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Electrical heat gain - Page 9

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Posted by Eeyore on April 15, 2009, 9:01 am

Mike wrote:

I think the heating industry hung on to BTUs for as long as possible to
put people off doing DIY central heating. If a radiator is rated at say
1kW, most people can relate to that.


Posted by RamRod Sword of Baal on April 14, 2009, 7:25 pm

Maybe, but this information as taken direct from the Carrier 'Handbook of
air conditioning design' and it was in Imperial measurements, and as most of
the people here seem to be from the USA it seem the best system to quote.

Here (Australia) we use the metric system, and I can use both. Using the
metric system to would be quite difficult for someone who was not familiar
with the metric system and if in the US you would need to transpose the
results into BTUs so as to select the equipment, as US equipment is shown as
capacity in BTUs not watts.

BTW the heat from the fluorescent fixture was a direct quote from this book
despite what Neon Joe or who ever says, so he had better tell Carrier they
are wrong :-)

Mind you Carrier say they invented air conditioning..............

In answer to the request for solar load, that is much more difficult as one
needs to know the design outdoor temperature, the design indoor temperature
and the orientations of the various walls and windows and if they have  sun
load, are the walls and ceiling insulated and if so how much, also the
construction of the walls and windows. It is much more involved than
electrical load.

Posted by Curbie on April 14, 2009, 10:51 pm

I don't want to speak for (Neon) John, but what I took away from his
point was a that I know of no device that converts energy with gains,
only losses. So I think something's amiss with that?

I have no problem with the metric system or the people who use or
"think" in it, they had no more choice what they learned in grade
school than I did.

Both systems are just different ways to describe the same thing. I
find easier to do a simple conversion than build a time-machine to
re-live 40 years of my life using the metric system. Just lazy I

I have a database containing Year, Month, Day, Hour, Temperature, Dew
Point, Relative Humidity, Atmospheric Pressure, and other valuable
climate readings, recorded hourly for ever day of a month, each month
is selected (some how?) from a 30 year data set recorded by 239
weather stations scattered around the US.  The data is collected,
compiled, and distributed by the US government with the title TMY
(Typical Meteorological Year) 2 or 3, I would imagine the Australian
government does something similar, but I don't really know.

I'm working on a spread-sheet to tie together resources (wind & solar)
and a climate summary based on proximity to a each TMY reporting
station to drive spread-sheet calculations for heating (space &
water), cooling, and electrical usage, generation. & storage needs.
I've already written functional sheets on electrical usage,
generation, & storage, also a household design sheet which functions
for calculating heat loss for a basic "box", but I've added N-S-E-W
parameters and I'm now pushing to add, heat gain for human, electrical
and solar heat gain. This is what has kicked off my questions about
heat gain and solpos.c.

Thanks for your time and help.


Posted by Eeyore on April 15, 2009, 9:02 am

Curbie wrote:

Energy is never lost, it just changes form, ending up as thermal.


Posted by daestrom on April 17, 2009, 1:09 am

The real question is, "When a flourescent fixture says it is rated for 40
watts, is that the power at the input to the entire fixture, or the input to
the bulb?"  Obviously if it is the input to the bulb, then the total power
input to the fixture is somewhat higher than that.  He's just applying a
'rule-of-thumb' saying the fixture is only 80% efficient.  To get 40 watts
at the bulb, you need to put in 50 watts to the fixture.  And of course all
50 watts eventually end up as heat in the room.

It's really not much different than saying a 1 hp (746 watt) motor draws
more than 746 watts of power from the line.  Motor ratings are given for the
output of the shaft, not the line input.  That's why to find the power draw
you divide the motor rating by the projected efficiency.  A small 1 hp motor
that is 75% efficient would draw about 995 watts of power.

Nobody claims you get more heat out than electricity input, it's just using
the output rating of the equipment and dividing by efficiency to get what
the nominal input would be.


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