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By removing energy by heat transfer from a room, a window air conditioner maintains the room at \(18^{\circ} \mathrm{C}\) on a day when the outside temperature is \(42^{\circ} \mathrm{C}\) (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 \(T_{\mathrm{C}}=21^{\circ} \mathrm{C}\) and \(T_{\mathrm{H}}=48^{\circ} \mathrm{C}\), and compare with the value found in part (a).

Step by step solution

01

- Understanding the Problem

An air conditioner cools a room to 18°C on a day when the outside temperature is 42°C. We need to determine the minimum theoretical power required per kW of cooling.

02

- Formulate the Carnot Efficiency Expression

For part (a), use the Carnot efficiency for a refrigerator between temperatures 18°C (inside) and 42°C (outside). The COP (Coefficient of Performance) for a Carnot refrigerator is given by \[\text{COP}_{\text{Carnot}} = \frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}}}\]where temperatures must be in Kelvin.

03

- Convert Temperatures to Kelvin

Convert the given temperatures to Kelvin. \[T_{\text{C}} = 18 + 273 = 291 \text{ K}\]\[T_{\text{H}} = 42 + 273 = 315 \text{ K}\]

04

- Calculate COP for Part (a)

Using the converted temperatures, calculate the COP:\[\text{COP}_{\text{Carnot}} = \frac{291}{315 - 291} = \frac{291}{24} \approx 12.13\]

05

- Determine Minimum Theoretical Power for Part (a)

The minimum theoretical power required per kW of cooling is the reciprocal of COP:\[\frac{1}{\text{COP}_{\text{Carnot}}} = \frac{1}{12.13} \approx 0.0824 \text{ kW per kW of cooling}\]

06

- Consider New Temperatures for Part (b)

Now consider the effect of new temperatures: \[T_{\text{C}} = 21^{\text{C}} = 294 \text{ K}\]\[T_{\text{H}} = 48^{\text{C}} = 321 \text{ K}\]

07

- Calculate COP for New Conditions in Part (b)

Using the new temperatures, calculate the new COP:\[\text{COP}_{\text{Carnot}} = \frac{294}{321 - 294} = \frac{294}{27} \approx 10.89\]

08

- Determine Minimum Theoretical Power for Part (b)

The minimum theoretical power required per kW of cooling for the new conditions is the reciprocal of the new COP:\[\frac{1}{\text{COP}_{\text{Carnot}}} = \frac{1}{10.89} \approx 0.0918 \text{ kW per kW of cooling}\]

09

- Compare the Results

Compare the power requirements for part (a) and part (b):For part (a): 0.0824 kW per kW of coolingFor part (b): 0.0918 kW per kW of coolingThe power required increases when cooling and heating temperatures are closer to the operational limits of the air conditioner.

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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 (COP) is a key concept in refrigeration and air conditioning. It measures the efficiency of a refrigeration system. Specifically, in the context of a Carnot refrigerator, the COP is given by the formula:
\[ \text{COP}_{\text{Carnot}} = \frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}}} \]
where:
- \( T_{\text{C}} \) is the temperature of the cooled space
- \( T_{\text{H}} \) is the temperature of the surroundings
All temperatures must be converted to Kelvin before applying the formula. This ratio tells us how much cooling effect is achieved per unit of work input. A higher COP indicates a more efficient system.

Heat Transfer

Heat transfer is the process of energy transfer from one body or system to another, because of a temperature difference. In refrigeration, heat is transferred from the interior of the refrigerator (or room) to the outside environment. This process is governed by the second law of thermodynamics.
There are three main modes of heat transfer:

• Conduction
• Convection
• Radiation

In the context of an air conditioner, the device absorbs heat from the indoor environment and releases it to the outside, maintaining the desired indoor temperature.

Theoretical Power Calculation

The theoretical power calculation in refrigeration involves determining the minimum power needed by the system to achieve a certain cooling effect. This is where the COP comes into play. Once the COP is known, the minimum power requirement (in kW) per kW of cooling can be calculated as:
\[ \text{Power} = \frac{1}{\text{COP}} \]
For example, in the exercise, once the COP of 12.13 was found for the initial condition, the power required was calculated as:
\[ \text{Power} = \frac{1}{12.13} \]This results in approximately 0.0824 kW per kW of cooling. Similarly, for the adjusted temperatures, the theoretical power calculation would yield different values due to the change in COP.

Temperature Conversion to Kelvin

In thermodynamics, calculations involving temperature differences require converting Celsius to Kelvin. Kelvin is the SI unit for temperature and it is essential for consistency in thermodynamic equations.
The conversion formula is:
\[ T(\text{K}) = T(\text{°C}) + 273.15 \]
Applying this to the exercise:
\[ T_{\text{C}} = 18 + 273 = 291 \text{ K} \]
\[ T_{\text{H}} = 42 + 273 = 315 \text{ K} \]
This ensures valid inputs to COP calculations and other thermodynamic equations.

Engineering Thermodynamics

Engineering thermodynamics deals with energy conversion and the principles governing such processes. It focuses on understanding and applying laws such as:

• The First Law of Thermodynamics - conservation of energy
• The Second Law of Thermodynamics - direction of energy transfer processes

For optimization, engineers use insights from thermodynamics to design systems like refrigerators and air conditioners that maximize efficiency. This includes minimizing theoretical power requirements while maintaining desired temperatures. Knowledge of concepts like COP and heat transfer helps in designing more effective thermal systems.