A rigid container with a volume of 2 m3 contains twenty kilograms of refrigerant 134aat an initial pressure of 240 kPa. The container is cooled until the pressure of therefrigerant is 100 kPa. Determine:a) The initial temperature of the refrigerant; (5 marks)b) The final state (i.e., compressed liquid, saturated liquid, saturated mixtureetc.) of the refrigerant, giving your reasoning; and (3 marks)c) The heat transfer associated with the process. (14 marks)d) Sketch the process on a P-v diagram.
Question
A rigid container with a volume of 2 m3 contains twenty kilograms of refrigerant 134aat an initial pressure of 240 kPa. The container is cooled until the pressure of therefrigerant is 100 kPa. Determine:a) The initial temperature of the refrigerant; (5 marks)b) The final state (i.e., compressed liquid, saturated liquid, saturated mixtureetc.) of the refrigerant, giving your reasoning; and (3 marks)c) The heat transfer associated with the process. (14 marks)d) Sketch the process on a P-v diagram.
Solution
To answer this question, we need to use the properties of refrigerant 134a and the ideal gas law.
a) The initial temperature of the refrigerant can be found using the ideal gas law, which is PV = mRT, where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and T is the temperature. The specific gas constant for refrigerant 134a is 0.0815 kJ/kg.K.
So, T = PV / mR = (240 kPa * 2 m3) / (20 kg * 0.0815 kJ/kg.K) = 294.5 K.
b) The final state of the refrigerant can be determined by looking at the pressure-temperature (P-T) diagram for refrigerant 134a. At 100 kPa, the saturation temperature is about 246 K. Since the initial temperature is higher than this, the refrigerant is initially in the superheated vapor state. As the container is cooled and the pressure drops, the refrigerant will move towards the saturated vapor line. If the cooling is stopped before it reaches the saturated vapor line, it will remain a superheated vapor. If the cooling continues until it reaches the saturated vapor line, it will become a saturated vapor. If the cooling continues beyond this point, it will become a saturated mixture of liquid and vapor.
c) The heat transfer associated with the process can be found by using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, there is no work done because the container is rigid. So, the heat transfer is equal to the change in internal energy of the refrigerant. This can be found by using the specific heat capacity at constant volume (Cv) for refrigerant 134a, which is 0.717 kJ/kg.K.
So, Q = m * Cv * (Tfinal - Tinitial) = 20 kg * 0.717 kJ/kg.K * (246 K - 294.5 K) = -697.26 kJ.
d) On a P-v diagram, the process would start at the initial pressure and volume and move horizontally to the left as the volume remains constant and the pressure decreases. The final point would be at the final pressure and the same volume. The area under the curve represents the work done, which is zero in this case because the container is rigid. The process would appear as a vertical line moving downwards.
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