Question 12.1: Work Needed for Isothermal and Adiabatic Compression Conside......
Work Needed for Isothermal and Adiabatic Compression
Consider a piston-cylinder system containing 0.5 m³ of air at 100 kPa and 20°C, which is to be compressed to 0.1 m³ volume. Determine the work needed when the compression is isothermal and when it is isentropic (reversible adiabatic).
Given: \gamma (air) = 1.4, p_{1} = 100 kPa, V_{1} = 0.5 m^{3} and V_{2} = 0.1 m^{3}
Figure: See Figure 12.1.
Assumptions: Compression process is in quasiequilibrium, ideal gas assumption for air.
Find: W_{isot} and W_{isen}
From Table 12.1, for isothermal compression,
W_{isot} = p_{1} V_{1} \ln \left\lgroup \frac{V_{2}}{V_{1}} \right\rgroup= 100 kPa \times 0.5 m^{3} \times \ln \left\lgroup \frac{0.1}{0.5} \right\rgroup \frac{1 kJ}{1 kPa . m^{3}} = -80.47 kJ
For isentropic compression,
W_{isen} = \frac{p_{1} V_{1}^{\gamma} (V_{2}^{1 – \gamma} – V_{1}^{1 – \gamma})}{(1 – \gamma)}= \frac{100 kPa \times (0.5 m^{2})^{1.4} \times (0.1^{1-1.4} – 0.5^{1-1.4})}{(1 – 1.4)} \frac{1 kJ}{1 kPa . m^{3}}
= -113.0 kJ
Comments
The work is negative implying that work has to be done on the system. The work needed for the isentropic compression is greater than that needed for isothermal compression (in this example, by about 40%). This increase in work is signicant, and impacts many HVAC components such as refrigerant compressors or air compressors. Hence, when the compression ratio is large, it is common design practice to use two stages of compression (i.e., a low-pressure compressor and a high-pressure compressor) with intercooling between the two stages.
| TABLE 12.1 | |||
| Work Output for Some Important Non ow Processes for Ideal Gases | |||
| Terminology | Type of Reversible Process |
Exponent n | Work Output |
| 1 Polytropic | General | – | W = \int pdV = \frac{(p_{2} V_{2} – p_{1}V_{1})}{1 – n} |
| 2 Isometric | Constant volume | – | W = 0 |
| 3 Isobaric | Constant pressure | n = 0 | W = p(V_{2} – V_{1}) |
| 4 Isothermal | Constant temperature | n = 1 | W = p_{1} V_{1} \ln \left\lgroup \frac{V_{2}}{V_{1}} \right\rgroup |
| 5 Isenthalpic | Constant enthalpy | – | W = 0 |
| 6 Isentropic (reversible adiabatic) | Constant entropy | n = γ | W = \frac{p_{1} V_{1}^{\gamma} (V_{2}^{1 – \gamma} – V_{1}^{1 – \gamma})}{(1 – \gamma)} |
The final and initial states are represented by subscripts 2 and 1, respectively.