Find expert advice and community support for all your questions on IDNLearn.com. Discover the information you need quickly and easily with our reliable and thorough Q&A platform.
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
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
