Thermodynamics

384 | ThermodynamicsT,°C4501ThrottlingprocessP 1 = 7 MPaT 1 = 450°C2h = const.FIGURE 7–66Schematic and T-s diagram forExample 7–18.P 2 = 3 MPas 1s 2sAnalysis We take the throttling valve as the system (Fig. 7–66). This is acontrol volume since mass crosses the system boundary during the process.We note that there is only one inlet and one exit and thus ṁ 1 ṁ 2 ṁ.Also, the enthalpy of a fluid remains nearly constant during a throttlingprocess and thus h 2 h 1 .The entropy of the steam at the inlet and the exit states is determined fromthe steam tables to beState 1:State 2:P 1 7 MPaT 1 450°C fh 1 3288.3 kJ>kgs 1 6.6353 kJ>kg # KP 2 3 MPafsh 2 h 2 7.0046 kJ>kg # K1Then the entropy generation per unit mass of the steam is determined fromthe entropy balance applied to the throttling valve,S # in S # out S # gen dS system >dt123Rate of net entropytransfer by heatand mass123Rate of entropygenerationm # s 1 m # s 2 S # gen 0S # gen m # 1s 2 s 1 2Dividing by mass flow rate and substituting gives(steady)¡0123Rate of changein entropys gen s 2 s 1 7.0046 6.6353 0.3693 kJ/kg # KThis is the amount of entropy generated per unit mass of steam as it isthrottled from the inlet state to the final pressure, and it is caused byunrestrained expansion. The increase of entropy principle is obviouslysatisfied during this process since the entropy generation is positive.