Heat transfer and combustion correlations

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The discussion revolves around the simplification of a heat transfer correlation for natural convection from a horizontal pipe to the atmosphere. Participants explore the relationship between Nusselt number (Nu), Grashof number (Gr), and Prandtl number (Pr), and how to derive the heat transfer coefficient (h). There is confusion regarding the algebraic manipulation of the terms, particularly how to handle the powers of d and the constants involved. The importance of unit consistency and the correct application of algebraic principles is emphasized throughout the conversation. Ultimately, the participants express a collaborative effort to clarify the calculations and arrive at the correct solution.
  • #31
Chestermiller said:
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?
 
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  • #32
##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.
 
  • #33
Rogue said:
##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?
 
  • #34
Chestermiller said:
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?
 
  • #35
##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?
 
  • #36
Rogue said:
##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.
 
  • #37
Chestermiller said:
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. :)
 
  • #38
Rogue said:
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.
 
  • #39
Chestermiller said:
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.
 

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