Explanation Draw the T − s diagram for the Brayton cycle with regeneration as shown in Figure (1). Write the pressure ratio relation for the process 1-2. P r 2 = P 2 P 1 P r 1 (I) Here, relative pressure at state 1 is P r 1 and relative pressure at state 2 is P r 2 . Write the pressure ratio relation for the process 3-4. P r 4 = P 4 P 3 P r 3 (II) Here, relative pressure at state 3 is P r 3 and relative pressure at state 4 is P r 4 . Write the expression for the efficiency of the compressor ( η C ) to calculate h 2 . η C = h 2 s − h 1 h 2 − h 1 h 2 = h 1 + h 2 s − h 1 η C (III) Here, enthalpy on isentropic condition 2 s is h 2 s . Write the expression for the efficiency of the turbine ( η T ) to calculate h 4 . η T = h 3 − h 4 h 3 − h 4 s h 4 = h 3 − η T ( h 3 − h 4 s ) (IV) Here, enthalpy on isentropic condition 4 s is h 4 s . Write the expression for the heat rejected by the regenerator ( q regen ) . q regen = ε ( h 4 − h 2 ) (V) Here, the effectiveness of the regenerator is ε and enthalpy at state 4 is h 4 . Write the expression to calculate the heat input to the gas turbine ( q in ) . q in = ( h 3 − h 2 ) − q regen (VI) Write the expression to calculate the net work output by the gas turbine ( w net,out ) . w net,out = w T,out − w C,in = ( h 3 − h 4 ) − ( h 2 − h 1 ) (VII) Here, the work done by the turbine is w T,out , and the work input to the compressor is w C,in . Write the thermal efficiency of the given turbine ( η t h ) as η t h = w net,out q in (VIII) Write the expression to calculate mass flow rate m ˙ . m ˙ = W ˙ net w net,out (IX) Here, mass flow rate is m ˙ , net work output is w net,out . Conclusion: From Table A-17, “Ideal-gas properties of air”, obtain the following properties at the temperature of 303 K . h 1 = 121.88 Btu / lbm P r 1 = 1.1369 Here, enthalpy at state 1 is h 1 and relative pressure at state 1 is P r 1 . From Table A-17, “Ideal-gas properties of air”, obtain the following properties at the temperature of 1561 K . h 3 = 549.35 Btu / lbm P r 3 = 230.12 Here, enthalpy at state 3 is h 3 and relative pressure at state 3 is P r 3 . Substitute 4 for P 2 P 1 , and 1.1369 for P r 1 in Equation (I). P r 2 = ( 4 ) ( 1.1369 ) = 4.548 From the Table A-17E, “Ideal-gas properties of air”. Obtain the value of enthalpy on isentropic state ( h 2 s ) at the relative pressure ( P r 2 ) of 4.548 by using interpolation method. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (X) Here, the variables denoted by x and y are relative pressure and enthalpy. Show relative pressure and enthalpy values from the Table A-17E. Relative pressure ( P r 2 ) Enthalpy ( h 2 ) , in Btu / lbm 4.193 177.23 4.548 ? 4.607 182.08 Substitute 4.193 for x 1 , 4.548 for x 2 , 4.607 for x 3 , 177.23 for y 1 , and 182.08 for y 3 in Equation (X). y 2 = ( 4.548 − 4.193 ) ( 182.08 − 177.23 ) ( 4.607 − 4.193 ) + 177.23 = 181.388 = 181.39 The value of enthalpy on isentropic state ( h 2 s ) at the relative pressure ( P r 2 ) of 4

9th Edition

ISBN: 9781259822674

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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