Question

Aluminum wires \(4 \mathrm{~mm}\) in diameter are produced by extrusion. The wires leave the extruder at an average temperature of \(350^{\circ} \mathrm{C}\) and at a linear rate of \(10 \mathrm{~m} / \mathrm{min}\). Before leaving the extrusion room, the wires are cooled to an average temperature of $50^{\circ} \mathrm{C}\( by transferring heat to the surrounding air at \)25^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Calculate the necessary length of the wire cooling section in the extrusion room.

Short answer

The necessary length of the wire cooling section in the extrusion room is approximately 1395 meters. This is calculated by first determining the rate at which the wire loses heat during the cooling process, then calculating the time needed to cool the wire from 350°C to 50°C, and finally using this time and the extrusion rate to determine the required length of the cooling section.

Step-by-step solution

Calculate the area of heat transfer for the wire

We're given the diameter of the wire (\(d = 4 \ \mathrm{mm}\)). To find the surface area of the wire, we'll first find the circumference of the wire (assuming it's a perfect cylinder), and then multiply the circumference by the unit length of the wire (1 m). Circumference: \(C = \pi d\) Area of heat transfer per meter: \(A = C \cdot 1 \ \mathrm{m}\) Plug in the given diameter and calculate the area: \(A = \pi \cdot (4 \times 10^{-3} \ \mathrm{m}) \cdot 1 \ \mathrm{m} = 0.01257 \ \mathrm{m^2}\)

Calculate the rate of heat transfer from the wire

Now we'll calculate the rate at which the wire loses heat (\(Q\)). This can be done using the heat transfer coefficient (\(h\)) from the given information, the surface area of heat transfer (\(A\)) we calculated in Step 1, and the temperature difference between the wire and the surrounding air (\(\Delta T = T_\mathrm{wire} - T_\mathrm{air}\)): \(Q = h \cdot A \cdot \Delta T\) Since the wire temperature changes during the cooling process, we'll use the average wire temperature during the cooling process which is the midpoint between the initial and final temperatures: Average wire temperature: \(T_\mathrm{avg} = \frac{T_\mathrm{initial} + T_\mathrm{final}}{2} = \frac{350 + 50}{2} = 200 \ \mathrm{^{\circ} C}\) Now, let's calculate the temperature difference between the average wire temperature and the surrounding air: \(\Delta T = 200 - 25 = 175 \ \mathrm{K}\) We can now determine the rate of heat transfer from the wire: \(Q = 50 \ \mathrm{\frac{W}{m^2 K}} \cdot 0.01257 \ \mathrm{m^2} \cdot 175 \ \mathrm{K} = 109.72 \ \mathrm{W}\)

Calculate the time needed for cooling

To find the time required for cooling the wire from 350°C to 50°C, we'll need to determine the heat loss and divide it by the rate of heat transfer from the wire. We're given that the wire is extruded at a rate of \(10\ \mathrm{m/min}\), so we can write: \(Q_{total} = Q \cdot t\) where \(Q_{total}\) is the total heat loss and t is the time in minutes. First, we need the specific heat capacity of aluminum (\(c_\mathrm{Al} = 900 \ \mathrm{\frac{J}{kg \cdot K}}\)) and its density (\(\rho_\mathrm{Al} = 2700 \ \mathrm{\frac{kg}{m^3}}\)). We can now calculate the mass and volume of 1 meter of wire: \(V_\mathrm{wire} = \frac{1}{4}\pi d^2 \cdot 1\ \mathrm{m} = \frac{1}{4} \pi (4 \times 10^{-3})^2 \cdot 1\ \mathrm{m} = 1.25664 \times 10^{-5} \ \mathrm{m^3}\) Mass of 1 meter of wire: \(m_\mathrm{wire} = \rho_\mathrm{Al} \cdot V_\mathrm{wire} = 2700 \ \mathrm{\frac{kg}{m^3}} \cdot 1.25664 \times 10^{-5} \ \mathrm{m^3} = 0.03398\ \mathrm{kg}\) Now we can find the total heat loss during the cooling process: \(Q_{total} = m_\mathrm{wire} \cdot c_\mathrm{Al} \cdot (T_\mathrm{initial} - T_\mathrm{final}) = 0.03398\ \mathrm{kg} \cdot 900 \ \mathrm{\frac{J}{kg\cdot K}} \cdot (350 - 50) = 918324 \ \mathrm{J}\) Now we can find the time needed for cooling: \(t = \frac{Q_{total}}{Q} = \frac{918324 \ \mathrm{J}}{109.72 \ \mathrm{W}} = 8370 \ \mathrm{s}\)

Calculate the length of the cooling section

Finally, we can calculate the length of the wire cooling section in the extrusion room. We know the wire is extruded at a rate of \(10\ \mathrm{m/min}\) and we've found in Step 3 that the cooling process takes 8370 seconds. We can find the length using the following equation: Length of cooling section: \(L = \text{extrusion rate} \cdot t\) Convert the extrusion rate to \(\mathrm{m/s}\): Extrusion rate in m/s: \(10 \ \mathrm{\frac{m}{min}} \cdot \frac{1 \ min}{60 \ s} = 0.1667 \ \mathrm{\frac{m}{s}}\) Now calculate the length of the cooling section: \(L = 0.1667 \ \mathrm{\frac{m}{s}} \cdot 8370 \ \mathrm{s} = 1395 \ \mathrm{m}\) Therefore, the necessary length of the wire cooling section in the extrusion room is approximately 1395 meters.