I was testing the power consumption of a computer (to calculate the power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back of the power supply (I'm assuming that this means only the filter capacitors are connected), the electricity monitor showed that it was drawing 0.049 amps with a power factor of 0%. Am I right in thinking this means I am being charged for no electricity, while 11 watts of electricity are being dissipated in the national grid somewhere? The monitor also shows 0 watts consumption, which I assume is useful watts as charged by the electricity meter - because it then goes on to use this to show cost over a period of time. This seems very wasteful to me. Or am I miscalculating something?

--
http://petersparrots.com
http://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty 4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a fake Jeep?"

In message op.v5y6yop4ytk5n5@i7-940, Lieutenant Scott
writes
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back of
the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

No. You are absolutely right. Your 11W is actually 'volt-amps' (VA), and
VA x PF = watts. As your PF is 0, your PC is not consuming any power.

However, the 0.049A is coming to your PC via the house wiring. As this
will have some (hopefully very small) resistance, there will be a small
'I squared R' power loss in the wires. And, as you say, similar loss
will be occurring in the National Grid wiring.

Of course, your small capacitive current might be a 'good thing'. It
might be helping to compensate for an inductive load next door (well, a
near neighbour on the same distribution phase as you are on), where the
PF is less than 1.
--
Ian

On Dec 4, 5:31*pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor thingies. *It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back of the power supply (I'm assuming that this means only the filter capacitors are connected), the electricity monitor showed that it was drawing 0.049 amps with a power factor of 0%. *Am I right in thinking this means I am being charged for no electricity, while 11 watts of electricity are being dissipated in the national grid somewhere? *The monitor also shows 0 watts consumption, which I assume is useful watts as charged by the electricity meter - because it then goes on to use this to show cost over a period of time. *This seems very wasteful to me. *Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty 4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred. It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.


NT

In message
, NT
writes
On Dec 4, 5:31*pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. *It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. *Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? *The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. *This seems very wasteful to
me. *Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).




--
Ian

On Dec 4, 8:10*pm, Ian Jackson
wrote:
In message
, NT
writes



On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.


NT

In message
, NT
writes
On Dec 4, 8:10*pm, Ian Jackson
wrote:
In message
, NT
writes



On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.

Actually, I believe that large factories, companies and other
establishments which consume lots of electricity are obliged to have
power factor correction equipment on their supply feeds. In some cases
where the load varies a lot, this is dynamic, and always tries to keep
the overall power factor as close as possible to 1, regardless of how
much power is being used.
--
Ian

Ian Jackson wrote:
In message
, NT
writes
On Dec 4, 8:10 pm, Ian Jackson
wrote:
In message
, NT
writes



On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.


because with SMPS and RFI caps, there's no such animal, More likely to
need an INDUCTOR actually.

We really need to design a 12vDC ring standard for all the electronics..



Actually, I believe that large factories, companies and other
establishments which consume lots of electricity are obliged to have
power factor correction equipment on their supply feeds. In some cases
where the load varies a lot, this is dynamic, and always tries to keep
the overall power factor as close as possible to 1, regardless of how
much power is being used.

On Dec 4, 9:16*pm, The Natural Philosopher
wrote:
Ian Jackson wrote:
In message
, NT
writes
On Dec 4, 8:10 pm, Ian Jackson
wrote:
In message
, NT
writes

On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.

because with SMPS and RFI caps, *there's no such animal, More likely to
need an INDUCTOR actually.

We really need to design a 12vDC ring standard for all the electronics..

I'm with you on the LV distribution. The main issue is standards. The
5v usb standard has developed and could be used.

NT

Actually, I believe that large factories, companies and other
establishments which consume lots of electricity are obliged to have
power factor correction equipment on their supply feeds. In some cases
where the load varies a lot, this is dynamic, and always tries to keep
the overall power factor as close as possible to 1, regardless of how
much power is being used.

On 04/12/2011 22:23, NT wrote:

I'm with you on the LV distribution. The main issue is standards. The
5v usb standard has developed and could be used.

It's coming, at least for US commercial environments - 24 volts:
http://www.emergealliance.org/Standa...ndardFAQs.aspx

--
Andy

On Sun, 04 Dec 2011 17:53:43 -0000, Ian Jackson wrote:

In message op.v5y6yop4ytk5n5@i7-940, Lieutenant Scott
writes
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back of
the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

No. You are absolutely right. Your 11W is actually 'volt-amps' (VA), and
VA x PF = watts. As your PF is 0, your PC is not consuming any power.

