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An electrical current of 15 A flows in an 18 gage copper wire ( \(1.02 \mathrm{~mm}\) diameter). If the wire has an electrical resistance of \(0.0209 \Omega / \mathrm{m}\), calculate (i) the rate of heat generation per meter length of wire. (ii) the rate of heat generation per unit volume of copper. (iii) the heat flux across the wire surface at steady state.

Short Answer

Expert verified
(i) 4.7025 W/m; (ii) 5.755 x 10^6 W/m^3; (iii) 1467.58 W/m^2.

Step by step solution

01

Rate of Heat Generation Per Meter

The rate of heat generation (power dissipation) in a resistive wire is given by the formula for electrical power: \( P = I^2 R \) where \( I \) is the current and \( R \) is the resistance. Here, the resistance per meter (\( R \)) is \( 0.0209 \Omega / \text{m} \). So, we calculate the power per meter of wire as follows:\[ P = (15 \, \text{A})^2 \times 0.0209 \, \Omega/\text{m} = 4.7025 \, \text{W/m} \].
02

Calculate Wire Volume Per Meter

The volume of the wire per meter length can be calculated using the formula for the volume of a cylinder, \( V = A \times L \), where \( A \) is the cross-sectional area. The area \( A \) can be calculated using the equation: \( A = \pi \times \left(\frac{d}{2}\right)^2 \). Given \( d = 1.02 \, \text{mm} = 0.00102 \, \text{m} \), the area becomes:\[ A = \pi \times \left(\frac{0.00102}{2}\right)^2 = 8.1724 \times 10^{-7} \, \text{m}^2 \]. Since the length \( L \) is 1 meter, the volume \( V \) is:\[ V = 8.1724 \times 10^{-7} \, \text{m}^3 \].
03

Rate of Heat Generation Per Unit Volume

The rate of heat generation per unit volume is calculated by dividing the power generated by the volume of the wire. From the previous steps, we know the power per meter \( P = 4.7025 \, \text{W/m} \) and the volume \( V = 8.1724 \times 10^{-7} \, \text{m}^3 \).Thus, the heat generation per unit volume is:\[ \frac{P}{V} = \frac{4.7025}{8.1724 \times 10^{-7}} = 5.755 \times 10^{6} \, \text{W/m}^3 \].
04

Calculate the Surface Area of the Wire

The surface area per meter length of cylindrical wire is given by \( A_s = \pi \times d \times L \). With \( d = 0.00102 \, \text{m} \) and \( L = 1 \, \text{m} \), the calculation is:\[ A_s = \pi \times 0.00102 \, \text{m} \times 1 \, \text{m} = 3.2041 \times 10^{-3} \, \text{m}^2 \].
05

Calculate Heat Flux Across the Wire Surface

Heat flux across the surface is the rate of heat transfer per unit area. It is calculated by dividing the power dissipation per meter by the surface area of the wire. Using \( P = 4.7025 \, \text{W/m} \) from Step 1 and the surface area \( A_s = 3.2041 \times 10^{-3} \, \text{m}^2 \) from Step 4:\[ \text{Heat flux} = \frac{4.7025}{3.2041 \times 10^{-3}} \approx 1467.58 \, \text{W/m}^2 \].

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Key Concepts

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

Electrical Resistance
Electrical resistance is a fundamental concept in understanding how currents flow through materials. It is the opposition that a material presents to the flow of electric current. Resistance is determined by the material's properties and its dimensions. For example, longer and thinner materials typically have higher resistance compared to shorter and thicker ones.
In our given exercise, the electrical resistance of the copper wire is specifically measured per meter, standing at \(0.0209 \Omega / \mathrm{m}\). This means every meter of wire contributes this amount of resistance to the flow of current.
The significance of resistance in our problem is apparent when we use it alongside the current to determine the heat generated. According to Ohm's Law, increasing resistance or current will result in more heat generation. This clearly highlights why understanding resistance is crucial when analyzing electrical systems, particularly for those involving heat generation as a byproduct.
Heat Flux
Heat flux is an essential concept when studying heat transfer characteristics of materials. It describes the rate of heat energy transfer per unit area, typically measured in \(\text{W/m}^2\). It's a useful metric for understanding how efficiently heat dissipates across a surface.
In the exercise, we see how the heat generated by the resistance of the copper wire affects the surrounding environment. The heat flux is determined by dividing the rate of heat generation by the surface area of the wire. This lets us know how much heat flows outward per square meter of the wire's surface.
By figuring this out, we can understand how much heat the wire will dissipate under normal conditions. This can be vital in applications where temperature control is important, preventing overheating which might damage the wire or any surrounding insulation.
Copper Wire
Copper is a common choice for electrical wiring due to its excellent electrical conductivity and ductility. It offers low resistance, facilitating effective current flow, which is why it's often preferred for making wires.
In this exercise, we are dealing with an 18-gauge copper wire, which has a diameter of \(1.02 \text{ mm}\). The gauge of a wire affects its ability to conduct current and its total resistance. Generally, higher gauge numbers tend to mean thinner wires with higher resistance.
The properties of copper, such as its resistance per meter, directly impact the rate of heat generation when current flows through the wire. Understanding these physical characteristics helps in predicting how the wire performs under different electrical loads and how well it handles thermal stresses.
Thermal Analysis
Thermal analysis involves examining the temperature distribution within or around an object and understanding how thermal energy is transferred. It's critical in designing systems prone to heat generation, such as electrical wires.
In the context of the exercise, thermal analysis allows us to predict where heat might collect and how it can exit the system. The steady state condition assumed in the problem implies that energy input and energy dissipation balance out, and the temperature profile stabilizes over time.
This sort of analysis can ensure safety and reliable operation, as it helps anticipate and mitigate issues like overheating. By evaluating heat distribution and dissipation through thermal analysis, engineers can optimize components for efficiency and longevity.

