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Homework help:Heat transfer coefficient of flow over a flat plate

  1. Jun 16, 2012 #1
    A 2-m x 3-m flat plate is suspended in a room, and is subjected to air flow parallel to its surfaces along its 3-m-long side. The free stream temperature and velocity of the air are 20oC and 7m/s. the total drag force acting on the plate is measured to be 0.86N. Determine the average heat transfer coefficient for the plate. Answer:h=12.7W/m2K



    Properties of air at 300K(from my textbook
    ρ=1.1763
    c=1.007e03
    μ=1.862e-05
    v=1.5e-05
    Pr=0.717
    k=2.614e-02




    Attempt at a solution
    ReL=UL/v=7*3/1.5e-05=1.5e06 >5e05
    xcr=(Recrv)/U=1.07

    Nux,Laminar=0.332Pr1/3Rex1/2
    Nux,Turbulent=(0.0296Re0.8Pr)/(1+2.11Rex-0.1(Pr-1))

    Avg heat coeff
    =(1/L){0xcrhx, Laminardx + xcr3hx, Turbulentdx}
    = (1/L){0xcrNux, Laminar(k/x)dx + xcr3Nux, Turbulent(k/x)dx}
    = ...
    =14.39


    I did approximate 1+2.11Rex-0.1(Pr-1) to be equal to 1 in my working because I didn't know how to integrate it. Is that acceptable? And I don't know how to use the drag force provided in this question. :x

    I've been pondering over this question for a few days and I still can't get the answer please help check if my approach is correct. Thank you.
     
  2. jcsd
  3. Jun 18, 2012 #2
    Hint: The problem states the force on the plate. That implies they want you to approach it from a Reynold's Analogy standpoint.
     
  4. Jun 18, 2012 #3
    Thanks, I'll give it a shot again.
     
  5. Jun 19, 2012 #4
    Okay thanks for the hint once again. I tried again using Reynold's Analogy.

    qx/(ρcu(T-Tw))=τwx/(ρu2)


    hx/c=τwx/(u)

    τwx = (hxu)/(c)

    AτwxdA = 0.86

    A(hxu)/(c)dA = 0.86

    (u/c)∫AhxdA = 0.86

    (u/c)∫AhxdA = 0.86

    AhxdA = (0.86c)/u

    avg h
    = (1/A)∫AhxdA
    = (1/A) (0.86c)/u

    Is this approach correct now? I subbed in the values of A=12, c = 1.007e03 and u=7 but the answer I got is 10.3 instead of the given 12.7.
    I think the difference might be attributed to the value of c.

    Thanks in advance
     
  6. Jun 19, 2012 #5
    I started off with the basic relationship:

    St*Pr^2/3 = Cf/2

    I don't see the Prndtl number anywhere in your computations.
     
  7. Jun 19, 2012 #6
    Oops, I misread my notes, the formula I used earlier was only for Pr=1 >.<

    I did as you told me and I got the answer.(at least reasonably close to it) ^^ Thanks a lot!(Sorry if I annoyed you with a lot of stupid mistakes and questions)
    Here is my working if anyone is interested.

    Stx = Nux/(RexPr) = hx/cpU

    StxPr2/3= Cfx/2=тwx/(pU2)

    (hx/cpU)Pr2/3= тwx(pU2)

    hx = (c/U)(Pr-2/3wx

    AhxdA = (c/U)(Pr-2/3AтwxdA

    Avg heat coeff
    =(1/A)∫AhxdA
    =(1/A)(c/U)(Pr-2/3AтwxdA

    sub A=12, c=1.007e03, U=7, Pr=0.717, ∫AтwxdA = 0.86

    Avg heat coeff=12.86
     
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