Refrigerant 134-a enters the condenser of a heat pump at 800kPa and 35°C at a rate of 0.12kg/s and leaves at 800kPa as a saturated liquid. If the compressor consumes 3.5kW of power, determine: (a) Draw the schematic diagram of the above heat pump (b) the COP of the heat pump (c) the rate of heat absorption from the outside air.
Refrigerant 134-a enters the condenser of a heat pump at 800kPa and 35°C at a rate of 0.12kg/s and leaves at 800kPa as a saturated liquid. If the compressor consumes 3.5kW of power, determine: (a) Draw the schematic diagram of the above heat pump (b) the COP of the heat pump (c) the rate of heat absorption from the outside air.
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Question
![**Heat Pump Analysis**
**Problem Statement:**
Refrigerant 134-a enters the condenser of a heat pump at 800 kPa and 35°C at a rate of 0.12 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 3.5 kW of power, determine:
(a) Draw the schematic diagram of the above heat pump
(b) The COP (Coefficient of Performance) of the heat pump
(c) The rate of heat absorption from the outside air
---
**Solution Steps:**
1. **Schematic Diagram:**
- The schematic would typically illustrate the components of a heat pump system, including the compressor, condenser, expansion valve, and evaporator. The flow direction of the refrigerant and heat exchange processes would be labeled.
2. **COP Calculation:**
- COP (Coefficient of Performance) is calculated using the formula:
\[
\text{COP} = \frac{\text{Heat output}}{\text{Work input}}
\]
- Heat output can be determined from the energy balance of the condenser, while the work input is given as 3.5 kW.
3. **Rate of Heat Absorption:**
- The rate of heat absorption from the outside air can be calculated using the properties of Refrigerant 134-a and the mass flow rate, taking into account the energy balance in the evaporator.
**Note:**
- Detailed thermodynamic properties of Refrigerant 134-a at given conditions can be referenced from refrigerant tables or software.
- The necessary calculations require the use of enthalpy values before and after condensation.
This problem demonstrates the principles of thermodynamics in real-world applications of heat pumps and can enhance the understanding of energy systems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F448d285c-e61e-464b-87b7-b0cab95f7d42%2Ffac432a5-8b94-4f46-b742-9bedd9553149%2F9v24t0q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Heat Pump Analysis**
**Problem Statement:**
Refrigerant 134-a enters the condenser of a heat pump at 800 kPa and 35°C at a rate of 0.12 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 3.5 kW of power, determine:
(a) Draw the schematic diagram of the above heat pump
(b) The COP (Coefficient of Performance) of the heat pump
(c) The rate of heat absorption from the outside air
---
**Solution Steps:**
1. **Schematic Diagram:**
- The schematic would typically illustrate the components of a heat pump system, including the compressor, condenser, expansion valve, and evaporator. The flow direction of the refrigerant and heat exchange processes would be labeled.
2. **COP Calculation:**
- COP (Coefficient of Performance) is calculated using the formula:
\[
\text{COP} = \frac{\text{Heat output}}{\text{Work input}}
\]
- Heat output can be determined from the energy balance of the condenser, while the work input is given as 3.5 kW.
3. **Rate of Heat Absorption:**
- The rate of heat absorption from the outside air can be calculated using the properties of Refrigerant 134-a and the mass flow rate, taking into account the energy balance in the evaporator.
**Note:**
- Detailed thermodynamic properties of Refrigerant 134-a at given conditions can be referenced from refrigerant tables or software.
- The necessary calculations require the use of enthalpy values before and after condensation.
This problem demonstrates the principles of thermodynamics in real-world applications of heat pumps and can enhance the understanding of energy systems.
Expert Solution
Step 1
Given,
mass flow rate is,
power consumption by the compressor is,
For refrigerant 134-a
It enters in heat pump at, and temperature is .
So enthalpy using the table of refrigerant 134-a will be,
And it leaves from heat pump at, as a saturated liquid.
So quality will be, x = 0
And enthalpy will be,
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