Question

Long aluminum wires of diameter $3 \mathrm{~mm}\left(\rho=2702 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, and \(\left.\alpha=9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) are extruded at a temperature of \(350^{\circ} \mathrm{C}\) and exposed to atmospheric air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine how long it will take for the wire temperature to drop to \(50^{\circ} \mathrm{C}\). (b) If the wire is extruded at a velocity of 10 \(\mathrm{m} / \mathrm{min}\), determine how far the wire travels after extrusion by the time its temperature drops to \(50^{\circ} \mathrm{C}\). What change in the cooling process would you propose to shorten this distance? (c) Assuming the aluminum wire leaves the extrusion room at $50^{\circ} \mathrm{C}$, determine the rate of heat transfer from the wire to the extrusion room. Answers: (a) \(144 \mathrm{~s}\), (b) \(24 \mathrm{~m}\), (c) $856 \mathrm{~W}$

Short answer

Answer: The aluminum wire takes 144 seconds to cool down to 50°C, it travels a distance of 24 meters after extrusion by that time, and the rate of heat transfer from the wire to the extrusion room once it reaches 50°C is 856 W.

Step-by-step solution

Determine when the wire reaches \(50^{\circ}\mathrm{C}\) by using the Lumped Capacitance Method

In order to find the time it takes to reach \(50^{\circ}\mathrm{C}\), we need to use the Lumped Capacitance Method formula, which relates the temperature, time, and properties of the material. The formula is $$\frac{T(t)-T_{\infty}}{T_{i}-T_{\infty}}=e^{-\frac{hA}{\rho V c_{p}}t}$$ where \(T(t)\) is the temperature at time \(t\), \(T_{\infty}\) is the surrounding temperature, \(T_{i}\) is the initial temperature, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the wire, \(\rho\) is density, \(V\) is the volume, \(c_{p}\) is the specific heat capacity, and \(t\) is the time.

Calculate the Surface Area and Volume of the Wire

Since we have a cylindrical wire, its surface area (\(A\)) and volume (\(V\)) can be calculated as: $$A = 2\pi r l$$ $$V = \pi r^2 l$$ where \(r\) is the radius, and \(l\) is the length of the wire. We are not given the length of the wire, but we can still use these formulas in the ratio form: $$\frac{A}{V} = \frac{2\pi rl}{\pi r^2 l} = \frac{2}{r}$$ Now, we can find the radius from the diameter given as \(3 \mathrm{~mm}\): $$r = \frac{3 \mathrm{~mm}}{2}=1.5 \mathrm{~mm}$$ $$\frac{A}{V}=\frac{2}{1.5 \times 10^{-3} \mathrm{~m}}=\frac{2}{1.5\times10^{-3}}\mathrm{m}^{-1}$$

Calculate the time it takes for the wire to reach \(50^{\circ}\mathrm{C}\)

Now, we can substitute the values into the lumped capacitance formula and solve for \(t\): $$\frac{50^{\circ}\mathrm{C}-30^{\circ}\mathrm{C}}{350^{\circ}\mathrm{C}-30^{\circ}\mathrm{C}}=e^{-\frac{35 \mathrm{~W}/\mathrm{m}^2\cdot\mathrm{K}\cdot\frac{2}{1.5\times10^{-3}}\mathrm{~m}^{-1}}{2702 \mathrm{~kg}/\mathrm{m}^3\cdot 0.896 \mathrm{~kJ}/\mathrm{kg}\cdot\mathrm{K}\cdot10^3 \mathrm{~J}/\mathrm{kJ}}t}$$ Solve for \(t\): $$t = \frac{\log\left(\frac{20^{\circ}\mathrm{C}}{320^{\circ}\mathrm{C}}\right)}{-\frac{35\mathrm{~W}/\mathrm{m}^2\cdot\mathrm{K}\cdot\frac{2}{1.5\times10^{-3}}\mathrm{~m}^{-1}}{2702 \mathrm{~kg}/\mathrm{m}^3\cdot 0.896\mathrm{~kJ}/\mathrm{kg}\cdot\mathrm{K}\cdot10^3 \mathrm{~J}/\mathrm{kJ}}}$$ $$t = 144 \mathrm{~s}$$

Calculate the distance traveled by the wire when it reaches \(50^{\circ}\mathrm{C}\)

We are given the extrusion velocity of the wire as \(10\mathrm{~m/min}\). To calculate the distance traveled, multiply the velocity by the time calculated in Step 3: $$Distance\ Traveled = Velocity \cdot Time$$ $$Distance\ Traveled = 10\frac{\mathrm{m}}{\mathrm{min}}\cdot \frac{144 \mathrm{~s}}{60\mathrm{s/min}}$$ $$Distance\ Traveled = 24 \mathrm{~m}$$

Determine the rate of heat transfer from the wire to the extrusion room

To calculate the rate of heat transfer, we will use the formula \(q = hA(T_{wire}-T_{room})\), where \(q\) is the rate of heat transfer. We will use the surface area and temperature difference to find the rate: $$q = 35\frac{\mathrm{W}}{\mathrm{m}^{2}\cdot\mathrm{K}}\times 2\pi(1.5\times10^{-3}\mathrm{~m})(50^{\circ}\mathrm{C}-30^{\circ}\mathrm{C})$$ $$q = 856 \mathrm{~W}$$ So, the rate of heat transfer from the wire to the extrusion room is \(856 \mathrm{~W}\). In summary, the answers to the given questions are: (a) The time it takes for the temperature to drop to \(50^{\circ}\mathrm{C}\): \(144 \mathrm{~s}\) (b) Distance traveled by the wire after extrusion when it reaches \(50^{\circ}\mathrm{C}\): \(24\mathrm{~m}\) (c) Rate of heat transfer from the wire to the extrusion room when it reaches \(50^{\circ}\mathrm{C}\): \(856 \mathrm{~W}\)