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 \(100^{\circ} \mathrm{C}\) is (a) \(0.28 \mathrm{~h}\) (b) \(0.99 \mathrm{~h}\) (c) \(2.05 \mathrm{~h}\) (d) \(3.55 \mathrm{~h}\) (e) \(5.33 \mathrm{~h}\)

Step by step solution

01

List down the given parameters

We are given: - Thermal conductivity \((k)\): \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Thermal diffusivity \((\alpha)\): \(3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) - Plate thickness \((d)\): \(0.40 \mathrm{~m}\) - Initial temperature of plate \((T_\text{initial})\): \(750^{\circ} \mathrm{C}\) - Temperature of water bath \((T_\text{bath})\): \(20^{\circ} \mathrm{C}\) - Heat transfer coefficient \((h)\): \(600 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) - Final surface temperature of plate \((T_\text{final})\): \(100^{\circ} \mathrm{C}\)

02

Calculate the Biot number

To find the Biot number \((\text{Bi})\), we will use the formula \(\text{Bi} = \frac{h \cdot d}{k}\) Plugging in the given values, we get: \(\text{Bi} = \frac{600 \mathrm{~W} \text{/} \mathrm{m}^2 \cdot \mathrm{K} \cdot 0.40 \mathrm{~m}}{15 \mathrm{~W} \text{/} \mathrm{m} \cdot \mathrm{K}} = 16\)

03

Use the dimensionless temperature definition

Now, we will define dimensionless temperature (\(\Theta\)) as \(\Theta = \frac{T_\text{final} - T_\text{bath}}{T_\text{initial} - T_\text{bath}}\). Plugging in the given temperatures, we get: \(\Theta = \frac{100^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}}{750^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}} = \frac{80}{730} = 0.1096\)

04

Use the Newton's Law of Cooling formula

Newton's Law of Cooling states that \(\Theta = e^{-\text{Bi} \cdot \text{Fo}}\), where the Fourier number (Fo) is defined as \(\text{Fo} = \frac{\alpha \cdot t}{d^2}\) (with \(t\) being the time). We are trying to find \(t\), so we will rearrange the formula to find \(\text{Fo}\) and then solve for \(t\): 1. \(\text{Fo} = -\frac{1}{\text{Bi}} \cdot \ln{(\Theta)}\) 2. \(t = \frac{\text{Fo} \cdot d^2}{\alpha}\)

05

Calculate the Fourier number

Using the formula \(\text{Fo} = -\frac{1}{\text{Bi}} \cdot \ln{(\Theta)}\): \(\text{Fo} = -\frac{1}{16} \cdot \ln{(0.1096)} \approx 0.0982\)

06

Calculate the time

Now, we will use the formula \(t = \frac{\text{Fo} \cdot d^2}{\alpha}\) to calculate the time taken for the surface temperature to drop to \(100^{\circ} \mathrm{C}\). \(t = \frac{0.0982 \cdot (0.40 \mathrm{~m})^2}{3.91 \times 10^{-6} \mathrm{~m}^2 \text{/} \mathrm{s}} \approx 4010 \mathrm{s}\) Lastly, convert the time in seconds to hours: \(t = \frac{4010 \mathrm{s}}{3600 \mathrm{s/h}} \approx 1.11 \mathrm{~h}\) The closest answer is (b) \(0.99 \mathrm{~h}\).

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