Cost of running a PC?

  • Thread starter Thread starter terry smith
  • Start date Start date
If you are air-conditioning, it takes at least 2 watts to get rid
of one watt of heat. So, for cost estimates, at least triple the
display power usage.
Click to expand...

???

It takes far less than 1:1 even with cheap, poor efficiency
units.

For example, a 12,000 BTU unit with an EER of 12.
12,000 / 12 = 1000 Watt used
12,000 BTU * .293 BTU/Watt = 3516 Watt removed

3.5:1 Energy removed:used
 
kony said:
.... snip ...

???

It takes far less than 1:1 even with cheap, poor efficiency
units.

For example, a 12,000 BTU unit with an EER of 12.
12,000 / 12 = 1000 Watt used
12,000 BTU * .293 BTU/Watt = 3516 Watt removed
Click to expand...

You are mixing units, to start with. 1 kilowatt _hour_ represents
about 3500 BTUs.

I am basing my comment on the basic efficiencies of heat engines,
which an air conditioner is. It uses the Carnot cycle, and the
efficiency is highly dependant on the ratio of source and sink
temperatures, in absolute units. I don't know what the EER
actually represents.

The Carnot cycle is reversible, and this is the basis of heat
pumps, which can be used for both heating and cooling. Now I have
just about run out of what I remember from thermodynamics classes
untold moons ago.
 
You are mixing units, to start with. 1 kilowatt _hour_ represents
about 3500 BTUs.
Click to expand...

1000 / .293 = 3412, not too far off of 3500.

I am basing my comment on the basic efficiencies of heat engines,
which an air conditioner is. It uses the Carnot cycle, and the
efficiency is highly dependant on the ratio of source and sink
temperatures, in absolute units. I don't know what the EER
actually represents.
Click to expand...

EER (Energy Efficiency Ratio) -
A ratio calculated by dividing the cooling capacity in Btu's
per hour (Btuh) by the power input in watts at any given set
of rating conditions, expressed in Btuh per watt
(Btuh/watt).

The calculation is based on a test of the system at constant
operation with an outdoor temperature of 95°F. and 50%
relative humidity. The indoor temperature used is 80°F and
50% relative humidity. The total of the electricity used for
the entire system with a specific indoor coil is the divisor
for the formula.
The Carnot cycle is reversible, and this is the basis of heat
pumps, which can be used for both heating and cooling. Now I have
just about run out of what I remember from thermodynamics classes
untold moons ago.
Click to expand...

I just take them at their word, never tried measuring AC
consumption but it "seems" far enough beyond 1:1 that even
if the testing were done in optimal conditions, real-life
results should still be well above 2:1 Input:Removed
(energy). I don't even recall seeing any AC units below an
EER of 10, but that might be the (legal minimum EER in
USA?).
 
kony said:
1000 / .293 = 3412, not too far off of 3500.


EER (Energy Efficiency Ratio) -
A ratio calculated by dividing the cooling capacity in Btu's
per hour (Btuh) by the power input in watts at any given set
of rating conditions, expressed in Btuh per watt
(Btuh/watt).

The calculation is based on a test of the system at constant
operation with an outdoor temperature of 95°F. and 50%
relative humidity. The indoor temperature used is 80°F and
50% relative humidity. The total of the electricity used for
the entire system with a specific indoor coil is the divisor
for the formula.


I just take them at their word, never tried measuring AC
consumption but it "seems" far enough beyond 1:1 that even
if the testing were done in optimal conditions, real-life
results should still be well above 2:1 Input:Removed
(energy). I don't even recall seeing any AC units below an
EER of 10, but that might be the (legal minimum EER in USA?).
Click to expand...

There is something fundamentally wrong in your calculations. They
seem to imply that all the heat energy installed in a room by 1000
watts can be pumped out with another 1000 watts. Since Carnot
engines are reversible, that implies that the original 1000 watts
can be generated by that engine and that ambient/sink difference.
That is a perpetual motion machine, and I don't believe a word of
it.

Aha - we have the original 1000 watts, plus the 1000 watts to pump
it out. That is 2000 watts. Thus the power cost to run something
dissipating x watts is 2x watts, assuming we maintain constant
temperature. Meanwhile we also have to pump out at least some
fraction of the 1000 watts of pump power, which raises the pump
power needed, etc. The maximum efficiency of 50% I can believe.

So my original statement of adding 2x power should be adding at
least 1x power. So if you save 1 watt in an air conditioned
environment, you are actually saving at least 2 watts.
 
There is something fundamentally wrong in your calculations. They
seem to imply that all the heat energy installed in a room by 1000
watts can be pumped out with another 1000 watts. Since Carnot
engines are reversible, that implies that the original 1000 watts
can be generated by that engine and that ambient/sink difference.
That is a perpetual motion machine, and I don't believe a word of
it.
Click to expand...

The difference is that the AC only moves the heat energy
outside the room, it is a two-part system, each part fairly
insulated from the other. The inside gets cooler but the
outside hotter. There is no perpetuality to it, just
movement of energy.


Aha - we have the original 1000 watts, plus the 1000 watts to pump
it out. That is 2000 watts. Thus the power cost to run something
dissipating x watts is 2x watts, assuming we maintain constant
temperature. Meanwhile we also have to pump out at least some
fraction of the 1000 watts of pump power, which raises the pump
power needed, etc. The maximum efficiency of 50% I can believe.

So my original statement of adding 2x power should be adding at
least 1x power. So if you save 1 watt in an air conditioned
environment, you are actually saving at least 2 watts.
Click to expand...

There is no doubling of the "1000 watts".

AC units are tested and certified to be of a certain EER
value. That value is the BTU/Watts efficiency, and in my
example the EER was 12 (common EER for even the cheapest $80
AC unit is going to be around 10).


Given an AC with an EER of 12, and giving it 1000 watt hour
input, it will move 12,000 BTU of energy out of the room.
Next we need determine the wattage equivalent of 12000 BTU.

Btu/hour x 0.293 = watts (standard conversion)

12000 x 0.293 = 3516

So, the AC will need 1000W input to move 3516 Watt of heat
out of the room at given interval, to return room to same
temp prior to the introduction of this 3516W device(s). The
number of hours, rate of cooling needed would determine
needed BTU rating of the AC but given the EER of 12 for this
example, as a fixed constant, all that is needed is to buy
the right size AC unit, that variable can be ignored... not
in real life could it be ignored, but for this specific
scenario of 3512W @ 80F inside, 95F outside, it can be, as
it is the definition of EER.

The reaining issue is how much of that 1000W used by the AC,
is relased into the room rather than outside the room. For
the sake of keeping the model simple I will suggest that it
is a perfectly insulated window unit and that ALL of the
input power to the AC, is heat that never enters the room,
is removed by the blower outside.
 
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