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Sagot :
By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.
Given the following data:
Surface temperature = 15°C
Bulk temperature = 75°C
Side length of plate = 150 mm to m = 0.15 meter.
How to calculate the average heat transfer coefficient.
Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:
[tex]T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45[/tex]
Film temperature = 45°C to K = 273 + 45 = 318 K.
For the coefficient of thermal expansion, we have:
[tex]\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}[/tex]
From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:
- Kinematic viscosity, v = [tex]1.750 \times 10^{-5}[/tex] m²/s.
- Thermal conductivity, k = 0.02699 W/mk.
- Thermal diffusivity, α = [tex]2.416 \times 10^{-5}[/tex] m²/s.
- Prandtl number, Pr = 0.7241.
Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:
[tex]R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10} }\\\\R_{aL}=1.48 \times 10^{7}[/tex]
Also take note, g(Pr) is given by this equation:
[tex]g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }[/tex]
g(Pr) = 0.430
For GrL, we have:
[tex]G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7[/tex]
Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:
[tex]N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times [\frac{4}{3} \times (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k[/tex]
Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:
[tex]N_{uL}=\frac{h_{L}L}{k} =0.68 + \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19} } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 + \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19} })\\\\h_{L}=4.87\;W/m^2k[/tex]
Read more on heat transfer here: https://brainly.com/question/10119413
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