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Loss due to inefficiency... 
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#1
Mar2705, 03:59 PM

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I am just double checking an idea that I've always figured was true, but never really asked someone who knows about. In basically any machine that is not 100% efficient, the only "place" the lost energy (due to the inefficiency) can go is into its surroundings as heat energy, or radiation, correct? If this is so, how easily/effectively can radiation be contained and transferred back into heat energy? How is this done? (I assume just very reflective surfaces surrounding the thing that is emitting radiation)
Thanks. 


#2
Mar2705, 04:50 PM

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AM 


#3
Mar2805, 07:59 AM

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#4
Mar2805, 10:05 AM

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Loss due to inefficiency...
The energy loss due to radiation is: [tex]E = \sigma T^4[/tex] where [itex]\sigma = 5.67051 \times 10^{8} W/m^2K^4[/itex] So if the temperature (eg steam) is in the order of 400K, you are looking at a maximum of 1.5 kilowatts/m^2 of radiation loss, or about 2 horsepower. AM 


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Mar2805, 10:29 AM

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Mar2805, 10:31 AM

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#7
Mar2805, 12:18 PM

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So basically there's only sound, heat, and infrared to take into consideration?



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