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A 2.1-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at \(20^{\circ} \mathrm{C}\). Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be \(180^{\circ} \mathrm{C}\) in steady operation. Also, the voltage drop and electric current through the wire are measured to be \(110 \mathrm{~V}\) and \(3 \mathrm{~A}\), respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: \(156 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short Answer

Expert verified
Based on the given step by step solution, the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room is calculated to be 156 W/m²·K.

Step by step solution

01

Calculate the power generated by resistance heating

To calculate the power generated by resistance heating, we can use the following formula: \(P = VI\) where P is the power, V is the voltage drop, and I is the electric current. In this case, V = 110 V, and I = 3 A. Therefore, the power generated is: \(P = (110 \mathrm{~V})(3 \mathrm{~A}) = 330 \mathrm{~W}\)
02

Calculate the surface area of the wire

The surface area (A) of a cylinder can be calculated using the following formula: \(A = 2 \pi rL\), where r is the radius, and L is the length of the wire. In this case, the diameter of the wire is 0.2 cm, which means the radius is 0.1 cm or 0.001 m, and the length is 2.1 m. Therefore, the surface area of the wire is: \(A = 2\pi(0.001\mathrm{~m})(2.1\mathrm{~m})=0.0132\mathrm{~m}^2\)
03

Calculate total heat transfer by convection

Now that we have the power generated (P) and the surface area of the wire (A), we can calculate the total heat transfer (Q) by convection using the following formula: \(Q = P - Q_{rad}\) Since we're disregarding any heat transfer by radiation (Q_rad = 0), the total heat transfer by convection (Q) is equal to the power generated (P): \(Q = 330 \mathrm{~W}\)
04

Calculate the convection heat transfer coefficient

Finally, we can calculate the convection heat transfer coefficient (h) using the following formula: \(h = \frac{Q}{A(T_s - T_{room})}\) where \(T_s\) is the surface temperature of the wire, and \(T_{room}\) is the room temperature. In this case, T_s = 180°C, and T_room = 20°C, so \(T_s - T_{room} = 180 - 20 = 160 \mathrm{~K}\). Therefore, the convection heat transfer coefficient is: \(h = \frac{330 \mathrm{~W}}{(0.0132\mathrm{~m}^2)(160\mathrm{~K})}=156\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) The convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room is 156 W/m²·K.

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Key Concepts

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

Resistance Heating
Resistance heating is a process where electrical energy is converted into heat as an electrical current passes through a resistant material. In the context of our exercise, an electrical wire acts as the resistor. When current flows through the wire, the electrical resistance of the wire causes it to heat up.

The power generated can be calculated using the formula:
\(P = VI\), where \(P\) is the power in watts (W), \(V\) is the voltage in volts (V), and \(I\) is the current in amperes (A). To put this into perspective, if a wire in our exercise has a voltage drop of 110 V and carries a current of 3 A, the power generated by resistance heating is \(P = 110 V \times 3 A = 330 W\). This amount of power consequently raises the wire's temperature, which can be observed by its surface temperature.
Heat Transfer by Convection
Heat transfer by convection occurs when heat is carried away from the surface of an object by a fluid such as air or water. In our exercise, the wire is surrounded by air at a lower temperature, and heat flows from the wire into the air.

To describe this process quantitatively, we introduce the convection heat transfer coefficient (\(h\)), which is measured in watts per square meter per kelvin (\(W/m^2\cdot K\)). The coefficient represents the efficiency of heat transfer from the wire's surface to the surrounding air.

Our goal is to determine the value of \(h\) for our specific setup. We already know that radiation is disregarded, simplifying our calculation. We use the formula:
\(h = \frac{Q}{A(T_s - T_{\text{room}})}\), where \(A\) is the wire's surface area, \(T_s\) is the wire's surface temperature, and \(T_{\text{room}}\) is the room's air temperature. By calculating the values of \(A\) and \(Q\) from the wire's dimensions and power generated, we can solve for \(h\). In the given problem, the convection heat transfer coefficient is found to be 156 \(W/m^2\cdot K\), indicating the rate of heat loss per unit area per degree of temperature difference between the wire and the surrounding air.
Thermal Analysis of Electrical Wire
The thermal analysis of electrical wire involves understanding how the wire's temperature changes under the influence of current flow and the subsequent heat transfer to its surroundings. Our thermal analysis is crucial for determining the safe operating conditions of the wire and ensuring it does not overheat and cause damage or fire.

Such an analysis will consider factors such as the wire's material, its physical dimensions (diameter and length, which determine the surface area), the ambient temperature, and the properties of the surrounding medium (usually air). In the context of the given exercise, the wire attains a steady-state temperature of \(180^\circ C\) because the generated heat by resistance heating is balanced by the heat lost through convection.

The thermal analysis allows us to calculate essential parameters, like the convection heat transfer coefficient (\(h\)), which serves as an indicator of the wire's ability to dissipate heat into the environment. Ensuring the wire does not exceed certain temperature thresholds is vital for safety and maintaining the integrity of electrical systems.

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Most popular questions from this chapter

On a still clear night, the sky appears to be a blackbody with an equivalent temperature of \(250 \mathrm{~K}\). What is the air temperature when a strawberry field cools to \(0^{\circ} \mathrm{C}\) and freezes if the heat transfer coefficient between the plants and air is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) because of a light breeze and the plants have an emissivity of \(0.9\) ? (a) \(14^{\circ} \mathrm{C}\) (b) \(7^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-3^{\circ} \mathrm{C}\)

An aluminum pan whose thermal conductivity is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has a flat bottom with diameter \(15 \mathrm{~cm}\) and thickness \(0.4 \mathrm{~cm}\). Heat is transferred steadily to boiling water in the pan through its bottom at a rate of \(1400 \mathrm{~W}\). If the inner surface of the bottom of the pan is at \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

An AISI 316 stainless steel spherical container is used for storing chemicals undergoing exothermic reaction that provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner diameter of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~cm}\). For safety reason to prevent thermal burn on individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). If the ambient temperature is \(23^{\circ} \mathrm{C}\), determine the necessary convection heat transfer coefficient to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Is the necessary convection heat transfer coefficient feasible with free convection of air? If not, discuss other option to prevent the container's outer surface temperature from causing thermal burn.

The north wall of an electrically heated home is 20 \(\mathrm{ft}\) long, \(10 \mathrm{ft}\) high, and \(1 \mathrm{ft}\) thick, and is made of brick whose thermal conductivity is \(k=0.42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}\). On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about \(62^{\circ} \mathrm{F}\) and \(25^{\circ} \mathrm{F}\), respectively, for a period of \(8 \mathrm{~h}\). Determine \((a)\) the rate of heat loss through the wall that night and \((b)\) the cost of that heat loss to the home owner if the cost of electricity is \(\$ 0.07 / \mathrm{kWh}\).

A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -hour period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\).

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