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Zigmanuir [339]
3 years ago
10

Two 60 cm parallel disks are separated by 40 cm and are aligned directly on top of each other. Both disks are black surfaces wit

h temperatures of 450 K. The backs of the disks are insulated and the surroundings may be treated as a black body at 300 K. Determine the net radiation heat transfer from the disks to the environment.
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

775.48 W

Explanation:

given,

diameter of disk = 0.6 cm

length of the disk = 0.4 m

T₁ = 450 K         T₂ = 450 K      T₃ = 300 K

\dfrac{d}{r_1}=\dfrac{0.4}{0.3} = 1.33

now,

the value of view factor (F₁₂)corresponding to 1.33

F₁₂ = 0.265

F₁₃ = 1 - 0.265 = 0.735

now,

net rate of radiation heat transfer from the disk to the environment:

=\dot{Q_{1-3}+Q_{2-3}} = 2 \dot{Q_{1-3}}

       = 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)

       = 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)

       = 775.48 W

Net radiation heat transfer from the disks to the environment = 775.48 W

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