^{2})$$
Therefore, the general equation is
$$k_{0}(T + \frac {\beta}{2}T^{2}) = \frac {k_{0}}{L}(T_{2} + \frac {\beta}{2}T_{2}^{2} - T_{1} - \frac {\beta}{2}T_{1}^{2})x + k_{0}(T_{1} + \frac {\beta}{2} T_{1}^{2})$$
The $k_{0}$ all cancel, leaving
$$T(1 + \frac {\beta}{2} T) = (T_{2} + \frac {\beta}{2}T_{2}^{2} - T_{1} - \frac {\beta}{2}T_{1}^{2}) \frac {x}{L} + T_{1} + \frac {\beta}{2}T_{1}^{2}$$

So for $\beta = 0$, this is
$$T = (T_{2} - T_{1}) \frac {x}{L} + T_{1}$$
So I think I've convinced myself a number of times that the temperature profile for $\beta = 0$ is going to be a straight negative slope. Now...How the heck will I make a temperature profile when beta has a nonzero value? This is a nonlinear equation so I cant just do a plot of T vs. x on MATLAB, or at least not sure how to since the synthax will only allow me to define a variable, and I can't define it in terms of itself. How would you think of it if you just did it by hand? I should know how to do it by hand, but I like using computer software to analyze these things too

Last edited: Oct 11, 2014
22. Oct 11, 2014

### Staff: Mentor

I would solve for x as a function of T, and then plot x as the abscissa and T as the ordinate.
Chet

23. Oct 11, 2014

### Maylis

Alright, well I just ran this through matlab, and it seems like unless $\beta$ is on the order of 100 or 1000, then it hardly makes a difference. The temperature profile is basically a straight line for all of them. Even when $\beta$ is around 1000 or 10,000, it is still straight with just some slight curvature.

Here is an example of my code for when $\beta = 0$, the other code cells are identical except I change the value of $\beta$
Code (Text):

k0 = 18;
beta = 0;
L = 0.15; % meters
T1 = 500;
T2 = 350;
A = 0.9;

k = k0*(1+beta*T);
T = linspace(T1,T2,1000);
x = (T.*(1+(beta/2).*T)-T1-(beta/2)*T1^2)*(L/(T2+(beta/2)*T2^2-T1-(beta/2)*T1^2));
Qdot = ((-k0*A)/L)*(T2 + (beta/2)*T2^2 - T1 - (beta/2)*T1^2);
qdot = Qdot/A;

subplot(3,2,1)
plot(x,T)
xlabel('distance (meters)')
ylabel('Temperature (K)')
title('Temperature vs. distance (\beta = 0)')

subplot(3,2,2)
plot(x,qdot)
xlabel('distance (meters)')
ylabel('heat flux')
title('heat flux vs. distance (\beta = 0)')

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Last edited: Oct 11, 2014
24. Oct 12, 2014

### Staff: Mentor

The temperature profile approaches a limiting curve as $\beta$ becomes infinite. You should be able to derive the equation for that limiting profile from your final equation. What is it, and how does it compare with what you got for $\beta = 10000$? Does this explain why the profile doesn't change much over the full range of beta values?

Chet

25. Oct 12, 2014

### Maylis

Okay, well I am trying for $\beta \rightarrow {\infty}$, and the equation is

$$lim_{\beta \to \infty} T = \frac {\beta[(\frac {1}{\beta} T_{2} + \frac {1}{2} T_{2}^2 - \frac {1}{\beta} T_{1} - \frac {1}{2} T_{1}^2) \frac {x}{L} + \frac {1}{\beta}T_{1} + \frac {1}{2} T_{1}^{2}]}{\beta(\frac {1}{\beta} + \frac {1}{2}T)}$$

I would think to use L'hôpital's rule here, but I don't even think its applicable, all those terms with $\frac {1}{\beta}$ will vanish, but I wonder if I can cancel out the $\beta$ that I pulled out.

Last edited: Oct 13, 2014