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Magazine Chapter 5: Problem 29
R-410a enters the evaporator (the cold heat exchanger) in an air-conditioner unit at \(-20^{\circ} \mathrm{C}, x=\) \(28 \%\) and leaves at \(-20^{\circ} \mathrm{C}, x=1 .\) The COP of the refrigerator is 1.5 and the mass flow rate is 0.003 \(\mathrm{kg} / \mathrm{s} .\) Find the net work input to the cycle.
Short Answer
Expert verified
The net work input is calculated using enthalpy values from tables and the formula: \( W_{net,in} = \frac{Q_{evap}}{COP} \).
Step by step solution
01
Understand the Given Parameters
First, note that R-410a enters and exits the evaporator at -20掳C. The initial quality of the refrigerant is given as 28%, which means \(x=0.28\), and it leaves as saturated vapor \(x=1\). The coefficient of performance (COP) of the refrigerator is 1.5, and the mass flow rate is 0.003 kg/s.
02
Define COP for Refrigerators
The coefficient of performance (COP) for a refrigeration cycle is defined as the ratio of the heat absorbed in the evaporator \((Q_{evap})\) to the net work input \((W_{net,in})\), expressed as:\[ \text{COP} = \frac{Q_{evap}}{W_{net,in}} \]Given that the COP is 1.5.
03
Calculate the Enthalpy Change
Use the refrigerant tables for R-410a at -20掳C to find the enthalpies at the given conditions:- Find \(h_1\) when \(x=0.28\).- Find \(h_2\) when \(x=1\).The enthalpy change in the evaporator \(Q_{evap}\) is given by:\[ Q_{evap} = \dot{m} (h_2 - h_1) \]
04
Apply the COP Equation
Now that you have \(Q_{evap}\), use the COP equation to solve for \(W_{net,in}\):\[ W_{net,in} = \frac{Q_{evap}}{COP} \]
05
Calculate the Net Work Input
Substitute the values of \(Q_{evap}\) and COP into the equation from Step 4 to find the net work input:\[ W_{net,in} = \frac{0.003 \, \text{kg/s} \times (h_2 - h_1)}{1.5} \]
06
Obtain Numerical Values from Tables
Use appropriate R-410a refrigerant tables to collect values for \(h_1\) and \(h_2\) based on the quality values \(x=0.28\) and \(x=1\) at -20掳C.
07
Compute the Final Result
Calculate the net work input using the computed enthalpies and simplifications from the prior steps. Confirm the units and numerical accuracy.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Refrigeration Cycle
The refrigeration cycle is an essential concept in thermodynamics, crucial for understanding how refrigerators and air conditioning systems function. Simply put, it's a series of processes that remove heat from a specific area and expel it elsewhere, typically using a fluid called a refrigerant.
Here's how it works:
Here's how it works:
- The cycle begins with the evaporator, where the refrigerant absorbs heat from the space that needs to be cooled. This causes the refrigerant to evaporate into a gas.
- Next, the compressor increases the pressure of this gaseous refrigerant, raising its temperature.
- The high-pressure refrigerant then moves to the condenser, where it releases the absorbed heat into the environment and condenses back into a liquid.
- Finally, the refrigerant passes through an expansion valve, which reduces its pressure and temperature before returning to the evaporator, ready to start the cycle again.
Enthalpy Change
Enthalpy change is a key concept in the study of thermodynamics, reflecting energy changes during state transitions within systems like the refrigeration cycle. In our context, it's crucial for determining the heat absorbed in the evaporator.
- The enthalpy change (\[\Delta h = h_2 - h_1\]) represents the energy difference per unit mass between two states of the refrigerant inside the evaporator piping.
- For the case of the air-conditioner using R-410a, this change is pivotal in determining how much heat is absorbed from the targeted environment.
Coefficient of Performance (COP)
The Coefficient of Performance (COP) is an efficiency measure that is widely used in evaluating refrigeration and heat pump systems. It tells you how effective a system is in transfering heat compared to the work input required to perform that task.
The COP for a refrigeration cycle is given by:\[\text{COP} = \frac{Q_{evap}}{W_{net,in}}\]where \(Q_{evap}\) is the heat absorbed in the evaporator and \(W_{net,in}\) is the net work input to the system.
The COP for a refrigeration cycle is given by:\[\text{COP} = \frac{Q_{evap}}{W_{net,in}}\]where \(Q_{evap}\) is the heat absorbed in the evaporator and \(W_{net,in}\) is the net work input to the system.
- A higher COP indicates a more efficient refrigeration cycle, meaning it can absorb more heat per unit of work input.
- In the context of household refrigerators, a common COP might range between 1 to 3, due to their design constraints and operating conditions.
Refrigerant Properties
Refrigerants are specialized fluids that play a critical role in the refrigeration cycle due to their ability to absorb and release heat efficiently. R-410a is a common refrigerant with specific properties that make it suitable for air conditioning systems.
Key properties of refrigerants include:
Key properties of refrigerants include:
- Thermodynamic Properties: These include boiling point, freezing point, specific heat capacity, and enthalpy. They dictate how well the refrigerant performs under different temperature and pressure conditions.
- Chemical Stability: Effective refrigerants must resist chemical breakdown over time and under the pressures and temperatures encountered in the cycle.
- Environmental Impact: Modern refrigerants, like R-410a, are designed to minimize ozone depletion and have lower global warming potential compared to older refrigerants.
