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krek1111 [17]
3 years ago
5

Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ___

_________ of the pipe.
Engineering
1 answer:
Nesterboy [21]3 years ago
8 0

Answer:

Inlet

Explanation:

Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.

This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.

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Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
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Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

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According to the first thermodynamic law, the energy must be conserved so:

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This equation can be solved by integration between an initial and a final state:

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\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

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