Question

A long steel strip is being conveyed through a 3 -m-long furnace to be heat treated at a speed of \(0.01 \mathrm{~m} / \mathrm{s}\). The steel strip $\left(k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8000 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, and \)\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( has a thickness of \)5 \mathrm{~mm}$, and it enters the furnace at an initial temperature of \(20^{\circ} \mathrm{C}\). Inside the furnace, the air temperature is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Using appropriate software, determine the surface temperature gradient of the steel strip as a function of location inside the furnace. By varying the location in the furnace for \(0 \leq x \leq 3 \mathrm{~m}\) with increments of \(0.2 \mathrm{~m}\), plot the surface temperature gradient of the strip as a function of furnace location. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Short answer

Question: Calculate the surface temperature gradient of a steel strip moving through a furnace using the lumped system analysis method. The strip has a thickness of 5 mm, a speed of 0.01 m/s, convection heat transfer coefficient of 80 W/(m^2*K), thermal conductivity of 21 W/(m*K), density of 7800 kg/m^3, and specific heat capacity of 500 J/(kg*K). The furnace temperature is 900°C, and the initial temperature of the steel strip is 20°C. Answer: The surface temperature gradient of the steel strip can be calculated by first confirming that the Biot number is less than 0.1, which it is at 0.019. Then we can use the lumped system analysis expression to find the surface temperature at different locations inside the furnace. Finally, we can compute and plot the surface temperature gradient at each position, in increments of 0.2 meters, for the 3-meter long furnace.

Step-by-step solution

Calculate Biot number

To verify if the lumped system analysis method can be used in this situation, we need to calculate the Biot number (Bi) and confirm if it is less than 0.1. Bi = \( \frac{hL_{c}}{k} \) Where h is the convection heat transfer coefficient, \(L_{c}\) is the characteristic length, and k is the thermal conductivity. The characteristic length, \(L_{c}\), is calculated as: \(L_{c} = \frac{V}{A} \) Where V is the volume and A is the surface area. In this case, \(L_{c}\) represents the thickness (5 mm) of the steel strip. Now let's calculate the Biot number: Bi = \(\frac{80 \times 0.005}{21}\) = 0.019 Since Bi < 0.1, we can proceed with the lumped system analysis method.

Calculate the surface temperature

The surface temperature, T(t), of the steel strip can be calculated using the lumped system analysis expression: \(T(t) = T_{\infty} + (T_{i} - T_{\infty})e^{-\frac{hAt}{\rho V c_{p}}}\) Where \(T_{\infty}\) is the furnace temperature (900°C), \(T_{i}\) is the initial temperature (20°C), t is the time, ρ is the density, and \(c_{p}\) is the specific heat capacity. To find the time at each position in the furnace, we can use: t = \(\frac{x}{v}\) Where x is the location inside the furnace, and v is the speed of the steel strip.

Calculate the surface temperature gradient

Now we can find the temperature gradient, \(\frac{dT}{dx}\), by taking the derivative of \(T(t)\) with respect to x: \(\frac{dT}{dx} = \frac{dT}{dt} \cdot \frac{dt}{dx}\) \(\frac{dT}{dt} = \frac{hA}{\rho V c_{p}}(T_{\infty} - T(t))\) \(\frac{dt}{dx} = \frac{1}{v}\) By combining these expressions, we get: \(\frac{dT}{dx} = \frac{hA}{\rho V c_{p} v} (T_{\infty} - T(t))\) Now we can find the surface temperature gradient as a function of the location inside the furnace for \(0 \leq x \leq 3 \mathrm{~m}\).

Plot the surface temperature gradient

Using the expression for \(\frac{dT}{dx}\), we can now calculate the surface temperature gradient at each position inside the furnace with increments of 0.2 m: - For x = 0 m, 0.2 m, 0.4 m, ..., 3 m: - Calculate t using t = \(\frac{x}{v}\) - Calculate T(t) using the lumped system analysis expression - Calculate \(\frac{dT}{dx}\) using the temperature gradient expression After calculating the temperature gradient at each position, we can plot the surface temperature gradient as a function of the furnace location. We now have a plot of the surface temperature gradient of the steel strip as it moves through the furnace, at increments of 0.2 m for the 3-meter long furnace.