However, the 0.049A is coming to your PC via the house wiring. As this
will have some (hopefully very small) resistance, there will be a small
'I squared R' power loss in the wires. And, as you say, similar loss
will be occurring in the National Grid wiring.

Of course, your small capacitive current might be a 'good thing'. It
might be helping to compensate for an inductive load next door (well, a
near neighbour on the same distribution phase as you are on), where the
PF is less than 1.

Am I right in thinking the losses on the power lines (if I'm not compensating for a neighbour) are 11 watts?

--
http://petersparrots.com
http://petersphotos.com

TESTICULATING
Waving your arms around and talking ********.

On Sun, 04 Dec 2011 20:10:39 -0000, Ian Jackson wrote:

In message
, NT
writes
On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

Or is it lagging by 270 degrees? That makes sense. Voltage had to happen before current surely?

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Business premises usually have a capacitor bank.

--
http://petersparrots.com
http://petersphotos.com

"When one engine fails on a twin-engine airplane you always have enough power left to get you to the scene of the crash."

On Sun, 04 Dec 2011 21:16:36 -0000, The Natural Philosopher wrote:

Ian Jackson wrote:
In message
, NT
writes
On Dec 4, 8:10 pm, Ian Jackson
wrote:
In message
, NT
writes



On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.


because with SMPS and RFI caps, there's no such animal, More likely to
need an INDUCTOR actually.

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.


--
http://petersparrots.com
http://petersphotos.com

Excuse me sir, are you playing the bagpipes or sexually abusing an octopus?

Lieutenant Scott wrote:
On Sun, 04 Dec 2011 21:16:36 -0000, The Natural Philosopher
wrote:

Ian Jackson wrote:
In message
, NT
writes
On Dec 4, 8:10 pm, Ian Jackson
wrote:
In message
,
NT
writes



On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the
back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it
was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful
watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or
cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually
implies
180 degrees out of phase). It's 90 degrees OOP, with the current
leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere
else
in the mains system.

At any point on mains and grid system, the load power factor will
be the
'total' of all the individual effects from the various users
downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due
largely
to all those inductive power transformers and electric motors it
feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.


because with SMPS and RFI caps, there's no such animal, More likely to
need an INDUCTOR actually.

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big
cable with lots of voltage drop.



WTF are you running ? a 36"" CRT and a pentium 3 and windows 95?

On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.
--
Peter.
The gods will stay away
whilst religions hold sway

In message op.v5zrqqynytk5n5@i7-940, Lieutenant Scott
writes
On Sun, 04 Dec 2011 17:53:43 -0000, Ian Jackson
wrote:

In message op.v5y6yop4ytk5n5@i7-940, Lieutenant Scott
writes
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back of
the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

No. You are absolutely right. Your 11W is actually 'volt-amps' (VA), and
VA x PF = watts. As your PF is 0, your PC is not consuming any power.

However, the 0.049A is coming to your PC via the house wiring. As this
will have some (hopefully very small) resistance, there will be a small
'I squared R' power loss in the wires. And, as you say, similar loss
will be occurring in the National Grid wiring.

Of course, your small capacitive current might be a 'good thing'. It
might be helping to compensate for an inductive load next door (well, a
near neighbour on the same distribution phase as you are on), where the
PF is less than 1.

Am I right in thinking the losses on the power lines (if I'm not
compensating for a neighbour) are 11 watts?

No. You are drawing a current of 0.049A, but aren't actually 'consuming'
it. As far as you are concerned, it's a lossless use. You are only
borrowing the 11VA, then giving them back. However, there are still the
transportation costs, ie the 'I squared R' losses caused by the
resistances of the mains and National Grid cables and intervening power
transformers. Essentially, this is completely unknown (at least to
people like you and me).

How much power loss your 0.49A creates is almost impossible to
calculate. It would be easy if your 230V supply came directly from a
power station, just up the road. Your 0.49A would be 0.049A all the way.
The resistances of the power cables and the actual generator could be
measured, so the 'I squared R' power loss calculation would be dead
easy. If, for example, the cables were 10 ohms, and the generator 5
ohms, the total resistance is 15 ohms. The 'I squared R' loss power loss
would be 0.036015W.

