Question A cold air chamber is proposed for quenching steel ball bearings of diameter D = 0.2 m and initial temperature Ti = 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 within the chamber is 1000 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: and c = 450 J/kg ∙ K. A soda lime glass sphere of diameter D1 25 mm is encased in a bakelite spherical shell of thickness L 10 mm. The composite sphere is initially at a uniform temperature, Ti = 40°C, and is exposed to a fluid at T∞ = 10°C with h = 30 W/m2 ∙ K. Determine the center temperature of the glass at t = 200 s. Neglect the thermal contact resistance at the interface between the two materials.

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A cold air chamber is proposed for quenching steel ball bearings of diameter D = 0.2 m and initial temperature Ti = 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 within the chamber is 1000 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: and c = 450 J/kg ∙ K. A soda lime glass sphere of diameter D1 25 mm is encased in a bakelite spherical shell of thickness L 10 mm. The composite sphere is initially at a uniform temperature, Ti = 40°C, and is exposed to a fluid at T∞ = 10°C with h = 30 W/m2 ∙ K. Determine the center temperature of the glass at t = 200 s. Neglect the thermal contact resistance at the interface between the two materials.

Transcribed Image Text: A cold air chamber is proposed for quenching steel ball bearings of diameter D = 0.2 m and initial temperature Ti = 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 within the chamber is 1000 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: and c = 450 J/kg ∙ K. A soda lime glass sphere of diameter D1 25 mm is encased in a bakelite spherical shell of thickness L 10 mm. The composite sphere is initially at a uniform temperature, Ti = 40°C, and is exposed to a fluid at T∞ = 10°C with h = 30 W/m2 ∙ K. Determine the center temperature of the glass at t = 200 s. Neglect the thermal contact resistance at the interface between the two materials.

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Calculate the Biot number.\mathrm{Bi}=\frac{h L_{c}}{k}=\frac{h r_{0}}{3 k}=\frac{1000 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}} \times \frac{0.2}{2} \mathrm{~m}}{3 \times 50 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}}=0.667As the Biot number is greater than 0.1 , the lumped capacitance method is not going to be used and use the approximate solution. For approximate solution, calculate Biot number.\mathrm{Bi}=\frac{h r_{0}}{k}=\frac{1000 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}} \times \frac{0.2}{2} \mathrm{~m}}{50 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}}=2From the coefficients used in the one-term approximation to the series solutions for transient one-dimensional conduction, The coefficients for the Biot number 2 are given by C_{1}=1.4793 and \zeta_{1}=2.0288 (For sphere).Using the approximate solution considering total energy transfer,\begin{aligned}\theta_{o}^{*} & =\left(1-\frac{Q}{Q_{0}}\right) \frac{\zeta_{1}^{3}}{3\left[\sin \left(\zeta_{1}\right)-\zeta_{1} \cos \left(\zeta_{1}\right)\right]} \\& =(1-0.7) \frac{2.0288^{3}}{3[\sin (2.0288)-2.0288 \times \cos (2.0288)]} \\& =0.4655\end{aligned}Calculate the Fourier number.\mathrm{Fo}=-\frac{1}{\zeta_{1}^{2}} \ln \left(\frac{\theta_{o}^{*}}{C_{1}}\right)=-\frac{1}{2.0288^{2}} \ln \left(\frac{0.4655}{1.4793}\right)=0.281Calculate the residence time of the balls within the chamber.t=\mathrm{Fo}_{\mathrm{o}} \frac{r_{0}^{2}}{\alpha}=0.281 \times \frac{\left(\frac{0.2}{2} \mathrm{~m}\right)^{2}}{2 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{sec}}=140.5 \mathrm{sec}Calculate the drive velocity.V=\frac{L}{t}=\frac{5 \mathrm{~m}}{140 \mathrm{sec}}=0.036 \mathrm{~m} / \mathrm{sec} ...