Question

The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to \(6^{\circ} \mathrm{C}\) at an average rate of \(10 \mathrm{kg} / \mathrm{h}\). If the COP of this refrigerator is \(3.1,\) the required power input to this refrigerator is \((a) 197 \mathrm{W}\) (b) \(612 \mathrm{W}\) \((c) 64 \mathrm{W}\) \((d) 109 \mathrm{W}\) \((e) 403 \mathrm{W}\)

Short answer

Question: A refrigerator is used to cool water from an initial temperature of \(23^\circ \mathrm{C}\) to a final temperature of \(6^\circ \mathrm{C}\) at a cooling rate of \(10 \ \mathrm{kg/h}\). If the Coefficient of Performance (COP) of the refrigerator is 3.1, determine the power required to provide the cooling effect. Answer: (c) \(64 \ \mathrm{W}\)

Step-by-step solution

Given Values

Identify the given values from the problem statement: - Initial temperature of water, \(T1 = 23^\circ \mathrm{C}\) - Final temperature of water, \(T2 = 6^\circ \mathrm{C}\) - Cooling rate, \(\frac{dm}{dt} = 10 \mathrm{kg} / \mathrm{h}\) - Coefficient of performance, \(\mathrm{COP} = 3.1 \)

Calculate the Mass Flow Rate

Convert the mass flow rate from \(\mathrm{kg}/\mathrm{h}\) to \(\mathrm{kg}/\mathrm{s}\): $$ \frac{dm}{dt} = 10 \frac{\mathrm{kg}}{\mathrm{h}} \times \frac{1\ \mathrm{h}}{3600\ \mathrm{s}} = \frac{1}{360}\ \mathrm{kg/s} $$

Calculate the Cooling Effect

Use the specific heat capacity of water (\(c_p = 4.18 \times 10^3 \ \mathrm{J/(kg \cdot K})\)) and the mass flow rate to find the cooling effect (\(Q_c\)): $$ Q_c = \frac{dm}{dt} \times c_p \times (T1-T2) = \frac{1}{360} \ \mathrm{kg/s} \times 4.18\times 10^3 \ \mathrm{J/(kg \cdot K)}\times (23-6) \ \mathrm{K} $$

Calculate the Cooling Effect in Watts

Simplify the equation to find the cooling effect in Watts: $$ Q_c = \frac{1}{360} \times 4.18\times 10^3 \times 17 = 197.1 \ \mathrm{W} $$

Calculate the Power Input

Use the Coefficient of Performance (COP) formula to find the power input (\(W\)): $$ \mathrm{COP} = \frac{Q_c}{W} $$ Rearrange the equation to solve for power input \(W\): $$ W = \frac{Q_c}{\mathrm{COP}}= \frac{197.1 \ \mathrm{W}}{3.1} = 63.6 \ \mathrm{W} $$ Looking at the answer choices, the closest option is: (c) \(64 \ \mathrm{W}\)

Key concepts

Coefficient of Performance

The coefficient of performance (COP) is a measure of a refrigerator's efficiency. It is defined as the ratio of the cooling effect produced by the refrigerator to the work (or power) input. Mathematically, we express it as:
\[\text{COP} = \frac{Q_c}{W}\]where \(Q_c\) is the total heat removed from the refrigerated space (cooling effect), and \(W\) is the work done by the refrigerator (power input). In simple terms, a higher COP means that the refrigerator is more efficient, as it provides more cooling for each unit of energy consumed. When calculating COP, it’s crucial to use consistent units to ensure an accurate result.

Mass Flow Rate

The mass flow rate is a key concept in thermodynamics and engineering that indicates how much mass of a substance passes through a given surface per unit time. It's typically denoted as \(\frac{dm}{dt}\) or \(\dot{m}\), where \(dm\) represents an infinitesimal amount of mass and \(dt\) represents an infinitesimal amount of time. This rate is crucial in the analysis of systems like refrigerators, where the cooling load depends on both the amount of substance (in this case, water) and how quickly it's cooled. For refrigeration calculations, converting the mass flow rate into a consistent unit like kilograms per second (kg/s) is often necessary, as it directly relates to the rate at which heat is removed from the substance being cooled.

Cooling Effect Calculation

To calculate the cooling effect, we need to understand the amount of energy removed from the substance as it cools. This is determined by using the specific heat capacity of the substance, the mass flow rate, and the temperature change. The specific heat capacity (\(c_p\)) is a property that denotes how much energy is needed to raise the temperature of a unit mass of substance by one degree Celsius (or Kelvin).
The formula to calculate the cooling effect (\(Q_c\)) in a refrigeration process is:
\[Q_c = \frac{dm}{dt} \times c_p \times (T1 - T2)\]where \(T1\) and \(T2\) are the initial and final temperatures, respectively. The result gives the energy per time unit, which is the power (\(W\)). The proper understanding of this concept enables one to determine the necessary power input for a refrigeration system based on its desired cooling capacity.

Specific Heat Capacity

Specific heat capacity (\(c_p\)) is a property that describes how much heat energy is required to raise the temperature of a given mass of a substance by one degree. It's expressed in units of energy per mass per temperature, such as joules per kilogram per Kelvin (\(\text{J/kg}\cdot\text{K}\)). Water, for instance, has a high specific heat capacity, meaning it can absorb a lot of heat before it gets significantly warmer. This property is vital to calculations involving heating or cooling substances because it directly affects how much energy is needed for a temperature change. In the context of refrigeration, knowing the specific heat capacity of the substance being cooled (like water in our exercise) is a critical piece of information for computing the cooling effect and, subsequently, the efficiency of the cooling system.