Refrigerant 134-a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa w quality of 20 percent and leaves at 100kPa and -26 C as shown in Figure 2. If the compressor consumes 600 W of power and t COP of the refrigerator is 1.2 determine: The mass flow rate of the refrigerant The rate of heat rejected to the kitchen air Condenser W. Expansion 100 kPa x-0.2 Compressor Evaporator e 100 kPa 26°C
Refrigerant 134-a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa w quality of 20 percent and leaves at 100kPa and -26 C as shown in Figure 2. If the compressor consumes 600 W of power and t COP of the refrigerator is 1.2 determine: The mass flow rate of the refrigerant The rate of heat rejected to the kitchen air Condenser W. Expansion 100 kPa x-0.2 Compressor Evaporator e 100 kPa 26°C
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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![**Refrigeration Cycle Analysis of R-134a in a Household Refrigerator**
Refrigerant 134-a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa with a quality of 20 percent and leaves at 100 kPa and -26°C, as shown in Figure 2. If the compressor consumes 600 W of power and the Coefficient of Performance (COP) of the refrigerator is 1.2, determine:
- The mass flow rate of the refrigerant
- The rate of heat rejected to the kitchen air
### System Description and Diagram
Figure 2 illustrates the refrigeration cycle with key components: the evaporator, compressor, condenser, and expansion valve. The refrigerant flow path and energy interactions are shown in the schematic.
1. **Evaporator**:
- **Inlet Conditions**: 100 kPa, 20% quality
- **Outlet Conditions**: 100 kPa, -26°C
- **Heat Absorption**: \( Q_L \)
2. **Compressor**:
- Consumes electrical power \( W_c = 600 \text{ W} \)
3. **Condenser**:
- **Heat Rejection**: \( Q_H \)
4. **Expansion Valve**:
- Expansion process is assumed to be isenthalpic.
### COP and Performance Calculations
The COP of the refrigerator is given as 1.2. The COP for a refrigeration cycle is defined as:
\[ \text{COP} = \frac{Q_L}{W_c} \]
Where \( Q_L \) is the heat absorbed by the refrigerant in the evaporator and \( W_c \) is the work input to the compressor.
### Calculation Steps
1. **Determine Heat Absorbed \( Q_L \)**:
Using the given COP and compressor work:
\[ Q_L = \text{COP} \times W_c \]
\[ Q_L = 1.2 \times 600 \text{ W} \]
\[ Q_L = 720 \text{ W} \]
2. **Mass Flow Rate**:
The mass flow rate can be calculated using the refrigerant properties and the heat absorbed in the evaporator. The enthalpy values at the specific states are needed (which can be obtained from refrigerant tables for](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24a60763-065d-4deb-b582-ca07220062f2%2F9c636db4-f1d1-4183-8bbf-705ab53ec56d%2Fgrv5ge7_processed.png&w=3840&q=75)
Transcribed Image Text:**Refrigeration Cycle Analysis of R-134a in a Household Refrigerator**
Refrigerant 134-a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa with a quality of 20 percent and leaves at 100 kPa and -26°C, as shown in Figure 2. If the compressor consumes 600 W of power and the Coefficient of Performance (COP) of the refrigerator is 1.2, determine:
- The mass flow rate of the refrigerant
- The rate of heat rejected to the kitchen air
### System Description and Diagram
Figure 2 illustrates the refrigeration cycle with key components: the evaporator, compressor, condenser, and expansion valve. The refrigerant flow path and energy interactions are shown in the schematic.
1. **Evaporator**:
- **Inlet Conditions**: 100 kPa, 20% quality
- **Outlet Conditions**: 100 kPa, -26°C
- **Heat Absorption**: \( Q_L \)
2. **Compressor**:
- Consumes electrical power \( W_c = 600 \text{ W} \)
3. **Condenser**:
- **Heat Rejection**: \( Q_H \)
4. **Expansion Valve**:
- Expansion process is assumed to be isenthalpic.
### COP and Performance Calculations
The COP of the refrigerator is given as 1.2. The COP for a refrigeration cycle is defined as:
\[ \text{COP} = \frac{Q_L}{W_c} \]
Where \( Q_L \) is the heat absorbed by the refrigerant in the evaporator and \( W_c \) is the work input to the compressor.
### Calculation Steps
1. **Determine Heat Absorbed \( Q_L \)**:
Using the given COP and compressor work:
\[ Q_L = \text{COP} \times W_c \]
\[ Q_L = 1.2 \times 600 \text{ W} \]
\[ Q_L = 720 \text{ W} \]
2. **Mass Flow Rate**:
The mass flow rate can be calculated using the refrigerant properties and the heat absorbed in the evaporator. The enthalpy values at the specific states are needed (which can be obtained from refrigerant tables for
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