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o-na [289]
3 years ago
8

2.4 kg of nitrogen at an initial state of 285K and 150 kPa is compressed slowly in an isothermal process to a final pressure of

600 kPa. Determine the work.
Engineering
1 answer:
Neko [114]3 years ago
8 0

Answer:

W=-280.67 KJ

Explanation:

Given that

Initial pressure = 150 KPa

Final pressure = 600 KPa

Temperature T= 285 K

Mass m=2.4 Kg

We know that ,work in isothermal process given as

W=mRT\ ln\dfrac{P_1}{P_2}

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

W=mRT\ ln\dfrac{P_1}{P_2}

W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}

W=-280.67 KJ

Negative sign indicates that it is compression process and work is done on the gas.

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Xelga [282]

Answer:

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(b) Total current I is 3.257 A

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Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

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(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

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I=\int {\textbf{J} \, \textbf{ds}

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I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

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