Heat transfer and combustion correlations

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SUMMARY

The discussion centers on the correlation for heat transfer by natural convection from a horizontal pipe to the atmosphere, specifically the equation Nu = 0.53Gr^0.25Pr^0.25. Participants worked through the derivation of the heat transfer coefficient, h, and simplified the equation to h ≈ 1.34 (Ts-Tf / d)^0.25. Key constants such as α = 3.077 x 10^-3, ρ = 1.086, Cp = 1.0063, k = 2.816 x 10^-5, and μ = 1.962 x 10^-5 were utilized in calculations. The conversation highlighted the importance of unit consistency and algebraic manipulation in achieving the correct results.

PREREQUISITES
  • Understanding of Nusselt number (Nu) and its significance in heat transfer.
  • Familiarity with Grashof number (Gr) and Prandtl number (Pr) calculations.
  • Basic algebraic manipulation skills, particularly with exponents.
  • Knowledge of thermal properties such as thermal conductivity (k), specific heat (Cp), and dynamic viscosity (μ).
NEXT STEPS
  • Study the derivation of the Nusselt number in natural convection scenarios.
  • Learn about the significance of the Grashof and Prandtl numbers in fluid dynamics.
  • Explore advanced heat transfer topics, including forced convection and heat exchanger design.
  • Review algebraic techniques for simplifying complex equations in engineering contexts.
USEFUL FOR

Engineers, physicists, and students involved in thermal analysis, heat transfer applications, and fluid dynamics will benefit from this discussion. It is particularly relevant for those working with natural convection systems and thermal management in engineering designs.

  • #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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