500 kg of water is in a 10 m³ boiler under 5 Mpa. The temperature, enthalpy, and the mass of each phase of water are needed to monitor the vaporization process.

(1) Determining the state of water

The specific volume of the water is

      v_av = V/m = 10/500 = 0.02 m³/kg

From the saturated water table, find the specific volume for saturated liquid vapor at 5 Mpa.

      v_f = 0.001286 m³/kg

      v_g = 0.03944 m³/kg

       v_f = 0.001286 < v_av = 0.02< v_g = 0.03944 m³/kg

Since v_f  < v_av< v_g, the water is in saturated mixture state.

(2) Temperature

Since the water is a saturated mixture, the temperature is the saturation temperature. From the saturated water table, the saturation temperature at 5 Mpa is 263.99^oC.

(3) Quality x

The quality of the saturated mixture is given by

      x = (v_av - v_f)/v_fg

      v_fg = v_g - v_f = 0.038154 m³/kg

      x = (0.02 - 0.001286)/0.038154 = 49%

(4) Enthalpy

Once again, from the saturated water table, h_f and h_g can be found at P = 5 Mpa.

      h_f = 1,154.23 kJ/kg

      h_g = 2,794.3 kJ/kg

     h_fg = h_g - h_f = 1,640.1 kJ/kg

     h_av = h_f + xh_fg

           = 1,154.23 + (49%)(1,640.1) = 1,958 kJ/kg

(5) Mass of each phase

The definition of quality is given by

      x = m_g/m_total

The mass for each phase can be calculated as

      m_g = x m_total = ( 49%)(500) = 245 kg

      m_l = m_total - m_g = 500 - 245 = 255 kg