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Explain how you can evaluate the outlet temperatures of the cold and hot fluids in a heat exchanger after its effectiveness is determined.

Short Answer

Expert verified
Answer: The method to evaluate the outlet temperatures of the cold and hot fluids after determining the effectiveness of a heat exchanger involves the following steps: 1) Understanding the effectiveness and inlet temperatures, 2) Calculating the heat capacity rates of both fluids, 3) Finding the minimum and maximum heat capacity rates, 4) Calculating the heat transfer using effectiveness, and 5) Determining the outlet temperatures of the hot and cold fluids using the actual heat transfer and heat capacity rates.

Step by step solution

01

Understanding Effectiveness and Temperatures in Heat Exchanger

Effectiveness (ε) measures the performance of a heat exchanger by comparing its actual heat transfer to the maximum possible heat transfer. It ranges between 0 (no heat transfer) and 1 (maximum heat transfer). The actual heat transfer depends on the inlet temperatures (T_h,in and T_c,in) of both the hot and cold fluids.
02

Calculate the heat capacity rates

The heat capacity rate of each fluid is defined as the product of its mass flow rate and its specific heat capacity. Calculate the heat capacity rates (C_h and C_c) for both the hot and cold fluids using the given specific heat capacities (c_ph and c_pc) and mass flow rates (ṁ_h and ṁ_c): C_h = ṁ_h * c_ph C_c = ṁ_c * c_pc
03

Find the minimum and maximum heat capacity rates

Determine the minimum (C_min) and maximum (C_max) heat capacity rates from the calculated heat capacity rates (C_h and C_c): C_min = min(C_h, C_c) C_max = max(C_h, C_c)
04

Calculate the heat transfer using effectiveness

Using the effectiveness (ε) and the minimum heat capacity rate (C_min), calculate the actual heat transfer (Q_actual) in the heat exchanger: Q_actual = ε * C_min * (T_h,in - T_c,in)
05

Determine the outlet temperatures

Use the actual heat transfer (Q_actual) and the heat capacity rates (C_h and C_c) to determine the outlet temperatures of the hot and cold fluids (T_h,out and T_c,out): T_h,out = T_h,in - (Q_actual / C_h) T_c,out = T_c,in + (Q_actual / C_c) By following these steps, we can evaluate the outlet temperatures of the cold and hot fluids in a heat exchanger after its effectiveness is determined.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Outlet Temperature Calculation
In a heat exchanger, knowing the outlet temperatures of both the hot and cold fluids is crucial for understanding how effectively thermal energy is being transferred. Once we have established the heat exchanger's effectiveness, the next step involves determining these temperatures. Heat exchanger effectiveness (\( \varepsilon \)) reveals how efficiently a heat exchanger transfers heat. It plays a central role in outlet temperature calculations.

To start, consider the actual heat transfer, \(Q_{\text{actual}}\), in the exchanger. It is calculated using the effectiveness formula: \(Q_{\text{actual}} = \varepsilon \times C_{\text{min}} \times (T_{h,in} - T_{c,in})\). Here, \(T_{h,in}\) and \(T_{c,in}\) represent the inlet temperatures of the hot and cold fluids.

With \(Q_{\text{actual}}\) determined, the outlet temperature of the hot fluid, \(T_{h,out}\) is found by reducing the inlet temperature, \(T_{h,in}\) by a fraction of the transferred heat: \(T_{h,out} = T_{h,in} - \frac{Q_{\text{actual}}}{C_h}\).

Similarly, to find the outlet temperature of the cold fluid, \(T_{c,out}\), add the corresponding fraction of the heat transfer to its inlet temperature: \(T_{c,out} = T_{c,in} + \frac{Q_{\text{actual}}}{C_c}\). By following these calculations, one can precisely determine the outlet temperatures of both fluids, offering insights into the performance of the system.
Heat Capacity Rate
The heat capacity rate (\(C\)) of a fluid within a heat exchanger is a key factor in determining how much heat can be transferred. It's essentially a measure of the fluid's ability to absorb or release heat for a given temperature difference.

Calculating the heat capacity rate requires knowledge of both the mass flow rate (\(\dot{m}\)) and the specific heat capacity (\(c_p\)) of the fluid. The formula for heat capacity rate is straightforward: \(C = \dot{m} \times c_p\). Both the hot and cold fluids will have their respective heat capacity rates. For example, the rate for the hot fluid is \(C_h = \dot{m}_h \times c_{ph}\) and for the cold fluid, it's \(C_c = \dot{m}_c \times c_{pc}\).

Understanding the heat capacity rates is crucial because they directly influence the overall temperature change of the fluids, impacting the effectiveness and efficiency of the heat exchanger. Heat exchanges involve maximum and minimum capacity rates. The minimum heat capacity rate, \(C_{min}\), is the lower of the two capacity rates in the system. This value is pivotal because it will determine how much heat can practically be transferred between the two fluids under set conditions.

In practice, recognizing which fluid has the minimum heat capacity rate helps in optimizing the design and operation of heat exchangers to achieve desired thermal performance.
Heat Transfer Calculation
Heat transfer calculation is an essential aspect of understanding heat exchanger performance. It involves determining not just how much heat is transferred, but also the conditions and factors affecting it.

The calculation starts with the concept of heat exchanger effectiveness (\(\varepsilon\)). It quantifies the ratio of the actual heat transfer to the maximum possible heat transfer in an ideal scenario.

The actual heat transferred, \(Q_{\text{actual}}\), is given by the expression: \(Q_{\text{actual}} = \varepsilon \times C_{\text{min}} \times (T_{h,in} - T_{c,in})\). This formula underscores the influence of inlet temperatures, highlighting the importance of initial thermal differences alongside the system's effectiveness.

The minimum heat capacity rate, \(C_{\text{min}}\), serves as a limiting factor for how much heat can be transferred. It reflects the weaker link in the heat transfer chain as determined by the fluid flow properties and thermal capacity.

This value is crucial because it ensures feasibility in calculations, preventing scenarios where more heat transfer is predicted than is physically possible. Thus, heat transfer calculation is critical for the design, evaluation, and optimization of a heat exchanger system, ensuring efficiency and reliability across varying operating conditions.

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Most popular questions from this chapter

A cross-flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine (a) the exit temperature of the hot fluid and \((b)\) the rate of heat transfer in the heat exchanger.

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools \(5000 \mathrm{~kg} / \mathrm{h}\) of air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) which enters the air handler at \(25^{\circ} \mathrm{C}\). If these streams are in counter-flow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Glycerin \(\left(c_{p}=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(20^{\circ} \mathrm{C}\) and \(0.3 \mathrm{~kg} / \mathrm{s}\) is to be heated by ethylene glycol \(\left(c_{p}=2500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(60^{\circ} \mathrm{C}\) and the same mass flow rate in a thin-walled double-pipe parallel-flow heat exchanger. If the overall heat transfer coefficient is \(380 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area is \(5.3 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer and \((b)\) the outlet temperatures of the glycerin and the glycol.

Reconsider Prob. 11-131. Using EES (or other) software, plot the number of tube passes as a function of water velocity as it varies from \(1 \mathrm{~m} / \mathrm{s}\) to \(8 \mathrm{~m} / \mathrm{s}\), and discuss the results.

Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell side at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). If the correction factor is \(F=\) \(0.96\), determine the outlet temperatures of the oil and air.

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