Consider a double-pipe parallel-flow heat exchanger of length \(L\) The inner and outer diameters of the inner tube are \(D_{1}\) and \(D_{2}\), respectively, and the inner diameter of the outer tube is \(D_{3}\). Explain how you would determine the two heat transfer surface areas \(A_{i}\) and \(A_{o}\). When is it reasonable to assume \(A_{i} \approx A_{o} \approx A_{s}\) ?

Short Answer

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Answer: The heat transfer surface area of the inner and outer tubes in a double-pipe parallel-flow heat exchanger can be determined using the following formulas: - For the inner tube: \(A_i = 2 * \pi * \frac{D_2}{2} * L\) - For the outer tube: \(A_o = 2 * \pi * \frac{D_3}{2} * L\) It is reasonable to assume that the heat transfer surface areas are roughly equal when the ratio of the tube diameters is close to 1, which is when \(\frac{D_2}{D_3} \approx 1\). This generally occurs when the wall thickness between the two tubes is very small.

Step by step solution

01

Determine the surface area of the inner tube, \(A_{i}\)

To find the surface area of the inner tube, we need to consider the length of the inner tube and its outer diameter. The surface area of the inner tube, \(A_{i}\), can be calculated using the formula for the lateral surface area of a cylinder: \(A_i = 2 * \pi * \frac{D_2}{2} * L\) where \(\frac{D_2}{2}\) is the radius of the inner tube.
02

Determine the surface area of the outer tube, \(A_{o}\)

To find the surface area of the outer tube, we need to consider the length of the outer tube and the inner diameter of the outer tube, which is in contact with the surface of the inner tube. The surface area of the outer tube, \(A_o\), can be calculated using the formula for the lateral surface area of a cylinder: \(A_o = 2 * \pi * \frac{D_3}{2} * L\) where \(\frac{D_3}{2}\) is the radius of the outer tube.
03

Determine when it is reasonable to assume \(A_{i} \approx A_{o} \approx A_{s}\)

In some cases, the wall thickness between the two tubes is very small, meaning that there is little difference between \(A_i\) and \(A_o\). In these situations, it is reasonable to assume that \(A_{i} \approx A_{o} \approx A_{s}\). This is generally the case when the ratio of the tube diameters is close to 1: \(\frac{D_2}{D_3} \approx 1\) In summary, we have determined the heat transfer surface areas of the inner and outer tubes \(A_{i}\) and \(A_{o}\), and discussed when it is reasonable to assume that the heat transfer surface areas are roughly equal.

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Key Concepts

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

Heat Transfer Surface Area Calculation
Understanding how to calculate the heat transfer surface area is essential when analyzing a double-pipe heat exchanger. The ability to transfer heat effectively across a surface area is fundamental to the exchanger's efficiency.

As detailed in the solution, the surface area (\(A_i\) for the inner tube and \(A_o\) for the outer tube) is vital for the heat transfer process. For the inner tube's surface area \(A_i\), the key formula is \(A_i = \pi \cdot D_2 \cdot L\), where \(D_2\) is its outer diameter and \(L\) represents the length of the tube. Similarly, the outer tube's surface area \(A_o\) uses \(A_o = \pi \cdot D_3 \cdot L\), with \(D_3\) being the inner diameter of the outer tube.

In practical applications, the surfaces through which heat is exchanged are often measured to determine how quickly and efficiently heat can be transferred from one fluid to another. Ideal conditions for the assumption that \(A_i\) is approximately equal to \(A_o\) occur when the double-pipe structure is designed with minimal space between the two tubings, making the ratios of their diameters close to unity. When designing such heat exchangers, engineers need to ensure that the calculated surface areas will provide an adequate rate of heat transfer for the intended operation.
Parallel-Flow Heat Exchanger
The double-pipe heat exchanger mentioned in the exercise operates on the principle of parallel flow. In a parallel-flow heat exchanger, both the hot and cold fluids move in the same direction. This alignment presents a high temperature difference at one end of the exchanger, but this difference typically decreases along the length of the exchanger as the fluids approach thermal equilibrium.

