Explanation The figure below shows the system of valve and flash chamber as the control volume. Figure-(1) Write the expression of power of turbine. W ˙ t = m ˙ 2 ( h 2 − h 4 ) (I) Here, the mass of saturated vapor at entry of turbine is m ˙ 2 , the specific enthalpy at 4 is h 4 , the specific enthalpy at 2 is h 2 . Write the expression for steady state equation for the control volume. W ˙ c v + m ˙ 2 h 2 + m ˙ 3 h 3 = Q ˙ c v + m ˙ 1 h 1 (II) Here, the rate of work flow through the control volume is W ˙ c v , the mass flow rate at 3 is m ˙ 3 , the specific enthalpy at 3 is h 3 , the mass flow rate at 1 is m ˙ 1 the specific enthalpy at 1 is h 1 and the rate of heat transfer through the control volume is Q ˙ c v . The heat flow through the control volume is negligible that is Q ˙ c v is zero. The rate of work flow through the control volume is W ˙ c v = 0 . Write the mass balance expression. m ˙ 1 = m ˙ 2 + m ˙ 3 m ˙ 3 = m ˙ 1 − m ˙ 2 Substitute m ˙ 1 − m ˙ 2 for m ˙ 3 , 0 for W ˙ c v , 0 for Q ˙ c v in Equation (II). 0 + m ˙ 2 h 2 + ( m ˙ 1 − m ˙ 2 ) h 3 = 0 + m ˙ 1 h 1 m ˙ 2 = m ˙ 1 ( h 1 − h 3 ) ( h 2 − h 3 ) (III) Write the expression for the specific enthalpy at exit of turbine. h 4 = h f + x ( h g − h f ) (IV) Here, the specific enthalpy at saturated vapor is h f , the specific enthalpy at saturated liquid is h g and the dryness fraction is x . Conclusion: Refer to Table A-2 “Properties of Saturated Water (Liquid–Vapor): Temperature Table” to obtain specific enthalpy at saturated vapor as 763.22 kJ / kg with respect to 15 bar pressure and 180 ° C . Refer to Table A-3 “Properties of Saturated Water (Liquid–Vapor): Pressure Table” to obtain specific enthalpy at saturated vapor as 604.74 kJ / kg and specific enthalpy at saturated vapor as 2738.6 kJ / kg with respect to 4 bar pressure. Substitute 5 kg / s for m ˙ 1 , 763.22 kJ / kg for h 1 , 604.74 kJ / kg for h 3 and 2738.6 kJ / kg for h 2 in Equation (III)

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 4.12, Problem 105P

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