The electric power generator for an orbiting satellite is...

The electric power generator for an orbiting satellite is composed of a long, cylindrical uranium heat source that is housed within an enclosure of square cross section. The only way for heat that is generated by the uranium to leave the enclosure is through four rows of the thermoelectric modules of Example 3.13. The thermoelectric modules generate electric power and also radiate heat into deep space characterized by $T_{\text {sur }}=4 \mathrm{~K}$. Consider the situation for which there are 20 modules in each row for a total of $M=4 \times 20=80$ modules. The modules are wired in series with an electrical load of $R_{e, \text { load }}=250 \Omega$, and have an emissivity of $\varepsilon=0.93$. Determine the electric power generated for $\dot{E}_{g}=1,10$, and $100 \mathrm{~kW}$. Also determine the surface temperatures of the modules for the three thermal energy generation rates.

Key Concepts

Energy Balance in Thermal Systems

The energy balance in thermal systems requires that the thermal energy produced, converted, and lost is in equilibrium. For a satellite power generator, this means that the nuclear heat produced must be balanced by the energy converted into electricity and the energy lost through radiative cooling. Achieving this balance is essential for controlling the system’s operating temperatures and ensuring reliable performance of both the thermoelectric modules and the satellite as a whole.

Series Wiring and Electrical Load

When thermoelectric modules are wired in series, their electrical outputs add up while the same current flows through each module. This configuration affects the overall voltage and current delivered to an external electrical load. Matching the output characteristics of the generator with the electrical load, including considerations like resistance, is crucial for maximizing power transfer and overall system efficiency.

Thermoelectric Conversion

This concept involves converting heat energy directly into electrical energy using the thermoelectric effect. Materials that exhibit this effect generate a voltage when subjected to a temperature gradient, allowing for the harvesting of waste heat or, in this case, a nuclear heat source. The efficiency of this conversion depends on material properties and design parameters, making it a critical factor in energy generation systems in environments such as space.

Radiative Heat Transfer

Radiative heat transfer refers to the process by which energy is emitted from a surface in the form of electromagnetic radiation. In space, where convective and conductive modes of heat transfer are negligible, radiation is the primary mechanism for dissipating heat. The Stefan-Boltzmann law governs this process, stating that the power radiated per unit area is proportional to the fourth power of the surface temperature, modulated by the emissivity of the surface.

Transcript

00:01 So here in this problem for part a we have the intensity of a spherical black body as a function of the wavelength.

00:10 So here i as a function of lambda d lambda just equal to 2 pi times h c squared.

00:28 We're here h is blanks constant and c is the speed of light divided by lambda power 5 multiplied by e power hc over lambda k t and here k is a bultzman constant and t is the temperature of this black body minus 1 d lambda but we also know that the energy is related to the wavelength through this relation hc over lambda so from here is hc divided by e and now the absolute value of d lambda is just equal to hc divided by e squared de so now i of lambda d lambda is just equal to 2 pi hc squared divided by e squared e power e over kt minus 1 multiplied by e power 5 over h power 5 c bar 5 times hc divided by e squared de and this is equal to 2 pi e cubed divided by h cubed divided by h cubed c cubed c squared multiplied by e -power e divided by k t minus 1 d e and this is the intensity as a function of the energy.

02:54 So now the intensity as a function of the energy adjust 2 pi e cubed divided by h cubed c squared times e power e over k t minus 1.

03:05 This is for part a.

03:07 Now for part b, we have here i as a function of e is just equal to alpha times e cubed divided by e power e over k t minus 1.

03:32 We're here alpha is just equal to 2x divided by h cubed c square.

03:48 So now to get the maximum value of this intensity as a function of the energy i, so here we need to have d .i by d .e to be equal to zero.

04:05 So now d .i.

04:13 By d .e.

04:15 Is just equal to alpha times 3e squared times e power e over k t minus 1 minus a cube divided by k t multiplied by e power e over k t this is all divided by e power e over k t minus 1 squared so now this must be equal to 0 so now 3 minus e over k t times e power e over k t divided by a power e over k t minus 1 is just equal to 0 we now let x to be equal to e over k t so now we would have 3 minus x is just equal to 3 times e power negative x and now the solution is x is equal to 2 .8 to 1.

06:26 So now e is just equal to 2 .8 to 1 times kt and this is the value at which the intensity is the maximum and we call it e0...

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