Explanation Given: Pressure at state 1 is 100 kPa . Pressure at state 2 is 1000 kPa . Pressure at state 3 is 100 kPa . The volume of boiler tank is 100 L . Concept used: Write the expression for energy balance equation. Q T = m 2 ⋅ u 2 − m 1 ⋅ u 1 + m 0 ⋅ h 0 …… (1) Here, u i is specific internal energy at entry, m 1 is initial mass, h 0 specific enthalpy at exit u 2 specific internal energy at exit, m 2 mass at stage 2, Q T is heat transfer through tank and m 0 is mass at exit. From saturated water table B 1.2 corresponding to 100 kPa the exiting specific enthalpy is 267.46 kJ / kg From saturated water table B 1.2. corresponding to 100 kPa the obtained liquid and vapor specific volume and internal energy: The specific internal energy for liquid is 417.33 kJ / kg . The specific enthalpy for liquid is 0 .001043 kJ / kg . Write the expression for initial mass. m 1 = V v f …… (2) Here, V is volume of tank and v f is specific volume. From saturated water table B 1.2. Corresponding to 1000 kPa the obtained stage 2 liquid and vapor specific volume and internal energy: The specific internal energy for gas is 2583.64 kJ / kg . The specific enthalpy for gas is 2776.08 kJ / kg . The specific volume for gas is 0 .19444 m 3 / kg . Write the expression for mass at stage 2 m 2 = V v g …… (3) Here, V is volume of tank, v g specific volume of gas. Write the expression for exit mass. m o = m 1 − m 2 …… (4) Write the expression for heat transfer through heater. W = m o ( h 0 − h i ) …… (5) Here, W is turbine work, h i is inlet specific enthalpy. Calculation: Substitute 100 L for V and 0 .001043 kJ / kg for v f equation (2) m 1 = 100 L ( 1 m 3 1000 L ) 0

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Author: Sonntag

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Chapter 4, Problem 4.115P

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