# Homework Help: Lumped Capitance Method (Calculating temp. + heat transfer)

1. Apr 19, 2013

### watsup1993

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

Problem is in attachment pic.

2. Relevant equations

3. The attempt at a solution

My attempt is on the pics 1 and 2.
Firstly, is it valid to assume length as 0.075m?
- For part (i), from the question it infers that one of the cylinders is applicable to the lumped capacitance method, for this Bi should be less than 0.1, although only cylinder is 0.15 which is the lowest. Bt I will assume that this is the one because the question asks to use the cylinder in which lumped capacitance works.
- In my working for part (i), I have calculated the temperature after 30mins, but the question asks for the temperature at the AXIS and at the SURFACE of the cylinder. So I am a little confused on how to go about with this
- For part (ii), is the way I calculated amount of energy per unit length released by cylinder in 30mins of cooling correct? Also it says there are two methods, has anyone got any idea on what the second method could be

Thanks alot!

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2. Apr 21, 2013

### Staff: Mentor

I wasn't able to see very well what you had written, but it seemed that the first part looked OK.

The length of the cylinder is irrelevant in this problem. The characteristic length in this problem is either the radius or the diameter. If the characteristic length in your book's definition of the Biot number is the radius of the cylinder, then using 0.075m was correct to do. If the characteristic length in your book's definition was the diameter, then you should have used 0.15m. I don't have your book, so I don't know, but in many books, the radial temperature profile within a cylinder is given graphically as a function of dimensionless time at specific values of the Biot number.

Regarding which cylinder is the appropriate one for the lumped capacitance method to be valid, you have to ask yourself "in which cylinder would you expect the temperature profile to be more uniform?"

Chet

3. Apr 21, 2013

### watsup1993

Okay, thanks for your reply. Yes, indeed in my book it says that the characteristic length is the radius, so what I have done is correct. Cylinder 1 with a Biot number of 0.15 would be appropriate. But then in my calculations, I calculated the temperature for that. But my question is how to temp. calculations for AXIS and SURFACE?

I mean, the temperature I calculated is it even correct, and is that for AXIS or SURFACE
-------------------------------------------------------------------------------------------------------------------
I actually did some thinking, so assuming Cylinder 1 has to be the one chosen for the temperature calculations. Since it has a low biot number of 0.15, is it safe to assume there is a relatively small temperature difference between the axis and surface of the cylinder (uniform temperature distribution) So its a trick question? The value for the temp. I got is for axis and surface?

Can you shed some light on part (ii), I am quite sure the method I have used is not scientifically correct, because first I calculated Q. Then I simply divided it by the length, to get energy per length (is that allowed?) Also what other way could be used

Last edited: Apr 21, 2013
4. Apr 21, 2013

### Staff: Mentor

Let's stick with part (i) for now. Your assessment regarding the small temperature difference is correct, but it is still possible to determine the temperature difference you are looking for. It is not a trick question. As I said, many books have plots of dimensionless temperature as a function of dimensionless time, with dimensionless radial location as a parameter. These plots are for specific values of biot number. Does your book have such plots? If not, you can find them in McAdams (I think). I can also show you another simple way of getting this answer without having to solve the unsteady state heat conduction equation. This approximation is applicable specifically to cases where the biot number is small. But it takes a little explaining, and it probably would be better if you could lay your hands on some of those plots.

Chet

5. Apr 21, 2013

### watsup1993

These plots of dimensionless temperature as a function of dimensionless time, they are not in my lecture handouts but I did look online and got some graphs. I just want to add this question is from a Past Exam Paper and all the information on the question given was only that, meaning no extra data (regarding plot value, or anything else) was given. So knowing that, if we were not meant to use these plot values. What would have been the correct approach.

6. Apr 21, 2013

### Staff: Mentor

Unfortunately, I don't have time to describe what I would have done right now. But I will get back with you as soon as I can (next day or so). In the meantime, see if you can use your solution to determine the time constant for exponential decay of the temperature difference. Are you able to come up with a number for this?

Chet

7. Apr 22, 2013

### watsup1993

'see if you can use your solution to determine the time constant for exponential decay of the temperature difference. Are you able to come up with a number for this'

The thing is i dont think I have to make it as complicated as your are suggesting, I am probably wrong. But I have no clue how this exponential decay of temp. diff. will help, moreover I have not come across this at university.

Please can you just tell me the value of temperature I got, what does that represent?

