The diagram below illustrates a simplified version of the Jet Propulsion Laboratory's 25-ft space simulator chamber. Within it, tests are run for deep space probes and their components. In this particular experiment, a sensor assembly with its housing is placed inside. The sensor-housing electronics create heat such that q=3181.0 W exits the assembly's top surface into the chamber. The chamber itself is under vacuum, and its walls are at cryogenic temperatures to simulate conditions in space. Surfaces 1 and 2 are flat. Assume they are flush. The top surface of the chamber, Surface 4, can be approximated as a perfectly insulated, flat ceiling. The exposed surface of the floor, Surface 2, is a black, donut- shaped surface. Additional surface information can be seen below, including various parameters and dimensions. Assume all surfaces are opaque and diffuse, and conditions are at steady-state. PARAMETERS & DIMENSIONS Surface 1: A 7.07 m²; diameter d; = 3 m; &₁ = 0.5; Surface 2: A45.74 m²; black surface, &₂ = 1; Surface 3: A=669.8 m²; diameter d3=8.2 m; Surface 4: A = 52.8 m²; diameter d. = 8.2 m; d4 = ds 7 7 } d1 q=3181.0 W T₂ = 100 K L=26m; &=0.9 790 K E=0.1: perfectly insulated

Elements Of Electromagnetics
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The diagram below illustrates a simplified version of the Jet Propulsion
Laboratory's 25-ft space simulator chamber. Within it, tests are run for deep space
probes and their components. In this particular experiment, a sensor assembly with its
housing is placed inside. The sensor-housing electronics create heat such that
q=3181.0 W exits the assembly's top surface into the chamber.
The chamber itself is under vacuum, and its walls are at cryogenic temperatures
to simulate conditions in space. Surfaces 1 and 2 are flat. Assume they are flush.
The top surface of the chamber, Surface 4, can be approximated as a perfectly
insulated, flat ceiling. The exposed surface of the floor, Surface 2, is a black, donut-
shaped surface.
Additional surface information can be seen below, including various parameters
and dimensions. Assume all surfaces are opaque and diffuse, and conditions are at
steady-state.
PARAMETERS & DIMENSIONS
Surface 1: A = 7.07 m²; diameter d; = 3 m; &₁ = 0.5;
Surface 2: A₂ = 45.74 m²; black surface, E₂ = 1; T₂ = 100 K
Surface 3: A = 669.8 m²; diameter ds = 8.2 m;
Surface 4: A₁ = 52.8 m²; diameter d = 8.2 m;
d4 = ds
7
6
LN
di
q=3181.0 W
L= 26 m; &=0.9 ; T3=90 K
E= 0.1 perfectly insulated
Transcribed Image Text:The diagram below illustrates a simplified version of the Jet Propulsion Laboratory's 25-ft space simulator chamber. Within it, tests are run for deep space probes and their components. In this particular experiment, a sensor assembly with its housing is placed inside. The sensor-housing electronics create heat such that q=3181.0 W exits the assembly's top surface into the chamber. The chamber itself is under vacuum, and its walls are at cryogenic temperatures to simulate conditions in space. Surfaces 1 and 2 are flat. Assume they are flush. The top surface of the chamber, Surface 4, can be approximated as a perfectly insulated, flat ceiling. The exposed surface of the floor, Surface 2, is a black, donut- shaped surface. Additional surface information can be seen below, including various parameters and dimensions. Assume all surfaces are opaque and diffuse, and conditions are at steady-state. PARAMETERS & DIMENSIONS Surface 1: A = 7.07 m²; diameter d; = 3 m; &₁ = 0.5; Surface 2: A₂ = 45.74 m²; black surface, E₂ = 1; T₂ = 100 K Surface 3: A = 669.8 m²; diameter ds = 8.2 m; Surface 4: A₁ = 52.8 m²; diameter d = 8.2 m; d4 = ds 7 6 LN di q=3181.0 W L= 26 m; &=0.9 ; T3=90 K E= 0.1 perfectly insulated
Given the information you have for Surface 2, choose the equation
necessary to help solve for the radiosity J2.
(A.) 92=0W = A₂F21 U2 −J₁) + A₂F2302-J3) + A₂F2402-J4)
(B.) (OT2 -√₂)E₂A₂ = A2F21(J2 − J1) + A2F23(J2 − J3) + A₂F24(J2-J4)
1-E₂
(C.)/₂=0T₂
(D.) OT = A₂F2102-1) + A₂F2302-/3) + A₂F24(√2-J4)
(E.) 92 = 07₂
Transcribed Image Text:Given the information you have for Surface 2, choose the equation necessary to help solve for the radiosity J2. (A.) 92=0W = A₂F21 U2 −J₁) + A₂F2302-J3) + A₂F2402-J4) (B.) (OT2 -√₂)E₂A₂ = A2F21(J2 − J1) + A2F23(J2 − J3) + A₂F24(J2-J4) 1-E₂ (C.)/₂=0T₂ (D.) OT = A₂F2102-1) + A₂F2302-/3) + A₂F24(√2-J4) (E.) 92 = 07₂
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