Question

A refrigeration cycle having a coefficient of performance of 3 maintains a computer laboratory at \(18^{\circ} \mathrm{C}\) on a day when the outside temperature is \(30^{\circ} \mathrm{C}\). The thermal load at steady state consists of energy entering through the walls and windows at a rate of \(30,000 \mathrm{~kJ} / \mathrm{h}\) and from the occupants, computers, and lighting at a rate of \(6000 \mathrm{~kJ} / \mathrm{h}\). Determine the power required by this cycle and compare with the minimum theoretical power required for any refrigeration cycle operating under these conditions, each in \(\mathrm{kW}\).

Short answer

3.33 kW vs 0.41 kW

Step-by-step solution

Understand the Coefficient of Performance (COP)

The Coefficient of Performance (COP) of a refrigeration cycle is given by \[ \text{COP} = \frac{Q_L}{W} \]where \( Q_L \) is the heat removed from the cooled space, and \( W \) is the work input. The COP is given as 3 in this problem.

Calculate Total Thermal Load

The total thermal load, \( Q_L \), is the sum of energy entering through walls and windows and the energy from occupants, computers, and lighting. This is given as: \[ Q_L = 30000 \frac{\text{kJ}}{\text{h}} + 6000 \frac{\text{kJ}}{\text{h}} = 36000 \frac{\text{kJ}}{\text{h}} \]Convert this to kW: \[ Q_L = \frac{36000}{3600} \text{ kW} = 10 \text{ kW} \]

Determine Actual Power Required

Using the COP definition, the actual power required by the refrigeration cycle is given by \[ W = \frac{Q_L}{\text{COP}} \]Substituting the values, \[ W = \frac{10 \text{ kW}}{3} = 3.33 \text{ kW} \]

Calculate the Minimum Theoretical Power

The minimum theoretical power required for any refrigeration cycle is determined using the Carnot COP, which is given by \[ \text{COP}_{\text{Carnot}} = \frac{T_L}{T_H - T_L} \]where \( T_L \) is the absolute temperature of the cold reservoir (18°C) and \( T_H \) is the absolute temperature of the hot reservoir (30°C). Convert these temperatures to Kelvin: \[ T_L = 18 + 273 = 291 \text{ K} \]\[ T_H = 30 + 273 = 303 \text{ K} \]Then the Carnot COP is: \[ \text{COP}_{\text{Carnot}} = \frac{291}{303 - 291} = \frac{291}{12} = 24.25 \]The minimum theoretical power is \[ W_{\text{min}} = \frac{Q_L}{\text{COP}_\text{Carnot}} = \frac{10 \text{ kW}}{24.25} = 0.41 \text{ kW} \]

Compare the Powers

Compare the actual power required to the minimum theoretical power: - Actual power required: 3.33 kW - Minimum theoretical power: 0.41 kW

Key concepts

Coefficient of Performance (COP)

The Coefficient of Performance, or COP, is a crucial parameter in refrigeration systems. It measures the efficiency of a refrigeration cycle by comparing the amount of heat removed from the cooled space to the work input required to remove it. Mathematically, it is expressed as: \( \text{COP} = \frac{Q_L}{W} \)
where \( Q_L \) is the heat removed from the cooled space, and \( W \) is the work input. The higher the COP, the more efficient the refrigeration system.
In our problem, the COP is given as 3. This means that for every unit of work input, three units of heat are removed from the cooled space. This simple equation helps us understand how well the refrigeration system uses energy to maintain the desired temperature.

Thermal Load Calculation

Thermal load calculation involves determining the total amount of heat that needs to be removed from a space to maintain a specific temperature. In our exercise, the thermal load consists of energy entering through the walls and windows, as well as the energy generated by occupants, computers, and lighting.
To find the total thermal load \( Q_L \), we sum these individual heat loads:
\( Q_L = 30000 \frac{ \text{kJ}}{ \text{h}} + 6000 \frac{ \text{kJ}}{ \text{h}} = 36000 \frac{ \text{kJ}}{ \text{h}} \)
It's often useful to convert the thermal load from kJ/h to kW by dividing by 3600 (the number of seconds in an hour):
\( Q_L = \frac{36000}{3600} \text{ kW} = 10 \text{ kW} \)
This total thermal load is crucial for designing the cooling system and determining the power required to maintain the desired temperature.

Carnot COP

The Carnot Coefficient of Performance (COP) represents the highest possible efficiency for a refrigeration cycle operating between two temperatures. It's derived from the Second Law of Thermodynamics and is a benchmark for the theoretical limit of efficiency. The formula for Carnot COP is: \( \text{COP}_{\text{Carnot}} = \frac{T_L}{T_H - T_L} \)
Here, \( T_L \) is the absolute temperature of the cold reservoir and \( T_H \) is the absolute temperature of the hot reservoir. Remember to convert temperatures to Kelvin by adding 273. For instance, for our problem:

• \( T_L = 18^{\circ} \mathrm{C} + 273 = 291 \text{ K} \)
• \( T_H = 30^{\circ} \mathrm{C} + 273 = 303 \text{ K} \)

Plugging these values into the formula gives us: \( \text{COP}_{\text{Carnot}} = \frac{291}{303 - 291} = \frac{291}{12} = 24.25 \) This means that, theoretically, for every unit of work input, 24.25 units of heat can be removed from the cooled space. Using this ideal COP, we can calculate the minimum theoretical power required: \( W_{\text{min}} = \frac{ Q_L}{ \text{COP}_{\text{Carnot}} } = \frac{10 \text{ kW}}{24.25} = 0.41 \text{ kW} \)
Comparing this with the actual power required (3.33 kW) helps us understand the efficiency gap between practical and ideal systems.