The path of an isentropic process on p-v coordinates can be inferred from Eq (3.13):
The constant in this equation is determined by the properties of the initial state.The isentrope (Fig. 3.4) is of the form p x 1/v7, while the isotherm (Fig. 3.2) varies as p x 1/v. Since 7 > 1 for all gases, the isentrope decreases more rapidly with v than does the isotherm.
The work done by isentropically expanding an ideal gas is obtained by substituting Eq (3.15) into Eq (3.1):
Problems 3.4 - 3.6 are examples of isentropic expansion calculations involving ideal gases. 3.5.2 Water
Although Eqs (3.9) and (3.10) could be employed to approximate entropy changes for processes in superheated steam, use of the steam table A.3 is preferred. Similarly, entropy changes of single-phase (compressed) liquid water are conveniently handled using Table A.4. Many practical problems require calculation of entropy changes when water is partially vaporized. Such calculations utilize the entropies of the saturated liquid and saturated vapor provided in steam tables A.1 and A.2. The average entropy of a two-phase mixture of quality x is:
The entropy of vaporization (sfg), although not listed in the steam tables, can be obtained from hfg. The vaporization process occurs at constant temperature and pressure. Under these restraints, the second term on the right of Eq (3.8a) is zero and the remainder of the equation can be directly integrated from the saturated liquid state to the saturated vapor state, yielding:
This equation applies to any pure substance, not just to water. The temperature in Eq (3.18) must be in Kelvins even though the steam tables use degrees Celsius.
Example: Verify Eq (3.18) from Steam-Table entries at 150oC (423 K)
At 150oC, Table A.1 gives f = 2114.3 kJ/kg. Applying Eq (3.18), sfg should be:
Table A.1 gives sfg = sg - sf = 6.838 - 1.842 = 5.00 kJ/kg-K, verifying the value obtained by Eq( 3.18).
Example: Superheated steam at 1 MPa and 300oC is isentropically expanded to 10 kPa. What are the temperature and quality of the final state?
From Table A.3, the entropy of the initial state is 7.123 kJ/kg-K. This is also the entropy in the final state. Assuming that the final state is a two-phase mixture, Table A.2 gives a final temperature of 45.8oC for the specified final pressure of 10 kPa. The quality is determined by solving Eq (3.17) for x:
3.6 More complicated processes
Table 3.2 summarizes the work and heat effects in an ideal gas in the four iso processes discussed above.
Many complex processes can be broken up into two or more "iso" steps. A simple example is changing the state of water from one (p,T) combination to another analyzed in Sect. 2.5.1. Here the prescribed change in state was achieved an isothermal step and an isobaric step.
Table 3.2 Iso processes in an ideal gas
Av = 0
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