Question

Consider a water-to-water counter-flow heat exchanger with these specifications. Hot water enters at \(95^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of hot water is \(15^{\circ} \mathrm{C}\) greater than that of cold water, and the mass flow rate of hot water is 50 percent greater than that of cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(1400 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the outlet temperature of the cold water, \((b)\) the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

Short answer

Answer: The outlet temperature of the cold water is 42°C, the effectiveness of the heat exchanger is approximately 0.293, the mass flow rate of the cold water is approximately 0.108 kg/s, and the heat transfer rate is approximately 997.92 W.

Step-by-step solution

Use energy balance equation for hot water

Since the mass flow rate of hot water is 50 percent greater than that of cold water, the mass flow rate of hot water can be represented as \(\dot{m}_{h} = 1.5\dot{m}_{c}\). From the energy balance equation for hot water, we have: \(Q = \dot{m}_{h}c_{p}(T_{h, in} - T_{h, out})\) Using \(\dot{m}_{h} = 1.5\dot{m}_{c}\), we get: \(Q = 1.5\dot{m}_{c}c_{p}(95 - (T_{c, out}+15))\) ... (1)

Use energy balance equation for cold water

From the energy balance equation for cold water, we have: \(Q = \dot{m}_{c}c_{p}(T_{c, out} - 20)\) ... (2)

Solve for outlet temperature of cold water

From (1) and (2), we can solve for the outlet temperature of the cold water: \(1.5\dot{m}_{c}c_{p}(95 - (T_{c, out}+15)) = \dot{m}_{c}c_{p}(T_{c, out} - 20)\) Solving for \(T_{c, out}\), we get: \(T_{c, out} = 42^{\circ} C\)

Calculate the effectiveness of the heat exchanger

The effectiveness of the heat exchanger is defined as: \(\epsilon = \frac{Q}{\dot{m}_{c}c_{p}(T_{h, in} - T_{c, in})}\) Using the results from Step 2 and Step 3, we get: \(\epsilon = \frac{\dot{m}_{c}c_{p}(42 - 20)}{\dot{m}_{c}c_{p}(95 - 20)} = \frac{22}{75} \approx 0.293\)

Determine the mass flow rate of cold water

We are given that the product of heat transfer surface area and the overall heat transfer coefficient is \(U = 1400 \frac{W}{K}\). We can write the following equation: \(Q = U\Delta{T}\) We can use the results from Step 1 to obtain the value of \(\Delta{T}\): \(\Delta{T} = \frac{1.5\dot{m}_{c}c_{p}(60)}{1400} = 22\) Finally, we can solve for the mass flow rate of the cold water: \(\dot{m}_{c} = \frac{U\Delta{T}}{c_{p}(T_{c, out} - T_{c, in})} = \frac{1400 \times 22}{4180\times(42-20)} \approx 0.108 \frac{kg}{s}\)

Calculate the heat transfer rate

Now that we have determined the mass flow rate of the cold water, we can calculate the heat transfer rate: \(Q = \dot{m}_{c}c_{p}(T_{c, out} - T_{c, in}) = 0.108\times4180\times(42-20) \approx 997.92 W\) #Summary# In conclusion, we have determined the following values for this counter-flow heat exchanger problem: (a) Outlet temperature of cold water: \(T_{c, out} = 42^{\circ} C\) (b) Effectiveness of the heat exchanger: \(\epsilon \approx 0.293\) (c) Mass flow rate of cold water: \(\dot{m}_{c} \approx 0.108 \frac{kg}{s}\) (d) Heat transfer rate: \(Q \approx 997.92 W\)

Key concepts

Counter-Flow Heat Exchanger

A counter-flow heat exchanger is a device used to transfer heat between two fluids moving in opposite directions. In our example, the hot water and cold water in the heat exchanger enter from opposite ends, which creates a situation where temperature gradients drive heat transfer throughout the entire length of the exchanger. This configuration is highly efficient compared to other types, such as parallel-flow, because it allows the maximum change in temperature for the fluids involved.
The temperature gradient, a measure of the difference in temperature between the hot and cold fluids, is more consistent across the length of a counter-flow exchanger. This consistency leads to improved heat transfer, as the heat transfer rate is directly proportional to the temperature difference. Understanding how counter-flow heat exchangers work is vital because they are widely used in industries like power plants, refrigeration, and chemical processing. These systems maximize energy efficiency and ensure a better match between the requirements and capabilities of the equipment.

Thermal Effectiveness

Thermal effectiveness (\(\epsilon\)) is a measure of how efficiently a heat exchanger performs compared to an ideal case. It is defined as the ratio of the actual heat transfer to the maximum possible heat transfer under the same conditions. In our problem, the effectiveness is calculated using the formula:
\[\epsilon = \frac{Q}{\dot{m}_{c}c_{p}(T_{h, in} - T_{c, in})}\]
The numerator represents the actual heat transfer, while the denominator represents the maximum energy that could be transferred if the system were 100% effective. For this specific heat exchanger, the effectiveness is approximately 0.293, indicating that it transfers about 29.3% of the maximum possible heat. While this number might seem low, effectiveness can vary greatly depending on system design, operating conditions, and fluid properties. It's essential to have an understanding of what is reasonable or achievable in real-world applications, and knowing how to calculate effectiveness can aid in designing or adjusting systems for better performance.

Energy Balance Calculation

Energy balance calculations are key to understanding the dynamics within a heat exchanger. They involve calculating the energy transferred between the entering and exiting fluids. For the hot and cold water streams in our example, the energy balance equations are:- For hot water:\[Q = \dot{m}_{h}c_{p}(T_{h, in} - T_{h, out})\]- For cold water:\[Q = \dot{m}_{c}c_{p}(T_{c, out} - T_{c, in})\]These equations ensure that the energy lost by the hot fluid is equal to the energy gained by the cold fluid, a fundamental principle often referred to as conservation of energy. By using these equations, we can derive relationships between variables such as mass flow rate and temperature change, and solve for unknown values like the outlet temperature of cold water or the mass flow rate of cold water. Energy balance calculations are integral to ensuring every part of the system functions as expected, leading to precise control over heat exchange processes.

Heat Transfer Rate

The heat transfer rate (\(Q\)) is an important parameter that quantifies how much energy is being transferred from the hot fluid to the cold fluid per unit time in a heat exchanger. In our exercise, the heat transfer rate is calculated using one of the energy balance equations:\[Q = \dot{m}_{c}c_{p}(T_{c, out} - T_{c, in})\]This formula allows us to determine how effectively the heat exchanger is doing its job. The answer of approximately 997.92 watts in our exercise tells us that nearly 1000 joules of energy are transferred every second between the hot and cold streams.Understanding the heat transfer rate is vital for evaluating the performance of a heat exchanger. It helps determine whether the exchanger is suitable for a given application or if adjustments are necessary to meet specific energy requirements. Additionally, knowing the heat transfer rate aids in calculating the size and capacity of the equipment, which is crucial for both design and real-life usage.