However, your 0.049A won't stay at 0.049A for long. At your local
substation transformer, the 230V becomes (say?) 11,000V. From there to
the next upstream transformer, your 0.049A will be 0.00102A, so the 'I
squared R' loss will be a lot less. Eventually, your original relatively
massive 0.049A will be a relatively minuscule 0.000575A flowing in the
400kV supergrid, and the 'I squared R' losses will be very small indeed.
[Note that, as the 'I squared R' losses are proportional to the square
of the current, it is much more efficient to carry the electricity at as
high a voltage as possible.]

But it would take someone with a complete knowledge of the specification
and characteristics of the whole power system to calculate what the sum
total of all the individual 'I squared R' your original 0.049A creates.
--
Ian

On Dec 4, 8:39*pm, Ian Jackson
wrote:
In message
, NT
writes



On Dec 4, 8:10*pm, Ian Jackson
wrote:
In message
, NT
writes

On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.

Actually, I believe that large factories, companies and other
establishments which consume lots of electricity are obliged to have
power factor correction equipment on their supply feeds. In some cases
where the load varies a lot, this is dynamic, and always tries to keep
the overall power factor as close as possible to 1, regardless of how
much power is being used.
--
Ian- Hide quoted text -

- Show quoted text -

Power factor correction equipment exists for industrial sized
installations that need it.
Large consumers are encouraged to have it by being charged by the KVa
rather then the Kw. Kw=KVa if power factor is unity ie volts and amps
in phase. If they have lots of electric motors, the PF will be
lagging, iecurrent will lag behind voltage.
This means a larger current than a unity power factor will flow
necesitating larger cables, switchgear and transformers.

The alternative is to fit a capacitor to each inductive load.
Fluorescent lights with a choke/ballast usually have one fitted.
Capacitors have an opposite effect to inductors/motors,(they cause the
current to leadt he voltage)

On Dec 4, 11:22*pm, Andy Wade wrote:
On 04/12/2011 22:23, NT wrote:

I'm with you on the LV distribution. The main issue is standards. The
5v usb standard has developed and could be used.

It's coming, at least for US commercial environments - 24 volts:http://www.emergealliance.org/Standa...ndardFAQs.aspx

--
Andy

There's a lot of gross over simplification there.

On 05/12/2011 08:49, Ian Jackson wrote:

How much power loss your 0.49A creates is almost impossible to
calculate. It would be easy if your 230V supply came directly from a
power station, just up the road. Your 0.49A would be 0.049A all the way.
The resistances of the power cables and the actual generator could be
measured, so the 'I squared R' power loss calculation would be dead
easy. If, for example, the cables were 10 ohms, and the generator 5
ohms, the total resistance is 15 ohms. The 'I squared R' loss power loss
would be 0.036015W.

Also consider the hypothetical situation of someone on the same phase
further down the road with an inductive load drawing an out of phase
current matching your capacitive load. The affected current will bounce
backwards and forwards between the pair of you without drawing anything
from the substation and the I^2R losses only arise in the cable in your
street.

--
Mike Clarke

On Mon, 05 Dec 2011 08:16:48 -0000, The Natural Philosopher wrote:

Lieutenant Scott wrote:
On Sun, 04 Dec 2011 21:16:36 -0000, The Natural Philosopher
wrote:

Ian Jackson wrote:
In message
, NT
writes
On Dec 4, 8:10 pm, Ian Jackson
wrote:


Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.


because with SMPS and RFI caps, there's no such animal, More likely to
need an INDUCTOR actually.

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big
cable with lots of voltage drop.



WTF are you running ? a 36"" CRT and a pentium 3 and windows 95?

How would windows 95 use 350 watts?

--
http://petersparrots.com
http://petersphotos.com

I think car alarms should be set of explode after two minutes.
That way, we either take out a car thief, or deprive a noise-polluting jerk of his wheels.

On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:

On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.

What is it a ZX spectrum?

--
http://petersparrots.com
http://petersphotos.com

A chicken crossing the road is poultry in motion.

On Mon, 05 Dec 2011 09:00:00 -0000, harry wrote:

On Dec 4, 8:39 pm, Ian Jackson
wrote:
In message
, NT
writes



On Dec 4, 8:10 pm, Ian Jackson
wrote:
In message
, NT
writes

On Dec 4, 5:31 pm, "Lieutenant Scott" wrote:
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back
of the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

--http://petersparrots.comhttp://petersphotos.com

A guy bought his wife a beautiful diamond ring for Christmas.
A friend of his said, "I thought she wanted one of those sporty
4-Wheel drive vehicles."
"She did," he replied. "But where in the hell was I gonna find a
fake Jeep?"