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Most popular questions from this chapter

A \(5 \mathrm{~mm} \times 2 \mathrm{~mm} \times 1 \mathrm{~mm}\)-thick semiconductor laser is mounted on a \(1 \mathrm{~cm}\) cube copper heat sink and enclosed in a Dewar flask. The laser dissipates \(2 \mathrm{~W}\), and a cryogenic refrigeration system maintains the copper block at a nearly uniform temperature of \(90 \mathrm{~K}\). Estimate the top surface temperature of the laser chip for the following models of the dissipation process: (i) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer underneath the top surface of the laser. (ii) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer at the midplane of the chip. (iii) The energy is dissipated uniformly through the chip. Take \(k=170 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the chip, and neglect parasitic heat gains from the Dewar flask.

A thin-wall, spherical stainless steel vessel has an outside diameter of \(40 \mathrm{~cm}\) and contains biochemical reactants maintained at \(160^{\circ} \mathrm{C}\). The tank is located in a laboratory where the air is maintained at \(20^{\circ} \mathrm{C}\). A layer of insulation is to be added to prevent workers from being burned by accidental contact with the vessel. The insulation chosen has a thermal conductivity of \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). How thick should the layer be if the threshold for a skin burn on a nonmetallic surface can be taken as \(55^{\circ} \mathrm{C}\) ? Also calculate the heat loss. The combined convective and radiative heat transfer coefficients on the outside of the insulation is estimated to be \(9 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

The absorber of a simple flat-plate solar collector with no coverplate consists of a \(2 \mathrm{~mm}\)-thick aluminum plate with \(6 \mathrm{~mm}\)-diameter aluminum water tubes spaced at a pitch of \(10 \mathrm{~cm}\), as shown. On a clear summer day near the ocean, the air temperature is \(20^{\circ} \mathrm{C}\), and a steady wind is blowing. The solar radiation absorbed by the plate is calculated to be 680 \(\mathrm{W} / \mathrm{m}^{2}\), and the convective heat transfer coefficient is estimated to be \(14 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If water at \(20^{\circ} \mathrm{C}\) enters the collector at \(7 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\) per meter width of collector, and the collector is \(3 \mathrm{~m}\) long, estimate the outlet water temperature. For the aluminum take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and \(\varepsilon=0.20\). (Hint: Evaluate the heat lost by reradiation using Eq. (1.19) with a constant value of \(h_{r}\) corresponding to a guessed average plate temperature. Then apply the steady-flow equation to an elemental length of the collector, and so derive a differential equation governing the water temperature increase along the collector.)

Consider a heat barrier consisting of a \(2 \mathrm{~mm}\)-thick brass plate to which \(3 \mathrm{~mm}\) copper tubing is soldered. The tubes are spaced \(10 \mathrm{~cm}\) apart. Cooling water passed through the tubes keeps them at approximately \(315 \mathrm{~K}\). The underside of the brass wall is insulated with a \(1.5 \mathrm{~cm}\)-thick asbestos layer, which in turn contacts a hot wall at \(600 \mathrm{~K}\). Assuming that the heat transfer from the cold side of the brass plate is negligible, estimate the temperature of the hottest spot on the brass wall. Take \(k=0.16 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\) for the asbestos and \(111 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the brass.

A steel heat exchanger tube of \(2 \mathrm{~cm}\) outer diameter is fitted with a steel spiral annular fin of \(5 \mathrm{~cm}\) outer diameter and \(0.2 \mathrm{~mm}\) thickness, wound at a pitch of 3 \(\mathrm{mm}\). If the outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the bare tube and 15 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\) on the finned tube, determine the reduction in the outside thermal resistance achieved by adding the fins. Take \(k_{\text {steel }}=42 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

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