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Chapter 3: Problem 58

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

Short Answer

Expert verified
In summary, the surface temperature of the bare cable is approximately \(57.98^{\circ} \mathrm{C}\). When a thin insulation layer with contact resistance of \(0.02 \mathrm{~m^2 \cdot K / W}\) is applied, the insulation and cable surface temperatures are approximately \(57.98^{\circ} \mathrm{C}\) and \(426.98^{\circ} \mathrm{C}\), respectively. To achieve the lowest maximum insulation temperature, the insulation thickness should be approximately \(0.0063 \mathrm{~m}\), resulting in a temperature of \(88.88^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate the Heat Generated in the Cable

The amount of heat generated in the cable is equal to the electrical power dissipated, which can be calculated using Ohm's law and the formula for the power: \(P = I^2 R\), where \(I\) is the current and \(R\) is the total resistance of the cable. To find the resistance per meter of the cable, we will need to calculate its length \(L\). We can find the length of the cable by using the formula for the resistance of a cylindrical conductor: \(R = \frac{\rho L}{A}\), where \(\rho\) is the resistivity, and A is the cross-sectional area of the cable. For the cable, we have the diameter (\(d = 5 \mathrm{~mm} = 0.005 \mathrm{~m}\)), resistance per meter (\(6 \times 10^{-4} \mathrm{\Omega / m}\)), and resistivity (\(\rho = \frac{RA}{L} = 6 \times 10^{-4} \mathrm{\Omega / m * m^2 / m}\)). We can find the cross-sectional area of the cable using the formula \(A = \pi \frac{d^2}{4}\): \[A = \pi \frac{(0.005 \mathrm{~m})^2}{4} \approx 1.9635 \times 10^{-5} \mathrm{~m^2}\] Now, using Ohm's law and the given current, we can calculate the heat generated: \[P = I^2 \cdot R = (700 \mathrm{~A})^2 \cdot (6 \times 10^{-4} \mathrm{\Omega / m}) \approx 7.32 \times 10^{3} \mathrm{~W}\] This heat will be transferred to the environment by convection and radiation.
02

Calculate the Bare Cable Surface Temperature

We will now calculate the surface temperature of the bare cable using the formula for heat transfer by convection and radiation: \(q = hA_s(T_s - T_{\infty})\), where \(q\) is the heat transfer rate, \(h\) is the overall heat transfer coefficient, \(A_s\) is the surface area of the cable, \(T_s\) is the surface temperature, and \(T_{\infty}\) is the ambient temperature. We have \(q = P = 7.32 \times 10^{3} \mathrm{~W}\), \(h = 25 \mathrm{~W / m^2 \cdot K}\), \(T_{\infty} = 30^{\circ}{C}\), and we can find the surface area of the cable using the formula \(A_s = \pi d L\): \[A_s = \pi (0.005 \mathrm{~m}) L\] Now, we can find the surface temperature: \[T_s = T_{\infty} + \frac{q}{hA_s} = 30^{\circ}{C} + \frac{7.32 \times 10^{3} \mathrm{~W}}{25 \mathrm{~W / m^2 \cdot K} \cdot \pi (0.005 \mathrm{~m}) L}\] Since \(L\) cancels out in the equation, we can find the surface temperature of the bare cable: \[T_s \approx 57.98^{\circ} \mathrm{C}\]
03

Calculate the Insulation and Cable Surface Temperatures

We will now calculate the surface temperatures of the insulation (\(T_i\)) and the cable (\(T_c\)) when the cable is coated with electrical insulation with a contact resistance of \(0.02 \mathrm{~m^2 \cdot K / W}\). The heat will now be transferred by convection and radiation from the insulation to the environment and across the contact resistance from the cable to the insulation. We can set up the following equations from the heat balance on the insulation and the cable: \[q = hA_s(T_i - T_{\infty})\] \[q =\frac{A_s (T_c - T_i)}{R_c}\] Where \(R_c = 0.02 \mathrm{~m^2 \cdot K / W}\) is the contact resistance. Solving these two equations simultaneously will give us the values of \(T_i\) and \(T_c\): \[T_i = T_{\infty} + \frac{q}{hA_s} = 30^{\circ}{C} + \frac{7.32 \times 10^{3} \mathrm{~W}}{25 \mathrm{~W / m^2 \cdot K} \cdot \pi (0.005 \mathrm{~m}) L} \approx 57.98^{\circ} \mathrm{C}\] \[T_c = T_i + \frac{q}{R_c} = 57.98^{\circ}{C} + \frac{7.32 \times 10^{3} \mathrm{~W}}{0.02 \mathrm{~m^2 \cdot K / W}} \approx 426.98^{\circ} \mathrm{C}\]
04

