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Chapter 5: Problem 60

A heat pump is under consideration for heating a research station located on an Antarctica ice shelf. The interior of the station is to be kept at \(15^{\circ} \mathrm{C}\). Determine the maximum theoretical rate of heating provided by a heat pump, in \(\mathrm{kW}\) per kW of power input, in each of two cases: The role of the cold reservoir is played by (a) the atmosphere at \(-20^{\circ} \mathrm{C}\), (b) ocean water at \(5^{\circ} \mathrm{C}\).

Short Answer

Expert verified
8.23 kW/kW for atmosphere at \(-20^{\circ}C\), 28.82 kW/kW for ocean water at \(5^{\circ}C\).

Step by step solution

01

Understand the Coefficient of Performance (COP)

The Coefficient of Performance (COP) for a heat pump is given by \[ \text{COP}_{\text{HP}} = \frac{T_{h}}{T_{h} - T_{c}} \] where \(T_{h}\) is the temperature of the hot reservoir (interior of the station) and \(T_{c}\) is the temperature of the cold reservoir (either the atmosphere or ocean water).
02

Convert temperatures to Kelvin

To use the formula for COP, convert the temperatures from Celsius to Kelvin. For case (a): \(T_{h} = 15^{\circ}C + 273.15 = 288.15K \) \(T_{c} = -20^{\circ}C + 273.15 = 253.15K \)For case (b): \(T_{h} = 15^{\circ}C + 273.15 = 288.15K \) \(T_{c} = 5^{\circ}C + 273.15 = 278.15K \)
03

Compute the COP for Case (a)

Plug the values into the COP formula for case (a): \[\text{COP}_{\text{HP,a}} = \frac{288.15}{288.15 - 253.15} = \frac{288.15}{35} \approx 8.23 \]
04

Compute the COP for Case (b)

Plug the values into the COP formula for case (b): \[\text{COP}_{\text{HP,b}} = \frac{288.15}{288.15 - 278.15} = \frac{288.15}{10} \approx 28.82 \]
05

Interpret the COP values

The COP represents the maximum theoretical rate of heating provided by the heat pump per kW of power input. Thus, the theoretical maximum rate of heating for case (a) is 8.23 kW/kW and for case (b) is 28.82 kW/kW.

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Key Concepts

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

Heat Pump
A heat pump is a device that transfers heat energy from a colder area (cold reservoir) to a warmer area (hot reservoir) using mechanical work. In simple terms, it makes a space warmer by absorbing heat from a source like the atmosphere or water. Heat pumps are essential in regions with extreme temperatures like Antarctica, as they help maintain a stable and comfortable indoor environment even when it's freezing outside. The efficiency of a heat pump is often measured using the Coefficient of Performance (COP). It's crucial to understand that the COP indicates how efficiently the device uses energy to transfer heat.
Thermal Efficiency
Thermal efficiency refers to how well a system converts input energy into useful output energy, often expressed as a percentage. For heat pumps, COP is a key indicator of this efficiency. Higher COP means better efficiency, which is ideal for reducing energy consumption and lowering costs. In the context of the exercise, finding the COP helps identify how effectively the heat pump can warm the research station. This ensures that the least amount of input energy provides the most heat, making the system as efficient as possible. Efficiency is a critical consideration, especially in remote areas like Antarctica, where resources are limited.
Temperature Conversion
Converting temperatures is an essential step in calculating the COP. The COP formula requires temperatures to be in Kelvin, not Celsius. This conversion is vital as Kelvin is the standard unit for absolute temperature in thermodynamic calculations. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example, converting the station's indoor temperature from 15掳C:
  • Indoor Temperature,
    - In Celsius: 15 + 273.15 = 288.15K
Similarly, we convert the cold reservoirs:
  • Atmosphere at -20掳C: -20 + 273.15 = 253.15K
  • Ocean water at 5掳C: 5 + 273.15 = 278.15K
Conversion ensures the accuracy needed for thermodynamic formulas.
Energy Transfer
Energy transfer in a heat pump involves moving heat from a cold reservoir to a hot reservoir. This process requires energy, commonly supplied by electric power. The higher the COP, the more efficient this energy transfer is, meaning less energy input for more heat output. For instance, a COP of 8.23 would mean that 1 kW of electrical energy input can generate 8.23 kW of heating power. Understanding this transfer helps in designing systems that are energy-efficient and cost-effective for heating. Efficient energy transfer is particularly vital in extreme conditions, ensuring comfort without excessive energy use.
Thermodynamic Cycles
The operation of a heat pump is based on thermodynamic cycles, specifically the refrigeration cycle. This cycle involves several steps: evaporation, compression, condensation, and expansion. Each step plays a critical role in transferring heat from the cold space to the warm space. Key Steps in the Cycle:
  • Evaporation: Absorbs heat from the cold reservoir.
  • Compression: Increases the temperature and pressure of the refrigerant.
  • Condensation: Releases heat into the warm reservoir.
  • Expansion: Lowers the temperature and pressure of the refrigerant.
    • Understanding these steps is crucial for comprehending how a heat pump works. This helps to better grasp how efficiency and COP are influenced by the different components and phases of the cycle.

