/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 A two-fluid heat exchanger has i... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 45

A two-fluid heat exchanger has inlet and outlet temperatures of 65 and \(40^{\circ} \mathrm{C}\) for the hot fluid and 15 and \(30^{\circ} \mathrm{C}\) for the cold fluid. Can you tell whether this exchanger is operating under counterflow or parallelflow conditions? Determine the effectiveness of the heat exchanger.

Short Answer

Expert verified
The heat exchanger is operating under counterflow conditions, as the hot fluid outlet temperature (\(40^{\circ} C\)) is colder than the cold fluid outlet temperature (\(30^{\circ} C\)), and the hot fluid inlet temperature (\(65^{\circ} C\)) is hotter than the cold fluid inlet temperature (\(15^{\circ} C\)). The Log Mean Temperature Difference (LMTD) is calculated as \(29.52^{\circ} C\). Due to the lack of full data, the exact effectiveness value cannot be determined, but the heat exchanger is comparatively more effective in counterflow conditions than parallel flow conditions.

Step by step solution

01

Recall the definition of counterflow and parallel flow heat exchangers

In counterflow heat exchangers, the two fluids flow in opposite directions, while in parallel flow heat exchangers, they flow in the same direction.
02

Compare the given temperatures to determine the flow conditions

We are given the following temperatures: Hot fluid inlet temperature: \(65^{\circ} C\) Hot fluid outlet temperature: \(40^{\circ} C\) Cold fluid inlet temperature: \(15^{\circ} C\) Cold fluid outlet temperature: \(30^{\circ} C\) For counterflow heat exchangers, the hot fluid outlet temperature should be colder than the cold fluid outlet temperature, and the hot fluid inlet temperature should be hotter than the cold fluid inlet temperature. Conversely, for parallel flow heat exchangers, the hot fluid outlet temperature must be hotter than the cold fluid inlet temperature. Comparing the temperatures, we find that the hot fluid outlet temperature (\(40^{\circ} C\)) is colder than the cold fluid outlet temperature (\(30^{\circ} C\)), and the hot fluid inlet temperature (\(65^{\circ} C\)) is hotter than the cold fluid inlet temperature (\(15^{\circ} C\)). Therefore, the heat exchanger is operating under counterflow conditions.
03

Determine the Log Mean Temperature Difference (LMTD)

The LMTD is given by the formula: \[LMTD = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\] Where \(\Delta T_{1}\) is the temperature difference between the hot fluid inlet temperature and the cold fluid outlet temperature, and \(\Delta T_{2}\) is the temperature difference between the hot fluid outlet temperature and the cold fluid inlet temperature. Let's calculate the LMTD: \[\Delta T_{1} = 65^{\circ} C - 30^{\circ} C = 35^{\circ} C\] \[\Delta T_{2} = 40^{\circ} C - 15^{\circ} C = 25^{\circ} C\] Then, \[LMTD = \frac{35^{\circ} C - 25^{\circ} C}{\ln(\frac{35^{\circ} C}{25^{\circ} C})} = 29.52^{\circ} C\]
04

Determine the effectiveness of the heat exchanger

The effectiveness of a heat exchanger is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate. It can be calculated using the following formula: \[Effectiveness = \frac{q_{actual}}{q_{max}}\] Where \(q_{actual}\) is the actual heat transfer rate and \(q_{max}\) is the maximum possible heat transfer rate. We can calculate the actual heat transfer rate (\(q_{actual}\)) using the hot-side heat capacity rate (\(\dot{C}_{h}\)), the hot fluid inlet temperature (\(T_{hi}\)), and the hot fluid outlet temperature (\(T_{ho}\)): \[q_{actual} = \dot{C}_{h} (T_{hi} - T_{ho})\] Unfortunately, we don't have the hot-side heat capacity rate (\(\dot{C}_{h}\)) given in this problem, so we cannot determine the exact effectiveness value. Nonetheless, we can deduce that the heat exchanger is more effective in counterflow conditions than it would be in parallel flow conditions due to the observed temperature differences. In conclusion, this heat exchanger is operating under counterflow conditions, and the calculated LMTD is \(29.52^{\circ} C\). Without the full data, we cannot determine the exact effectiveness, but we can claim that it is more effective in counterflow conditions than parallel flow conditions.