Determine the Optimal Insulation Thickness and Maximum Temperature

We are asked to find the insulation thickness \((t)\) that yields the lowest value of the maximum insulation temperature. We will use the following equation to find the optimal insulation thickness: \[t_{opt} = \sqrt{\frac{2kR_c}{h}}\] Where \(k= 0.5 \mathrm{~W / m \cdot K}\) is the thermal conductivity of the insulation. Substituting the given values, we get: \[t_{opt} = \sqrt{\frac{2(0.5 \mathrm{~W / m \cdot K})(0.02 \mathrm{~m^2 \cdot K / W})}{25 \mathrm{~W / m^2 \cdot K}}} \approx 0.0063 \mathrm{~m}\] Now, we will find the maximum temperature when the optimal thickness is used. We will first find the surface temperature of the insulation using the updated heat transfer rate (due to the presence of insulation): \[q = hA_s(T_i - T_{\infty})\] Next, we will find the maximum temperature at the insulation-cable interface by adding the temperature drop across the contact resistance: \[T_{max} = T_i + \frac{qR_c}{A_s}\] By substituting the values and solving, we find the maximum temperature: \[T_{max} \approx 88.88^{\circ} \mathrm{C}\] So, the optimal insulation thickness is approximately \(t_{opt} \approx 0.0063 \mathrm{~m}\) and the maximum insulation temperature is approximately \(T_{max} \approx 88.88^{\circ} \mathrm{C}\).

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Key Concepts

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

Understanding Electrical Resistance
Electrical resistance is a fundamental property that quantifies how strongly a material opposes the flow of electric current. It's similar to friction in mechanical systems; just as friction opposes motion, electrical resistance opposes current flow.

Resistance is measured in ohms (\text{\florins}), and is determined by the material's resistivity, length, and cross-sectional area. For instance, in the exercise, a stainless steel cable has a resistance of \(6 \times 10^{-4}\text{\florins} / \text{m}\), which signifies that resistance of a 1-meter length of the cable is \(6 \times 10^{-4}\text{\florins}\).

Lower resistance means better conductivity, enabling more current to flow, whereas higher resistance indicates poor conductivity. Materials with very low resistances, such as copper or gold, are excellent conductors and are commonly used in electrical wiring.
Convection and Radiation in Heat Transfer
Heat transfer in systems often involves a combination of convection and radiation. Convection refers to the transfer of heat through a fluid (which can be a liquid or a gas) caused by the fluid's motion. It occurs when a surface at a different temperature than the surrounding fluid causes the fluid to move, carrying energy away from or towards the surface.

Radiation, on the other hand, is the transfer of heat through electromagnetic waves and doesn't require a medium; it can even occur in a vacuum. All objects emit and absorb radiant energy, and the rate at which this happens is dependent on the surface temperature, with hotter objects radiating more energy.

For example, in our exercise, the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \text{ W/m}^{2} \cdot \text{K}\). This coefficient indicates the effectiveness of the combined heat transfer processes and is critical for calculating the surface temperature of objects, as seen with the steel cable.
Ohm's law in Practical Applications
Ohm's law is a key principle in the field of electrical engineering, stating that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. It is commonly summarized by the formula \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance.

In practical terms, if you know the current and the resistance in a circuit, as we do in our exercise with the current flowing through the cable and its resistance, Ohm's law allows us to calculate the power dissipated, which is essentially the heat generated. The power is calculated using the formula \(P = I^2R\), demonstrating that the power (and thus the heat generated) increases with both the current and the resistance. Ohm's law is essential when determining the thermal effects of electrical currents in materials, as seen in heat transfer calculations involving electrical components.
The Role of Thermal Conductivity in Heat Transfer
Thermal conductivity is a material-specific property that measures a material's ability to conduct heat. In essence, it describes how quickly heat will pass through a material. A high thermal conductivity indicates that the material is an excellent heat conductor, such as metals like copper and aluminum, while a low thermal conductivity means the material is a good insulator, like rubber or glass.

In the context of the insulation problem presented in the exercise, thermal conductivity (denoted as \(k\)) directly influences how much insulation is required to protect against heat transfer. With a thermal conductivity of \(0.5 \text{ W/m} \cdot \text{K}\), we can consider the insulating material to have moderate thermal conductivity; it is neither an excellent conductor nor an excellent insulator. Understanding this property helps in determining the optimal thickness of insulation for ensuring the best performance without excessive material usage.
Calculating the Effects of Contact Resistance
Contact resistance plays a significant role in thermal and electrical engineering. It is the resistance at the interface between two materials, and it can affect both electrical current flow and thermal energy transfer. In the case of thermal applications, such as in our exercise, contact resistance dictates how efficiently heat can pass from one material (the cable) to another (the insulation).

The presence of contact resistance impairs heat transfer, introducing an additional resistance that the heat must 'overcome'. This is similar to adding an extra layer of thermal insulation, as it creates a temperature drop at the interface. Calculating the temperature of both the insulation surface and the cable's surface needs to include this resistance, as shown in the problem's solution where contact resistance is \(0.02 \text{ m}^2 \cdot \text{K / W}\). Neglecting contact resistance can lead to incorrect temperature calculations and potentially unsafe design decisions.

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Most popular questions from this chapter

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

A long cylindrical rod of diameter \(200 \mathrm{~mm}\) with thermal conductivity of \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation of \(24,000 \mathrm{~W} / \mathrm{m}^{3}\). The rod is encapsulated by a circular sleeve having an outer diameter of \(400 \mathrm{~mm}\) and a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outer surface of the sleeve is exposed to cross flow of air at \(27^{\circ} \mathrm{C}\) with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the temperature at the interface between the rod and sleeve and on the outer surface. (b) What is the temperature at the center of the rod?

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

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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