In a vapor compression refrigeration cycle, the refrigerant enters the compressor in saturated vapor state at evaporator pressure, with specific enthalpy equal to 250 kJ/kg. The exit of the compressor is superheated at condenser pressure with specific enthalpy equal to 300 kJ/kg. At the condenser exit, the refrigerant is throttled to the evaporator pressure. The coefficient of performance (COP) of the cycle is 3. If the specific enthalpy of the saturated liquid at evaporator pressure is 50 kJ/kg, then the dryness fraction of the refrigerant at entry to evaporator is _________.

The correct answer is 0.25.

1. Analysis of the T-s Diagram:
Based on the provided temperature-entropy ($T-s$) diagram of the vapor compression refrigeration cycle:
• State 1 represents the refrigerant entering the compressor as a saturated vapor at the evaporator pressure.
• State 2 represents the superheated vapor exiting the compressor at the condenser pressure.
• State 3 represents the refrigerant exiting the condenser at the condenser pressure.
• State 4 represents the refrigerant entering the evaporator after passing through the throttling (expansion) valve at the evaporator pressure.

2. Given Data:
• Specific enthalpy at compressor inlet (saturated vapor at evaporator pressure), $h_1=250\text{kJ/kg}$
• Specific enthalpy at compressor exit, $h_2=300\text{kJ/kg}$
• Coefficient of performance (COP) of the cycle, $\text{COP}=3$
• Specific enthalpy of saturated liquid at evaporator pressure, $h_f=50\text{kJ/kg}$

3. Step-by-Step Derivation:

Step 3.1: Work input to the compressor ($w$)
The work input to the compressor is given by the difference in enthalpy between the inlet and outlet of the compressor:

$w=h_2-h_1$

Substituting the given values:

$w=300-250=50\text{kJ/kg}$

Step 3.2: Refrigerating effect ($\text{RE}$)
Using the definition of Coefficient of Performance ($\text{COP}$):

$\text{COP}=\frac{\text{Refrigerating Effect (RE)}}{\text{Work Input (w)}}$

Rearranging the equation to solve for $\text{RE}$:

$\text{RE}=\text{COP}\times w$

$\text{RE}=3\times 50=150\text{kJ/kg}$

Step 3.3: Specific Enthalpy at Evaporator Entry ($h_4$)
The refrigerating effect in the evaporator is the difference in enthalpy between the exit and entry of the evaporator:

$\text{RE}=h_1-h_4$

Substituting the values:

$150=250-h_4$

$h_4=250-150=100\text{kJ/kg}$

Step 3.4: Dryness Fraction ($x$) at Evaporator Entry
State 4 lies in the wet region at the evaporator pressure. The enthalpy of the mixture can be expressed as:

$h_4=h_f+x\times (h_g-h_f)$

Where:
• $h_f$ is the specific enthalpy of the saturated liquid at evaporator pressure ($50\text{kJ/kg}$).
• $h_g$ is the specific enthalpy of the saturated vapor at evaporator pressure (which is $h_1=250\text{kJ/kg}$).
• $x$ is the dryness fraction at state 4.

Substituting these values:

$100=50+x\times (250-50)$

$100=50+200x$

$200x=50$

$x=\frac{50}{200}=0.25$

Thus, the dryness fraction of the refrigerant at the entry to the evaporator is 0.25.