Answer:
T1 = 0.61°C and h1 = 54.43 KJ/Kg
T2 = 21.55 °C and h2 = 84.58 KJ/Kg
Explanation:
We are given;
Volume; V = 14L = 0.014m³
Mass; m = 10.08kg
Let's calculate the total specific volume of the mixture;
α_tot = V/m = 0.014/10.08 = 0.00139 m³/kg
Looking at the table first table i attached, at pressure of 300KPa we have;
α_vap = 0.0678
α_liq = 0.000773
Now formula for quality is given as;
q1 = (α_tot - α_liq)/(α_vap - α_liq)
Thus, plugging in the relevant values to obtain;
q1 = (0.00139 - 0.000773)/(0.0678 - 0.000773)
q1 = 0.009
Now, from the second table i attached, and by interpolation, we can find specific enthalpies. So,at 300 KPa,
h_liquid = 52.65 KJ/kg
And h_evap = 198.195 KJ/Kg
Now,
Enthalpy is; h1 = h_liquid + q1•h_evap
Thus, h1 = 52.65 + 0.009(198.195) = 54.43 KJ/Kg
Also, from the tables attached, by interpolation, at P = 300 KPa, T1 = 0.61°C
And also at P = 600 KPa, T2 = 21.55°C
Now, at 600 KPa, from the first table attached,
α_vap = 0.03433
α_liq = 0.00082
Now formula for quality is given as;
q2 = (α_tot - α_liq)/(α_vap - α_liq)
Thus, plugging in the relevant values to obtain;
q2 = (0.00139 - 0.00082)/(0.03433 - 0.00082)
q2 = 0.017
Now, from the second table i attached, and by interpolation, we can find specific enthalpies. So,at 600 KPa,
h_liquid = 81.5 KJ/kg
And h_evap = 180.95 KJ/Kg
Now,
Enthalpy is; h1 = h_liquid + q2•h_evap
Thus, h2 = 81.5 + 0.017(180.95) = 84.58 KJ/Kg
