A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7. Saturated water vapor at 260 o C is slowly withdrawn through a pressure-regulating valve at the top of the tank as energy is transferred by heat to maintain the pressure constant in the tank. This continues until the tank is filled with saturated vapor at 260 o C. Determine the amount of heat transfer, in kj

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 = 0.85 m^3 / 0.02993 m^3 / kg

= 28.4 kg

The mass finally contained in the tank is

m2 = V2/v

= 0.85 m^3 / 0.0422 m^3 / kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 - m1u1 - he (m2 - m1)

Qcv = (20.14) (2599) - (28.4) (2158) - (2797) (20.14 - 28.4) = 14,160 kJ