Thermo Question - Fridge cycles

[Jacob87411]

Consider a 300 kJ/min refrigeration system that operates on an ideal vapor-compression cycle with refrigerant-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 140 kPa and is compressed to 800 kPa. Determine the (a) quality of the refrigerant at the end of the throttling process, (b) the coefficient of performance, and (c) the power input into the compressor.

Homework Equations

In a refigeration cycle I am considering state 1 after the evaporator, 2 after the compressor, 3 after the condenser and 4 before the evaporator.

P1=140 kPa, x (quality)=1 since its a saturated vapor.
P2=800 kPa, x=1

W=m(h2-h1)

The Attempt at a Solution

300 kJ/min = 5KW

In an ideal refigeration cycle isn't the only work considered the work for the compressor, so isn't the answer to (c) 5KW?

I'm having the most problems with (a) because I am unsure to begin. I can find other states such as enthalpies with the pressure I have been given, is that the correct approach?

 

[Nino MMP]

(a) The quality of the refrigerant at the end of the throttling process can be determined using the ideal gas law. Since the pressure is known (P2 = 800 kPa), the specific volume can be calculated using the ideal gas law:v2 = R * T2/P2, where R is the gas constant for refrigerant-134a, and T2 is the temperature at state 2. The quality is then given by x = (v_g - v_f)/(v_g - v_f), where v_g and v_f are the specific volumes of the saturated vapor and saturated liquid, respectively. (b) The coefficient of performance is defined as the ratio of the net work output to the input energy required by the compressor. In this case, the net work output is 300 kJ/min, while the input energy required by the compressor is 5 kW. Therefore, the COP is 60 kJ/kW = 0.6.(c) The power input into the compressor is 5 kW.

 

[Blueshift5]

Yes, the work input into the compressor is 5 kW, as it is the only part of the cycle that requires work. To determine the quality of the refrigerant at the end of the throttling process (state 4), we can use the fact that the throttling process is isenthalpic (meaning there is no change in enthalpy). This means that the enthalpy at state 4 will be equal to the enthalpy at state 3 (after the condenser). Using the pressure and enthalpy values at state 3, we can determine the enthalpy at state 4. From there, we can use the saturated vapor table for refrigerant-134a to find the corresponding temperature and quality at state 4.

The coefficient of performance (COP) for a refrigeration cycle is defined as the ratio of the desired output (cooling effect) to the required input (work input). In this case, the COP can be calculated as the ratio of the cooling effect (300 kJ/min) to the work input (5 kW), giving a COP of 60.

To summarize:
(a) To find the quality at state 4, use the isenthalpic condition and the saturated vapor table for refrigerant-134a.
(b) The COP can be calculated as the ratio of the cooling effect to the work input.
(c) The power input into the compressor is the only work input in an ideal refrigeration cycle, so it is equal to 5 kW.