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Loss due to inefficiency... |
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| Mar27-05, 03:59 PM | #1 |
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Loss due to inefficiency...
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)
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| Mar27-05, 04:50 PM | #2 |
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Recognitions:
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AM |
| Mar28-05, 07:59 AM | #3 |
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| Mar28-05, 10:05 AM | #4 |
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Recognitions:
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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 |
| Mar28-05, 10:29 AM | #5 |
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| Mar28-05, 10:31 AM | #6 |
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| Mar28-05, 12:18 PM | #7 |
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So basically there's only sound, heat, and infrared to take into consideration?
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