Explanation The flow of steam through the nozzle is steady and isentropic. Write the expression for the efficiency of nozzle. η N = h 01 − h 2 h 01 − h 2 s (I) Here, efficiency of nozzle is η N , stagnation enthalpy at the inlet is h 01 , enthalpy at exit is h 2 , and superheated enthalpy at exit is h 2 s . Write the expression of energy balance equation for the converging-diverging nozzle. h 1 + V 1 2 / 2 = h 2 + V 2 2 / 2 Inlet velocity is equal to zero V 1 = 0 . V 2 = 2 ( h 1 − h 2 ) (II) Here, velocity of steam at the inlet of nozzle is V 1 , and velocity of steam at exit of nozzle is V 2 . Write the expression to calculate the exit area of the nozzle. A 2 = m ˙ v 2 V 2 (III) Here, mass flow rate of steam is m ˙ , specific volume of the steam at the exit is v 2 . Write the expression to calculate the velocity of sound through the steam at the exit of nozzle. c 2 = ( ∂ P ∂ ρ ) s 1 / 2 = ( Δ P Δ ( 1 / v ) ) s 1 / 2 (IV) Here, pressure drop in the nozzle is Δ P , and drop in the specific volume of the steam is Δ ( 1 / v ) . Write the expression to calculate the Mach number for the steam at the exit of nozzle. Ma 2 = V 2 c 2 (V) Here, Mach number of the steam at the exit is Ma 2 . Conclusion: Refer Table A-6, “Superheated water”, obtain the values of h 01 , and s 2 s at a pressure of 1 MPa and temperature of 500 ° C as 3479 .1 kJ / kg , and 7.7642 kJ / kg ⋅ K respectively. Here, at superheated condition the entropy of saturated steam is s 2 s . Refer Table A-6, “Superheated water”, obtain the values of h 2 s at an entropy of 7.7642 kJ / kg ⋅ K and a pressure of 0.2 MPa as 3000 kJ / kg . Substitute 3000 kJ / kg for h 2 s , 3479 .1 kJ / kg for h 01 , and 0.90 for η N in Equation (I). 0.90 = 3479 .1 kJ / kg − h 2 3479 .1 kJ / kg − 3000 kJ / kg h 2 = 3047.91 kJ / kg The stagnation enthalpy of steam at the inlet is equal to the actual enthalpy at the inlet ( h 01 = h 1 ) . Substitute 3479 .1 kJ / kg for h 1 and 3047.91 kJ / kg for h 2 in Equation (II). V 2 = 2 ( 3479 .1 kJ / kg − 3047.91 kJ / kg ) = 2 ( 3479

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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Chapter 17.7, Problem 112P

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Chapter 17 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

Ch. 17.7 - A high-speed aircraft is cruising in still air....Ch. 17.7 - What is dynamic temperature?Ch. 17.7 - Prob. 3P Ch. 17.7 - Prob. 4P Ch. 17.7 - Prob. 5P Ch. 17.7 - Prob. 6P Ch. 17.7 - Calculate the stagnation temperature and pressure...Ch. 17.7 - Prob. 8P Ch. 17.7 - Prob. 9P Ch. 17.7 - Prob. 10P

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The exit area and the Mach number at the exit of nozzle.

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