# Heat transfer and combustion

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1. Oct 15, 2014

### Mitch1

Hi am a bit stuck on understanding how these two correlations can be similar I am wondering if anyone can ahead any light
Can anyone solve

Correlation for heat transfer by natural convection from a horizontal pipe to atmosphere is;

Nu = 0.53Gr^0.25Pr^0.25
Nu = hd / k

Where, Gr = αρ²d³(Ts-Tf)g / μ²
And Pr = Cpμ / k

Show that the above correlation can be simplified to;
h ≈ 1.34 (Ts-Tf / d)^0.25

α=3.077 x 10^-3
ρ=1.086
Cp=1.0063
k=2.816 x 10^-5
μ=1.962 x 10^-5

2. Oct 15, 2014

### RUber

Plug in the information:
$Nu = 0.53(\frac{αρ²d³(Ts-Tf)g}{ μ²})^{0.25}(\frac{Cpμ }{ k})^{0.25} = hd / k$
$h = \frac{k}{d} 0.53(\frac{αρ²d³(Ts-Tf)g}{ μ²})^{0.25}(\frac{Cpμ }{ k})^{0.25}$
Distribute through the exponents across multiplication, and compute the constants, and you should get an answer similar to
$a \left( \frac{(Ts-Tf)}{d}\right)^{.25}$

3. Oct 15, 2014

### RUber

Note: When I carried out the calculation, I did not get 1.34 as a coefficient. You may want to verify all you units to make sure they are balanced.

4. Oct 15, 2014

### Mitch1

Hi ruber
All units are correct as stated I am also unsure how to get 1.34 I have tried various ways.
When I follow your advise how would d^3 disappear in a sense as this is a unknown
Thanks again

5. Oct 15, 2014

### RUber

The $d^3$ doesn't disappear, you have $\frac{(d^3)^{.25}}{d}=\frac{1}{d^{.25}}$.
What is g? That constant was not given in your original post.

6. Oct 15, 2014

### Mitch1

I unstandard that now thanks a lot!
And g is 9.81 it is just gravity

7. Oct 25, 2014

### blitzman

did you try:

(d^4)/d

I`m working on same question.

8. Oct 26, 2014

### Mitch1

Hi, would that not result in d^3

9. Nov 6, 2014

### Big Jock

Anyone get any further with this question? Its really becoming a bug bear of mine I cant seem to get it any further along....

10. Nov 6, 2014

### Staff: Mentor

RUber gave the solution essentially completely in post #2. I don't see what the problem is.

Chet

11. Nov 7, 2014

### Big Jock

your 100% correct Chet I have got the correct answer my post was in haste

12. Nov 11, 2014

### Mitch1

Hi again I have completed this question however I am unsure how we get from
(D^3)^.25 /d. = (1/d^.25)

13. Nov 11, 2014

### Staff: Mentor

It's just basic algebra.

$$\frac{(d^3)^{0.25}}{d}=\frac{d^{0.75}}{d}=\frac{1}{d^{0.25}}$$
Chet

14. Nov 11, 2014

### Mitch1

Hi Chet
Yes I realised once I posted it
Thanks anyway

15. May 20, 2015

### rjc45y

Am I allowed to ask what "Distribute through the exponents across multiplication, and compute the constants" means for mere mortals struggling along. Or is that too much of an embarrassing admission?

16. May 21, 2015

### RUber

What I meant by "distribute through across multiplication" was:
$(ab^2c^a)^x = a^x b^{2x} c^{ax}$
Then, combine all your like terms.

17. May 23, 2015

### rjc45y

Many thanks, got the correct answer but at the mo' it's x10^-3 out, will preserver

18. Oct 11, 2016

### sootybeau

19. Oct 12, 2016

### Mitch1

Yes you are correct on both counts

20. Oct 12, 2016

### sootybeau

Do I need to multiply 0.915 that I found by μ2? and also 0.53 by μ2