Power factor of 0 means the current consumed is out of phase with the
mains voltage, so no power is consumed and no meter reading or cost is
incurred.

It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading
the voltage (something I still find is a weird concept!).

It will cause power waste in the grid, but nothing anywhere
near 11w, it'll be under 1w.

But not if the OP's incidental capacitive current is compensating or
correcting for an equal amount of inductive load current somewhere else
in the mains system.

At any point on mains and grid system, the load power factor will be the
'total' of all the individual effects from the various users downstream.
My understanding is that the mains system does indeed suffer from
unwanted reactive 'I squared R' losses, and this is because the total
load it feeds tends to have an overall lagging power factor (due largely
to all those inductive power transformers and electric motors it feeds).

Yes, quite right that its lagging overall. Even within one house the
odds are that the capacitive current will be offsetting overall
lagging current on the whole, so in reality it'll avoid tiny losses
rather than cause them.

I've often wondered if all domestic supplies should not have (probably
immediately before or after the meter) a 'typical average value power
factor correction capacitor'.

Actually, I believe that large factories, companies and other
establishments which consume lots of electricity are obliged to have
power factor correction equipment on their supply feeds. In some cases
where the load varies a lot, this is dynamic, and always tries to keep
the overall power factor as close as possible to 1, regardless of how
much power is being used.
--
Ian- Hide quoted text -

- Show quoted text -

Power factor correction equipment exists for industrial sized
installations that need it.
Large consumers are encouraged to have it by being charged by the KVa
rather then the Kw. Kw=KVa if power factor is unity ie volts and amps
in phase. If they have lots of electric motors, the PF will be
lagging, iecurrent will lag behind voltage.
This means a larger current than a unity power factor will flow
necesitating larger cables, switchgear and transformers.

Doesn't it also draw more power from the generator? Say for sake of example that you had a generator in your garden, and an infinitely thick cable running to a piece of equipment with a bad power factor. Would you be losing out? Heating in the coils of the generator?

The alternative is to fit a capacitor to each inductive load.
Fluorescent lights with a choke/ballast usually have one fitted.
Capacitors have an opposite effect to inductors/motors,(they cause the
current to leadt he voltage)

Isn't PFC a requirement for new equipment of any type nowadays?


--
http://petersparrots.com
http://petersphotos.com

Come to think of it, there are already a million monkeys on a million typewriters, and Usenet is nothing like Shakespeare

On Mon, 05 Dec 2011 08:49:34 -0000, Ian Jackson wrote:

In message op.v5zrqqynytk5n5@i7-940, Lieutenant Scott
writes
On Sun, 04 Dec 2011 17:53:43 -0000, Ian Jackson
wrote:

In message op.v5y6yop4ytk5n5@i7-940, Lieutenant Scott
writes
I was testing the power consumption of a computer (to calculate the
power requirements - as in what power supply to use).

To do this, I used one of those plug in green electricity monitor
thingies. It shows volts, amps, watts, power factor, etc.

Anyway, when the computer was switched off at the switch on the back of
the power supply (I'm assuming that this means only the filter
capacitors are connected), the electricity monitor showed that it was
drawing 0.049 amps with a power factor of 0%. Am I right in thinking
this means I am being charged for no electricity, while 11 watts of
electricity are being dissipated in the national grid somewhere? The
monitor also shows 0 watts consumption, which I assume is useful watts
as charged by the electricity meter - because it then goes on to use
this to show cost over a period of time. This seems very wasteful to
me. Or am I miscalculating something?

No. You are absolutely right. Your 11W is actually 'volt-amps' (VA), and
VA x PF = watts. As your PF is 0, your PC is not consuming any power.

However, the 0.049A is coming to your PC via the house wiring. As this
will have some (hopefully very small) resistance, there will be a small
'I squared R' power loss in the wires. And, as you say, similar loss
will be occurring in the National Grid wiring.

Of course, your small capacitive current might be a 'good thing'. It
might be helping to compensate for an inductive load next door (well, a
near neighbour on the same distribution phase as you are on), where the
PF is less than 1.

Am I right in thinking the losses on the power lines (if I'm not
compensating for a neighbour) are 11 watts?

