Problem: Air at 100 kPa and 20 °C is compressed in a continuous process in an adiabatic compressor to a pressure of 400 kPa. Determine the exit temperature and the work required per kilogram of air. Find: Exit temperature [latex]T_2[/latex] and work required per kilogram of air [latex]w_{12}[/latex] for compression. Known: Initial pressure [latex]P_1[/latex] = 100 kPa, initial temperature [latex]T_1[/latex] = 20 °C = 293.15 K, final pressure [latex]P_2[/latex] = 400 kPa, adiabatic process so [latex]Q_{12}[/latex] = 0. Assumptions: Air is an ideal gas with constant specific heat, the process is reversible and adiabatic so ∆s = 0, changes in kinetic and potential energy are negligible in the compressor. Governing equations: Isentropic process (ideal gas) [latex]\frac{T_2}{T_1}= \left\lgroup\frac{P_2}{P_1}\right\rgroup ^{(γ-1)/γ}[/latex] First law (control volume) [latex]w_{12}=h_2-h_1=c_p(T_2-T_1)[/latex] Properties: Air at 293.15 K (interpolation) has specific heat [latex]c_p [/latex] = 1.004 kJ / kgK (Appendix 4), specific heat ratio of air at 293.15 K (interpolation) γ = 1.400 (Appendix 4).

The exit temperature is [latex]T_2=T_1\left\lgroup\frac{P_2}{P_1}\right\rgroup ^{(γ-1)/γ} , \\ T_2 = 293.15 \ K\left\lgroup \frac{400 \ kPa}{100 \ kPa}\right\rgroup ^{(1.400-1)/1.400}=435.66 \ K [/latex] The work done per kilogram of gas is [latex]w_{12}=c_p(T_2-T_1)=1.004 \ kJ/kgK \times (435.66 \ K -293.15 \ K)=143.19 \ kJ/kg[/latex] Answer: The exit temperature is 435.7 K and the work done per unit mass of air is 143.2 kJ / kg.

Question 6.6: Problem: Air at 100 kPa and 20 °C is compressed in a continu...

Energy, Entropy and Engines An Introduction to Thermodynamics

Problem: Air at 100 kPa and 20 °C is compressed in a continuous process in an adiabatic compressor to a pressure of 400 kPa. Determine the exit temperature and the work required per kilogram of air.

Find: Exit temperature $T_2$ and work required per kilogram of air $w_{12}$ for compression.

Known: Initial pressure $P_1$ = 100 kPa, initial temperature $T_1$ = 20 °C = 293.15 K, final pressure $P_2$ = 400 kPa, adiabatic process so $Q_{12}$ = 0.

Assumptions: Air is an ideal gas with constant specific heat, the process is reversible and adiabatic so ∆s = 0, changes in kinetic and potential energy are negligible in the compressor.

Governing equations:
Isentropic process (ideal gas) $\frac{T_2}{T_1}= \left\lgroup\frac{P_2}{P_1}\right\rgroup ^{(γ-1)/γ}$

First law (control volume) $w_{12}=h_2-h_1=c_p(T_2-T_1)$

Properties: Air at 293.15 K (interpolation) has specific heat $c_p$ = 1.004 kJ / kgK (Appendix 4), specific heat ratio of air at 293.15 K (interpolation) γ = 1.400 (Appendix 4).

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