A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 °C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed in an isobaric process to the initial state. Determine the boundary work for each process and the net work for the cycle?

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oeerivona oeerivona · Answer 1 · 2020-03-05T06:50:15+01:00

Answer:

Isothermal expansion W₁ =-37198.9 J

Polytropic Compression W₂ =-34872.82 J

Isobaric Compression W₃ = -6974.566 J

The net work for the cycle = -79046.29 J

Explanation:

Mass of air = 0.15 kg = 150 g

Molar mass = 28.9647 g/mol

Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air

PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³

For isothermal expansion we have

P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³

Therefore work done

W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J

Stage 2

Compression polytropically we have

[tex]\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n[/tex] where P₃ = 2 MPa

Therefore V₃ = [tex](\frac{1}{4} )^{\frac{1}{1.2} }*V_2[/tex] = 1.6904×10⁻² m³

Work = W₂ = [tex]\frac{P_2V_2-P_3V_3}{n-1}[/tex] = -34872.82 J

[tex]\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}[/tex] or T₃ = [tex]T_2*(\frac{P_3}{P_2})^\frac{n-1}{n}[/tex] = 785.12 K

Isobaric compression we have thus

Work done W₃ = P(V₁ -V₃) = -6974.566 J

Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J