# Homework Help: Heat transfer and combustion correlations

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1. 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

2. 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?

3. Dec 31, 2017

### Rogue

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

4. 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.

5. Dec 31, 2017

### Rogue

My algebraic expression:

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

6. Dec 31, 2017

### Rogue

Is the above ok?

7. 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.

8. Dec 31, 2017

### Staff: Mentor

So, what is the problem?

9. 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?

10. 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?

11. Dec 31, 2017

### Staff: Mentor

Yes. This is basic high school algebra.

12. 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. :)

13. 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.

14. Jan 1, 2018

### Rogue

No worries.

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