Explanation Write the expression for mass rate balance for the streams m ˙ 1 = m ˙ 2 − m ˙ 3 m ˙ 3 = m ˙ 1 − m ˙ 2 (I) Here, the mass flow rate at inlet of section (1) is m ˙ 1 , mass flow rate at the exit of section (2) is m ˙ 2 , mass flow rate at the exit of section (3) is m ˙ 3 . Write the expression for steady flow energy equation for energy balance. − Q ˙ C V − m ˙ 1 h 1 + m ˙ 2 h 2 + m ˙ 3 h 3 = 0 Q ˙ C V = m ˙ 2 h 2 + m ˙ 3 h 3 − m ˙ 1 h 1 (VI) Here, heat added to the steam separator is Q ˙ C V , enthalpy at entry for stream at section (1) is h 1 , enthalpy at entry for stram at section (2) is h 2 , enthalpy at exit for stream at section (3)is h 3 , mass flow rate for air at section (1) is m ˙ 1 , mass flow rate for air at section (2) is m ˙ 2 , and mass flow rate for air at section (3) is m ˙ 3 . Write the expression for enthalpy at section (1). h 1 = h f 1 + x 1 h f g 1 (VIII) Here, liquid enthalpy at state (1) is h f 1 , enthalpy of liquid vapor mixture at section (1) is h f g 1 , and dryness fraction at state (1) is x 1 . Conclusion: Substitute 10 kg / s for m ˙ 1 , 6.9 kg / s for m ˙ 2 in Equation (V). m ˙ 3 = 10 kg / s − 6.9 kg / s = 3.1 kg / s Refer to table A-3 “Properties of Saturated Water” to obtain the values of enthalpy 467.11 kJ / kg for h f 1 and 2226.5 kJ / kg for h f g 1 at pressure 1.5 bar . Substitute 0.7 for x 1 , 467

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Chapter 4.12, Problem 88P

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Chapter 4.12, Problem 87P

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