# Heat Equation and Energy Balance

1. Apr 19, 2016

### joshmccraney

Hi PF!

For the longest time I thought an energy balance and the heat equation were identical procedures. However, recently I saw an example of a steady state, constant property, laminar flow of fluid between two flat surfaces where the top surface moves in the $x$ direction at $V_1$ and we assume fully developed hydrodynamical and thermally. The distance between plates is $a$ and the top surface is insulated with temperature $T_o$ and bottom surface has constant heat flux $q''$.

After an energy balance I arrive at $$\frac{\partial T}{\partial x} = \frac{q''}{a V_1 \rho c}$$
But the energy equation gives (after simplification) $$\frac{\partial T}{\partial x} = \frac{\alpha}{V_1}\frac{\partial^2 T}{\partial y^2}$$
And clearly the right hand side of these two equations, while equal, are represented quite differently. Ultimately we find $$ak\frac{\partial^2 T}{\partial y^2}=q''$$ but I thought flux was equal to $-k \nabla T=-k\partial_yT$ for this problem.

Thanks so much!

2. Apr 19, 2016

### Staff: Mentor

The "energy balance equation" is incorrect. The V1 should be the average velocity in the x direction, T should be the "mixing cup average" temperature, and the derivative should be an ordinary derivative. You can obtain the correct "energy balance equation" by multiplying both sides of the "energy equation" by $V_1(y) \rho c$ and integrating with respect to y between the two boundaries.

3. Apr 19, 2016

### joshmccraney

You're totally correct, sorry for this. However, if we are thermally fully developed isn't it true $\partial_xT_{mean} = \partial_x T$?
So is the equality I've concluded with wrong? Namely, $akT_{yy} \neq q''$ since the velocity should be an average in the energy balance but an actual velocity profile in the energy equation?

4. Apr 19, 2016

### Staff: Mentor

Ah. I now see what this problem is about that you are trying to solve. You are trying to determine the asymptotic temperature variation in a constant heat flux problem at large values of x. I have lots of experience with this kind of problem, and have solved many like it during my career.

The key to getting a solution is to a problem like this is to recognize the general mathematical form that the solution must take, to wit:
$$T=T_0+Kx+f(y)$$
Note that it is a function of x plus a function of y. By making use of the energy differential equation and the boundary conditions, you need to determine the constant K and the function f(y). The function f(y) must be such that its mixing cup average is zero.

chet

5. Apr 20, 2016

### MexChemE

Hi, Chet and Josh! The way I'm interpretting the constant heat flux statement is as a boundary condition for y = 0, not that the heat flux for all y is constant. Taking this approach, the energy balance can be written as
$$\Delta y W (\rho v_x \hat{H} |_x - \rho v_x \hat{H} |_{x+\Delta x}) + \Delta x W (q'' |_y - q'' |_{y+\Delta y}) = 0$$
Rearranging and taking limits
$$\rho v_x \frac{d \hat{H}}{dx} + \frac{d q''}{dy} = 0$$
Inserting the definition of enthalpy and Fourier's Law
$$\rho v_x c \frac{\partial T}{\partial x} - k \frac{\partial^2 T}{\partial y^2} = 0$$
$$\frac{\partial T}{\partial x} = \frac{k}{\rho c v_x} \frac{\partial^2 T}{\partial y^2} = \frac{\alpha}{v_x} \frac{\partial^2 T}{\partial y^2}$$
Taking this approach, the energy balance now agrees with the energy equation. However, as Chet pointed out, vx is the average velocity, not the velocity of the upper plate.

6. Apr 20, 2016

### Staff: Mentor

Actually, in your equations, $v_x$ is the local velocity at y, $v_x(y)$.

The problem that Josh is solving is analogous to the problem of laminar flow in a pipe with a constant wall heat flux at all axial locations x > 0. For long distances down the pipe, the temperature approaches an asymptotic profile in which the radial temperature variation stops changing with x. This problem is solved in BSL.

7. Apr 20, 2016

### joshmccraney

Hi MexChemE!

Regarding you energy balance:
I don't think $q''$ is evaluated at both endpoints of the energy balance (since this flux is imposed on the base plate. My balance looks like $$W \Delta x q'' + V_{mean} \rho c a W T(x) = V_{mean} \rho c a W T(x + \Delta x) \implies\\ q'' = \rho c a V_{mean} \frac{d T}{d x}$$
Is this correct?

8. Apr 20, 2016

### Staff: Mentor

Yes. dT/dx is the value of the constant K in my post #4. So you've already determined that. Now, all you need to do is determine the function f(y).

9. Apr 20, 2016

### joshmccraney

Awesome, but I have a few questions. First, when computing this infinitesimal mass balance technically $T(x,y)$ is actually $T(x,y-mean)$, where $y-mean$ implies the average temperature in $y$, right?

Secondly, if this mass balance is correct, why isn't it recognizably identical to the energy equation (if my understanding is correct, the energy equation is a mass balance once you apply the divergence theorem to the flux); is this correct? Please feel free to elaborate on this point.

And lastly, how would anyone know to come up with the ansatz you proposed for the solution form?

