Question

Find the temperature, mass and total enthalpy of the liquid and vapor phases of 5 kg of water in a container with a pressure of 150 kPa and a constant volume of 200L.

Answer 1

$m=5\; kg \\ p= 150 \; kPa = 150 \times 10^3 \; Pa \\ V = 200 \;L = 0.2 \; m^3$

The specific volume:

$v = \frac{V}{m} = \frac{0.2 \;m^3}{5 \;kg}=0.04 \;m^3/kg$

The specific volume of saturated liquid:

$v_f = 0.00105273 \; m^3/kg$

The specific volume of saturated vapor:

$v_g = 1.1593 \;m^3/kg \\ v=v_f +x(v_g -v_f)$

x = dryness fraction

$0.04 \; m^3/kg = 0.00105273 \;m^3/kg + x(1.1593 -0.00105273) \\ x = \frac{0.03894727 \; m^3/kg}{1.15824727 \;m^3/kg} \\ x = 0.0336$

The mass of water:

$x \times m = 0.0336 \times 5 = 0.168 \; kg$

The mass of vapor:

$(1-x) \times m = 4.832 \;kg$

The specific enthalpy of saturated liquid

$h_f = 467.13 \; kJ/kg$

The specific enthalpy of saturated vapor:

h_g = 2693.1 \; kJ/kg

Specific enthalpy:

$h =h_f +x(h_g-h_f) \\ h = 467.13 \;kJ/kg + 0.0336(2693.1 -467.13) \; kJ/kg \\ h = 541.92 \;kJ/kg$

The enthalpy:

$H = m \times h \\ H = 5 \times 541.92 = 2709.6 \; kJ$