1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat transfer and combustion correlations

  1. Oct 15, 2014 #1
    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. jcsd
  3. Oct 15, 2014 #2

    RUber

    User Avatar
    Homework Helper

    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}##
     
  4. Oct 15, 2014 #3

    RUber

    User Avatar
    Homework Helper

    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.
     
  5. Oct 15, 2014 #4
    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
     
  6. Oct 15, 2014 #5

    RUber

    User Avatar
    Homework Helper

    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.
     
  7. Oct 15, 2014 #6

    I unstandard that now thanks a lot!
    And g is 9.81 it is just gravity
     
  8. Oct 25, 2014 #7
    did you try:

    (d^4)/d

    I`m working on same question.
     
  9. Oct 26, 2014 #8
    Hi, would that not result in d^3
     
  10. Nov 6, 2014 #9
    Anyone get any further with this question? Its really becoming a bug bear of mine I cant seem to get it any further along....
     
  11. Nov 6, 2014 #10
    RUber gave the solution essentially completely in post #2. I don't see what the problem is.

    Chet
     
  12. Nov 7, 2014 #11
    your 100% correct Chet I have got the correct answer my post was in haste
     
  13. Nov 11, 2014 #12
    Hi again I have completed this question however I am unsure how we get from
    (D^3)^.25 /d. = (1/d^.25)
     
  14. Nov 11, 2014 #13
    It's just basic algebra.

    [tex]\frac{(d^3)^{0.25}}{d}=\frac{d^{0.75}}{d}=\frac{1}{d^{0.25}}[/tex]
    Chet
     
  15. Nov 11, 2014 #14
    Hi Chet
    Yes I realised once I posted it
    Thanks anyway
     
  16. May 20, 2015 #15
    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?
     
  17. May 21, 2015 #16

    RUber

    User Avatar
    Homework Helper

    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.
     
  18. May 23, 2015 #17
    Many thanks, got the correct answer but at the mo' it's x10^-3 out, will preserver
     
  19. Oct 11, 2016 #18
     
  20. Oct 12, 2016 #19
    Yes you are correct on both counts
     
  21. Oct 12, 2016 #20
    Do I need to multiply 0.915 that I found by μ2? and also 0.53 by μ2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Heat transfer and combustion correlations
  1. Heat transfer (Replies: 5)

Loading...