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Chapter 3: Problem 163

Consider the conditions of Problem \(3.149\) but now allow for a tube wall thickness of \(5 \mathrm{~mm}\) (inner and outer diameters of 50 and \(60 \mathrm{~mm}\) ), a fin-to-tube thermal contact resistance of \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the fact that the water temperature, \(T_{w}=350 \mathrm{~K}\), is known, not the tube surface temperature. The water-side convection coefficient is \(h_{w}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer per unit tube length \((\mathrm{W} / \mathrm{m})\) to the water. What would be the separate effect of each of the following design changes on the heat rate: (i) elimination of the contact resistance; (ii) increasing the number of fins from four to eight; and (iii) changing the tube wall and fin material from copper to AISI 304 stainless steel \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ?

Short Answer

Expert verified
The rate of heat transfer per unit tube length is 93,729 W/m. Eliminating contact resistance increases the heat rate to 115,242 W/m, increasing the number of fins to eight increases the heat rate to 103,305 W/m, and changing the tube wall and fin material to AISI 304 stainless steel results in a heat rate of 50,351 W/m.

Step by step solution

01

1. Determine the tube geometry

First, let's find the radius of the inner and outer tube surfaces: \[r_i = \frac{D_i}{2} = \frac{50}{2} = 25 \mathrm{~mm} \] \[r_o = \frac{D_o}{2} = \frac{60}{2} = 30 \mathrm{~mm} \]
02

2. Calculate the thermal resistance due to conduction of the tube

We need to find the thermal resistance of the copper tube. The formula for the resistance is: \[R_{cond} = \frac{\ln(\frac{r_o}{r_i})}{2 \pi k} \] Assuming the thermal conductivity of copper is \(k = 400 \ \mathrm{W/m \cdot K}\), the thermal resistance of the tube is: \[R_{cond} = \frac{\ln(\frac{30}{25})}{2 \pi (400)} = 1.15 \times 10^{-4} \ \mathrm{m^2K/W} \]
03

3. Calculate the thermal resistance due to convection of the water

Next, we need to find the thermal resistance due to convection from the water at temperature \(T_w = 350 \ \mathrm{K}\). The formula for the convective resistance is: \[R_{conv} = \frac{1}{h_w A} \] The water-side convection coefficient is given as \(h_w = 2000 \ \mathrm{W/m^2 \cdot K} \). The area of the inner surface of the tube is: \[A = 2 \pi r_i L = 2 \pi (25) L = 50 \pi L \] Thus, the convective resistance is: \[R_{conv} = \frac{1}{2000 (50 \pi)}= 3.18 \times 10^{-4} \ \mathrm{m^2K/W} \]
04

4. Calculate the total thermal resistance

Now we consider the fin-to-tube thermal contact resistance (\(R_{contact}=10^{-4} \ \mathrm{m^2 \cdot K / W}\)). To find the total thermal resistance, we sum up all threes resistances: \[R_{total} = R_{conv} + R_{cond} + R_{contact} = 3.18 \times 10^{-4} + 1.15 \times 10^{-4} + 10^{-4} = 5.33 \times 10^{-4} \ \mathrm{m^2K/W} \]
05

5. Determine the rate of heat transfer

We need to determine the rate of heat transfer (\(Q_{total}\)) per unit tube length. To do so, we use Newton's Law of Cooling: \[Q_{total} = \frac{T_w - T_\infty}{R_{total}} \] We are given that \(T_\infty = 300 \ \mathrm{K}\), so: \[Q_{total} = \frac{350 - 300}{5.33 \times 10^{-4}} = 93,729 \ \mathrm{W/m} \] Now we will analyze the separate effects of the mentioned design changes on the heat rate.
06

6. Eliminating the contact resistance

If we eliminate the contact resistance, the total thermal resistance becomes: \[R'_\text{total} = R_{conv} + R_{cond} = 4.33 \times 10^{-4} \ \mathrm{m^2 }\cdot \mathrm{K/W}\] The new rate of heat transfer becomes: \[Q'_{total} = \frac{350 - 300}{4.33 \times 10^{-4}} = 115,242 \ \mathrm{W/m}\]
07

7. Increasing the number of fins

Increasing the number of fins will directly impact the contact resistance. Let's assume the new contact resistance with eight fins is half of the original: \[R'_{contact} = \frac{R_{contact}}{2} = 5 \times 10^{-5} \ \mathrm{m^2K/W}\] New total thermal resistance is given as: \[R''_{total} = R_{conv} + R_{cond} + R'_{contact} = 4.83 \times 10^{-4} \ \mathrm{m^2K/W}\] The heat transfer rate with twice the number of fins is: \[Q''_{total} = \frac{350 - 300}{4.83 \times 10^{-4}} = 103,305 \ \mathrm{W/m} \]
08

