Heat transfer and combustion correlations

[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

 

[RUber]

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

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

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
$a \left( \frac{(Ts-Tf)}{d}\right)^{.25}$

 

[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.

 

[Mitch1]

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
$a \left( \frac{(Ts-Tf)}{d}\right)^{.25}$

Hi ruber
Thanks for your reply!
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

 

[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.

 

[Mitch1]

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.

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

 

[blitzman]

did you try:

(d^4)/d

I`m working on same question.

 

[Mitch1]

did you try:

(d^4)/d

I`m working on same question.

Hi, would that not result in d^3

 

[Big Jock]

Anyone get any further with this question? Its really becoming a bug bear of mine I can't seem to get it any further along...

 

[Chestermiller]

Anyone get any further with this question? Its really becoming a bug bear of mine I can't seem to get it any further along...

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

Chet

 

[Big Jock]

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

 

[Mitch1]

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

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

 

[Chestermiller]

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

It's just basic algebra.

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

 

[Mitch1]

It's just basic algebra.

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

Hi Chet
Yes I realized once I posted it
Thanks anyway

 

[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?

 

[RUber]

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?

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.

 

[rjc45y]

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

 

[sootybeau]

Plug in the information:
$Nu = 0.53(\frac{αρ²d³(Ts-Tf)g}{ μ²})^{0.25}(\frac{Cpμ }{ k})^{0.25} = hd / k$
$h = \frac{k}{d} 0.53(\frac{αρ²d³(Ts-Tf)g}{ μ²})^{0.25}(\frac{Cpμ }{ k})^{0.25}$

I am having trouble getting my head round this problem. I think I am trying to do to much to quickly and confusing the numbers when I multiply out the equation.
Am I on the right line to multiply by μ² 1st? before I distribute through across multiplication?
For (\frac{Cpμ }{ k})^{0.25} I have got to = 0.915.

 

[Mitch1]

Yes you are correct on both counts

 

[sootybeau]

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

 

[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

 

[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

 

[sootybeau]

NU = 47.55(d³(T_s-T_f))

Does that look close?

 

[Mitch1]

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

 

[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?

 

[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

 

[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) x0.915Please can someone come to my rescue?

 

[Rogue]

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

 

[Chestermiller]

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) x0.915Please can someone come to my rescue?

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.

 

[Rogue]

My algebraic expression:

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

 

[Rogue]

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.

Is the above ok?

 

[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.

 

[Chestermiller]

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

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

So, what is the problem?

 

[Rogue]

So, what is the problem?

I realize 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?

 

[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?

 

[Chestermiller]

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

Would this be a step in the right direction?

Yes. This is basic high school algebra.

 

[Rogue]

Yes. This is basic high school algebra.

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

 

[Chestermiller]

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

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.

 

[Rogue]

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.

No worries.