Question

Aluminum wires, \(3 \mathrm{~mm}\) 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

Answer: To find the required length of the wire cooling section, use the formula: \(L = \frac{\dot{m} \cdot c_{Al} \cdot \Delta T}{h \cdot \pi \cdot d \cdot \Delta T_{air-wire}}\) where \(\dot{m}\) is the mass flow rate, \(c_{Al}\) is the specific heat capacity of aluminum, \(\Delta T\) is the temperature difference, \(h\) is the heat transfer coefficient, \(d\) is the diameter of the wire, and \(\Delta T_{air-wire}\) is the temperature difference between the wire and surrounding air. Plug in the values and calculations from Step 1 and Step 2 to determine the length \(L\) of the wire cooling section in the extrusion room.

Step-by-step solution

Calculate the heat loss required to cool the wire

To calculate the heat loss required to decrease the wire's temperature from \(350^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), we need to find the mass flow rate of the aluminum wire, its specific heat capacity, and the temperature difference. The mass flow rate can be determined using the volumetric flow rate, the density of aluminum, and the wire's cross-sectional area. Given the diameter of the aluminum wire, \(d = 3 \mathrm{~mm}\), we can find the cross-sectional area, \(A\) using the formula: \(A = \frac{\pi d^2}{4}\) The linear rate at which the wire is extruded is given as \(10 \mathrm{~m} / \mathrm{min}\) or \(10/60 \mathrm{~m} / \mathrm{s}\), which represents the volumetric flow rate per unit time. The density of aluminum, \(\rho_{Al} = 2700 \mathrm{~kg} / \mathrm{m}^3\). Therefore, the mass flow rate, \(\dot{m}\) can be obtained as: \(\dot{m} = \rho_{Al} \cdot A \cdot \frac{10}{60}\) The specific heat capacity of aluminum, \(c_{Al} = 900 \mathrm{~J} / (\mathrm{kg} \cdot \mathrm{K})\). The temperature difference, \(\Delta T\), required is also given as \((350 - 50) \mathrm{~K}\) or \(300 \mathrm{~K}\). Therefore, the heat loss required, \(Q\) can be represented as: \(Q = \dot{m} \cdot c_{Al} \cdot \Delta T \) Plug the values to find the heat loss required to cool the wire.

Calculate the heat transfer rate

We are given the heat transfer coefficient, \(h = 50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the temperature difference between the wire and surrounding air, \(\Delta T_{air-wire} = (50 - 25) \mathrm{~K} = 25 \mathrm{~K}\). The heat transfer rate, \(\dot{Q}_{conv}\), can be calculated using the formula: \(\dot{Q}_{conv} = h \cdot A_s \cdot \Delta T_{air-wire}\) where \(A_s\) is the surface area of the wire. For a cylindrical wire of length \(L\), the surface area can be represented as: \(A_s = \pi \cdot d \cdot L\)

Calculate the required length of the wire cooling section

To find the required length of the wire cooling section, we need to equate the heat loss required to cool the wire and the heat transfer rate: \(Q = \dot{Q}_{conv}\) \(\dot{m} \cdot c_{Al} \cdot \Delta T = h \cdot \pi \cdot d \cdot L \cdot \Delta T_{air-wire}\) Rearranging to find the length, \(L\), we get: \(L = \frac{\dot{m} \cdot c_{Al} \cdot \Delta T}{h \cdot \pi \cdot d \cdot \Delta T_{air-wire}}\) Now, plug in the values and calculations from Step 1 and Step 2 to find the length of the wire cooling section, \(L\), in the extrusion room.

Key concepts

Heat Loss Calculation

To understand the cooling process in wire extrusion, it's crucial to start with the basics of heat loss calculation. Heat loss refers to the amount of heat that's transferred from the hot wire to the surrounding environment, which, in this case, is air. The amount of heat that needs to be removed to cool the wire from one temperature to another can be calculated by considering the wire's mass flow rate, the material's specific heat capacity, and the temperature drop the wire must undergo.

Using the formula Q = \( \dot{m} \cdot c_{Al} \cdot \Delta T \), you're essentially multiplying the mass of the wire passing a point per unit of time, the material's ability to store heat (specific heat capacity), and the difference in temperature before and after cooling. The specific heat capacity is like the thermal 'capacity' of a material—how much heat it can hold before its temperature rises by 1°C. For an extruded aluminum wire, lower temperatures mean that the specific heat capacity will cause significant transfer of energy, and thus, more consideration to heat loss is necessary to design an efficient cooling process.

Heat Transfer Coefficient

The heat transfer coefficient is a measure of how easily heat passes from the wire to its surroundings. A high coefficient indicates that heat can transfer quickly, while a lower coefficient means heat moves more slowly. The coefficient depends on material properties, surface conditions, and the flow characteristics of the cooling medium, which in the case of wire extrusion is typically air.

Given by the symbol h and expressed in terms of W/(m²·K), this coefficient is used in the formula \( \dot{Q}_{conv} = h \cdot A_s \cdot \Delta T_{air-wire} \) to calculate the rate at which heat is being transferred via convection. It's a crucial variable because it directly affects the length of the cooling section needed—the higher the coefficient, the shorter the cooling section required to achieve the desired temperature drop.

Thermal Conductivity

Thermal conductivity, denoted by the symbol k, is a property of materials that quantifies the ability to conduct heat. Materials with high thermal conductivity can transfer heat rapidly, which in manufacturing processes such as wire extrusion, is an essential consideration. As heat is conducted from the hot core of the wire to its cooler surface, it then gets convected away by the surrounding air.

The significance of thermal conductivity in the wire extrusion process becomes apparent when ensuring that the wire's temperature is uniformly reduced throughout its cross-section by the time it reaches the end of the cooling section. This uniform cooling helps maintain the structural integrity and desired properties of the wire.

Mass Flow Rate

Mass flow rate is at the heart of the cooling equation. It represents the amount of mass passing through a given section of the wire per unit time and is usually denoted as \( \dot{m} \). In wire extrusion, knowing the mass flow rate is essential to determining the heat loss calculation because it impacts how much heat energy is carried away from the wire per unit time. The mass flow rate depends on the wire's density, cross-sectional area, and its linear extrusion rate.

To compute the mass flow rate for an extruded wire, you'd use the cross-sectional area of the wire and multiply it by the linear rate at which the wire is extruded and the density of the material. The result gives you a concrete value that, when plugged into the heat loss formula, contributes to finding the amount of cooling necessary.

Cooling Wire Extrusion Process

In the cooling wire extrusion process, properly calculating the length of the cooling section is pivotal to ensuring the final quality of the wire. As the extruded wire leaves the die, it's at a high temperature and needs to be cooled to a manageable level before it can be spooled or further processed. The rate of cooling must be sufficient to solidify the wire but not so fast as to induce stresses or uneven cooling which may cause warping or weakness.

The process involves transferring heat from the wire to the surrounding cooling medium, which is often air or sometimes liquid if rapid cooling is required. To optimize this process, one must understand the concepts of mass flow rate, thermal conductivity, and heat transfer coefficient deeply as they dictate the heat transfer rates and, ultimately, the required cooling length within the extrusion room. An efficient cooling process minimizes energy loss, prevents material waste, and delivers a high-quality end product, which is why these calculations are so important in industrial settings.