Also, I know you want me to first go through part (i) which is understandable. Bt does part (ii), look reasonable to you?

Thanks!

8. Apr 22, 2013

### Staff: Mentor

The temperature variation you got is only a first order approximation to the solution, but adequate for getting the average temperature. In fact, at this level of approximation, you were correct in deducing that the temperature you got is approximately representative of the average, the temperature at the surface, and the temperature at the center. But now the problem asks you to estimate the temperature at the center and at the surface. That means that you need to need to go to a slightly higher order of approximation. This can be done pretty easily, once you figure out how to reason things out.

I understand your reluctance to evaluate the time constant for exponential decay of the temperature without some guarantee that it will lead to a result reasonably quickly. So, what I'm going to do is lay out the whole thing for you now.

Please feel free to ask any questions about what I have discussed above. Also, if you would like me to lead you through this in small bite-size steps, I would be glad to do so.

In part ii there are two ways to do it.
1. Integrate the heat flux with respect to time, starting at time zero
2. Recognize that the total heat is just equal to the increase in enthalpy, based only on the initial and final temperatures.

9. Apr 24, 2013

### Staff: Mentor

Suppose you have a cylinder of diameter D that has a constant heat flux q applied at its surface. What is the rate of increase of the cross-section average temperature of the cylinder?
$$\rho C_p\frac{\pi D^2}{4}\frac{d\overline{T}}{dt}=\pi Dq$$
or, equivalently
$$\frac{d\overline{T}}{dt}=\frac{ 4q}{\rho C_pD}$$
After the radial temperature profile has had the opportunity to develop within the cylinder (i.e., at times long compared to D2/α, where α is the thermal diffusivity), the temperature within the cylinder can be expressed as:
$$T =\overline{T}+ f(r)$$
The transient heat conduction equation within the cylinder is given by:
$$\frac{\partial T}{\partial t}=\alpha \frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial T}{\partial r})$$
where $$\alpha = \frac{k}{\rho C_p}$$

If we combine the previous equations, we obtain:

$$\frac{ 2q}{\rho C_pR}=\frac{k}{\rho C_p} \frac{1}{r}\frac{d}{dr}(r\frac{df}{d r})$$
or equivalently
$$\frac{1}{r}\frac{d}{dr}(r\frac{df}{d r})=\frac{ 2q}{kR}$$
Integrating once with respect to r, we obtain:
$$\frac{df}{d r}=\frac{ qr}{kR}$$
Integrating again with respect to r, we obtain:
$$f=\frac{ qr^2}{2kR}+C$$
where C is a constant of integration. The constant of integration is determined by making using of the constraint that the average value of the radial variation f over the cross section is equal to zero:

$$\int_0^R{2\pi r f dr }= 0$$
From this, if follows that $$C=-\frac{qR}{4k}$$
and that:
$$f=\frac{ qR}{4k}(2(\frac{r}{R})^2-1)$$
Thus,$$T=\overline{T}+\frac{ qR}{4k}(2(\frac{r}{R})^2-1)$$
At the center of the cylinder, the temperature is $T=\overline{T}-\frac{ qR}{4k}$.
At the surface of the cylinder, the temperature is $T=\overline{T}+\frac{ qR}{4k}$.
If we reverse the sign on q so that the cylinder is cooling rather than heating, we get
At the center of the cylinder, the temperature is $T=\overline{T}+\frac{ qR}{4k}$.
At the surface of the cylinder, the temperature is $T=\overline{T}-\frac{ qR}{4k}$.

10. Apr 26, 2013

### watsup1993

Chet, my man. Thanks a billion for all your input, it is very much appreciated. I did not reply for the past few days because I was still working on it. I finished my working, realised my mistakes and even got it verified by a phD student. I will show the correct working, so if anyone come across this thread they may use it.

Firstly, the biot number I calculated for both cylinders were incorrect because I was using the wrong value for the length. The L(c) is always Volume/Area. So if we use the equations for a cylinder: (pi*r^2*h)/2pi*r*h = r/2.

Now using a length of r/2 = 0.0375, for cylinder 1 I get a biot number of 0.075 (less than 0.1 which satisfies the lumped capacitance rule)

So in my initial thumbnail, showing my calculation of the temperature giving 519.6K (is correct) And since, it is lumped capacitance it is safe to assume that this the temperature at the surface and the axis. (so in reality, it was straight forward)

Now for part (ii), as shown in the second thumbnail of my working. For the first method, I used integration of the lumped capacitance equation to get a value of Q. Now after getting this, I simply divided the value of Q by the length, (3039/0.075) = 40524 J/m. Now the guy said its fine, bt I doubt it. Chet, can you verify that what I have done is acceptable.

