Explanation The following figure shows the heat exchanger for applying steady flow energy equation. Figure-(1) Write the expression for steady flow energy equation for energy balance. m 1 ( h 1 − h 2 ) + m 3 ( h 3 − h 4 ) = 0 m 1 ( h 1 − h 2 ) = m 3 ( h 4 − h 3 ) m 3 = m 1 ( h 1 − h 2 ) ( h 4 − h 3 ) (I) Here, enthalpy at entry for steam is h 1 , enthalpy at exit for steam is h 2 , enthalpy at entry for cooling water is h 3 , enthalpy at exit for cooling water is h 4 , mass flow rate for steam is m ˙ 1 and mass flow rate for cooling water is m ˙ 3 . Write the expression for enthalpy at state (1). h 1 = h 2 + x 1 ( h g − h 2 ) (II) Here, the dryness fraction at section (1) is x 1 . Conclusion: Refer to table A-3 ,“Properties of saturated water” to obtain the values of specific enthalpy of saturated liquid and saturated vapour at 0.08 bar as 173.88 kJ / kg and 2576.98 kJ / kg respectively. Substitute 173.88 kJ / kg for h 2 at pressure 0.08 bar and 2576.98 kJ / kg for h g in Equation (II). h 1 = 173.88 kJ / kg + 0.932 × ( 2576.98 kJ / kg − 173.88 kJ / kg ) = 173

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Chapter 4.12, Problem 74P

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The mass flow rate of cooling water.

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Chapter 4.12, Problem 73P

Chapter 4.12, Problem 75P

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