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Part Two 1. (15 pts) In the manufacturing process of small ball bearings, spherical steel balls of 0.25-cm diameter are heat-treated to obtain a particular

Part Two 1. (15 pts) In the manufacturing process of small ball bearings, spherical steel balls of 0.25-cm diameter are heat-treated to obtain a particular metal structure. They are then submerged in a moving engine oil at a temperature of T = 60C. The cooling process lasts 2 seconds, when the ball bearings' temperature drops from 300C to 140C. Determine the oil stream velocity, Voor required in this process. Assume that the Lumped-capacitance Model is valid in your calculations. Use the following Nusselt-Reynolds correlation equation for a sphere: Nup = 0.3 Re06 Pri, where Nup hD VooD = k Use the following for the properties of steel: p = 7801, k = 43, Cp = 473 K W and Rep , a = 1.172 x 10-5 m Use the following for the properties of engine oil: k = 0.133- W v = 6.60 10-6 m, Pr = 98.3. m.K BSE
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1. (15 pts) in the manufacturing process of small ball bearings, spherical steel balls of \\( 0.25-\\mathrm{cm} \\) diameter are heat-treated to obtain a particular metal structure. They are then submerged in a moving engine oil at a temperature of \\( T_{c o}=60^{\\circ} \\mathrm{C} \\). The cooling process lasts 2 seconds, when the \\( V_{\\infty} \\), required in this process. Assume that \\( 300^{\\circ} \\mathrm{C} \\) to \\( 140^{\\circ} \\mathrm{C} \\). Determine the oil stream velocity, calculations. Use the following Nusselt-Reynolds correlation equation for a sphere: \\( N u_{D}=0.3 R e_{D}^{0.6} P r^{\\frac{1}{3}} \\), where \\( N u_{D}=\\frac{\\hbar D}{k} \\), and \\( R e_{D}=\\frac{v_{\\infty} D}{v} \\). Use the following for the properties of steel: \\( \ ho=7801 \\frac{\\mathrm{kg}}{\\mathrm{m}^{3}}, k=43 \\frac{\\mathrm{w}}{\\mathrm{m} \\cdot \\mathrm{K}}, c_{p}=473 \\frac{\\mathrm{l}}{\\mathrm{kgk}}, \\alpha=1.172 \\times 10^{-5} \\frac{\\mathrm{m}^{2}}{\\mathrm{~s}} \\). Use the following for the properties of engine oil: \\( k=0.133 \\frac{\\mathrm{w}}{\\mathrm{m} \\cdot \\mathrm{K}}, v=6.60 \\times 10^{-6} \\frac{\\mathrm{m}^{2}}{\\mathrm{~s}}, \\mathrm{Pr}=98.3 \\)

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