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

Short Answer

Expert verified
Yes, COP for a refrigeration cycle can be less than or greater than one. For heat pump cycles, COP is generally greater than one.

Step by step solution

01

Understand the Coefficient of Performance (COP)

The coefficient of performance (COP) is a measure of efficiency for refrigeration and heat pump cycles. For a refrigeration cycle, it is defined as the ratio of the heat removed from the cold reservoir to the work input. For a heat pump cycle, it is the ratio of the heat delivered to the hot reservoir to the work input.
02

Formula for COP of Refrigeration Cycle

The COP for a refrigeration cycle is given by COPref=QcW where Qc is the heat removed from the cold reservoir and W is the work input.
03

COP for Refrigeration Cycle Being Less Than One

For the COP to be less than one in a refrigeration cycle, it means that the work input W is greater than the heat removed Qc. This is possible in real-world applications, especially in less efficient systems or under certain conditions.
04

COP for Refrigeration Cycle Being Greater Than One

For the COP to be greater than one in a refrigeration cycle, it means that the heat removed Qc is greater than the work input W. This is common for most practical refrigeration systems under normal operating conditions.
05

Formula for COP of Heat Pump Cycle

The COP for a heat pump cycle is given by COPhp=QhW where Qh is the heat delivered to the hot reservoir and W is the work input.
06

COP for Heat Pump Cycle Being Less Than One

For the COP to be less than one in a heat pump cycle, the work input W is greater than the heat delivered Qh. While this is theoretically possible, it is generally not the case in practical heat pump systems.
07

COP for Heat Pump Cycle Being Greater Than One

For the COP to be greater than one in a heat pump cycle, the heat delivered Qh is greater than the work input W. This is usually the case for most practical heat pump systems.

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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 a process that removes heat from a space or substance to keep it cool.
It is essential for devices like refrigerators and air conditioners. The efficiency of this cycle is measured by the Coefficient of Performance (COP). To understand COP, let's dive into its formula and practicality.
The COP of a refrigeration cycle is given by the formula:
COPref=QcW where Qc represents the heat removed from the cold area, and W is the work done by the cycle.

**COP Less Than One**
For the COP to be less than one, the energy needed (work input) is more than the heat extracted. This scenario can happen in systems with low efficiency or under challenging conditions where more work is needed to remove less heat.

**COP Greater Than One**
When the COP is greater than one, the heat extracted from the space is more than the energy used. This is the norm for most refrigeration systems, making them effective for cooling purposes.
Heat Pump Cycle
A heat pump cycle works similarly to a refrigeration cycle but with a different goal.
Instead of cooling a space, it aims to heat a space. It transfers heat from a cold area to a warmer area. The COP of a heat pump cycle measures its efficiency and is described by the formula:
COPhp=QhW Here, Qh is the heat delivered to the warm area, and W is the work input.

**COP Less Than One**
When the COP is less than one in a heat pump cycle, it implies more energy is used than the heat delivered. This situation is usually rare in practical situations, because efficient heat pumps typically deliver more heat than the energy they consume.

**COP Greater Than One**
Most heat pump systems have a COP greater than one. This indicates that they deliver more heat than the work input, which is a standard requirement for efficient heating. Heat pumps are widely used for providing warmth in colder climates due to their high efficiency.
Thermodynamic Efficiency
Thermodynamic efficiency is a broader concept that applies to both refrigeration and heat pump cycles.
It helps us understand how effectively a system converts input energy into useful work. For both cycles, efficiency is about how well they use work input to move heat.

To break it down:
* In a refrigeration cycle, high efficiency means more heat removal with less input work.
* In a heat pump cycle, high efficiency means more heat delivery with less input work.

**Key Points**
* Efficient systems use less energy to achieve desired heating or cooling.
* The higher the COP, the more efficient the system.
* Real-world factors can influence efficiency, like system design and operating conditions.

By understanding these concepts, you can better appreciate the mechanisms behind everyday devices like refrigerators and heat pumps. Efficient designs help save energy and costs, making them crucial for sustainability.

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Most popular questions from this chapter

During January, at a location in Alaska winds at 30C can be observed. Several meters below ground the temperature remains at 13C, however. An inventor claims to have devised a power cycle exploiting this situation that has a thermal efficiency of 10%. Discuss this claim.

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at temperatures TH and TC, would it be better to increase TH or decrease TC by equal amounts?

Two kilograms of water execute a Camot power cycle. During the isothermal expansion, the water is heated until it is a saturated vapor from an initial state where the pressure is 40 bar and the quality is 15%. The vapor then expands adiabatically to a pressure of 1.5 bar while doing 491.5 kJ/kg of work. (a) Sketch the cycle on pv coordinates. (b) Evaluate the heat and work for each process, in kJ. (c) Evaluate the thermal efficiency.

Two reversible power cycles are arranged in series. The first cycle receives energy by heat transfer from a reservoir at temperature TH and rejects energy to a reservoir at an intermediate temperature T. The second cycle receives the energy rejected by the first cycle from the reservoir at temperature T. and rejects energy to a reservoir at temperature TC lower than T. Derive an expression for the intermediate temperature T in terms of TH and TC when (a) the net work of the two power cycles is equal. (b) the thermal efficiencies of the two power cycles are equal.

For each kW of power input to an ice maker at steady state, determine the maximum rate that ice can be produced, in kg/h, from liquid water at 0C. Assume that 333 kJ/kg of energy must be removed by heat transfer to freeze water at 0C, and that the surroundings are at 20C.

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