For the second method of obtaining the energy per unit length, he said to use: Q = m c(p) (Ti - Tend)
I am assuming Ti = 1000K, Tend from part (i) = 519.6K. c(p) is given as 450. Now I am a bit confused about mass, how do I get that. Is this even correct, and he said after I get 'Q', simply divide it by the length (0.075m) to get the energy per unit length.

Thanks!

P.S Sorry for the extra work, Chet!

Well actually, if the length is r/2, the height should also be r/2, correct?

Last edited: Apr 26, 2013
11. Apr 26, 2013

### Staff: Mentor

The length of the cylinder is irrelevant. You will see that if you write out the equations algebraically. In every case, the length will cancel out of the numerator and diameter. So it doesn't matter whether you use 0.075 or 75000, you will get the same answer. Check this out to make sure you are comfortable with it.

You also need to make sure that the two answers you get in part ii are exactly the same (as they should be).

I'm surprised you haven't used the final equations I derived in my previous post to estimate the temperature at the axis and at the surface, to satisfy yourself that they indeed differ from the average temperature and from one another by only a few degrees. In these equations,

q = h (519.6 - 293) W/m^2

Please try it out, and see what you get.

Chet

Last edited: Apr 26, 2013
12. Apr 26, 2013

### watsup1993

Okay, so using your equations I get temperature at center: 528.85K and at surface: 510.35. So yes, there is not a major difference.

'You also need to make sure that the two answers you get in part ii are exactly the same (as they should be).' <------------- This is bothering me now.

So in my working using the lumped capacitance, I got a value of Q as 3039J.
Second method: mc(Ti - Tend)
(im still assuming, height as 0.075 so its easy to compare to wht i have already done)
Mass = density*volume = 7500*pi*0.075^3 = 132.54 kg (because r = h = 0.075m)
Using value of mass, Q = 132.54*450*(1000-519.6) = 2,865,2497 J (obviously incorrect)
Third Method: Q = hA (Ti - Tend)
Q = 80*2*pi*0.075^2 * (1000-519.6) = 1358.29 J (incorrect)
---
How am I getting 3 different values for Q (energy), if they were the same I would just divide by length, and I would have 2 methods of calculating energy per unit length.

Is something wrong in what I am doing?

EDIT: All these Q values should technically be W or J/s, instead of J, right? So not only do I divide by length, I also have to multiply it by the time (30 min = 30*60s)?

13. Apr 26, 2013

### Staff: Mentor

For the original problem, I get

$$\rho C_p (\frac{\pi D^2}{4})L\frac{dT}{dt}=-h\pi DL (T-T_∞)$$

or, equivalently,

$$\frac{dT}{dt}=-\frac {4h}{\rho C_pD} (T-T_∞)$$

Notice, the L's have canceled.

The solution to the above equation is:

$$(T-T_∞)=(T_0-T_∞)e^{-\frac{4ht}{\rho C_pD}}$$

At 1800 seconds (30 min), this equation predicts T = 520 K

Method 1 for getting the total heat removed per unit length:

$$\frac{dQ}{dt}=πDh(T-T_∞)=h(T_0-T_∞)e^{-\frac{4ht}{\rho C_pD}}$$
Integrating from t = 0 to t = t, I get,

$$Q=\rho C_p (\frac{\pi D^2}{4})(T_0-T_∞)(1-e^{-\frac{4ht}{\rho C_pD}})$$

Method 2 for getting the heat removed per unit length:

$$Q=\rho C_p (\frac{\pi D^2}{4})(T_0-T)$$

Results from the two methods:

From both these equations, for t = 1800 sec (30 min), I get 28.6 million Joules heat removed per unit length

14. Apr 26, 2013

### watsup1993

Yes, you are correct. My equation is basically the same, i just havent simplified it as you. I also realised I forgot to enter a value into my equations, now I am getting the exact same result as you.
Once again, Thank you

Can you just have a quick look at the other thread regarding the heat exchangers.

15. Apr 27, 2013

### watsup1993

Actually for units, will it not be W/m with the equations you have given. Or is it J/m ?

16. Apr 27, 2013

### Staff: Mentor

The problem statement is asking for the cumulative amount of heat over 30 minutes, so it's J/m.