8. Changing the material from copper to AISI 304 stainless steel

Changing the material of tube wall and fins will affect the thermal conductance. The given thermal conductivity of AISI 304 stainless steel is 20 W/mK. The new thermal resistance due to conduction becomes: \[R'_\text{cond} = \frac{\ln\left(\frac{r_o}{r_i}\right)}{2\pi k'} = \frac{\ln\left(\frac{30}{25}\right)}{2\pi(20)}= 5.75 \times 10^{-4} \ \mathrm{m^2K/W} \] The new total thermal resistance with the new tube material is: \[R'''_ \mathrm{total} = R_{conv} + R'_\mathrm{cond} + R_\mathrm{contact} = 9.93 \times 10^{-4} \ \mathrm{m^2K/W} \] The heat transfer rate with the new AISI 304 stainless steel material is given as: \[Q'''_\mathrm{total} = \frac{T_w - T_\infty}{R'''_\mathrm{total}} = \frac{350 - 300}{9.93 \times 10^{-4}} = 50,351 \ \mathrm{W/m}\] In summary, the rate of heat transfer per unit tube length is 93,729 W/m. Eliminating contact resistance increases the heat rate to 115,242 W/m, increasing the number of fins to eight increases the heat rate to 103,305 W/m, and changing the tube wall and fin material to AISI 304 stainless steel results in a heat rate of 50,351 W/m.

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Key Concepts

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

Thermal Resistance
Thermal resistance is a critical concept in understanding how heat moves through materials. It acts like an obstacle that slows down the rate at which heat is transferred through a substance. Think of it like insulation in a house; it prevents heat from escaping.

In the context of the problem, thermal resistance in the tube is calculated for both conduction and convection processes, as well as for the contact between the fin and the tube wall. This total resistance ultimately determines how efficiently heat can travel from the water inside the tube to the surrounding environment.

Due to different materials or physical states, each component in the system has its specific thermal resistance. While conduction resistance depends on the material's properties and geometry, convection resistance revolves around the heat transfer coefficient and surface area. The concept also involves fin-to-tube contact resistance, highlighting how interfaces between dissimilar materials can change the heat transfer rates.
Convection
Convection is the process by which heat is transferred from one place to another through fluids like liquids or gases. This happens either by the bulk movement of the fluid itself or through thermal exchanges at the surface of a solid.

In this exercise, convection on the water side of the tube wall is a key player. The water flowing through the tube acts as a medium to transfer heat, driven by the water-side convection coefficient. This coefficient, denoted as \(h_w\), describes how effectively the heat moves from the fluid to the surface.

In the example problem, we use the formula for convective thermal resistance, \(R_{conv} = \frac{1}{h_w A}\), to determine how the convective process limits the rate of heat transfer. Understanding convection helps students make sense of how heat loss or retention can be optimized in real-life systems.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is a crucial property when analyzing the heat transfer through conduction. The higher the thermal conductivity, the better the material is at transferring heat.

In the provided exercise, the copper tube has a thermal conductivity of 400 W/m路K, making it a very good conductor of heat. Copper's high thermal conductivity is why it is often used in heat exchangers and other applications where efficient heat transfer is needed. Conversely, AISI 304 stainless steel has a substantially lower thermal conductivity of 20 W/m路K.

The switch from copper to stainless steel significantly modifies the thermal resistance due to conduction, as seen in the exercise. This change dramatically decreases the overall efficiency of heat transfer, which serves to illustrate how material selection is crucial in engineering design.
Newton's Law of Cooling
Newton's Law of Cooling gives us a way to estimate the rate of heat transfer between a body and its surroundings. It states that the rate of heat loss of a body is directly proportional to the difference in temperatures between the body and its environment.

In the problem, this law is applied to calculate the overall heat transfer rate from the water in the tube to the surroundings. Using the formula, \(Q_{total} = \frac{T_w - T_\infty}{R_{total}}\), we can see how the temperature difference, along with the total thermal resistance, affects the heat transfer rate.

This principle is fundamental in engineering because it helps predict how quickly an object will reach thermal equilibrium with its environment. By mitigating factors like thermal resistance and optimizing materials and surface designs, engineers can enhance heat exchange efficiency.

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Most popular questions from this chapter

A spherical Pyrex glass shell has inside and outside diameters of \(D_{1}=0.1 \mathrm{~m}\) and \(D_{2}=0.2 \mathrm{~m}\), respectively. The inner surface is at \(T_{s, 1}=100^{\circ} \mathrm{C}\) while the outer surface is at \(T_{s, 2}=45^{\circ} \mathrm{C}\). (a) Determine the temperature at the midpoint of the shell thickness, \(T\left(r_{m}=0.075 \mathrm{~m}\right)\). (b) For the same surface temperatures and dimensions as in part (a), show how the midpoint temperature would change if the shell material were aluminum.

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

Copper tubing is joined to the absorber of a flat-plate solar collector as shown. The aluminum alloy (2024-T6) absorber plate is \(6 \mathrm{~mm}\) thick and well insulated on its bottom. The top surface of the plate is separated from a transparent cover plate by an evacuated space. The tubes are spaced a distance \(L\) of \(0.20 \mathrm{~m}\) from each other, and water is circulated through the tubes to remove the collected energy. The water may be assumed to be at a uniform temperature of \(T_{w}=60^{\circ} \mathrm{C}\). Under steady-state operating conditions for which the net radiation heat flux to the surface is \(q_{\text {rad }}^{\prime \prime}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\), what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? Note that \(q_{\text {rad }}^{\prime \prime}\) represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. You may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).
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