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T, = 700K E 0.8 RI T, =900K 2 R2 E20.5 3. Two gray-diffuse spheres have...

Question:

T, = 700K E 0.8 RI T, =900K 2 R2 E20.53. Two gray-diffuse spheres have properties and temperatures shown below. R and R2 are 0.1 m and 0.3 m respectively. (i) Comp

T, = 700K E 0.8 RI T, =900K 2 R2 E20.5
3. Two gray-diffuse spheres have properties and temperatures shown below. R and R2 are 0.1 m and 0.3 m respectively. (i) Compute heat transfer rate of sphere 1, q W]. Also, to minimize the heat loss, you decide to put a radiation shield in between the two spheres. (ii) Where do you prefer to place the radiation shield, i.e., close to sphere 1 or sphere 2 or right in the middle? (ii) Somehow you place the radiation shield with R3 = 0.2, E3 = 0.1. What is the temperature of the shield? (iv) compare the heat transfer rate with and without the radiation shield. The Stefan-Boltzmann constant is 5.67 x 10-8 W/m2-K4 (40 pt)

Answers

Two concentric sphere with radius R1 and R2.

GAVEN DATA Te 700K E=0.8 T2 =900k E2 -0.5 RI=0-1M R2 O3m Solution:Cwihaut Shield) Sur fate area, Al 4TY 4 1X (D-12 A」ニ 012ちも2 Vm R21-2 1D.5 20.8849 1-13x0.5 A2 Eb, Eba (Ri-2) 11 Rtotal Rtotal RI+R2tR t.99 +08849+7.96) Vm2 Rbobal =lo83 4 =07-T2t) C@1

with shield Radiation Shield wi th radius (R2 o 2) 1-03 For Concentrie Sphore, F321 AYQA of Radiabion Shield 4 xO 5026 m A3=Q 1a Cuith Shield) = (Tt-T) Rtotal Ri+R2tRR4+ RS Rto tal with shiel = 199 +08849 +H 90 +T 961+1.9&96 = 30.7255 2 5 4TX l68 T0

4 -767-67 X27.85 troo-Tt) 57x108 Shie td Temperature T3 @f6. 34 K Haat trans fer reduction - Quith shield due to Tad tation SCtii) Temperature ot Shield,3886.34K Civ 1-2 li thut Shield)= 2171.95W. T67. 61w 01-2 (with shield) due to radiation shied he

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