Explanation The below figure shows the T − s diagram of the process. Figure-(1) Write the expression for the specific enthalpy at state 1. h 1 = h a + V a 2 2 (I) Here, the specific enthalpy at state 1 is h 1 , the specific volume at state a is V a and the specific enthalpy at state a is h a . Write the expression for the pressure at state 1. p 1 = ( p r 1 p r a ) p a (II) Here, the pressure at state 1 is p 1 , the pressure at state a is p a , the reduced pressure at state 1 is p r 1 and the reduced pressure at state a is p r a . Write the expression for the saturated pressure at state 2. p r 2 s = ( p 2 p 1 ) p r 1 (III) Here, the pressure at state 2 is p 2 and the saturated pressure at state a is p r 2 s . Write the expression for the specific enthalpy in terms of compressor efficiency at state 2. h 2 = h 1 + ( h 2 s − h 1 η c ) (IV) Here, the compressor efficiency is η c , the specific enthalpy at state 2 is h 2 and the saturated specific enthalpy at state 2 is h 2 s . Write the expression for the specific enthalpy at state 4. h 4 = h 3 − η t ( h 3 − h 4 s ) (V) Here, the specific enthalpy at state 4 is h 4 and the specific enthalpy at state is h 3 . Write the expression for the saturated specific enthalpy in terms of turbine efficiency at state 4. h 4 s = h 3 − ( h 2 − h 1 η t ) (VI) Here, the turbine efficiency is η t , the saturated specific enthalpy at state 4 is h 4 s . Write the expression for the pressure at state 4. p 4 = ( p r 4 s p r 3 ) p 3 (VII) Here, the pressure at state 4 is p 4 , the pressure at state 3 is p 3 , the reduced pressure at state 3 is p r 3 and the saturated reduced pressure at state 4 is p r 4 s . Write the expression for the reduced pressure at state 5. p r 5 = ( p 5 p 4 ) p r 4 (VIII) Here, the pressure at state 5 is p 5 , the reduced pressure at state 5 is p r 5 and the reduced pressure at state 4 is p r 4 . Write the expression for the pressure at state 4. p 4 = ( p r 4 s p r 3 ) p 3 (IX) Here, the pressure at state 4 is p 4 , the pressure at state 3 is p 3 , the reduced pressure at state 3 is p r 3 and the saturated reduced pressure at state 4 is p r 4 s . Write the expression for the velocity at the nozzle exit. ( V 5 ) 1 = 2 ( h 4 − h 5 s ) (X) Here, the velocity at the nozzle exit is ( V 5 ) 1 . Write the expression for force developed. ( F x ) 1 = m 1 ( ( V 5 ) 1 − ( V a ) 1 ) (XI) Here, the initial velocity is ( V a ) 1 . Write the expression for thrust force. ( F t h r u s t ) 1 = ( − F x ) 1 (XII) Write the expression for the reduced pressure at state 5. p ′ r 5 = ( p 5 p 4 ) p ′ r 4 (XIII) Here, the pressure at state 4 is p 4 , the pressure at state 5 is p 5 , the pressure at state 5 is p 5 , the reduced pressure at state 5 is p ′ r 5 and the reduced pressure at state 4 is p ′ r 4 . Write the expression for isentropic efficiency of nozzle. η N = h ′ 4 − h 5 h ′ 4 − h 5 s (XIV) Write the expression for the velocity at the nozzle exit. ( V 5 ) 2 = 2 ( h ′ 4 − h 5 ) (XV) Here, the velocity at the nozzle exit is ( V 5 ) 2 , the specific enthalpy at state 4 is h ′ 4 and the specific enthalpy at state 5 is h ′ 5 . Write the expression for force developed. ( F x ) 2 = m 2 ( ( V 5 ) 2 − ( V a ) 2 ) (XVI) Here, the initial velocity is V a . Write the expression for thrust force. ( F t h r u s t ) 2 = ( − F x ) 2 (XVII) Conclusion: Refer to table A-22E “ideal gas properties of air” to obtain the specific enthalpy and reduced pressure at temperature T a = 420 ° R as 100.32 Btu / lb and 0.5760 . Substitute 100.32 Btu / lb for h a , 750 ft / s for V a in Equation (I). h 1 = ( 100.32 Btu / lb ) + ( 750 ft / s ) 2 2 = ( 100.32 Btu / lb ) + 281250 ( ft / s ) 2 ( 1 lbf 32.2 lb ⋅ ft / s 2 ) ( 1 Btu 778 ft ⋅ lbf ) = 111.5468 Btu / lb ≈ 111.55 Btu / lb Refer to table A-22E “ideal gas properties of air” to obtain the reduced pressure at specific enthalpy h 1 = 111.55 Btu / lb as 0.8349 . Substitute 0.8349 for p r 1 , 0.5760 for p r a and 9 lbf / in 2 for p a in Equation (II). p 1 = ( 0.8349 0.5760 ) ( 9 lbf / in 2 ) = 1.4494 ( 9 lbf / in 2 ) ≈ 13.05 lbf / in 2 Substitute 12 for p 2 p 1 and 0.8349 for p r 1 in Equation (III). p r 2 s = ( 0.8349 ) ( 12 ) = 10.01 Refer to table A-22E “ideal gas properties of air” to obtain the saturated specific enthalpy at reduced pressure p r 2 s = 10.01 as 227.29 Btu / lb . Substitute 227.29 Btu / lb for h 2 s , 111.55 Btu / lb 254.22 kJ / kg for h 1 and 0.88 for η c in Equation (IV). h 2 = 111.55 Btu / lb + ( 227.29 Btu / lb − 111.55 Btu / lb 0.88 ) = 111.55 Btu / lb + 131.52 Btu / lb = 243.07 Btu / lb Calculate the ratio of pressure at state 1 and 2. p 2 p 1 = 12 p 2 = 12 ( 13.05 lbf / in 2 ) = 156.6 lbf / in 2 Refer to table A-22E “ideal gas properties of air” to obtain the specific enthalpy and reduced pressure at temperature T 3 = 2400 ° R as 617.22 Btu / lb and 367.6 . Since, the pressure remains constant for the process 2 and 3. Hence, p 2 = p 3 . Substitute 243.07 Btu / lb for h 2 , 617.22 Btu / lb for h 3 , 111.55 Btu / lb for h 1 and 0.90 for η t in Equation (VI). h 4 s = 617.22 Btu / lb − ( 243

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Chapter 9.14, Problem 107P

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