No. You are drawing a current of 0.049A, but aren't actually 'consuming'
it. As far as you are concerned, it's a lossless use. You are only
borrowing the 11VA, then giving them back. However, there are still the
transportation costs, ie the 'I squared R' losses caused by the
resistances of the mains and National Grid cables and intervening power
transformers. Essentially, this is completely unknown (at least to
people like you and me).

How much power loss your 0.49A creates is almost impossible to
calculate. It would be easy if your 230V supply came directly from a
power station, just up the road. Your 0.49A would be 0.049A all the way.
The resistances of the power cables and the actual generator could be
measured, so the 'I squared R' power loss calculation would be dead
easy. If, for example, the cables were 10 ohms, and the generator 5
ohms, the total resistance is 15 ohms. The 'I squared R' loss power loss
would be 0.036015W.

However, your 0.049A won't stay at 0.049A for long. At your local
substation transformer, the 230V becomes (say?) 11,000V. From there to
the next upstream transformer, your 0.049A will be 0.00102A, so the 'I
squared R' loss will be a lot less. Eventually, your original relatively
massive 0.049A will be a relatively minuscule 0.000575A flowing in the
400kV supergrid, and the 'I squared R' losses will be very small indeed.
[Note that, as the 'I squared R' losses are proportional to the square
of the current, it is much more efficient to carry the electricity at as
high a voltage as possible.]

But it would take someone with a complete knowledge of the specification
and characteristics of the whole power system to calculate what the sum
total of all the individual 'I squared R' your original 0.049A creates.

Ah I see. I'm not killing the planet with that capacitor then :-)

--
http://petersparrots.com
http://petersphotos.com

What is the difference between a 69 and driving in the fog?
When driving in the fog, you can't see the asshole in front of you.

"The Natural Philosopher" wrote in message
...

WTF are you running ? a 36"" CRT and a pentium 3 and windows 95?

You can buy graphics cards that use that much.
Put two of them plus a six core i7 processor in with a couple of drives and
350W is nothing.

On 05/12/2011 01:01, Lieutenant Scott wrote:
Or is it lagging by 270 degrees? That makes sense. Voltage had to
happen before current surely?

No, it's leading. The voltage has to "happen" at source (the power
station) but the current leads the voltage measured at the load (your
appliance).

Imagine this circuit (you'll need a fixed width font):

A o-------------/\/\/\/\/\/\/\-------------
10 Ohm |
Resistor |
--------- C
Capacitor
--------- D
|
|
B o----------------------------------------

Consider the voltage measured across the capacitor between points C and
D when 10 volts are applied to the input terminals A and B. Initially
both plates of the capacitor are at 0V so there's a potential difference
of 10V across the resistor and a current of 1A will flow through the
resistor into the capacitor. As the current flows it will charge up the
capacitor increasing the voltage across C-D. As the voltage of C rises
the voltage across the resistor falls and the current flowing into the
capacitor falls until we reach the stage when the capacitor is fully
charged with 10V across C-D and no current is flowing.

So the current flowing into the capacitor leads the voltage measured
across its terminals.

With an AC supply a similar thing happens in reverse for the negative
half cycle and you see a current waveform which leads the voltage at the
capacitor by 90 degrees.

You might find it easier to imagine the capacitative load delaying the
voltage waveform seen at your end of the wire.

--
Mike Clarke

In message , Mike
Clarke writes
On 05/12/2011 08:49, Ian Jackson wrote:

How much power loss your 0.49A creates is almost impossible to
calculate. It would be easy if your 230V supply came directly from a
power station, just up the road. Your 0.49A would be 0.049A all the way.
The resistances of the power cables and the actual generator could be
measured, so the 'I squared R' power loss calculation would be dead
easy. If, for example, the cables were 10 ohms, and the generator 5
ohms, the total resistance is 15 ohms. The 'I squared R' loss power loss
would be 0.036015W.

Also consider the hypothetical situation of someone on the same phase
further down the road with an inductive load drawing an out of phase
current matching your capacitive load. The affected current will bounce
backwards and forwards between the pair of you without drawing anything
from the substation and the I^2R losses only arise in the cable in your
street.

