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sashaice [31]
3 years ago
14

. (20 pts) A horizontal cylindrical pipe (k = 10 W/m·K) has an outer diameter of 15 cm and a wall thickness of 5 cm. The pipe is

situated in a stationary air, where the air and surrounding temperature is 27°C. The outer surface temperature of the pipe is 127°C, and the pipe surface has an emissivity of 0.5. Determine the inner surface temperature of the pipe. Use the following air properties for the analysis: k = 0.03 W/m∙K, ν = 20.92 × 10−6 m2 /s, α = 29.90 × 10−6 m2 /s, Pr = 0.70

Engineering
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Which process is a from of mechanical weathering
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3 years ago
A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of
lara [203]

Answer:

T_C = 27+273.15 = 300.15 K

T_H = 477+273.15 = 750.15 K

And replacing in the Carnot efficiency we got:

e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%

W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

e = 1 -frac{T_C}{T_H}

We have on this case after convert the temperatures in kelvin this:

T_C = 27+273.15 = 300.15 K

T_H = 477+273.15 = 750.15 K

And replacing in the Carnot efficiency we got:

e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%

And the maximum power output on this case would be defined as:

W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}

Where Q_H represent the heat associated to the deposit with higher temperature.

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Answer:

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Answer:

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(Lift greater than weight, thrust greater than drag) accelerate upwards and forwards.

4 0
3 years ago
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