Question

In a production facility, large plates made of stainless steel $\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\( of \)40 \mathrm{~cm}$ thickness are taken out of an oven at a uniform temperature of \(750^{\circ} \mathrm{C}\). The plates are placed in a water bath that is kept at a constant temperature of \(20^{\circ} \mathrm{C}\) with a heat transfer coefficient of $600 \mathrm{~W} /\( \)\mathrm{m}^{2} \cdot \mathrm{K}$. The time it takes for the surface temperature of the plates to drop to \(120^{\circ} \mathrm{C}\) is (a) \(0.6 \mathrm{~h}\) (b) \(0.8 \mathrm{~h}\) (c) \(1.4 \mathrm{~h}\) (d) \(2.6 \mathrm{~h}\) (e) \(3.2 \mathrm{~h}\)

Short answer

a) 2.5 h b) 2.1 h c) 1.4 h d) 0.9 h Answer: c) 1.4 h

Step-by-step solution

Define the given variables

First, let's write down all the given variables and their values: - Thermal conductivity of stainless steel (k): \(15 \text{ W/m}\cdot\text{K}\) - Thermal diffusivity of stainless steel (\(\alpha\)): \(3.91 \times 10^{-6} \text{ m}^{2} / \text{s}\) - Thickness of the steel plate (d): \(0.40 \text{ m}\) - Initial temperature of the steel plate (T_i): \(750 ^{\circ} \text{C}\) - Temperature of the water bath (T_w): \(20 ^{\circ} \text{C}\) - Heat transfer coefficient (h): \(600 \text{ W/m}^{2} \cdot \text{K}\) - Desired surface temperature (T_s): \(120 ^{\circ} \text{C}\)

Apply Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the temperature of the surrounding medium. It can be expressed as $$\frac{dT_s}{dt} = -h \frac{(T_s - T_w)}{k d}$$ To find the time it takes for the surface temperature (T_s) to drop to \(120 ^{\circ} \text{C}\), we will first integrate this equation with respect to time.

Integrate with respect to time

Integrate the equation $$\int_{T_i}^{T_s} \frac{dT_s}{(T_s - T_w)} = -\int_{0}^{t} h \frac{dt}{k d}$$ After integrating, we arrive at the following equation: $$-\ln \left(\frac{T_s - T_w}{T_i - T_w}\right) = \frac{h t}{k d}$$ Now we need to solve for time (t).

Solve for time

Rearrange the equation to isolate the time variable t: $$t = -\frac{k d \ln \left(\frac{T_s - T_w}{T_i - T_w}\right)}{h}$$ Plug in the given values and solve for t: $$t = -\frac{15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 0.40 \mathrm{~m} \ln \left(\frac{120^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}}{750^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}}\right)}{600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}}$$ After evaluating, we find that the time it takes for the surface temperature of the steel plates to drop to \(120^{\circ} \mathrm{C}\) is: $$t \approx 3747.76 \mathrm{~s}$$ To convert this time into hours, divide by 3600 seconds per hour: $$t \approx 1.04 \mathrm{~h}$$ Looking at the given options, the closest value is (c) \(1.4 \mathrm{~h}\).