Question

Consider a solid stainless steel wire with a thermal conductivity of $14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(. The wire has a diameter of \)1 \mathrm{~mm}\(, has a resistivity of \)45 \times 10^{-8} \Omega \cdot \mathrm{m}\(, and carries a current of \)120 \mathrm{~A}$. (a) Determine the rate of heat generated within the wire $\left(\mathrm{W} / \mathrm{m}^{3}\right.\( ), and \)(b)$ calculate the maximum temperature rise in the wire.

Short answer

Question: Determine the heat generated within the wire and the maximum temperature rise of the wire with the following given values: diameter \(d = 1 \times 10^{-3}\,\text{m}\), resistivity \(\rho = 45 \times 10^{-8}\, \Omega \cdot \mathrm{m}\), current \(I = 120\, \mathrm{A}\), and thermal conductivity \(k = 14\, \mathrm{W}/\mathrm{m}\cdot\mathrm{K}\).

Step-by-step solution

Calculate the resistance per unit length

Firstly, we have to calculate the resistance per unit length. To do this, we use the formula: \(\text{Resistance per unit length} = \rho\frac{L}{A}\) The area of the wire can be found using the diameter of the wire by the following formula: \(\text{Area} = A = \pi(\frac{d}{2})^2\)

Calculate the heat generated per unit length

Now that we have the resistance per unit length, we can find the heat generated per unit length using the formula: \(\text{Heat generated per unit length} = Q = I^2R\)

Calculate the heat generated per unit volume

After we have the heat generated per unit length, we can find the heat generated per unit volume using the formula: \(\text{Heat generated per unit volume} = q = \frac{Q}{V}\) We need to find the volume of the wire first: \(\text{Volume} = V = A \cdot L\)

Calculate the maximum temperature rise

Finally, we can find the maximum temperature rise using the formula: \(\text{Maximum temperature rise} = T = \frac{Q}{k}\) Now we will plug in the given values and solve for the different parameters. Given: Diameter \(d = 1 \times 10^{-3}\,\text{m}\) Resistivity \(\rho = 45 \times 10^{-8}\, \Omega \cdot \mathrm{m}\) Current \(I = 120\, \mathrm{A}\) Thermal conductivity \(k = 14\, \mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) Calculate: \(A = \pi(\frac{d}{2})^2 \Rightarrow A = 7.854 \times 10^{-7}\, \mathrm{m}^2\) \(\rho\frac{L}{A} \Rightarrow R = 5.731\times10^{-5}\, \Omega\cdot\mathrm{m}\) \(Q = (120\,\mathrm{A})^2\cdot(5.731\times10^{-5}\, \Omega\cdot\mathrm{m}) \Rightarrow Q = 822.99\, \mathrm{W}/\mathrm{m}\) \(V=L\cdot A = L\cdot7.854 \times 10^{-7}\, \mathrm{m}^2\) \(q = \frac{822.99\, \mathrm{W}/\mathrm{m}}{7.854 \times 10^{-7}\, \mathrm{m}^2\cdot L} \Rightarrow q=1.047\times10^{6}\, \mathrm{W}/\mathrm{m}^3\) \(T = \frac{822.99\, \mathrm{W}/\mathrm{m}}{14\, \mathrm{W}/\mathrm{m}\cdot\mathrm{K}} \Rightarrow T = 58.78 \,^\circ \mathrm{K}\) So, the heat generated within the wire is \(1.047 \times 10^6 \, \mathrm{W}/\mathrm{m}^{3}\), and the maximum temperature rise in the wire is \(58.78\,^\circ \mathrm{K}\).