# Homework Help: Heat transfer and combustion correlations

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1. Oct 12, 2016

### Mitch1

I am not sure what you are trying to do? Your 0.915 is essentially Pr but you can just plug the values into the equation you quoted in post #18

2. Oct 12, 2016

### Mitch1

Try working out the nusselt number with the values given and 0.915 you worked out.
Once you have nu you can then put this into the heat transfer coefficient equation

3. Oct 12, 2016

### sootybeau

NU = 47.55(d3(Ts-Tf))

Does that look close?

4. Oct 12, 2016

### Mitch1

Yes that looks correct but there needs to be a power of 0.25 at the end of your equation

5. Oct 12, 2016

### sootybeau

I missed that by mistake sorry.

Your help has been great and really got me past my mental block. I am 10-3 out but should I have used k in W not kW?

6. Oct 12, 2016

### Mitch1

Your thermal conductivity is given in kW m-1 k-1 this will give your answer x10-3 kW m-2 k-2 which when you covert to W you get your answer

7. Dec 31, 2017

### Rogue

Apologies for the old thread resurrection but I can't get to grips with this.

I've tried it my long complicated way and the way suggested above and am struggling solving the equation.

Once I've punched in the numbers I end up with:

h=((1.49248x10^-5)/d)x((0.43433xd^0.75x(Ts-Tf))/0.0044) x0.915

Please can someone come to my rescue?

8. Dec 31, 2017

### Rogue

It's like the 0.0044 is in my way which was formerly (u^2)^0.25

9. Dec 31, 2017

### Staff: Mentor

Any chance you can write this out using LaTex. https://www.physicsforums.com/help/latexhelp/

And please express your results first in terms of algebraic parameters, and then show how you substituted. I would like to see your whole derivation.

10. Dec 31, 2017

### Rogue

My algebraic expression:

$h= \frac {k}{d} 0.53 ( \frac { \alpha p^2 d^3 (T_s - T_f) g} {u^2})^{0.25} (\frac {Cpu}{k})^{0.25}$

11. Dec 31, 2017

### Rogue

Is the above ok?

12. Dec 31, 2017

### Rogue

$h= \frac {2.816 \times 10^-5}{d} 0.53 ( \frac { (3.077 \times 10^-3) 1.086^2 d^3 (T_s - T_f) 9.81} {(1.962 \times 10^-5)^2})^{0.25} (0.915)$

The 0.915 at the end is my calculated Pr number to the power of 0.25.

13. Dec 31, 2017

### Staff: Mentor

So, what is the problem?

14. Dec 31, 2017

### Rogue

I realise it's probably going to be something daft, and I've done the hard part (hopefully), but I just can't seem to progress it from here?

15. Dec 31, 2017

### Rogue

$0.0044h= \frac {(2.816\times 10^-5)}{d} 0.53 (0.2355 \times 1.0421 d^{0.75}(T_s - T_f)^{0.25} 1.7698 ) (0.915)$

Would this be a step in the right direction?

16. Dec 31, 2017

### Staff: Mentor

Yes. This is basic high school algebra.

17. Dec 31, 2017

### Rogue

Bit harsh there Chet, but thanks.
I had just solved this, guess I need to try keep thinking straightforward rather than over complicating it. :)

18. Dec 31, 2017

### Staff: Mentor

Sorry. I didn't mean to be harsh. I was just helping my grandson study for the SATs yesterday, and this kind of exponential algebra was one of the things the study guide touched upon.

19. Jan 1, 2018

No worries.