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Checking for applicability of lumped capacity model: Biot number (for sphere): B i=\frac{h r}{3 k} where h = 100 W/m2K r = 0.1 / 2 = 0.05 m k = 73 W/mK B i=\frac{h r}{3 k} B i=\frac{100 \times 0.05}{3 \times 73}=>B i=0.0228 Since Bi < 0.1, lumped capacitance model is valid. Given 70% energy to be lost during the amount of time in the chamber. Thus, Q_{\text {lost }}=0.7 \times Q_{\text {total }} =>m . c .\left(T_{\text {initial }}-T_{\text {final }}\right)=0.7 \times\left(\right. m.c. \left.\left(T_{\text {initial }}-T_{\text {ambient }}\right)\right) =>T_{\text {initial }}-T_{\text {final }}=0.7 \times\left(T_{\text {initial }}-T_{\text {ambient }}\right) where T(initial) = 400 C T(final) = final exit temp of balls = ? T(ambient) = -15 C T_{\text {initial }}-T_{\text {final }}=0.7 \times\left(T_{\text {initial }}-T_{\text {ambient }}\right) 400-T_{\text {final }}=0.7 \times(400-(-15))=>T_{\text {final }}=109.5^{\circ} \mathrm{C} Temperature variation with time: \frac{T_{\text {final }}-T_{\text {ambient }}}{T}=e^{-\frac{h A}{\rho V c} \times \tau}T_{\text {initial }}-T_{\text {ambient }} where A = surface area of balls = 4 \prod r^{2}=4 \prod \times 0.05^{2}=0.0314 \mathrm{~m}^{2} t r = radius of sphere = 0.05 m V = volume of sphere = \frac{4}{3} \prod r^{3}=\frac{4}{3} \prod \times 0.05^{3}=5.236 e-4 m^{3} \rho = density of the material = \frac{k}{\alpha c}=\frac{73}{2 e-5 \times 450}=8111.11 \mathrm{~kg} / \mathrm{m}^{3} t k = 73 W/mK \mathrm{C} = 2e-5 m2/s c = 450 J/kgK T = residence time =  time required to cool to this temp = ? \frac{T_{\text {final }}-T_{\text {ambient }}}{T}=e^{-\frac{h A}{\rho V c} \times \tau}T_{\text {initial }}-T_{\text {ambient }} \frac{109.5-(-15)}{400-(-15)}=e^{-\frac{100 \times 0.0314}{8111.11 \times 5.236 \varepsilon-4 \times 450} \times \tau}=>\tau=732.7898 \sec (s)( Ans. )   Suggestable velocity of conveyor:     --------------------------------------- Kindly upvote. :)     ...