Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

Step by step solution

01

1. Write down the Newton's Law of Cooling formula

Newton's Law of Cooling can be represented as: \(\frac{dT}{dt}=-hA\frac{(T-T_{\infty})}{\rho Vc_p}\), where \(\frac{dT}{dt}\) is the rate of change in temperature, \(h\) is the heat transfer coefficient, \(A\) is the area through which heat transfer occurs, \(T\) is the temperature of the object, \(T_{\infty}\) is the ambient temperature, \(\rho\) is the density of the material, \(V\) is the volume of the object, and \(c_p\) is the specific heat.

02

2. Rewrite the cooling formula in terms of diameter and temperature

We know the diameter of ball bearings, so we can find the surface area and volume expressions using the diameter \(d\). Surface area \(A = 4 \pi (\frac{d}{2})^2 = \pi d^2\) Volume \(V = \frac{4}{3} \pi (\frac{d}{2})^3 = \frac{1}{6} \pi d^3\) Now substitute these expressions of \(A\) and \(V\) into the cooling formula: \(\frac{dT}{dt} = -h\pi d^2 \frac{(T-T_{\infty})}{\rho (\frac{1}{6} \pi d^3)c_p}\)

03

3. Simplify the formula for the rate of change in temperature

Now we can simplify the formula for \(\frac{dT}{dt}\): \(\frac{dT}{dt} = -\frac{6h}{d} \frac{(T-T_{\infty})}{\rho c_p}\)

04

4. Solve the cooling formula for time

Now we need to find the time taken for the balls to cool from \(900^{\circ} \mathrm{C}\) to \(850^{\circ} \mathrm{C}\). Integrate both sides of the equation between the initial (900°C) and final (850°C) temperatures and their respective times (0 and t) as limits: \(\int_{900}^{850} \frac{dT}{(T-T_{\infty})} = -\int_{0}^{t} \frac{6h}{d\cdot \rho c_p} dt\) Solve the integral and we get: \(-\ln \frac{(850-T_{\infty})}{(900-T_{\infty})}= \frac{6ht}{d\rho c_p}\)

05

5. Solve the equation for t

Rearrange the equation and solve for time \(t\): \(t = \frac{d\rho c_p}{6h} \ln \frac{900-T_{\infty}}{850-T_{\infty}}\)

06

6. Plug in the given values

Now substitute all given values into the equation: \(t = \frac{0.012 \times 8085 \times 480}{6\times 125} \ln \frac{900-30}{850-30}\)

07

7. Calculate the time to cool down

Finally, calculate the time it takes for the balls to cool down from 900°C to 850°C: \(t = 0.0109 \mathrm{~hour}\) or \(t \approx 39.2\mathrm{~seconds}\) The stainless steel balls can remain in the air for approximately 39.2 seconds before being dropped into the water without their temperature falling below 850°C.

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