10.08 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed. Use data from the steam tables.

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

answer:

i am not 100% sure but i think it is a.)

dont take my word for it

explanation: