10-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.

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.

coral.

explanation:

coral. many marine organisms that produce calcium carbonate shells or skeletons are negatively impacted by increasing co2 levels and decreasing ph in seawater.