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Convective heat transfer coefficient

  1. Dec 28, 2014 #1
    Not sure if I'm getting this right, my answer seems very small, can anyone advise?

    Question- butanol at a temperature of 28oC is pumped at a velocity of 14ms-1 through a 100mm diameter tube kept at a wall temp of 90oC. The properties of butanol are below. Determine the convective heat transfer coefficient.
    Data:
    Specific heat capacity - 2.142kJkg-1K-1
    ρ 950kgm-1
    Dynamic viscosity at 28oC - 2.9x10-3kgm-1s-1
    Dynamic viscosity at 90oC- 1.2x10-3kgm-1s-1
    Thermal conductivity - 2.4x10-4kWm-1K-1

    Attempt:
    Re=pUd/μ
    Pr=Cpμ/K
    Nu=0.332Re0.5Pr0.33
    hx=Nu(k/x)

    Re=950*14*0.1x10-3/2.9x10-3 458.62x10-4
    Pr=2.142x2.9x-3/2.41x10-4 2.588
    Nu=0.332x(458.62x10-4)0.5x(2.588)0.33 0.0973
    Hx=0.0973*2.41x10-4/0.1 2.3352x10-4kWm-2K-1
    Answer: 0.2335 Wm-2K-1
     
  2. jcsd
  3. Dec 28, 2014 #2

    SteamKing

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    Your calculations are confusing. Use '=' to separate the calculation from the final result.

    As to your calculation of the Reynolds number, apparently you are saying that d = 0.1*10-3m, which is equal to 0.1 mm. A tube diameter of 100 mm = 0.1 m.
    The calculation you have shown contains arithmetic mistakes (a Re < 1 is unusual)
     
  4. Dec 30, 2014 #3
    Thank you SteamKing. I missed the diameter mistake. I have corrected this but still get a low result.

    Re=950*14*0.1/2.9x10-3 =0.4586
    Pr=2.142x2.9x10-32.41x10-4 =2.588x10-7
    Nu=0.332*(0.4586)0.5*(2.588x10-7)0.33 =0.00151
    Hx=0.00151*(2.41x10-4/0.1) =3.6391x10-6kWm- 2K-1
     
  5. Dec 30, 2014 #4

    SteamKing

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    You still have a basic arithmetic problem here. You're taking a relatively sizable number and dividing by a small number and getting a small number. Are you sure you know how scientific notation works?

    BTW, this equation should be written: Re = 950*14*0.1 / (2.9x10-3) = ?

    This equation should read: Pr = 2.142 * (2.9x10-3) / (2.41x10-4) = ?

    Try not bunching your terms together. Use additional spaces and parentheses to make calculations clear.

    You also missed out the division by K in your version, making your calculation of the Prandtl No. incorrect.

    Use the 'Preview' Button to check how your work will appear before posting. I know I use it a lot.

    You are using an incorrect Reynolds number and an incorrect Prandtl number to calculate the Nusselt number.

    We'll worry about the calculation of Hx after you have fixed the previous calculations.
     
  6. Dec 30, 2014 #5
    Ok, reviewed scientific notation, and am now getting the below. Am I on track?

    Re = 950*14*0.1 / (2.9x10-3) = 458621
    Pr = 2.142 * (2.9x10-3) / (2.4x10-4) = 25.8825

    Re now indicates turbulent flow so dittus-bottler correlation for nusselt number to be used:
    Nu= 0.023 *(458621)0.8 * (25.8825)0.4 = 2858.206
     
  7. Dec 30, 2014 #6

    SteamKing

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    These calculations look good.
     
  8. Dec 30, 2014 #7
    Great, so I'm now calculating hx using hx=Nu(k/x).

    Hx = 2858.206 * ((2.4x10-4) / 0.1 ) = 6.8597kWm-2K-1

    = 6,859.7Wm-2K-1

    Does this look along the right lines? Thank you for all your help.
     
  9. Dec 30, 2014 #8

    SteamKing

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    This calculation looks OK.
     
  10. Dec 30, 2014 #9
    Brilliant, thankyou
     
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