Question

A large iron slab $\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\(, and \)k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ was initially heated to a uniform temperature of \(150^{\circ} \mathrm{C}\) and then placed on a concrete floor $\left(\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=840 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\(, and \)\left.k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)$. The concrete floor was initially at a uniform temperature of \(30^{\circ} \mathrm{C}\). Determine \((a)\) the surface temperature between the iron slab and concrete floor and \((b)\) the temperature of the concrete floor at the depth of \(25 \mathrm{~mm}\), if the surface temperature remains constant after \(15 \mathrm{~min}\).

Short answer

(a) The surface temperature between the iron slab and the concrete floor (T_s) after 15 minutes can be determined using the formula: $$T_s = 30 + \frac{80.2 (150 - 30)}{d}\cdot \frac{d_{concrete}}{0.79}$$ where "d" is the thickness of the iron slab. Without knowing the thickness, we can't provide a numerical value for T_s, but with the thickness, we can substitute the value and find T_s. (b) The temperature of the concrete floor at a depth of 25mm (T_d) can be determined using the one-dimensional heat conduction equation: $$T_d = T_s + (30 - T_s) \cdot e^{-\frac{0.025}{\sqrt{4 \cdot 6.24 \times 10^{-7} \cdot 900}}}$$ Since we don't have a numerical value for T_s, we cannot definitively solve for T_d. However, if the thickness of the iron slab is known, then we can substitute the value of T_s and find the temperature T_d.

Step-by-step solution

Determine heat transfer from iron to concrete (Q)

As heat is transferred from the iron slab to the concrete floor, we will first determine the amount of heat loss by conduction through the iron slab. For this, we'll use the following formula: $$Q = \frac{k \cdot A \cdot(T_{iron} - T_{floor})}{d}$$ where: - \(Q\) = heat transfer (W) - \(k\) = thermal conductivity of iron (\(80.2 W/m\cdot K\)) - \(A\) = area of the interface between the iron slab and the concrete floor - \(T_{iron}\) = initial temperature of the iron slab (\(150^{\circ} \mathrm{C}\)) - \(T_{floor}\) = initial temperature of the concrete floor (\(30^{\circ} \mathrm{C}\)) - \(d\) = thickness of the iron slab Since we are not given the thickness of the iron slab and the area of contact, we can express the heat transfer in terms of \(Q/A\): $$ \frac{Q}{A} = \frac{k (T_{iron} - T_{floor})}{d} $$

Determine surface temperature between iron and concrete (T_s)

Now, we'll assume that the heat transfer from the iron slab to the concrete floor is in steady-state. Under steady-state conditions, the temperature gradient between the slab and the floor is constant. So we can determine the temperature \(T_s\) at the interface between the iron slab and the concrete floor using the thermal resistance concept (for conduction): $$T_s = T_{floor} + \frac{Q}{A} \cdot \frac{d_{concrete}}{k_{concrete}}$$ Substituting the values for the thermal conductivity of concrete \((k_{concrete} = 0.79 W/m\cdot K)\) and assuming contact at \(15\) minutes: $$T_s = 30 + \frac{80.2 (150 - 30)}{d}\cdot \frac{d_{concrete}}{0.79}$$ Since we don't have numerical values of thickness, we'll have to leave the solution in terms of the thickness "\(d\)" of the iron slab.

Determine the temperature of the concrete floor at depth of 25mm (T_d)

Finally, we'll determine the temperature profile within the concrete floor. We will use the one-dimensional heat conduction equation to find the temperature \(T_d\) at the specified depth (\(25mm\)). The equation is as follows: $$T_d = T_s + (T_{surface} - T_s) \cdot e^{-\frac{depth}{\sqrt{4 \cdot \alpha \cdot t}}}$$ where: - \(T_d\) = temperature at 25mm depth - \(T_s\) = surface temperature - \(T_{surface}\) = initial temperature of the concrete floor (\(30^{\circ} \mathrm{C}\)) - \(\alpha\) = thermal diffusivity of the concrete floor \(= \frac{k}{\rho \cdot c_p} = \frac{0.79}{1600 \cdot 840} \approx 6.24 \times 10^{-7} m^2/s\) - \(depth = 0.025m\) - \(t = 15 min = 900s\) Substituting the values into the above equation: $$T_d = T_s + (30 - T_s) \cdot e^{-\frac{0.025}{\sqrt{4 \cdot 6.24 \times 10^{-7} \cdot 900}}}$$ We cannot definitively solve for \(T_d\) since we have left \(T_s\) in terms of the thickness of the iron slab. However, if the thickness is known, then we can substitute the value of \(T_s\) and find the temperature \(T_d\). This completes our step-by-step solution to the problem.