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An inventor claims to have developed a refrigerator that at steady state requires a net power input of \(0.54 \mathrm{~kW}\) to remove \(12,800 \mathrm{~kJ} / \mathrm{h}\) of energy by heat transfer from the freezer compartment at \(-20^{\circ} \mathrm{C}\) and discharge energy by heat transfer to a kitchen at \(27^{\circ} \mathrm{C}\). Evaluate this claim.

Short Answer

Expert verified
The inventor's claim is invalid because the actual COP exceeds the Carnot COP.

Step by step solution

01

Understanding the Problem

First, identify the main components of the problem. We are given the power input to the refrigerator, the rate of heat removal from the freezer compartment, and the temperatures of the freezer and the kitchen. We need to evaluate the claim by comparing the actual performance to the theoretical maximum performance of the refrigerator.
02

Calculate the Coefficient of Performance (COP)

The coefficient of performance (COP) for a refrigerator is given by \[ COP_{actual} = \frac{Q_L}{W} \]where \( Q_L = 12,800 \frac{\mathrm{kJ}}{\mathrm{h}} \)converted to kilowatts (\( \mathrm{kW} \)) is \[ Q_L = \frac{12,800 \mathrm{~kJ}}{3600 \mathrm{~s}} = 3.56 \mathrm{~kW} \]and \( W = 0.54 \mathrm{~kW} \). Substituting in the values, \[ COP_{actual} = \frac{3.56 \mathrm{~kW}}{0.54 \mathrm{~kW}} = 6.59 \]
03

Calculate the Maximum Possible COP (Carnot COP)

The maximum possible COP for a refrigerator (Carnot COP) between two temperatures is given by \[ COP_{Carnot} = \frac{T_L}{T_H - T_L} \]where \( T_L = -20^{\circ} \mathrm{C} = 253 \mathrm{~K} \) and \( T_H = 27^{\circ} \mathrm{C} = 300 \mathrm{~K} \). Substituting in the values, \[ COP_{Carnot} = \frac{253 \mathrm{~K}}{300 \mathrm{~K} - 253 \mathrm{~K}} = \frac{253 \mathrm{~K}}{47 \mathrm{~K}} = 5.38 \]
04

Compare Actual COP with Carnot COP

Compare the actual COP to the Carnot COP. Here, \( COP_{actual} = 6.59 \) and \( COP_{Carnot} = 5.38 \). Since the actual COP exceeds the Carnot COP, which represents the maximum theoretical efficiency, the inventor's claim is not possible. Thus, the claim is not valid.

