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RideAnS [48]
4 years ago
7

Gas contained in a piston-cylinder assembly undergoes a polytropic process for which the relation between pressure and volume is

given by p∀ n = constant. The initial volume is 0.04 m3 and the initial pressure is 2 bar. The final volume of the gas is 0.1 m3 , For each of the following cases, determine the final pressure [bar] and the work for the process [kJ]. Show integration for each case. a. n = 0, b. n = 1, c. n = 1.4.
Engineering
1 answer:
Charra [1.4K]4 years ago
6 0

Answer:

a) P = 2 bar, W =  12 kJ

B)P = 0.8 bar, W =  7.33 kJ

C) P = 0.55 bar, W =  6.14 kJ

Explanation:

pV^n = constant.

The initial volume V_1 = 0.1 m3,

The final volume V_2 = 0.04 m3,

The initial pressure P_1 = 2 bar.

We know that

P_1 V_1^n = P_2 V_2^n,

P_2 = P_1\frac{(V_1}{V_2)^n}

      = 2\frac{(0.04}{0.1)^n}

`a) n = 0, P_2 = 2\frac{(0.04}{0.1)^0}  = 2 bar

W = P_2(V_2 - V_1) = 2*100 kPa * (0.06 m3) = 12 kJ

b) n = 1, P_2 = 2\frac{(0.04}{0.1)^1}  = 0.8 bar

W = P_2 V_2 ln\frac{(V2}{V1} = 7.33 kJ

c)  n = 1.4, P_2 = 2\frac{(0.04}{0.1)^1.3}  = 0.5542 bar

W = \frac{(P_2 V_2 - P_1 V_1)}{(1 - n)} = 6.14 kJ

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Answer:

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The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

Q_{in} = U_{2} - U_{1}

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State 1 - (Liquid-Vapor Mixture)

P = 101.42\,kPa

T = 100\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 675.761\,\frac{kJ}{kg}

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P = 476.16\,kPa

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m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }

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The heat transfer require to the process is:

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A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-v
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Answer:

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Explanation:

h₁ = 3755.39

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