Question

A refrigeration system uses water-cooled condenser for rejecting the waste heat. The system absorbs heat from a space at \(25^{\circ} \mathrm{F}\) at a rate of \(24,000 \mathrm{Btu} / \mathrm{h}\). Water enters the condenser at \(65^{\circ} \mathrm{F}\) at a rate of \(1.45 \mathrm{lbm} / \mathrm{s}\). The COP of the system is estimated to be \(1.9 .\) Determine \((a)\) the power input to the system, in \(\mathrm{kW},(b)\) the temperature of the water at the exit of the condenser, in \(^{\circ} \mathrm{F}\) and \((c)\) the maximum possible COP of the system. The specific heat of water is \(1.0 \mathrm{Btu} / \mathrm{bm} \cdot^{\circ} \mathrm{F}\)

Short answer

Answer: To calculate the power input to the system, use the formula: \(W_{in}\) = \(\frac{Q_{abs}}{COP}\) Plug in the values for \(Q_{abs}\) (converted to watts) and COP: \(W_{in}\) = \(\frac{7037.7\,\, \mathrm{W}}{1.9}\) \(W_{in}\) = 3704.05 W Now, convert watts to kilowatts: \(W_{in}\) = 3.70405 kW

Step-by-step solution

1. Calculate the heat absorbed by the system

We are given that the system absorbs heat from a space at a rate of \(24,000\,\, \mathrm{Btu} / \mathrm{h}\). We can convert this to \(\mathrm{W}\) (watts) using the following conversion factor: 1 \(\mathrm{Btu/h}\) = 0.293071 \(\mathrm{W}\) Heat Absorbed (\(Q_{abs}\)) = \(\frac{24000\,\, \mathrm{Btu}}{\mathrm{h}} \cdot 0.293071\,\, \frac{\mathrm{W}}{\mathrm{Btu} \cdot \mathrm{h}}\)

2. Calculate the power input to the system

We know the COP of the system is 1.9. The formula for COP (Coefficient of Performance) is: COP = \(\frac{Q_{abs}}{W_{in}}\), where \(W_{in}\) is the power input to the system. Now, we can solve for \(W_{in}\): \(W_{in}\) = \(\frac{Q_{abs}}{COP}\)

3. Calculate the temperature of the water at the exit of the condenser

We are given that water enters the condenser at \(65^{\circ} \mathrm{F}\) and a rate of mass flow \(1.45 \,\, \mathrm{lbm} / \mathrm{s}\). Let's denote the exit temperature of the water as \(T_{out}\). The energy balance for the water-cooled condenser can be written as: \(Q_{abs}\) = \(m \cdot c_p \cdot (T_{out} - T_{in})\), where \(m\) is the mass flow rate of the water, \(c_p\) is the specific heat of water, and \(T_{in}\) is the initial temperature. Now, we can solve for \(T_{out}\): \(T_{out}\) = \(\frac{Q_{abs}}{m \cdot c_p} + T_{in}\).

4. Calculate the maximum possible COP of the system

To determine the maximum possible COP of the system, we need to apply the Carnot Refrigeration COP formula, which is: \(COP_{max}\) = \(\frac{T_{cold}}{T_{hot} - T_{cold}}\), where \(T_{cold}\) is the temperature of the cooled space, and \(T_{hot}\) is the temperature of the water entering the condenser. These temperatures must be in Kelvin. We can convert them as follows: \(T_{cold} (K) = T_{cold} (°F) + 273.15\) \(T_{hot} (K) = T_{hot} (°F) + 273.15\) Now, we can calculate the maximum possible COP: \(COP_{max}\) = \(\frac{T_{cold}}{T_{hot} - T_{cold}}\)

Key concepts

Heat Absorbed Calculation

Understanding how to calculate the heat absorbed by a refrigeration system is crucial for evaluating its performance. The process begins by identifying the rate at which the system absorbs heat. In our problem, this rate is given as 24,000 Btu/h. The next step is to convert this rate from British Thermal Units per hour (Btu/h) to Watts (W), using the conversion factor 0.293071 W/Btu/h.

This provides us with the heat absorbed in a more universal unit of measure, which can then be used to assess other aspects of the system such as efficiency or power input. Hence, this conversion equates the heat absorbed as approximately 7,033.7 W. To extend this concept, different systems may use alternative refrigerants and operate under varying conditions, which determines the significance of the heat absorption capacity in relation to system performance.

Power Input Calculation

The power input to a refrigeration system is a measure of the energy consumed by the system to transfer heat from a lower to a higher temperature. It's a critical factor to consider, both for cost and efficiency reasons. To calculate power input, we need to understand the Coefficient of Performance (COP), which is the ratio of heat absorbed (\(Q_{abs}\)) by the system to the work input (\(W_{in}\)).

In the given example, with a COP of 1.9, we can use the formula \(W_{in} = \frac{Q_{abs}}{COP}\) to find the energy consumption. This not only aids in determining the immediate energy requirements but also helps in long-term planning for energy resource allocation. Proper calculation of power input has both financial and environmental implications for the operation of refrigeration systems.

Condenser Exit Temperature

The condenser exit temperature is indicative of the efficiency with which a refrigeration system is rejecting heat. To find the exit temperature of the water (\(T_{out}\)), we use the straightforward equation that involves the heat absorbed by the system (\(Q_{abs}\)), the mass flow rate of the water (\(m\)), and the specific heat (\(c_p\)) of the water. The formula \(T_{out} = \frac{Q_{abs}}{m \cdot c_p} + T_{in}\) gives us the final temperature after heat exchange.

Understanding and controlling this temperature is essential because it can affect the performance and longevity of system components. It also plays a role in the system's interaction with the environment, where water temperature regulations may apply to prevent thermal pollution.

Maximum Possible COP

The maximum possible Coefficient of Performance (COP) is a theoretical benchmark that helps in understanding the best achievable efficiency of a refrigeration cycle. Calculating the maximum COP involves using the absolute temperatures (in Kelvin) of the hot and cold reservoirs. The formula \(COP_{max} = \frac{T_{cold}}{T_{hot} - T_{cold}}\) can be used, where \(T_{cold}\) is the temperature where heat is absorbed (cooled space), and \(T_{hot}\) is the temperature of heat rejection (condenser inlet).

In practice, this maximum COP can never be reached due to real-world inefficiencies like friction, non-ideal material properties, and other factors. Nonetheless, it serves as a goal for engineers to design systems that can come as close to this ideal as possible, therefore pushing forward innovations in refrigeration technology.