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Consider an ideal refrigeration cycle that has a condenser temperature of \(45^{\circ} \mathrm{C}\) and an evaporator temperature of \(-15^{\circ} \mathrm{C}\). Determine the coefficient of performance of this refrigerator for the working fluids \(R-12\) and \(R-22\).

Short Answer

Expert verified
The coefficient of performance (COP) of the refrigerator for the working fluids R-12 and R-22, under ideal conditions, is 3.99

Step by step solution

01

Understanding Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of efficiency of a refrigerator or heat pump. It is calculated as the ratio of refrigeration effect to the amount of work input. In the case of reversed Carnot cycle, which is an ideal model for refrigeration cycle, the COP is inversely proportional to the difference in temperatures of the condenser and evaporator expressed in Kelvin.
02

Conversion of temperatures to Kelvin

Before proceeding with the calculations of COP, we need to convert the given Condenser (Tc) temperature from Celsius to Kelvin and the evaporator (Te) temperature from Celsius to Kelvin. The formula to convert Celsius temperature to Kelvin is K = C + 273.15. Thus, Tc = 45 + 273.15 = 318.15 K, Te = -15 + 273.15 = 258.15 K.
03

Calculation of the Coefficient of Performance of the refrigerator

The Coefficient of Performance (COP) for any fridge or heat pump operating under ideal conditions based on a reversal of Carnot cycle is given by the formula: COP = Te / (Tc - Te). Using this formula, we can calculate the COP for the given refrigerator. COP = Te / (Tc - Te) = 258.15 / (318.15 - 258.15) = 3.99. This is the ideal Coefficient of Performance of the refrigerator for both R-12 and R-22 refrigerants as the COP calculation under ideal conditions doesn't depend on the type of refrigerant used. In reality, the actual Coefficient of Performance would depend on the characteristics of the working fluid or refrigerant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Performance
The Coefficient of Performance, commonly known as the COP, is a key metric that describes how efficiently a refrigeration cycle operates. The COP is essentially a ratio of the useful refrigeration effect, or the heat removed from the cold space, to the work input required to achieve that effect. This means a higher COP indicates a more efficient system.

To calculate the COP, we use the formula:
  • COP = \(\frac{Q_c}{W} = \frac{T_e}{T_c - T_e}\)
This equation shows that the COP is inversely related to the temperature difference between the condenser (\(T_c\)) and the evaporator (\(T_e\)). When the temperature difference is smaller, the refrigeration system requires less work, thereby increasing its efficiency.

In real-world applications, the actual COP will vary depending on many factors including the type of refrigerant used and how closely the system operates to the ideal cycle.
Carnot Cycle
The Carnot Cycle serves as a fundamental theoretical model for understanding refrigeration systems. Named after French physicist Sadi Carnot, this ideal reversible cycle provides the maximum possible efficiency a refrigerator or a heat pump can achieve under given conditions.

In the Carnot cycle, the refrigeration process consists of:
  • Isothermal Expansion: Absorbs heat at a constant low temperature from the refrigerated space.
  • Adiabatic Compression: Elevates the temperature of the refrigerant without exchanging heat with its surroundings.
  • Isothermal Compression: Releases heat at a constant higher temperature to the environment.
  • Adiabatic Expansion: Lowers the refrigerant's temperature, preparing it to absorb more heat in the next cycle.
These stages form a loop that illustrates how an ideal cycle can convert work input into a refrigeration effect. Understanding the Carnot Cycle is crucial because it sets a benchmark for real systems—allowing engineers to gauge the efficiency of practical refrigeration cycles against this ideal standard.

Although no real system can achieve a Carnot cycle completely, it guides the design and optimization of refrigeration systems.
Refrigerants R-12 and R-22
Refrigerants are crucial in the operation of refrigeration and air conditioning systems. Among them, R-12 and R-22 have historically been popular choices, each with distinct properties that affect their application and efficiency.

  • R-12: This refrigerant was widely used in residential and automotive air conditioning systems for many years. It is known for its excellent thermodynamic properties, however, its use has been largely phased out due to environmental concerns, particularly its ozone-depleting potential.

  • R-22: Slightly different in composition, R-22 has also been a popular refrigerant, especially in air conditioning units. Despite its high efficiency and compatibility with various system components, R-22 is being replaced due to its contribution to ozone depletion and new regulations encouraging environmentally friendly alternatives.
In theoretical calculations, such as determining the COP under ideal conditions, the specific choice of refrigerant like R-12 or R-22 doesn't affect the outcome; the COP primarily depends on the temperatures of the system. However, in actual systems, the refrigerant type plays a significant role in overall performance and environmental impact.

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Most popular questions from this chapter

The power plant shown in Fig. 11.40 combines a gas-turbine cycle and a steam- turbine cycle. The following data are known for the gas-turbine cycle. Air enters the compressor at \(100 \mathrm{kPa}\) \(25^{\circ} \mathrm{C},\) the compressor pressure ratio is \(14,\) and the isentropic compressor efficiency is \(87 \%\); the heater input rate is \(60 \mathrm{MW}\); the turbine inlet temperature is \(1250^{\circ} \mathrm{C}\), the exhaust pressure is \(100 \mathrm{kPa},\) and the isentropic turbine efficiency is \(87 \%\); the cycle exhaust temperature from the heat exchanger is \(200^{\circ} \mathrm{C}\). The following data are known for the steam-turbine cycle. The pump inlet state is saturated liquid at \(10 \mathrm{kPa}\), the pump exit pressure is \(12.5 \mathrm{MPa}\), and the isentropic pump efficiency is \(85 \%\); turbine inlet temperature is \(500^{\circ} \mathrm{C}\), and the isentropic turbine efficiency is \(87 \% .\) Determine a. The mass flow rate of air in the gas-turbine cycle. b. The mass flow rate of water in the steam cycle c. The overall thermal efficiency of the combined cycle.

Consider an ideal Stirling-cycle engine in which the pressure and temperature at the beginning of the isothermal compression process are \(14.7 \mathrm{lbf} / \mathrm{in} .^{2}, 80 \mathrm{F}\), the compression ratio is \(6,\) and the maximum temperature in the cycle is 2000 F. Calculate the maximum pressure in the cycle and the thermal efficiency of the cycle with and without regenerators.

To approximate an actual spark-ignition engine, consider an air-standard Otto cycle that has a heat addition of \(1800 \mathrm{kJ} / \mathrm{kg}\) of air, a compression ratio of \(7,\) and a pressure and temperature at the beginning of the compression process of \(90 \mathrm{kPa}\) and \(10^{\circ} \mathrm{C}\). Assuming constant specific heat, with the value from Table A.5, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle, and the mean effective pressure.

A steam power plant operating in an ideal Rankine cycle has a high pressure of \(5 \mathrm{MPa}\) and a low pressure of 15 kPa. The turbine exhaust state should have a quality of at least \(95 \%\), and the turbine power generated should be \(7.5 \mathrm{MW}\). Find the necessary boiler exit temperature and the total mass flow rate.

A refrigerator in a laboratory uses \(R-22\) as the working substance. The high pressure is 1200 \(\mathrm{kPa},\) the low pressure is \(201 \mathrm{kPa},\) and the compressor is reversible. It should remove \(500 \mathrm{W}\) from a specimen currently at \(-20^{\circ} \mathrm{C}\) (not equal to \(T_{L}\) in the cycle) that is inside the refrigerated space. Find the cycle COP and the electrical power required.

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