In applications where temperature consistency is required quickly and with smaller surface areas, parallel-flow arrangements are often favorable. This design is also simpler and less expensive to build compared to counterflow configurations. However, it's worth noting that parallel-flow heat exchangers generally have a lower overall heat transfer efficiency than counterflow systems, as the temperature gradient, which drives the transfer, diminishes more quickly along the length of the exchanger.

The efficiency and effectiveness of heat transfer in parallel-flow exchangers directly depend on the geometry and the surface areas of the tubes, emphasizing the importance of accurate calculation of these areas to predict and optimize performance.
Lateral Surface Area of a Cylinder
The concept of lateral surface area of a cylinder is central to calculating the heat transfer surface areas in double-pipe heat exchangers. A cylinder's lateral surface area can be visualized as the 'unrolled' outer surface of its side. It excludes the top and bottom faces of the cylinder, focusing on the area around the central axis.

To calculate this, the formula \(A = 2 \cdot \pi \cdot r \cdot h\) is used, where \(r\) is the radius of the cylinder and \(h\) is its height or length. Applying this to our problem, for the inner and outer tubes in the double-pipe heat exchanger, their lateral surfaces are key to determining the heat transfer areas \(A_i\) and \(A_o\) respectively. This calculation assumes a perfect cylindrical shape and does not account for any deviations or irregularities in the physical structure of the tubes. Such irregularities could affect actual heat transfer rates.

Understanding the parameters influencing lateral surface area, including the impact of diameter and length variations, is crucial for accuracy in these calculations—crucial in designing effective heat exchangers for real-world applications.

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Most popular questions from this chapter

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(20^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(4.5 \mathrm{~kg} / \mathrm{s}\). Heat is supplied by hot oil \(\left(c_{p}=2300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(170^{\circ} \mathrm{C}\) at a rate of \(10 \mathrm{~kg} / \mathrm{s}\). For a tube-side overall heat transfer coefficient of \(350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area on the tube side.

Consider a closed loop heat exchanger that carries exit water \(\left(c_{p}=1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\rho=62.4 \mathrm{lbm} / \mathrm{ft}^{3}\right)\) of a condenser side initially at \(100^{\circ} \mathrm{F}\). The water flows through a \(500 \mathrm{ft}\) long stainless steel pipe of 1 in inner diameter immersed in a large lake. The temperature of lake water surrounding the heat exchanger is \(45^{\circ} \mathrm{F}\). The overall heat transfer coefficient of the heat exchanger is estimated to be \(250 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). What is the exit temperature of the water from the immersed heat exchanger if it flows through the pipe at an average velocity of \(9 \mathrm{ft} / \mathrm{s}\) ? Use \(\varepsilon-\mathrm{NTU}\) method for analysis.

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.

A shell-and-tube heat exchanger with 2-shell passes and 8 -tube passes is used to heat ethyl alcohol \(\left(c_{p}=2670 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(25^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(2.1 \mathrm{~kg} / \mathrm{s}\). The heating is to be done by water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(95^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger.

In a chemical plant, a certain chemical is heated by hot water supplied by a natural gas furnace. The hot water \(\left(c_{p}=\right.\) \(4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) is then discharged at \(60^{\circ} \mathrm{C}\) at a rate of \(8 \mathrm{~kg} / \mathrm{min}\). The plant operates \(8 \mathrm{~h}\) a day, 5 days a week, 52 weeks a year. The furnace has an efficiency of 78 percent, and the cost of the natural gas is \(\$ 1.00\) per therm ( 1 therm \(=105,500 \mathrm{~kJ})\). The average temperature of the cold water entering the furnace throughout the year is \(14^{\circ} \mathrm{C}\). In order to save energy, it is proposed to install a water-to-water heat exchanger to preheat the incoming cold water by the drained hot water. Assuming that the heat exchanger will recover 72 percent of the available heat in the hot water, determine the heat transfer rating of the heat exchanger that needs to be purchased and suggest a suitable type. Also, determine the amount of money this heat exchanger will save the company per year from natural gas savings.

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