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!

Homework Help: Archived Heat transfer calculation across a flat plate

  1. Mar 30, 2013 #1
    Heat Transfer Qts
    Engine oil at 60 oC flows over flat plate whose temperature is at 20 oC with a velocity of 2 m s-1. The length of the plate is 5m. What is the film temperature and the rate of heat transfer per unit width of the entire plate?
    Additional information (at the film temperature):
    Thermal conductivity of oil: Prandtl number:
    Dynamic viscosity:
    k1 = 0.144 W m-1 K-1
    Pr = 2870
    u = 0.21 kg m-1 s-1

    Can't get a Reynolds number without density..
    Is film temperature 40 degrees Celsius????
  2. jcsd
  3. Mar 24, 2016 #2
    First, the film temperature is
    [tex]T_f = \frac{T_{\infty} + T_S}{2} = 40 °C[/tex]
    Now, I'm almost certain this is a problem from Incropera's heat transfer book, so I got the density of the oil at Tf from table A.5 (Incropera). The values of k, Pr and μ also coincide with the values in the table for Tf. Now, calculating the average Reynolds number for a flat plate
    [tex]\textrm{Re} = \frac{\rho v L}{\mu} = \frac{\left(876.07 \frac{kg}{m^3} \right) \left(2 \frac{m}{s} \right)(5 \ m)}{0.21 \frac{kg}{m \cdot s}} = 41717.62[/tex]
    For a flat plate, Re < 500000 indicates laminar flow. Now we use the following correlation for the average Nusselt number for laminar flow over a flat plate (applies for all Pr ≥ 0.6)
    [tex]\textrm{Nu} = 0.664 \ \textrm{Re}^{\frac{1}{2}} \ \textrm{Pr}^{\frac{1}{3}} = 0.664 (41717.62)^{\frac{1}{2}} (2870)^{\frac{1}{3}} = 1927.326[/tex]
    Now we can calculate the average heat transfer coefficient using the Nusselt number
    [tex]h = \frac{\textrm{Nu} \ k}{L} = 1927.326 \frac{0.144 \frac{W}{m \cdot K}}{5 \ m} = 55.5 \frac{W}{m^2 \cdot K}[/tex]
    Finally, we calculate the rate of heat transfer per unit width of plate
    [tex]q' = hL(T_{\infty} - T_S) = \left(55.5 \frac{W}{m^2 \cdot K} \right) (5 \ m) (60 °C - 20 °C) = 11100 \frac{W}{m}[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted