Explanation Show the T-s diagram as in Figure (1). Express temperature at state 6 s . T 6 s = T 5 [ P 6 P 5 ] k − 1 k (I) Here, temperature at state 5 is T 5 , pressure at state 6 is P 6 , pressure at state 5 is P 5 and heat capacity ratio is k . Express temperature at state 6 using Carnot efficiency. T 6 = T 5 + T 6 s − T 5 η C , i (II) Here, temperature at state 6 is T 6 , temperature at state 6 s is T 6 s and initial isentropic efficiency of turbine is η C , i . Express temperature at state 8 s . T 8 s = T 7 [ P 8 P 7 ] k − 1 k (III) Here, temperature at state 7 is T 7 , pressure at state 8 and 7 is P 8 and P 7 respectively. Express temperature at state 8. T 8 = T 7 − η C , i ( T 7 − T 8 s ) (IV) Here, temperature at state 7 is T 7 . Express temperature at state 9. T 9 = T sat@800 psia + 50 R (V) Here, saturation temperature at pressure of 800 psia is T sat@800 psia . Express enthalpy at state 1. h 1 = h f @ 5 psia (VI) Here, enthalpy of saturation liquid at pressure of 5 psia is h f @ 5 psia . Express specific volume at state 1. v 1 = v f @ 5 psia (VII) Here, specific volume of saturation liquid at pressure of 5 psia is v f @ 5 psia . Express work done at inlet. w p,in = v 1 ( P 2 − P 1 ) (VIII) Express enthalpy at state 2. h 2 = h 1 + w p,in (IX) Here, enthalpy at state 1 is h 1 . Due to throttling process, write the entropy at state 3 is equal to entropy at state 4. s 4 = s 3 (X) Here, entropy at state 3 is s 3 and entropy at state 4 is s 4 . Express enthalpy at state 4. h 4 = h 3 − η T ( h 3 − h 4 s ) (XI) Here, enthalpy at state 3 is h 3 and isentropic efficiency of turbine is η T . Express the net work output of gas cycle. w net,gas cycle = c p ( T 7 − T 8 ) − c p ( T 6 − T 5 ) = c p ( T 7 − T 8 − T 6 + T 5 ) (XII) Here, specific heat constant is c p , temperature at state 5, 6, 7 and 8 is T 5 , T 6 , T 7 and T 8 respectively. Express the net work output of steam cycle. w net,steam cycle = ( h 3 − h 4 ) − w p,in (XIII) Express mass flow rate of water from an energy balance. m ˙ w = c p ( T 8 − T 9 ) h 3 − h 2 m ˙ a (XIV) Here, mass flow rate of air is m ˙ a . Express heat input to the system. q in = c p ( T 7 − T 6 ) (XV) Express heat supplied from the system. q out = c p ( T 9 − T 5 ) + m ˙ w m ˙ a ( h 4 − h 1 ) (XVI) Express the overall thermal efficiency of the combined cycle. η th = 1 − q out q in (XVII) Conclusion: Refer Table A-2E, “ideal gas specific heats of various common gases”, and write the value of specific heat constant of air and heat capacity of air. c p = 0.240 Btu / lbm ⋅ R k = 1.4 Substitute 80°F for T 5 , 10 for P 6 P 5 and 1.4 for k in Equation (I). T 6 s = ( 80°F ) ( 10 ) 1.4 − 1 1.4 = [ ( 80 + 460 ) R ] ( 1.9306 ) = ( 540 R ) ( 1.9306 ) = 1043 R Substitute 80°F for T 5 , 1043 R for T 6 s and 0.90 for η C , i in Equation (II). T 6 = ( 80°F ) + 1043 R − 80°F 0.90 = ( 540 R ) + 1043 R − 540 R 0.90 = 1099 R Substitute 2100°F for T 7 , 1 10 for P 8 P 7 and 1.4 for k in Equation (III). T 8 s = ( 2100°F ) [ 1 10 ] 1.4 − 1 1.4 = [ ( 2100 + 460 ) R ] ( 0.5179 ) = ( 2560 R ) ( 0.5179 ) = 1326 R Substitute 2100°F for T 7 , 0.90 for η C , i and 1326 R for T 8 s in Equation (IV). T 8 = ( 2100°F ) − ( 0.90 ) ( 2100°F − 1326 R ) = ( 2560 R ) − ( 0.90 ) ( 2560 R − 1326 R ) = 1449 R Refer Table A-5E, “saturated water-pressure table”, and write the saturation temperature at pressure of 800 psia . T sat@800 psia = 518 °F Substitute 518 °F for T sat@800 psia in Equation (V). T 9 = 518 °F + 50 R = ( 518 + 460 ) R + 50 R = 978 R + 50 R = 1028 R Refer Table A-5E, “saturated water-pressure table”, and write the enthalpy of saturation liquid at pressure of 5 psia . h f @ 5 psia = 130.18 Btu / lbm Substitute 130.18 Btu / lbm for h f @ 5 psia in Equation (VI). h 1 = 130.18 Btu / lbm Refer Table A-5E, “saturated water-pressure table”, and write the specific volume of saturation liquid at pressure of 5 psia . v f @ 5 psia = 0.01641 ft 3 / lbm Substitute 0.01641 ft 3 / lbm for v f @ 5 psia in Equation (VII). v 1 = 0.01641 ft 3 / lbm Substitute 0.01641 ft 3 / lbm for v 1 , 800 psia for P 2 and 5 psia for P 1 in Equation (VIII). w p,in = ( 0.01641 ft 3 / lbm ) ( 800 psia − 5 psia ) = ( 0.01641 ft 3 / lbm ) ( 795 psia ) [ Btu 5.404 psia ⋅ ft 3 ] = 2

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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ISBN: 9781260048636

Author: CENGEL

Publisher: MCG

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Chapter 10.9, Problem 105RP

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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