Explanation Given: The initial pressure of the nozzle ( P 1 ) is 45 psia. The initial temperature of the nozzle ( P 2 ) is 940 ° F . The initial velocity ( V 1 ) is low (0). The final velocity ( V 2 ) is 650 ft/s . The isentropic efficiency ( η ) is 80 % . Calculation: Refer to Table A-5E, “Ideal-gas properties of air”, obtain the below properties at the saturated pressure of 1400 R . h 1 = 342.90 Btu / lbm P r 1 = 42.88 Since, initial velocity V 1 is low velocity V 1 is equal to zero. Write the expression for the energy balance equation for closed system. E ˙ i n − E ˙ o u t = Δ E ˙ s y s t e m (I) Here, rate of net energy transfer in to the control volume is E ˙ i n , rate of net energy transfer exit from the control volume is E ˙ o u t and rate of change in internal energy of system is Δ E ˙ s y s t e m . The rate of change in internal energy of the system is zero at steady state. Substitute Δ E ˙ s y s t e m = 0 in Equation (I). E ˙ i n − E ˙ o u t = 0 E ˙ i n = E ˙ o u t m ˙ ( h 1 + V 1 2 2 ) = m ˙ ( h 2 + V 2 2 2 ) h 1 − h 2 = V 2 2 − V 1 2 2 342.90 Btu / lbm − h 2 = ( 650 ft / s ) 2 − 0 2 h 2 = 342.90 Btu / lbm − 211250 ft 2 / s 2 2 ( 1 Btu / lbm 25037 ft 2 / s 2 ) h 2 = 334.46 Btu / lbm Refer to Table A-17E, “Ideal-gas properties of air”, obtain the value of exit temperature ( T 2 a ) at the enthalpy ( h 2 ) of 334.46 Btu / lbm by using interpolation method. Show the enthalpy and temperature values from the Table A-13 as in below table. Enthalpy ( h ) ,in Btu / lbm Temperature ( R ) , in kJ/kg 332.48 1360 334.46 ? 342.9 1400 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 (I) Here, the variables denoted by x and y are enthalpy and temperature

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ISBN: 9780078027680

Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala

Publisher: McGraw-Hill Education

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Chapter 8, Problem 124P

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Chapter 8, Problem 123P

Chapter 8, Problem 126P

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Fundamentals of Thermal-Fluid Sciences

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