Question Cold air chamber is proposed for quenching steel ball bearings of diameter, D = 0.1 m and initial temperature T = 400°C. Air in the chamber is maintained at -15°C by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that 70% of the initial thermal energy content of the ball above -15°C be removed. Radiation effects may be neglected, and the convection heat transfer coefficient with in the chamber is 100 W/m².K. Estimate the residence time of the balls within the chamber, and recommend a drive velocity of the conveyor. The following properties may be used for the steel: k = 73 W/m.K, a = 2 x 10-m?/s, and cp = 450 J/kg.K. Steel ball Bearing in Cold air chember O

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Transcribed Image Text: Cold air chamber is proposed for quenching steel ball bearings of diameter, D = 0.1 m and initial temperature T = 400°C. Air in the chamber is maintained at -15°C by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that 70% of the initial thermal energy content of the ball above -15°C be removed. Radiation effects may be neglected, and the convection heat transfer coefficient with in the chamber is 100 W/m².K. Estimate the residence time of the balls within the chamber, and recommend a drive velocity of the conveyor. The following properties may be used for the steel: k = 73 W/m.K, a = 2 x 10-m?/s, and cp = 450 J/kg.K. Steel ball Bearing in Cold air chember O

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Solution; \begin{array}{l}\text { Diometer of ball bearings }=0.1 \mathrm{~m} \\\text { Radius }=0.1 / 2=0.05 \mathrm{~m} \\B_{i} \text { (Biot number) }=\frac{h L_{c}}{k}=\frac{h r_{0}}{3 k} \\=\frac{100 \mathrm{w} / \mathrm{m}^{2} \times 0.0 \mathrm{~s}}{3 \times 73 \mathrm{w} / \mathrm{mk}}=0.023 \\\text { Bi }<0.1 \\\end{array}therefore, No thermal energy generation within ppere.from, the coefficient used in one term approximation to series solution for tronsient one dimenpional conduction.The coefficient of bist number if 2 then,c_{1}=1.4793 \text { and } \rho_{1}=2.0288 \text {. }Using equation.\begin{array}{l}\frac{Q}{Q_{\max }}=1-3 Q_{\text {opphere }} \frac{\sin \rho_{1}-\rho_{1} \cos \rho_{1}}{\rho_{1}^{3}} \\\therefore Q_{0} \text { phere }=\left(1-\frac{Q}{Q_{\text {max }}}\right)\left(\frac{\rho_{1}^{3}}{3\left(\sin \rho_{1}-\rho_{1} \cos \rho_{1}\right)}\right) \\=(1-0.7)\left(\frac{2.0288^{3}}{3(\sin (2.0833)-2.083 \cos 2.083))}\right. \\=0.465 \\\text { fourier number }\left(F_{0}\right)=\frac{1}{\rho_{1}^{2}} \ln \left(\frac{Q_{0}}{c_{1}}\right) \\=\frac{1}{(2.0288)^{2}} \ln \left(\frac{0.465}{1.4793}\right) \\=0.281 \\\end{array}\begin{aligned} \text { Repidence fime } & =\frac{f_{0} r_{0}{ }^{2}}{d} \\ & =\frac{0.281 \times(0.05)^{2}}{2 \times 10^{-5}} \\ & =35.125 \mathrm{sec} \\ \text { Drive velocity } & =\frac{L}{t}=\frac{50 \mathrm{~m}}{35.125 \mathrm{~m}}=1.423 \mathrm{~m} / \mathrm{see}\end{aligned} ...