Question

A long homogeneous resistance wire of radius \(r_{o}=\) \(5 \mathrm{~mm}\) is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of \(5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) as a result of resistance heating. If the temperature of the outer surface of the wire remains at \(180^{\circ} \mathrm{C}\), determine the temperature at \(r=3.5 \mathrm{~mm}\) after steady operation conditions are reached. Take the thermal conductivity of the wire to be \(k=6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short answer

Answer: The temperature at a distance of 3.5 mm from the center of the wire is 178.86°C.

Step-by-step solution

The heat generated in the wire due to electric current

The rate at which heat is generated in the wire is given as \(q_g = 5\times 10^7~\text{W}/\text{m}^3\). This heat must be conducted outward through the wire radially.

Heat conduction in a radial direction

For radial heat conduction, the equation is given by: \(\frac{d}{dr}(r k \frac{dT}{dr}) + q_g = 0.\) Here, \(r\) is the distance from the center of the wire, \(T\) is the temperature, and \(k\) is the thermal conductivity of the wire.

Solving the differential equation

Substitute the value of \(q_g\) and \(k\) into the equation and solve for \(\frac{dT}{dr}\): \(\frac{d}{dr}(r k \frac{dT}{dr}) + 5\times 10^7 = 0\) Integration yields: \(r \frac{dT}{dr} = -\frac{5\times 10^7}{6}{r^2} + C_1\) Next, integrate with respect to \(r\): \(T(r) = -\frac{5\times 10^7}{12}{r^3} + C_1 \ln r +C_2\) Now, we need to apply boundary conditions to get the values of \(C_1\) and \(C_2\).

Boundary conditions

At the outer radius (\(r=r_o = 5 \text{ mm} = 5\times 10^{-3}~\text{m}\)), the temperature is \(180^\circ\text{C}\): \(T(r_o) = T(5\times 10^{-3})= 180^\circ\text{C} = 453.15\text{K}\) (converting Celsius to Kelvin) Substitute values into the equation: \(453.15 = -\frac{5\times 10^7}{12}(5\times 10^{-3})^3 + C_1 \ln (5\times 10^{-3}) +C_2\) There is no source of heat at the center of the wire (since it is homogenous). Hence, as \(r \rightarrow 0,\) \(\frac{dT}{dr} \rightarrow 0\): \(0 = -\frac{5\times 10^7}{6}(0)^2 + C_1\) This means \(C_1 = 0\). So we can rewrite temperature equation as: \(T(r) = -\frac{5\times 10^7}{12}{r^3} + C_2\) Now substitute the values again to get \(C_2\): \(453.15 = -\frac{5\times 10^7}{12}(5\times 10^{-3})^3 + C_2\) Solving for \(C_2\), we get \(C_2 = 453.15 + \frac{5\times 10^7}{12}(5\times 10^{-3})^3 = 453.9247\) The temperature equation can now be written as: \(T(r) = -\frac{5\times 10^7}{12}{r^3} + 453.9247\)

Calculate temperature at r = 3.5 mm

We are asked to find the temperature at \(r = 3.5\text{~mm} = 3.5\times10^{-3}~\text{m}\). Substitute this value into the temperature equation: \(T(3.5\times10^{-3}) = -\frac{5\times 10^7}{12}(3.5\times 10^{-3})^3 + 453.9247\) Evaluating the expression, we get the temperature in Kelvin: \(T(3.5\times10^{-3}) = 452.009\text{K}\) Now convert the temperature back to Celsius: \(T = 452.009\text{K} - 273.15 = 178.86^\circ\text{C}\) So, the temperature at a radius of \(3.5\text{~mm}\) from the center of the wire is \(178.86^\circ\text{C}\).

Key concepts

Radial Heat Conduction

Heat conduction in a radial direction involves the transfer of heat through cylindrical objects like wires or pipes. Unlike straight-line heat conduction, radial conduction occurs from the inner core of a cylinder moving outward. This means as you move from the center of the wire towards the surface, the heat spreads out radially. In our specific case, heat is generated inside the wire and then moves out towards the air in the room.
The governing equation for radial heat conduction is derived from Fourier's law but adapted for cylindrical coordinates. The equation is given by \[\frac{d}{dr}(r k \frac{dT}{dr}) + q_g = 0,\] where:

• \(r\) is the radial distance from the center of the cylinder.
• \(k\) is the thermal conductivity of the material.
• \(T\) is the temperature, and \(q_g\) is the volumetric heat generation rate.

This equation describes how the temperature changes as you move away from the center. It plays a crucial role in determining how heat moves through materials like the resistance wire in our problem.

Thermal Conductivity

Thermal conductivity is a material property that shows how well a material can conduct heat. Materials with high thermal conductivity can transfer heat more efficiently compared to those with low conductivity. In the context of our problem, the wire has a conductivity (\(k\)) of 6 W/m·K, which is a moderate value.
Thermal conductivity is important because it influences how quickly heat can move through a material. The higher the thermal conductivity, the quicker the heat can pass from the center to the outer surface. The effect of thermal conductivity in our analysis is represented in the differential equation for radial heat conduction. Each substance will have its unique conductive efficiency, represented by this value. Understanding thermal conductivity helps predict the heating behavior of different materials under similar conditions.

Steady State Heat Transfer

In steady state heat transfer, the temperature distribution in the material does not change over time. That is, once the system reaches a steady state, the amount of heat entering a specific part of the material is equal to the amount leaving it.
In our problem, when the steady state is reached, the surface temperature of the wire remains constant at \(180 \degree C\). Steady state conditions simplify our analysis because, after some time, the temperatures become stable, making calculations easier as we know the heat into and out of any part of the system is balanced.
When we carry out heat transfer calculations under steady state assumptions, we focus on how heat flows spatially rather than temporally. This means for the spherical coordinates in our problem, we're assured that after some time, the heat flowing through any point in the wire remains consistent.

Resistance Heating

Resistance heating is a process where electrical energy is converted into heat as current passes through a resistive material. This is common in devices like electric heaters or stoves. When electrical current flows through a resistive material, due to its resistance, heat is generated.
In our exercise, resistance heating is responsible for generating heat within the wire at a rate of \(5 \times 10^7 \text{W/m}^3\). This uniform generation of heat ensures that the wire steadily raises the room's air temperature.
Understanding resistance heating is important because it highlights how the generation of heat happens directly in the material with minimal energy loss. The efficiency of resistance heating leads to its wide application in household and industrial settings. Recognizing the rate at which the heat is generated helps in designing systems that need precise temperature control, ensuring they operate safely and effectively within desired limits.