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Answer :
14. The mass of the liquid is 1.0334 kg.(B)
15. The amount of heat required is 2796.9 kJ. (D)
16. The specific internal energy is 2959 kJ/kg. (A)
14. The mass of liquid in the vessel can be calculated by multiplying the volume of liquid by its density. The volume of the liquid can be calculated by subtracting the specific volume of the vapour from the specific volume of the mixture.
At 260°C, the specific volume of the saturated vapour is 1.467 m³/kg, and the specific volume of the saturated liquid is 0.001142 m³/kg. The specific volume of the mixture can be found using the quality, which is the mass of the vapour divided by the total mass.
At a quality of 0.05, the specific volume of the mixture is 0.0298 m³/kg. Therefore, the volume of the liquid is (0.0298 - 1.467 x 0.05) m³. The density of the liquid at 260°C is 206 kg/m³, so the mass of the liquid is (0.0298 - 1.467 x 0.05) x 206 kg = 1.0334 kg.(B)
15. The heat required to convert the saturated liquid to dry saturated steam is given by the specific enthalpy of vaporization. At 260°C, this is 2796.9 kJ/kg. Therefore, the total heat required to produce 1.0334 kg of dry saturated steam is 1.0334 x 2796.9 = 2796.9 kJ. (D)
16. The specific internal energy of the dry saturated steam can be found using the specific enthalpy and the specific volume. At 260°C, the specific enthalpy of the dry saturated steam is 3030.4 kJ/kg. The specific volume of the dry saturated steam is 0.138 m³/kg. Therefore, the specific internal energy is 3030.4 - 0.138 x 1000 = 2959 kJ/kg.(A)
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