Question

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

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

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: \(P = (110 \mathrm{~V})(3 \mathrm{~A}) = 330 \mathrm{~W}\)

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: \(A = 2\pi(0.001\mathrm{~m})(2.1\mathrm{~m})=0.0132\mathrm{~m}^2\)

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): \(Q = 330 \mathrm{~W}\)

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.

Key concepts

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.