Question

In terms of qo , sphere radius ro, surrounding convection coefficient h, surrounding temperature T... P4: 20% What is the amount of radiant power leaving the opening of a cylindrical cavity with the diameter of 16 mm and depth of 24 mm with the emissivity of .9 and temperature of 1200 K. The surrounding temperature is 300 K. P5: 10%

Answer 1

The radiation surface area is the bottom circular area and the side surface of cylindrical area. The cylindrical cavity is open from one end.

TD 1 open from above h -4) 2.06810 m 2 m² To determine, net radiated power (P) = ? Stefan-Boltzmann law: Pre GA (74-74) here ez emissivity = 0.94-{fadh)+()} A = radiating area = (0.0167.0.024.%*)=(12.06x10 T= temperature of radiator = 1200k (2.81x104) =(4.07 X10 Tc- surrounding temperature = 300k r = Stefan's constant = 5.6703 X 10 WM2k+ P= 0.9 X5-6703X10 X 14.07X10 x ((2004-(300)+) P = x 10'2x{1440000 - 90000} : P = P153 x 10-12 x 1350000. 96.93 wax100 watt . P= . P=96.93X100 watt * -8 71.80 7.80 74.8x1-35 x 10-6