Explanation Draw the T-s diagram of the simple ideal Rankine cycle as shown in figure (1). Write the expression for the work input: w p , i n = v 1 ( P 2 − P 1 ) (I) Here, specific volume at inlet is v 1 , pressure at state point 2 is P 2 , and pressure at state point 1 is P 1 . Write the expression for the enthalpy at state 2 using work input. h 2 = h 1 + w p , i n (II) Here, enthalpy at inlet is h 1 . Write the expression for the entropy relation at state 4. s 4 = s f 4 − x 4 s f g 4 (III) Here, entropy of saturated liquid is s f 4 , vapor quality at state 4 is x 4 , and entropy of vapor mixture is s f g 4 . Write the expression for the enthalpy relation at state 6. h 4 = h f − x 4 h f g (IV) Here, enthalpy of saturated liquid is h f and enthalpy of vapor mixture is h f g . Write the formula for the heat input for the turbine. q i n = h 3 − h 2 (V) Here, enthalpy at state 3 is h 3 and enthalpy at state 2 is h 2 . Write the formula for the heat output of the turbine. q o u t = h 4 − h 1 (VII) Here, enthalpy at state 4 is h 4 and enthalpy at state 1 is h 1 . Write the expression for the thermal efficiency η t h = ( 1 − q o u t q i n ) × 100 % (VI) Conclusion: Refer to Table A-5, “Saturated water—Pressure table”, obtain the entropy of saturated liquid at the pressure of 50 kPa . h 1 = h f @ 50 kPa = 340 .54 kJ / kg v 1 = v f @ 50 kPa = 0.001030 m 3 / kg Substitute 0.001030 m 3 / kg for v 1 , 6 MPa for P 2 , 5 MPa for P 1 in equation (I). w p , i n = ( 0.001030 m 3 / kg ) ( 6 000 kPa − 50 kPa ) ( 1 kJ 1 kPa ⋅ m 3 ) = 6.13 kJ / kg Substitute 340.54 kJ / kg for h 1 and 6.13 kJ / kg for w p , i n in equation (II). h 2 = 340.54 kJ / kg + 6.13 kJ / kg = 346.67 kJ / kg Refer to Table A-6, “Superheated water”, obtain the following properties at the pressure of 6000 kPa and temperature of 600 ° C . h 3 = 3658 .8 kJ / kg s 3 = 7 .1693 kJ / kg ⋅ K Refer to Table A-5, “Saturated water—Pressure table”, obtain the following properties of liquid at the pressure of 50 kPa . h f @ 50 kPa = 340 .54 kJ / kg h f g @ 50 kPa = 2304.7 kJ / kg s f @ 50 kPa = 1 .0912 kJ / kg ⋅ K s f g @ 50 kPa = 6.5019 kJ / kg ⋅ K Substitute 7.1693 kJ / kg ⋅ K for s 4 , 1.0912 kJ / kg ⋅ K for s f , and 6.5019 kJ / kg ⋅ K for s f g in equation (III) 7.1693 kJ / kg ⋅ K = 1.0912 kJ / kg ⋅ K − x 6 ( 6.5019 kJ / kg ⋅ K ) x 4 = ( 7.1693 − 1.0912 ) kJ / kg ⋅ K 6.5019 kJ / kg ⋅ K x 4 = 0.9348 Substitute 340.54 kJ / kg for h f , 0.9348 for x 6 , and 2304

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Chapter 10.9, Problem 91RP

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Chapter 10.9, Problem 90RP

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