Refrigerant 134A is held in a large storage tank at 100 lbfi

Refrigerant 134A is held in a large storage tank at 100 lbf/in^2 at 50 F. A technician refills a 3.5 ft^3 cylinder to take on a service call. Initially, the cylinder contains 5 lb of refrigerant at 70 F. The technician opens a valve and lets refrigerant flow into the cylinder until the amount of refrigerant in the cylinder is 30 lb handsets the cylinder in the van. After a while, the temperature in the cylinder is 70 F again. Neglecting kinetic and potential energy effects, determine the amount of energy transfer by heat in Btu.

The technician has a tank containing a mass of m₁ = 5 lb = 2.268 kg

By referring the refrigirant tables, we find out that the specific enthalpy of the gas stored at 70F

h₁ = 258.9 kJ/kg

Hence the total enthalpy of the gas present in the container H₁ = m₁xh₁ = 2.268 x 258.9 = 587.1852 kJ

The problem states that the final mass of the gas inside the container is 30 lb.

Hence mass of the gas added m₂ = 25lb = 11.34 kg

Specific enthalpy of gas being added h₂ = 252.9 kJ/kg

Total enthalpy added to the container H₂ = m₂xh₂ = 11.34x252.9 = 2867.866 kJ/kg

Therefore, the total enthalpy of the tank just after the addition of the mass H_initial = H₁ + H₂ = 587.1852 + 2867.866 = 3455.0712 kJ

But after the mass settles down into a stable state, the final condition states that the mass of 30 lb is maintained at 70 F.

The total enthalpy for gas in that state H_final = m_finalxh_{at t=70F} = 13.6x258.9 = 3521.04 kJ/kg

Thus, the difference in enthalpy is = 3521.04 kJ/kg - 3455.0712 kJ/kg = 65.97 kJ/kg

Converting it into imperial units, it gives us the answer 62.52 btu. Hence, 62.52 btu of heat energy is added to the cylinder from the surroundings.

Refrigerant 134A is held in a large storage tank at 100 lbf/in^2 at 50 F. A technician refills a 3.5 ft^3 cylinder to take on a service call. Initially, the cyl

Refrigerant 134A is held in a large storage tank at 100 lbfi

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