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A copper sphere is to be quenched in an oil bath whose

Chapter , Problem 7.57

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QUESTION:

A copper sphere is to be quenched in an oil bath whose temperature is \(50^{\circ} \mathrm{C}\). The sphere’s radius is 30 mm, and the convection coefficient is \(h=300 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\right)\). Assume the sphere and the oil properties are constant. These properties are given in the following table. The sphere’s initial temperature is \(400^{\circ} \mathrm{C}\).

Assume that the volume of the oil bath is large enough so that its temperature does not change when the sphere is immersed in the bath. Obtain the dynamic model of the sphere’s temperature T. How long will it take for T to reach \(130^{\circ} \mathrm{C}\)?

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QUESTION:

A copper sphere is to be quenched in an oil bath whose temperature is \(50^{\circ} \mathrm{C}\). The sphere’s radius is 30 mm, and the convection coefficient is \(h=300 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\right)\). Assume the sphere and the oil properties are constant. These properties are given in the following table. The sphere’s initial temperature is \(400^{\circ} \mathrm{C}\).

Assume that the volume of the oil bath is large enough so that its temperature does not change when the sphere is immersed in the bath. Obtain the dynamic model of the sphere’s temperature T. How long will it take for T to reach \(130^{\circ} \mathrm{C}\)?

ANSWER:

Step 1 of 5

Determine the dimension of the object \(L\).

\(L=\frac{V}{A}\)

Here \(V\) is the volume of the sphere and \(A\) is the area of the sphere.

\(\begin{aligned} L & =\frac{\frac{4}{3} \pi r^{3}}{4 \pi r^{2}} \\ & =\frac{r}{3} \end{aligned}\)

Substitute \(30 \times 10^{-3} \mathrm{~m}\) for \(r\)

\(\begin{aligned} L & =\frac{30 \times 10^{-3}}{3} \\ & =0.01 \mathrm{~m} \end{aligned}\)

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