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G6KUI > TECH 02.05.08 12:41l 65 Lines 2084 Bytes #999 (0) @ WW
BID : 22959_G6KUI
Read: GUEST
Subj: Re: Thermal conductivity question
Path: IZ3LSV<IQ0LT<IK2XDE<DB0RES<ON0AR<YU7R<HG8LXL<CX2SA<GB7CIP<GB7COV<
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Sent: 080502/1107z @:GB7DBY.#23.GBR.EU NPF2.54c [G6KUI PMS Alvaston Derby]
Andy G0FTD asked some very basic questions.
} Subject: Thermal conductivity question
}
}
} Thanks for reading.
}
} I have two questions related to heat transfer.
}
} a) How does heat loss work. For example, is it inverse square law, like
} RF losses. Or linear, or what ?
}
}
} b) I have a two boilers. One is 10 times bigger than the other. Do I
} simply need 10 times the power (kilowatts) *to raise the water to the
} same temperature ?*
}
Firstly, Heat is transferred by one , or more , of the following.
Conduction
Convection
Radiation
Molacular Conduction in Gases ( below a certain pressure )
Conduction - this is the usual way of heat transfer in a solid.
Depending on the material, and the temperature difference, this is
usually LINEAR to the temperature difference.
Convection - this is the usual way of heat transfer within a liquid or gas.
Things get rather complicated because it relies on the physical movement
of the liquid or gas to transport the heat. You can either have "streamined
flow" or "turbulent flow" or a combination of both. Streamlined flow is
more efficient at heat transfer than turbuent flow. At low temperature
differentials heat transfer will probably be by streamlined flow and
probably move to turbulent flow as the temperature differential increases.
Radiation - as the name suggests this is a means of heat transfer across a
gap from one object to another. The gap can have nothing in it ( a vacuum )
or be filled with a gas or liquid.
You probably have heard of "black body" radiation. This is the most efficent
and has a factor of 1. Everything else has something less than 1.
Assuming black bodies, the heat transfer is proportional to the 4th power
of the temperature on an absolute scale ( KELVIN ).
T1= temp of 1st object in Kelvin
T2= temp of 2nd object in Kelvin
Heat Transfer = Constant x ((T1xT1xT1xT1)-(T2xT2xT2xT2)).
Answering your second question.
No, you don't need 10 times the power , you just need 10 times the time
at the same power input ;-))
Hope that helps !
73, Pete G6KUI
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