Question

The walls of a furnace are made of \(1.5\)-ft-thick concrete $\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\( and \)\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)$. Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at $1800^{\circ} \mathrm{F}$ with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(3.0 \mathrm{~h}\)

Short answer

Answer: It will take approximately 3.0 hours for the outer surface temperature of the furnace walls to rise to 70.1°F.

Step-by-step solution

Identify Known Parameters

We start by listing all given parameters: - Concrete thickness: \(L = 1.5 \thinspace ft\) - Thermal conductivity: \(k = 0.64 \thinspace \frac{Btu}{h \cdot ft \cdot ^{\circ}F}\) - Thermal diffusivity: \(\alpha = 0.023 \thinspace \frac{ft^2}{h}\) - Initial temperature: \(T_i = 70^{\circ}F\) - Inner surface temperature after firing: \(T_{1f} = 1800^{\circ}F\) - Desired outer surface temperature: \(T_{2d} = 70.1^{\circ}F\)

Calculate Temperature Rise at the Outer Surface

In order to find the time it takes for the outer surface to reach the desired temperature, let's first find the temperature rise at the outer surface of the furnace walls: \(\Delta T = T_{2d} - T_i = 70.1 - 70 = 0.1^{\circ}F\)

Use Transient Heat Conduction Formula

Now, we will utilize the transient heat conduction formula for a semi-infinite wall with an initial temperature of \(T_i\) and a constant surface temperature of \(T_{1f}\) at the inner surface. The temperature at a distance \(L\) from the inner surface and after time \(t\) can be given as: $$ T(x=L, t) = T_i + (T_{1f} - T_i)[1 - \mathrm{erf}\left(\frac{L}{2\sqrt{\alpha t}}\right)]$$

Solve for the Time

Next, plug the known parameters and the temperature rise at the outer surface into the equation to solve for the time \(t\): $$ T_{2d} = T_i + (T_{1f} - T_i) \left[1 - \mathrm{erf}\left(\frac{L}{2\sqrt{\alpha t}}\right)\right]$$ Substituting the values: $$ 70.1 = 70 + (1800 - 70) \left[1 - \mathrm{erf}\left(\frac{1.5}{2\sqrt{0.023 t}}\right)\right]$$ After rearranging the equation and using a lookup table for the error function, we can find the value of \(t\). Solving for \(t\), we get: $$t \approx 3.0 \thinspace h$$ Thus, it will take about \(3.0\) hours for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ}F\).