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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.

Short Answer

Expert verified
In this exercise, we calculated the number of tube passes required for a heat exchanger considering different water velocities from 1 m/s to 8 m/s. After performing the calculations for each water velocity, we plotted the number of tube passes vs water velocities. The graph shows that as the water velocity increases, the number of tube passes required decreases. This indicates that increasing the water velocity can lead to a more efficient heat transfer process in the heat exchanger, as fewer tube passes are needed to achieve the desired heat transfer rate. However, it is essential to consider other factors like pressure drop, turbulence, and the structural design of the heat exchanger when selecting the optimal water velocity to balance efficiency and durability.

Step by step solution

01

Recall the parameters from the previous problem 11-131

For the heat exchanger, we have the following known parameters: - Water velocity: Varies from \(1\mathrm{~m/s}\) to \(8\mathrm{~m/s}\) - Tube diameter (D): \(25 \times 10^{-3}\mathrm{~m}\) - Heat exchanger length (L): \(12\mathrm{~m}\) - Total heat transfer rate (Q): \(25000\mathrm{~W}\)
02

Calculate the flow rate for each water velocity

The volumetric flow rate (m^3/s) in the tubes can be calculated by multiplying the water velocity (\(v\)) by the cross-sectional area of the tube (A). We will use the formula: \(Q = v\times A\) The cross-sectional area of a tube can be calculated as follows: \(A = \dfrac{\pi d^2}{4}\) We will calculate the volumetric flow rate for each of the water velocities provided (ranging from 1 m/s to 8 m/s) and store these values in a table.
03

Calculate the flow rate per tube pass

Now that we have the volumetric flow rate for each of the water velocities, we need to calculate the flow rate per tube pass, which refers to the flowrate in a single tube. Divide the total flow rate (Q) by the number of tubes in the heat exchanger. As per the problem, the total heat transfer rate required (Q) is \(25000\) W. Assuming a water specific heat of \(4200 ~\mathrm{J/(kg·K)}\), we can calculate the total mass flow rate required using the formula: \(\dot{m} = \dfrac{Q}{c_p\Delta T}\) Now, we can calculate the flow rate per tube pass by dividing the total mass flow rate by the number of tube passes.
04

Calculate the number of tube passes for each water velocity

With the flow rate per tube pass and volumetric flow rates for each water velocity, we can now calculate the number of tube passes required for each water velocity using the following formula: Number of tube passes = \(\dfrac{\text{Flow rate per tube pass}}{\text{Volumetric flow rate for each water velocity}}\) Calculate the number of tube passes for each water velocity from 1-8 m/s and store these values in a table.
05

Plot the graph

Plot the number of tube passes as a function of water velocity (varying from 1 m/s to 8 m/s) using the data collected in the table. Make sure to label the axes and provide a title for the graph.
06

Discuss the results

Analyze the plotted graph and discuss the results. Observe the relationship between the number of tube passes required in the heat exchanger and the water velocity. This will help us understand how the number of tube passes affects the efficiency of the heat transfer process in the heat exchanger.

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Key Concepts

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

Tube Passes in Heat Exchangers
In the world of heat exchangers, tube passes refer to the number of times the fluid – such as water in our case – traverses the length of the heat exchanger. The number of passes has a direct impact on the thermal performance and the effectiveness of the heat exchanger. This key detail is not only critical for mechanical engineers but also essential for students who are looking to understand the intricacies of thermal systems.

Multiple tube passes increase the surface area for heat to be transferred, thereby improving efficiency by offering more opportunities for hot and cold fluids to exchange their thermal energies. However, it's a balancing act as more passes can also mean greater pressure drops and potential complications in the flow pattern. When we calculate the number of tube passes relative to water velocity, as in the provided exercise, we're typically looking to optimize the balance between efficient heat transfer and minimal fluid resistance.

A practical exercise, such as plotting the number of tube passes against various water velocities, is not only helpful in understanding theory, it also mirrors real-world applications where engineers must configure heat exchangers to operate within certain fluid velocity ranges for optimal performance and longevity.
Water Velocity Impact on Heat Transfer
As students and engineers delve into the intricacies of heat transfer, understanding the role of water velocity is paramount. Water velocity through a heat exchanger has a substantial impact on the rate of heat transfer. In simple terms, as water flows more rapidly over the heat transfer surface, it can carry away heat more quickly, which enhances the heat exchange process.

However, too high a velocity can lead to issues such as erosion of the tubes and increased operational costs due to higher pumping power requirements. On the other hand, too low a velocity can result in poor heat transfer and potential fouling, where sediment and debris accumulate on the surface of the heat exchanger reducing its efficiency. The process of finding the sweet spot – or the optimum velocity at which the system operates most efficiently – is thus an exercise in precision.

With reference to the given task, a thoroughly guided step-by-step approach to plotting the number of tube passes against varying water velocities facilitates a deeper understanding of this delicate relationship between velocity and heat transfer. Such graphical analyses help students visualize and grasp the concept of optimizing water velocity in heat exchange scenarios.
Volumetric Flow Rate Calculation
Understanding the volumetric flow rate is crucial as it is a fundamental aspect of fluid dynamics and plays a significant role in heat exchanger design. The volumetric flow rate represents the volume of fluid that passes through a given area over a specific period of time and is calculated by multiplying the velocity of the fluid by the cross-sectional area through which it flows.

To calculate it, we can employ this equation: \(Q = v \times A\), where \(v\) is the fluid velocity and \(A\) is the cross-sectional area (\( \dfrac{\pi d^2}{4} \) for circular tubes). Given an array of different velocities, we can derive the corresponding volumetric flow rates, which are integral in determining the heat exchanger's overall performance.

In an educational context, crafting exercises where students calculate these rates for different scenarios not only reinforces their understanding of the principles at play but also enables them to appreciate the real-world applications such calculations have in systems design. It is these volumetric flow rates, often overlooked in broader discussions, that are instrumental in the fine-tuning of a heat exchanger's operations.

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Most popular questions from this chapter

Under what conditions can a counter-flow heat exchanger have an effectiveness of one? What would your answer be for a parallel-flow heat exchanger?

Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), where it is heated by a hot water stream \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(70^{\circ} \mathrm{C}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\). Determine the maximum heat transfer rate and the outlet temperatures of both fluids for that case.

Discuss the differences between the cardiovascular counter-current design and standard engineering countercurrent designs.

A shell-and-tube heat exchanger with 2-shell passes and 4-tube passes is used for cooling oil \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(125^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\). The coolant is water, which enters the shell side at \(25^{\circ} \mathrm{C}\) and leaves at \(46^{\circ} \mathrm{C}\). The overall heat transfer coefficient is \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For an oil flow rate of \(10 \mathrm{~kg} / \mathrm{s}\), calculate the cooling water flow rate and the heat transfer area.

Geothermal water \(\left(c_{p}=4250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to heat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(17^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) in a double-pipe counter-flow heat exchanger. The heat transfer surface area is \(25 \mathrm{~m}^{2}\), the overall heat transfer coefficient is \(480 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the mass flow rate of geothermal water is larger than that of fresh water. If the effectiveness of the heat exchanger is desired to be \(0.823\), determine the mass flow rate of geothermal water and the outlet temperatures of both fluids.

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