Heat transfer calculation across a flat plate

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SUMMARY

The discussion focuses on calculating heat transfer across a flat plate with engine oil flowing at 60 °C over a plate at 20 °C. The film temperature is determined to be 40 °C using the formula T_f = (T_∞ + T_S) / 2. The average Reynolds number is calculated as 41717.62, indicating laminar flow, and the average Nusselt number is found to be 1927.326. Consequently, the heat transfer coefficient is 55.5 W/m²·K, leading to a rate of heat transfer per unit width of the plate of 11100 W/m.

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tcaldwell
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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

Help!
Can't get a Reynolds number without density..
Is film temperature 40 degrees Celsius?
 
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First, the film temperature is
T_f = \frac{T_{\infty} + T_S}{2} = 40 °C
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
\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
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)
\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
Now we can calculate the average heat transfer coefficient using the Nusselt number
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}
Finally, we calculate the rate of heat transfer per unit width of plate
q&#039; = 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}
 

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