Answer:
Temperature = 0.605°C
Total enthalpy at 300kpa = 545.2 kJ
Total enthalpy at 600kpa = 846.45 kJ.
Explanation:
Checking the table for 134a pressure table. It is given that the specific volume of saturated liquid and the specific volume of the saturated vapor of 280kpa is 0.0007699 m^3/kg and 0.072352 m^3/kg respectively.
Also, the specific volume of saturated liquid and the specific volume of the saturated vapor of 320kpa is 0.0007772 m^3/kg and 0.063604 m^3/kg respectively.
The first thing to do is to determine the value for the specific volume of saturated liquid.
At 300 kpa, the specific volume of saturated liquid,n is given below as;
300 - 280/ 320 - 280 = (n - 0.0007699)/ 0.0007772 - 0.0007699.
Therefore, n = specific volume of saturated liquid = 0.0007735 m^3/kg.
300 - 280/ 320 - 280 = n - 0.072352/ (0.063604 - 0.072352).
n = 0.0679 m^3/kg.
The second thing to do is to determine the value of the specific volume.
Specific volume = 14 × 10^-3/ 10 = 0.0014 m^3/kg.
Determine the enthalpy of the mixture,b(I). This is given below as;
300 - 280/ 320 - 280 = b(I) - 199.54/ (196.7 - 199.54).
b(I) = 198.125 kJ/Kg.
Hence, b = [ 300 - 280/ 320 - 280 = j - 50.18 / 55.16 - 50.18] + [ ( 0.0014 - 0.00077735) / 0.067978 - 0.00077735] × 198.125.
b = 54.517 KJ/Kg.
Total enthalpy = 10 × 54.517 = 545.17 kJ.
Temperature can be Determine as below;
300 - 280/ 320 - 280 = T + 1.25 / 2.46 - 1.25.
Temperature = 0.605°C.
Hence, at 600kpa, the total enthalpy = [81.51 + ( 0.0014 - 0.0008199/ 0.034295 - 0.0008199) × 180.90] × 10
Total enthalpy at 600kpa = 846.45 kJ.