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Most popular questions from this chapter

The thermal efficiency of a reversible power cycle operating between hot and cold reservoirs is \(20 \%\). Evaluate the coefficient of performance of (a) a reversible refrigeration cycle operating between the same two reservoirs. (b) a reversible heat pump cycle operating between the same two reservoirs.

A heat pump cycle is used to maintain the interior of a building at \(21^{\circ} \mathrm{C}\). At steady state, the heat pump receives energy by heat transfer from well water at \(9^{\circ} \mathrm{C}\) and discharges energy by heat transfer to the building at a rate of \(120,000 \mathrm{~kJ} / \mathrm{h}\). Over a period of 14 days, an electric meter records that \(1490 \mathrm{~kW} \cdot \mathrm{h}\) of electricity is provided to the heat pump. Determine (a) the amount of energy that the heat pump receives over the 14-day period from the well water by heat transfer, in \(\mathrm{kJ}\). (b) the heat pump's coefficient of performance. (c) the coefficient of performance of a reversible heat pump cycle operating between hot and cold reservoirs at \(21^{\circ} \mathrm{C}\) and \(9^{\circ} \mathrm{C}\).

By removing energy by heat transfer from a room, a window air conditioner maintains the room at \(22^{\circ} \mathrm{C}\) on a day when the outside temperature is \(32^{\circ} \mathrm{C}\). (a) Determine, in \(\mathrm{kW}\) per \(\mathrm{kW}\) of cooling, the minimum theoretical power required by the air conditioner. (b) To achieve required rates of heat transfer with practicalsized units, air conditioners typically receive energy by heat transfer at a temperature below that of the room being cooled and discharge energy by heat transfer at a temperature above that of the surroundings. Consider the effect of this by determining the minimum theoretical power, in \(\mathrm{kW}\) per \(\mathrm{kW}\) of cooling, required when \(T_{\mathrm{C}}=18^{\circ} \mathrm{C}\) and \(T_{\mathrm{H}}=36^{\circ} \mathrm{C}\), and compare with the value found in part (a).

A power cycle operating at steady state receives energy by heat transfer at a rate \(\dot{Q}_{\mathrm{H}}\) at \(T_{\mathrm{H}}=1800 \mathrm{~K}\) and rejects energy by heat transfer to a cold reservoir at a rate \(\dot{Q}_{\mathrm{C}}\) at \(T_{C}=600 \mathrm{~K}\). For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible. (a) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=100 \mathrm{~kW}\) (b) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{W}_{\text {cycle }}=250 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=200 \mathrm{~kW}\) (c) \(\dot{W}_{\text {cyde }}=350 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=150 \mathrm{~kW}\) (d) \(\dot{Q}_{\mathrm{H}}=500 \mathrm{~kW}, \dot{Q}_{\mathrm{C}}=200 \mathrm{~kW}\)

A refrigeration cycle has a coefficient of performance equal to \(75 \%\) of the value for a reversible refrigeration cycle operating between cold and hot reservoirs at \(-5^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), respectively. For operation at steady state, determine the net power input, in kW per kW of cooling, required by (a) the actual refrigeration cycle and (b) the reversible refrigeration cycle. Compare values.
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