Question

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -30C by rejecting its waste heat to cooling water that enters the condenser at 18 C at a rate of 0.25 kg/s and leaves at 26C. The refrigerant enters the condenser at 1.2 MPa and 65 C and leaves at 42 C. The inlet state of the compressor is 60 kPa and 234 C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine:
a. the quality of the refrigerant at the evaporator inlet.
b. the refrigeration load.
c. the COP of the refrigerator
d. the theoretical maximum refrigeration load for the same power input to the compressor.

Answer 1

Answer:

a) x = 0.4795

b) QL = 5.85 KW

c) COP = 2.33

d) QL_max = 12.72 KW

Explanation:

Solution:-

- Assuming the steady state flow conditions for both fluids R-134a and water.

- The thermodynamic properties remain constant for respective independent intensive properties.

- We will first evaluate the state properties of the R-134a and water.

- Compressor Inlet, ( Saturated Vapor ) - Ideal R-134a vapor cycle

              P1 = 60 KPa, Tsat = -36.5°C  

              T1 = -34°C , h1 = hg = 230.03 KJ/kg

              Qin = 450 W - surrounding heat  

- Condenser Inlet, ( Super-heated R-134a vapor ):

              P2 = 1.2 MPa , Tsat = 46.32°C  

              T2 = 65°C   , h2 = 295.16 KJ/kg

- Condenser Outlet, ( Saturation R-134a point ):

             P3 = P2 = 1.2 MPa , Tsat = 46.32°C

             T3 = 42°C   , h3 = hf = 111.23 KJ/kg

- R-134a is throttled to the pressure of P4 = compressor pressure = P1 = 60 KPa by an "isenthalpic - constant enthalpy pressure reduction" expansion valve.

- Inlet of Evaporator - ( liquid-vapor state )

             P4 = P1 = 60 KPa, hf = 3.9 KJ/kg , hfg = 223.9 KJ/kg

             h4 = h3 = 111.23 KJ/kg

- The quality ( x ) of the liquid-vapor R-134a at evaporator inlet can be determined:

             x4 = ( h4 - hf ) / hfg

             x4 = ( 111.23 - 3.9 ) / 223.9

             x4 = 0.4795      Answer ( a )        

- Water stream at a flow rate flow ( mw ) = 0.25 kg/s is used to take away heat from the R-134a.

- Condenser Inlet, ( Saturated liquid water ):

             Ti = 18°C , h = hf = 75.47 KJ/kg  

- Condenser Outlet, ( Saturated liquid water ):

             To = 26°C , h = hf = 108.94 KJ/kg

- Since the heat of R-134a was exchanged with water in the condenser. The amount of heat added to water (Qh) is equal to amount of heat lost from refrigerant R-134a.

- Apply thermodynamic balance on the R-134a refrigerant in the condenser:

             Qh = flow (mr) * [ h2 - h3 ]

Where,

flow ( mr ) : The flow rate of R-134a gas in the refrigeration cycle

             flow ( mr ) = Qh / [ h2 - h3 ]

             flow ( mr ) = 8.3675 / [ 295.16 - 111.23 ]

             flow ( mr ) = 0.0455 kg/s

- The cooling load of the refrigeration cycle ( QL ) is determined from energy balance of the cycle net work input ( Compressor work input ) - "Win" and the amount of heat lost from R-134a in condenser ( Qh ).

- Apply the thermodynamic balance for the compressor:

           

            Win = flow ( mr )*[ h2 - h1 ] - Qin

            Win = 0.0455*[ 295.16 - 230.03] KW - 0.45 KW

            Win = 2.513 KW

- The cooling load ( QL ) for the refrigeration cycle can now be calculated. Apply thermodynamic balance for the refrigeration cycle:

            QL = Qh - Win

            QL = 8.3675 - 2.513

            QL= 5.85 KW  .... Refrigeration Load, Answer ( b )

- The COP of the refrigeration cycle is calculated as the ratio of useful work and total work input required:

           

             COP = QL / Win

             COP = 5.85 / 2.513

             COP =  2.33      Answer ( c )            

- For a compressor to be working at 100% efficiency or ideal then the maximum COP for the refrigeration cycle would be:

           

             COP_max = [ TL ] / [ Th - TL ]

Where,

            TL : The absolute temperature of heat sink, refrigerated space

            TH : The absolute temperature of heat source, water inlet

                 

            COP_max = [ -30+273 ] / [ (18+273) - (-30+273) ]          

            COP_max = 5.063

- The theoretical ideal refrigeration load ( QL max ) would be:

     

           COP_max = QL_max / Win

           QL_max = Win*COP_max

           QL_max = 2.513*5.063

           QL_max = 12.72 KW     Answer ( d )