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Chapter 11: Problem 10

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

Expert verified
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

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.

Consider a shell and tube heat exchanger in a milk be heated from \(20^{\circ} \mathrm{C}\) by hot water initially at \(140^{\circ} \mathrm{C}\) and flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The milk flows through 30 thin-walled tubes with an inside diameter of \(20 \mathrm{~mm}\) with each tube making 10 passes through the shell. The average convective heat transfer coefficients on the milk and water side are \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. In order to complete the pasteurizing process and hence restrict the microbial growth in the milk, it is required to have the exit temperature of milk attain at least \(70^{\circ} \mathrm{C}\). As a design engineer, your job is to decide upon the shell width (tube length in each pass) so that the milk exit temperature of \(70^{\circ} \mathrm{C}\) can be achieved. One of the design requirements is that the exit temperature of hot water should be at least \(10^{\circ} \mathrm{C}\) higher than the exit temperature of milk.

Consider two double-pipe counter-flow heat exchangers that are identical except that one is twice as long as the other one. Which heat exchanger is more likely to have a higher effectiveness?

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Consider a condenser unit (shell and tube heat exchanger) of an HVAC facility where saturated refrigerant \(\mathrm{R} 134 \mathrm{a}\) at a saturation pressure of \(1318.6 \mathrm{kPa}\) and a rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) flows through thin-walled copper tubes. The refrigerant enters the condenser as saturated vapor and it is desired to have a saturated liquid refrigerant at the exit. The cooling of refrigerant is carried out by cold water that enters the heat exchanger at \(10^{\circ} \mathrm{C}\) and exits at \(40^{\circ} \mathrm{C}\). Assuming initial overall heat transfer coefficient of the heat exchanger to be \(3500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the surface area of the heat exchanger and the mass flow rate of cooling water for complete condensation of the refrigerant. In practice, over a long period of time, fouling occurs inside the heat exchanger that reduces its overall heat transfer coefficient and causes the mass flow rate of cooling water to increase. Increase in the mass flow rate of cooling water will require additional pumping power making the heat exchange process uneconomical. To prevent the condenser unit from under performance, assume that fouling has occurred inside the heat exchanger and has reduced its overall heat transfer coefficient by \(20 \%\). For the same inlet temperature and flow rate of refrigerant, determine the new flow rate of cooling water to ensure complete condensation of the refrigerant at the heat exchanger exit.
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