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 )