Explanation Draw the T − s diagram for turboprop engine as shown in Figure (1). Write the temperature and pressure relation for the isentropic process 1-2. T 2 = T 1 ( r p ) ( k − 1 ) / k (I) Here. temperature at state 1 is T 1 , temperature at state 2 is T 2 , specific heat ratio is k and pressure ratio is r p . Write the temperature and pressure relation ratio for the process 3-5. T 5 = T 3 ( 1 r p ) ( k − 1 ) / k (II) Here, temperature at state 5 is T 5 and temperature at state 3 is T 3 . Write the temperature relation for the compressor and turbine. The work output produced by the expansion process 3-4 equals to work input of compression process 1-2. T 4 = T 3 − ( T 2 − T 1 ) (III) Here, temperature at state 4 is T 4 . During the expansion process 4-5 excess enthalpy is generated, it is used to increase the kinetic energy of the flow through propeller. Write the expression to calculate the exit velocity. m ˙ e c p ( T 4 − T 5 ) = m ˙ p V e x i t 2 − V i n l e t 2 2 V e x i t 2 = [ 2 m ˙ e m ˙ p c p ( T 4 − T 5 ) + V i n l e t 2 ] (IV) Here, inlet velocity is V i n l e t , the specific heat at constant pressure is c p , the mass flow rate through the propeller is m ˙ p and the mass flow rate through the compressor is m ˙ e . Write the expression to calculate the specific volume at state 1 v 1 . v 1 = R T 1 P 1 (V) Here, gas constant is R , temperature at state 1 is T 1 and pressure at state 1 is P 1 . Write the expression to calculate the mass flow rate of propeller m ˙ p . m ˙ p = A 1 V i n l e t v 1 = π D 2 4 V i n l e t v 1 (VI) Here, propeller diameter is D , area is A 1 and inlet velocity is V i n l e t . Write the expression to calculate the thrust force generated by the propeller ( F ) . F = m ˙ p ( V e x i t − V i n l e t ) (VII) Here, the thrust force produced by engine is F , mass flow rate of air is m ˙ , inlet velocity of air is V inlet , and outlet velocity of air is V exit . Conclusion. From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant ( R ) for air substance as 0.3704 psia ⋅ ft 3 / lbm ⋅ R . From Table A-2Ea, “Ideal-gas specific heats of various common gases”, obtain the following values for air at room temperature. k = 1.4 c p = 0.24 Btu / lbm ⋅ R Substitute − 10 ° F for T 1 , 10 for r p , and 1.4 for k in Equation (I). T 2 = ( − 10 ° F ) ( 10 ) 1.4 − 1 / 1.4 = ( − 10 + 460 ) R ( 10 ) 1.4 − 1 / 1.4 = 868.8 R Substitute 940 ° F for T 4 , 10 for r p and 1.4 for k in Equation (II). T 5 = ( 940 ° F ) ( 1 10 ) 1.4 − 1 / 1.4 = ( 940 + 460 ) R ( 1 10 ) 1.4 − 1 / 1.4 = 725.1 R Substitute 940 ° F for T 3 , − 10 ° F for T 1 , and 868.8 R for T 2 in Equation (III). T 4 = 940 ° F − ( 868

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Chapter 9.12, Problem 138P

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