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andrezito [222]
3 years ago
15

For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari

ation of the local heat transfer coefficient with position along the plate is
Engineering
1 answer:
Aliun [14]3 years ago
3 0

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = C(Re_x)^{0.5 (Pr)^{1/3

Nu = C(\frac{Vx}{v})^{0.5} (Pr)^{1/3

\frac{h_xx}{k} = C(\frac{V}{v})^{0.5}  (Pr)^{1/3  (x)^{0.5

h_x = \frac{1}{x^{1/2}}

Next we write down the expression for the local heat flux from the plate with  uniform surface temperature;

q = h_xA( T_s - T∞ )

q ∝ h_x

∴

q ∝  \frac{1}{x^{1/2}}

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

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A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co
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Answer:

a) T_{H} = 1.967\,^{\circ}F, b) COP_{R} = 9.105, c) T_{H} = 115.934\,^{\circ}F, d) COP_{R} = 6.995, e) T_{H} = 25.129\,^{\circ}C

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}

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b) The coefficient of performance is:

COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}

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c) The temperature of the hot reservoir can be determined with the help of the following relation:

\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}

\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}

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COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}

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e) The temperature of the cold reservoir is:

8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}

8.9\cdot T_{H} - 2386.535 = 268.15

T_{H} = 298.279\,K

T_{H} = 25.129\,^{\circ}C

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