Question

1. A plate is placed in a large vacuum chamber to control its cooling rate. To prevent cracking, the temperature gradient in the plate cannot exceed 15 K/mm at any point during cooling. The plate is initially at 900 K with an emissivity of 0.8 and thermal conductivity of 1.5 W/mK. Assume that one of the surface of the plate is insulated and the other is exposed to the chamber, and the chamber walls are considered as black surface. Apply the energy balance and determine the lowest temperature Tour allowed for the walls of the vacuum chamber? (10 pts) Plate Insulation

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Answer 1

Solution: Given that : Initial temperatene (Ti)= 900k. Emisrivity (t) = 0.8 Thermal conductivity (K)= 1.5 W W/mK. Ts = surface temperature of the plate. Given thar, the temperature gradiant in the plate respect km. Assume exceed is k/mm with to that can not dT dx < 15000 K maximum heat Flux allowed dt dn = 1.5 x 15 000 22500 W/m2 balance equation; Now, considering the energy CE ( Tsung - Ts") = -K DI = -22500 O dx minimum Surface temperatere in given by; Ts - To - 15000 L where L = length = 875k. mant yam 4 Taking L = Length - 1mm C Assume). Ts = 900-(15*1 ) substituting in ean oi (5.670x1008)60.8) ( Tsunj - 875x) = -22500 2554) 1/4 Trung? +(875 (5.670X10-8)x0.8 Tsurf = 547.95k -22500 - Am. CS Scanned with CamScanner (Thank you) (Upvore please)