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Key Concepts

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

Counterflow Heat Exchanger
Understanding how a counterflow heat exchanger operates is crucial when dealing with thermal systems. This type of exchanger is characterized by the two fluids moving in opposite directions, each with its own inlet and outlet. One of the main advantages of this setup is the potential for a greater temperature change compared to other designs, which leads to a higher heat transfer efficiency.

In the given problem, we see that the hot fluid cools from an inlet temperature of 65掳C to an outlet temperature of 40掳C, while the cold fluid heats up from 15掳C to 30掳C. The directional temperature profiles of the two fluids suggest that this heat exchanger is operating in a counterflow arrangement. This is because the hot fluid exiting the heat exchanger is cooler than the exiting cold fluid, a scenario typical of counterflow operations.
Parallel Flow Heat Exchanger
In contrast to the counterflow design, a parallel flow heat exchanger features both fluids moving in the same direction. Under these conditions, the temperature differences between the two fluids decrease along the flow path. This tends to limit the potential for heat transfer as the two fluids approach thermal equilibrium.

For the parallel flow heat exchanger, the maximum temperature difference that can be achieved is bounded by the inlet temperature of the hot fluid and the inlet temperature of the cold fluid. This often results in a lower Log Mean Temperature Difference (LMTD) and subsequently lower overall heat transfer rates compared to counterflow heat exchangers.
Log Mean Temperature Difference (LMTD)
The Log Mean Temperature Difference, or LMTD, is a critical concept in understanding heat exchangers. It represents an average temperature difference between the hot and cold fluids over the length of the heat exchanger. The LMTD is calculated with a specialized formula, which compensates for the varying temperature difference along the length of the heat exchanger.

The formula is expressed as: \[LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}\]Where \(\Delta T_1\) is the temperature difference at one end and \(\Delta T_2\) is the temperature difference at the other end. The LMTD method is particularly useful because it allows for a simplified calculation of the heat transfer rate in a heat exchanger, assuming a constant heat transfer coefficient. In the exercise, the calculated LMTD was 29.52掳C, which aids in evaluating the performance and effectiveness of the heat exchanger.
Heat Transfer Rate
The heat transfer rate is a measure of the thermal energy transferred per unit time in a heat exchanger. It is crucial for sizing and evaluating the performance of the heat exchanger. The actual heat transfer rate, \(q_{actual}\), is calculated by considering the heat capacity rates of the fluids involved and the temperature changes they undergo during the heat exchange process.

The general formula to calculate the actual heat transfer rate for the hot or cold fluid sides is given by: \[q_{actual} = \dot{C}(T_{in} - T_{out})\]where \(\dot{C}\) is the heat capacity rate of the fluid, and \(T_{in}\) and \(T_{out}\) are the inlet and outlet temperatures, respectively. Although the full data for calculating the actual heat transfer rate were not provided in the problem, understanding this formula is essential for evaluating the effectiveness of the heat exchanger, which is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate.

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Most popular questions from this chapter

The condenser of a steam power plant contains \(N=1000\) brass tubes \(\left(k_{\mathrm{t}}=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), each of inner and outer diameters, \(D_{i}=25 \mathrm{~mm}\) and \(D_{o}=\) \(28 \mathrm{~mm}\), respectively. Steam condensation on the outer surfaces of the tubes is characterized by a convection coefficient of \(h_{o}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If cooling water from a large lake is pumped through the condenser tubes at \(m_{c}=400 \mathrm{~kg} / \mathrm{s}\), what is the overall heat transfer coefficient \(U_{o}\) based on the outer surface area of a tube? Properties of the water may be approximated as \(\mu=9.60 \times\) \(10^{-4} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\mathrm{Pr}=6.6 .\) (b) If, after extended operation, fouling provides a resistance of \(R_{f, i}^{\prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), at the inner surface, what is the value of \(U_{o}\) ? (c) If water is extracted from the lake at \(15^{\circ} \mathrm{C}\) and \(10 \mathrm{~kg} / \mathrm{s}\) of steam at \(0.0622\) bars are to be condensed, what is the corresponding temperature of the water leaving the condenser? The specific heat of the water is \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Hot exhaust gases are used in a shell-and-tube exchanger to heat \(2.5 \mathrm{~kg} / \mathrm{s}\) of water from 35 to \(85^{\circ} \mathrm{C}\). The gases, assumed to have the properties of air, enter at \(200^{\circ} \mathrm{C}\) and leave at \(93^{\circ} \mathrm{C}\). The overall heat transfer coefficient is \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the effectiveness-NTU method, calculate the area of the heat exchanger.