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Key Concepts

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

Coefficient of Performance (COP)
The Coefficient of Performance, or COP, is a measure of a refrigerator's efficiency.
It tells us how well a refrigerator is using energy to transfer heat from the freezer to the kitchen. You can calculate COP using the formula:
\[ COP = \frac{Q_L}{W} \]where Q_L is the heat removed from the freezer compartment, and W is the work input to the refrigerator.
In our exercise, Q_L is given as 12,800 kJ/h. First, convert this to kW because power is usually measured in kilowatts:
\[ Q_L = \frac{12,800 \text{ kJ}}{3600 \text{ s}} = 3.56 \text{ kW} \]Next, the power input W is provided directly as 0.54 kW. By substituting these values into the COP formula, we get:
\[ COP_{actual} = \frac{3.56 \text{ kW}}{0.54 \text{ kW}} = 6.59 \]This COP of 6.59 will be compared to the theoretical maximum COP later on.
Carnot Refrigerator
The Carnot Refrigerator is a theoretical model for the most efficient refrigerator possible.
Named after the French physicist Sadi Carnot, it represents an ideal cycle where maximum efficiency is achieved with no energy losses.
To find the theoretical COP for a Carnot Refrigerator, we use the temperatures of the freezer and the kitchen expressed in Kelvin. This can be calculated using:
\[ COP_{Carnot} = \frac{T_L}{T_H - T_L} \]Here, T_L is the temperature of the freezer, and T_H is the temperature of the kitchen. Converting the temperatures from Celsius to Kelvin:
T_L = -20°C = 253 K
T_H = 27°C = 300 K
Substitute these temperatures into the Carnot COP formula:
\[ COP_{Carnot} = \frac{253 \text{ K}}{300 \text{ K} - 253 \text{ K}} = \frac{253}{47} = 5.38 \]The theoretical maximum COP, or COP_{Carnot}, is 5.38.
Thermal Efficiency
Thermal efficiency in refrigeration isn't directly about converting heat to work but rather about efficiently moving heat.
In refrigeration, we look at the COP to determine the efficiency. A higher COP means higher thermal efficiency as it implies more heat removal for less work input.
In practical scenarios, achieving the theoretical maximum COP like that of a Carnot refrigerator is impossible due to real-world inefficiencies such as friction and other non-ideal factors.
In our exercise, comparing the actual COP (6.59) and the Carnot COP (5.38) helps us understand the potential deviations in real systems versus ideal ones.
Second Law of Thermodynamics
The Second Law of Thermodynamics states that energy has quality as well as quantity.
Processes occur in a certain direction and energy tends to disperse. According to this law, no real process can be perfectly efficient.
This law sets the upper limit on the efficiency of any heat engine or refrigerator.
For refrigerators, this means that the actual COP can never exceed the Carnot COP. This explains why the claim of the inventor in our exercise cannot be valid: the actual COP of 6.59 exceeds the theoretical limit of 5.38, violating the second law. This indicates that the inventor is making an unrealistic claim.
Power Input Calculation
Calculating the power input is crucial for evaluating refrigerator efficiency.
The power input is the amount of electrical work required to remove a certain amount of heat from the fridge.
In our exercise, the inventor claims that the refrigerator requires a net power input of 0.54 kW. We used this value explicitly in our COP calculations.
If the COP is unrealistic as determined by our comparison with Carnot COP, the power input itself may need reevaluation.
Recalculating these values with more practical considerations and proper assumptions would help in assessing the true efficiency and power needs of refrigeration systems.

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Most popular questions from this chapter

A cyclic heat engine operates between a source temperature of \(700^{\circ} \mathrm{C}\) and a sink temperature of \(25^{\circ} \mathrm{C}\). What is the least rate of heat rejection per \(\mathrm{kW}\) net output of the engine?

Determine the minimum theoretical power, in \(\mathrm{kJ} / \mathrm{s}\), required at steady state by a refrigeration system to maintain a cryogenic sample at \(-126^{\circ} \mathrm{C}\) in a laboratory at \(21^{\circ} \mathrm{C}\), if energy leaks by heat transfer to the sample from its surroundings at a rate of \(0.09 \mathrm{~kJ} / \mathrm{s}\).

According to an inventor of a refrigerator, the refrigerator can remove heat from the freezer compartment at the rate of \(13,000 \mathrm{~kJ} / \mathrm{h}\) by net input power consumption of \(0.65 \mathrm{~kW}\). Heat is discharged into the room at \(23^{\circ} \mathrm{C}\). The temperature of freezer compartment is \(-15^{\circ} \mathrm{C}\). Evaluate this claim.

To maintain a dwelling at a temperature of \(20^{\circ} \mathrm{C}, 600 \mathrm{MJ}\) energy is required per day when the temperature outside the dwelling is \(4^{\circ} \mathrm{C}\). A heat pump is used to supply this energy. Determine the minimum theoretical cost to operate the heat pump, in \(\$ /\) day, if the cost of electricity is \(\$ 0.10\) per \(\mathrm{kW} \cdot \mathrm{h}\).

Is it possible for the coefficient of performance of a refrigeration cycle to be less than \(1 ?\) To be greater than \(1 ?\) Answer the same questions for a heat pump cycle.

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