10. Apr 20, 2016

### Staff: Mentor

T(x,y) is the temperature at the coordinates location x,y.
The "recognizable" heat balance equation is:$$\rho C_p v(y) \frac{\partial T}{\partial x}=k\frac{\partial^2 T}{\partial y^2}$$
where $v(y)=V_1\frac{y}{a}$
If we integrate this equation between y = 0 and y = a, we obtain:
$$\rho C_p\frac{d [\int_0^a{v(y)T(x,y)dy}]}{dx}=k\left(\frac{\partial T}{\partial y}\right)_a-k\left(\frac{\partial T}{\partial y}\right)_0\tag{1}$$
The mixing cup temperature at location x is defined as:
$$\bar{T}(x)=\frac{\int_0^a{v(y)T(x,y)dy}}{\int_0^a{v(y)dy}}=\frac{\int_0^a{v(y)T(x,y)dy}}{\bar{v}a}\tag{2}$$
So, combining Eqns. 1 and 2, we obtain:
$$\rho C_p\bar{v}a\frac{d\bar{T}}{dx}=q''\tag{3}$$So this equation follows directly as a result of integrating the "recognizable" heat balance.

From Eqn. 3, we see that the mixing cup temperature must be rising at a constant rate with x, irrespective of what the temperature is doing in the y direction. So, $$\bar{T}(x)=T_0+\frac{q''}{\rho C_p\bar{v}a}x\tag{4}$$
Once the temperature profile becomes fully developed, the temperature profile between y = 0 and y = a is no longer changing with x. So, once this region is reached, $T=\bar{T}(x)+f(y)$

11. Apr 21, 2016

### MexChemE

Oh, that makes more sense, I was totally misunderstanding the implications of the heat flux. Initially, the reason I suggested the balance I wrote was that it would agree with the energy equation, but it was the energy equation what had to be adapted to the problem, not the balance. I should've known better, I'm always more inclined to set up the balance personally instead of using the equations when solving this kind of problem.

By the way, I enjoyed your last post, specifically, the integration wrt y of the energy equation and how it ended up agreeing with Josh's balance. Awesome skills!

12. Apr 21, 2016

### Staff: Mentor

Having tons of practical experience helps a little.

Chet

13. Apr 21, 2016

### joshmccraney

Sorry, I was unclear in what I meant by $T(x,y-mean)$. I am asking if on this energy balance, since we are slicing this up pieces of $\Delta x$ and not $\Delta y$, the $y$ direction quantities must be averaged? So the temperature we use in the balance must be $1/a \int_0^a T(x,y) \, dy$ rather than simply $T(x,y)$; is this correct?

I do have one more question: when setting up the energy balance, initially I did not include a flux term for the stream-wise direction. Now I'm not sure why. Can you explain why there wouldn't also be flux term $-aWk\partial_xT(x,y)$ entering and $-aWk\partial_xT(x+\Delta x,y)$ leaving?

I agree with MexChemE, excellent explanations!

14. Apr 21, 2016

### Staff: Mentor

No. T(x,y) is what I said it was. If you are talking about the integrated average heat balance involving the mixing cup average temperature, this is not a function of y, and it is not the temperature at the average y either.
The effect of heat conduction in the x direction is usually neglected compared to the conduction in the thickness direction (in this case, y), because the temperature gradients in the x direction are much smaller. Dimensional analysis also elucidates this same result. Only in the case of liquid metals have I ever heard of conduction in the axial direction being included.

15. Apr 21, 2016

### joshmccraney

So if I were to include this flux term in the energy balance we would have
$$-aWk\partial_xT(x)+W \Delta x q'' + V_{mean} \rho c a W T(x) = -aWk\partial_xT(x+\Delta x,y)+V_{mean} \rho c a W T(x + \Delta x) \implies\\ q'' = \rho c a V_{mean} \frac{d T}{d x} - k\frac{d^2T}{dx^2}$$

Since $q''$ is a vertical flux, would we scale it as $k/y^2$ and conclude that it dominates over the $T_{xx}$ term since, by the channel geometry, $x>y\implies x^2 \gg y^2 \implies 1/y^2 \gg 1/x^2$ and thus $q''$ dominates over $T_{xx}$?

Also, I've been very lazy with partial and ordinary derivatives. Technically all should be partial derivatives in this post, right?

16. Apr 21, 2016

### Staff: Mentor

No. Start with the differential equation and integrate with respect to y. The last term here is incorrect.
Basically, yes.
Except in the equation obtained by averaging over y, where its dTbar/dx (for the case where the axial conduction is neglected).

17. Apr 21, 2016

### joshmccraney

Isn't it true that neither $q''$ nor $\rho c V_{mean} T_x$ are functions of $y$? If so, integrating gives an $a$ multiplier. If that balance is incorrect, how would I set up that flux term?

18. Apr 21, 2016

### Staff: Mentor

The balance is correct, but there should be an "a" in front of the T'' term. And, of course, this only applies to the region where the temperature profile is fully developed. Plus, that T'' term doesn't have any boundary conditions specified on either end.

19. Apr 21, 2016

### joshmccraney

You said earlier that the last term was incorrect. Was the $a$ what you were referring to?

20. Apr 21, 2016

yes.