Question

A 5 -mm-thick stainless steel strip $(k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ is being heat treated as it moves through a furnace at a speed of \(1 \mathrm{~cm} / \mathrm{s}\). The air temperature in the furnace is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the furnace length is \)3 \mathrm{~m}$ and the stainless steel strip enters it at \(20^{\circ} \mathrm{C}\), determine the temperature of the strip as it exits the furnace.

Short answer

Answer: The approximate temperature of the stainless steel strip as it exits the furnace is 458°C.

Step-by-step solution

Determine the heat transfer rate by convection

We need to find the rate of heat transfer by convection between the strip and the air inside the furnace. To do this, we will use the following formula: \(q_{conv} = hA(T_{air} - T_{strip})\) where - \(q_{conv}\) is the heat transfer rate by convection (W) - \(h\) is the convection heat transfer coefficient (\(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) - \(A\) is the surface area of the strip exposed to the air (m²) - \(T_{air}\) is the air temperature inside the furnace (\(900^{\circ} \mathrm{C}\)) - \(T_{strip}\) is the temperature of the strip (unknown) Since the steel strip is moving continuously through the furnace, we can assume that the surface area exposed to the air is equal to the width of the strip multiplied by the length of the furnace. Therefore, we can express the surface area as: \(A = w \cdot L\) where - \(w\) is the width of the strip (assumed to be 1 m for simplicity) - \(L\) is the length of the furnace (3 m) So we have: \(q_{conv} = h(w \cdot L)(T_{air} - T_{strip})\)

Apply energy conservation

The heat gained by the strip equals the mass of the strip multiplied by its specific heat capacity and the change in temperature: \(q = mc_p \Delta T_{strip}\) where - \(q\) is the heat gained by the strip (W) - \(m\) is the mass of the strip (kg) - \(c_p\) is the specific heat capacity of the stainless steel (\(570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)) - \(\Delta T_{strip}\) is the change in temperature of the strip (K) Since the mass of the strip is given by the product of its volume and density, we can express the mass as: \(m = \rho V\) where - \(\rho\) is the density of the stainless steel (\(8000 \mathrm{~kg} / \mathrm{m}^{3}\)) - \(V\) is the volume of the strip (m³) Now we can relate the heat transfer rate by convection to the heat gained by the strip by setting the two equal: \(h(w \cdot L)(T_{air} - T_{strip}) = mc_p \Delta T_{strip}\)

Solve for the final temperature

We will now solve for the final temperature of the strip, \(T_{final}\): \(T_{final} = T_{initial} + \Delta T_{strip}\) where - \(T_{initial}\) is the initial temperature of the strip (\(20^{\circ} \mathrm{C}\)) Using the previous equations, we can rewrite this as: \(T_{final} = T_{initial} + \frac{h(w \cdot L)(T_{air} - T_{final})}{mc_p}\) We can now input the given values and solve for \(T_{final}\): \(T_{final} = 20 + \frac{80(1 \cdot 3)(900 - T_{final})}{(8000)(570)}\) Solving for \(T_{final}\), we get: \(T_{final} \approx 458^{\circ} \mathrm{C}\) Thus, the temperature of the stainless steel strip as it exits the furnace is approximately \(458^{\circ} \mathrm{C}\).