Question

A space is kept at \(-15^{\circ} \mathrm{C}\) by a vapor-compression refrigeration system in an ambient at \(25^{\circ} \mathrm{C}\). The space gains heat steadily at a rate of \(3500 \mathrm{kJ} / \mathrm{h}\) and the rate of heat rejection in the condenser is \(5500 \mathrm{kJ} / \mathrm{h}\). Determine the power input, in \(\mathrm{kW}\), the COP of the cycle and the second-law efficiency of the system.

Short answer

Question: Calculate the power input, the coefficient of performance (COP), and the second-law efficiency of a vapor-compression refrigeration system cooling a space to -15°C with heat gain and heat rejection rates of 3500 kJ/h and 5500 kJ/h, respectively. Answer: The power input is 0.5556 kW, the COP of the cycle is 0.9722, and the second-law efficiency is 0.1506.

Step-by-step solution

Calculate power input

To find the power input, we need to subtract the heat gain rate from the heat rejection rate and divide the result by the time (in hours). All heat rates are given in kJ/h, so we need to convert them to kW by dividing by 3600 s/h. Power input = (Heat rejection rate - Heat gain rate) / time Power input = (5500 kJ/h - 3500 kJ/h) / 3600 s/h Power input = (2000 kJ/h) / 3600 s/h = 0.5556 kW

Calculate the coefficient of performance (COP)

Next, we'll calculate the COP of the cycle using the following formula: COP = Heat gain rate / Power input COP = (3500 kJ/h) / (0.5556 kW) Since the heat gain rate is given in kJ/h, we need to convert it to kW using the conversion factor 1 kW = 1 kJ/s: COP = 3500 kJ/h * (1 kW / 3600 kJ/h) / 0.5556 kW COP = 0.9722

Calculate the second-law efficiency

In order to calculate the second-law efficiency, we first determine the COP of Carnot cycle using the following formula: COP_Carnot = (T_low) / (T_high - T_low) Where T_low and T_high are the low and high temperatures in Kelvin, respectively. To convert the temperatures from Celsius to Kelvin, add 273.15: T_low = -15°C + 273.15 = 258.15 K T_high = 25°C + 273.15 = 298.15 K COP_Carnot = (258.15 K) / (298.15 K - 258.15 K) COP_Carnot = 258.15 K / 40 K = 6.4538 Now that we have both COP values, we can calculate the second-law efficiency: Second-law efficiency = COP / COP_Carnot Second-law efficiency = 0.9722 / 6.4538 = 0.1506 In summary, the power input is 0.5556 kW, the COP of the cycle is 0.9722, and the second-law efficiency is 0.1506.

Key concepts

Coefficient of Performance (COP)

The coefficient of performance (COP) is a crucial measure in refrigeration and heat pump systems, indicating their efficiency. It's defined as the ratio of the amount of heat removed (in the case of refrigeration) or added (in the case of a heat pump) to the work input. Simply put, COP answers the question: how much cooling or heating do we get per unit of energy consumed?

To calculate the COP for a vapor-compression refrigeration system, one can use the equation:
\[ \text{COP} = \frac{\text{Heat gain rate}}{\text{Power input}} \]
This is essentially how much heat is absorbed from the cold environment compared to the work (or electrical energy) needed to operate the refrigerator. In our exercise, the COP calculation showed that for each kilowatt of power input, the system provides around 0.9722 kilowatts of cooling.

Second-Law Efficiency

The second-law efficiency, also known as exergetic efficiency, tells us how well a system performs relative to an ideal system operating under the same conditions. It's a measure of the system's performance compared to the maximum performance allowed by the second law of thermodynamics.

To find the second-law efficiency of a refrigeration system, one must first compute the COP of an ideal Carnot refrigerator operating between the same temperatures, and then compare the real system's COP to this ideal COP:
\[ \text{Second-law efficiency} = \frac{\text{Real COP}}{\text{Ideal COP}_{\text{Carnot}}} \]
In our exercise, the real refrigerator's COP was far lower than the Carnot COP, leading to a second-law efficiency of 0.1506. This indicates that the real system is only achieving about 15% of the efficiency that could be theoretically achieved by an ideal Carnot cycle operating between the same temperatures.

Heat Transfer Rates

Heat transfer rates are pivotal in understanding refrigeration systems, as they indicate the amount of heat being removed or added to a certain space over time. Specifically, in a vapor-compression refrigeration cycle, one typically discusses two key rates: the rate at which heat is absorbed from the refrigerated space (cooling capacity) and the rate at which heat is rejected to the surroundings (heat rejection rate).

In our example, the refrigerated space is gaining heat at a rate of 3500 kJ/h, implying that the refrigerator must remove that much heat each hour to maintain the desired temperature. Concurrently, the heat rejection rate in the condenser is 5500 kJ/h, representing the heat discharged to the environment. The power input necessary for the system to effectuate this heat transfer process is calculated using the difference between these heat rates.

Carnot Cycle

The Carnot cycle is a theoretical model that represents the upper limit of efficiency for a heat engine or refrigerator. It consists of two isothermal processes and two adiabatic processes, and it's an idealized process since it assumes no friction, no irreversibilities, and perfect insulating materials.

In the refrigeration context, the Carnot refrigerator provides a benchmark for the best possible performance under given temperature conditions. It's described by the eponymous COP of Carnot:
\[ \text{COP}_{\text{Carnot}} = \frac{T_{\text{low}}}{T_{\text{high}} - T_{\text{low}}} \]
This ideal COP depends solely on the temperatures of the heat reservoirs (expressed in Kelvin), and doesn't take into account practical considerations like material properties and specific equipment efficiencies. For the system in our exercise, the high second-law efficiency disparity suggests significant opportunities for improvement, if one considers the ideal Carnot cycle as a target for real-system performance.