a) On the side of an orbiting space station is an instrumentation module whose
electronics dissipate 1 kW. This energy is dissipated into deep space from the surface of the package, which has an area of 1 m2. The emissivity of the surface is 0.8. Calculate the steady state temperature of the surface of the module under the
following conditions:
i) not exposed to solar radiation.
ii) exposed to a solar flux of 750 W m−2, if the absorptivity of the module surface to
solar radiation is 0.3.
b) Explain the physical significance of Laplace’s equation in the context of conduction heat transfer. You may illustrate your answer with sketches if you wish.
electronics dissipate 1 kW. This energy is dissipated into deep space from the surface of the package, which has an area of 1 m2. The emissivity of the surface is 0.8. Calculate the steady state temperature of the surface of the module under the
following conditions:
i) not exposed to solar radiation.
ii) exposed to a solar flux of 750 W m−2, if the absorptivity of the module surface to
solar radiation is 0.3.
b) Explain the physical significance of Laplace’s equation in the context of conduction heat transfer. You may illustrate your answer with sketches if you wish.
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Q=Ea∆T^4
1000= 5.67 x 10-8 x 0.8 x 1 x T^4
T^4=2.204586x10^10
T^5= 385k = 112 degrees celcius
ii)
Q= 1000+ 750x0.3
Q=Ea∆T
1225=5.67x10-8x0.8x1xT^4
T^4=2.700617x10^10
b) Laplace’s equations, d2t/dx2=0 , simply means that, while there may be a temperature gradient causing conduction heat flow (i.e. the first derivative is not zero), there is no difference in the heat flow that would cause accumulation (i.e. the second differential does equal zero , so there is no accumulation and therefore no temperature change. Laplace’s equation therefore defines steady state conditions.
1000= 5.67 x 10-8 x 0.8 x 1 x T^4
T^4=2.204586x10^10
T^5= 385k = 112 degrees celcius
ii)
Q= 1000+ 750x0.3
Q=Ea∆T
1225=5.67x10-8x0.8x1xT^4
T^4=2.700617x10^10
b) Laplace’s equations, d2t/dx2=0 , simply means that, while there may be a temperature gradient causing conduction heat flow (i.e. the first derivative is not zero), there is no difference in the heat flow that would cause accumulation (i.e. the second differential does equal zero , so there is no accumulation and therefore no temperature change. Laplace’s equation therefore defines steady state conditions.