An energy storage system is proposed to absorb thermal energy collected during the day with a solar collector and release thermal energy at night to heat a building. The key component of the system is a shelland-tube heat exchanger with the shell side filled with \(n\)-octadecane (see Problem 8.47). (a) Warm water from the solar collector is delivered to the heat exchanger at \(T_{h, i}=40^{\circ} \mathrm{C}\) and \(\dot{m}=2 \mathrm{~kg} / \mathrm{s}\) through the tube bundle consisting of 50 tubes, two tube passes, and a tube length per pass of \(L_{l}=2 \mathrm{~m}\). The thin-walled, metal tubes are of diameter \(D=25 \mathrm{~mm}\). Free convection exists within the molten \(n\)-octadecane, providing an average heat transfer coefficient of \(h_{o}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside of each tube. Determine the volume of \(n\) octadecane that is melted over a 12 -h period. If the total volume of \(n\)-octadecane is to be \(50 \%\) greater than the volume melted over \(12 \mathrm{~h}\), determine the diameter of the \(L_{j}=2.2\)-m-long shell. (b) At night, water at \(T_{c, i}=15^{\circ} \mathrm{C}\) is supplied to the heat exchanger, increasing the water temperature and solidifying the \(n\)-octadecane. Do you expect the heat transfer rate to be the same, greater than, or less than the heat transfer rate in part (a)? Explain your reasoning.

A cross-flow heat exchanger used in a cardiopulmonary bypass procedure cools blood flowing at \(5 \mathrm{~L} / \mathrm{min}\) from a body temperature of \(37^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) in order to induce body hypothermia, which reduces metabolic and oxygen requirements. The coolant is ice water at \(0^{\circ} \mathrm{C}\), and its flow rate is adjusted to provide an outlet temperature of \(15^{\circ} \mathrm{C}\). The heat exchanger operates with both fluids unmixed, and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The density and specific heat of the blood are \(1050 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. a) Determine the heat transfer rate for the exchanger. b) Calculate the water flow rate. c) What is the surface area of the heat exchanger? d) Calculate and plot the blood and water outlet temperatures as a function of the water flow rate for the range 2 to \(4 \mathrm{~L} / \mathrm{min}\), assuming all other parameters remain unchanged. Comment on how the changes in the outlet temperatures are affected by changes in the water flow rate. Explain this behavior and why it is an advantage for this application.

A liquefied natural gas (LNG) regasification facility utilizes a vertical heat exchanger or vaporizer that consists of a shell with a single-pass tube bundle used to convert the fuel to its vapor form for subsequent delivery through a land-based pipeline. Pressurized LNG is off-loaded from an oceangoing tanker to the bottom of the vaporizer at \(T_{c, i}=-155^{\circ} \mathrm{C}\) and \(\dot{m}_{\mathrm{LNG}}=150 \mathrm{~kg} / \mathrm{s}\) and flows through the shell. The pressurized LNG has a vaporization temperature of \(T_{f}=-75^{\circ} \mathrm{C}\) and specific heat \(c_{p l}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The specific heat of the vaporized natural gas is \(c_{p, v}=2210 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) while the gas has a latent heat of vaporization of \(h_{f g}=575 \mathrm{~kJ} / \mathrm{kg}\). The LNG is heated with seawater flowing through the tubes, also introduced at the bottom of the vaporizer, that is available at \(T_{h, i}=20^{\circ} \mathrm{C}\) with a specific heat of \(c_{\mu \mathrm{Sw}}=3985 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). If the gas is to leave the vaporizer at \(T_{c o}=8^{\circ} \mathrm{C}\) and the seawater is to exit the device at \(T_{\text {hot }}=10^{\circ} \mathrm{C}\), determine the required vaporizer heat transfer area. Hint: Divide the vaporizer into three sections, as shown in the schematic, with \(U_{\mathrm{A}}=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(U_{\mathrm{B}}=260 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(U_{\mathrm{C}}=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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