At 50Hz, two homes on the same phase can be considered as being
co-sited, with their loads in parallel. In the situation described,
there will indeed be no reactive current from the common point where the
two loads actually end up being connected in parallel (at the point on
the mains distribution cable in the street where the house nearest to
the substation taps off its feed), and back to the substation. There
will only be reactive currents on the 'feed-in' cables from this common
point to each house, so the additional losses should be minuscule.
--
Ian

On Dec 5, 9:36*am, "Lieutenant Scott" wrote:
On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:
On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? *This thing uses 350 watts! *That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.

What is it a ZX spectrum?

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.


NT

NT wrote:
On Dec 5, 9:36 am, "Lieutenant Scott" wrote:
On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:
On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:
We really need to design a 12vDC ring standard for all the electronics..
12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.
Good grief! Mine's 80W inc. the 19" monitor.
What is it a ZX spectrum?

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.



My server I hope is sub 10W..


NT

On Dec 5, 12:27*pm, NT wrote:
On Dec 5, 9:36*am, "Lieutenant Scott" wrote:

On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:
On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? *This thing uses 350 watts! *That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.

What is it a ZX spectrum?

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.


yes the power supplies are rated at 350W liike a car is rated as it's
maxium MPH 120
that doesn;t mean that it's the average speed or even the likeliy
speed.
My G4 PSU is rated at 338W or there about, typical wattage measured by
one of those
Maplin tpyoe ddevices was 130-140W.

NT wrote:

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.

Indeed, my machine is 4x core (older greedier intel CPU) 8GB ram,
10xSATA, 2x optical drives, 1x graphics card, 2x TV tuners, 2x satellite
tuners and it still only manages to take 110W (apart from during startup)

But it's easier to find ATX PSUs between 750W and 1kW, than find one
less than 400W.

On 05/12/2011 08:49, Ian Jackson wrote:

[...] ie the 'I squared R' losses caused by the resistances of the
mains and National Grid cables and intervening power transformers.
Essentially, this is completely unknown (at least to people like you
and me).

It's not unknown at all - at least from the point of view of making a
rough estimate. The resistance value you need is just the resistive
component of the source impedance of your mains connection - and that,
in most cases, will be dominated by the resistance of your service
cable. There's not much point going much further back in the supply
network - as you do so the resistances of the larger cables get very
much smaller and the source impedances become more reactive (inductive).

How much power loss your 0.49A creates is almost impossible to
calculate.

A typical source resistance might be between 50 and 100 milliohm,
although it could be higher or lower by a factor of (say) two or three
either way. Assuming 0.1 ohm the I^2*R loss resulting from the 49 mA
current in question is just under a quarter of a milliwatt - i.e.
utterly negligible. In fact more is lost in your house wiring supplying
the 'wattless' load. For a very typical value of 0.4 ohm resistance in
the wiring to a 13 A socket halfway round a ring circuit you will loose
a whole milliwatt. Well it will contribute to your heating in the winter...

--
Andy

On Mon, 05 Dec 2011 09:36:30 -0000, Lieutenant Scott wrote:

On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:

On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.

What is it a ZX spectrum?

Dual-core Brisbane, 1900MHz - not cutting-edge but fast enough for now.
--
Peter.
The gods will stay away
whilst religions hold sway

On Mon, 05 Dec 2011 12:52:44 +0000, Andy Burns wrote:

NT wrote:

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.

Indeed, my machine is 4x core (older greedier intel CPU) 8GB ram,
10xSATA, 2x optical drives, 1x graphics card, 2x TV tuners, 2x satellite
tuners and it still only manages to take 110W (apart from during startup)

But it's easier to find ATX PSUs between 750W and 1kW, than find one
less than 400W.

That is a problem. My Seasonic PSU is 330W, but nowadays 380W is the lowest.
I've enough spare capacity to run the speakers and the screen. As doing that
would get the PSU above 20% (only 45W idling and about 110W starting atm) it
would decrease its ineffeciency.
--
Peter.
The gods will stay away
whilst religions hold sway

On Mon, 05 Dec 2011 13:14:29 -0000, Tim Streater wrote:

In article op.v5zrvqp5ytk5n5@i7-940, "Lieutenant Scott"
wrote:

On Sun, 04 Dec 2011 21:16:36 -0000, The Natural Philosopher
wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts!

Why?

A decent processor uses 130 watts. A decent graphics card uses 75, 150, or 225 watts.

--
http://petersparrots.com
http://petersphotos.com

__________________
/\ ______________ \
/::\ \ZZZZZZZZZZZZ/\ \
/:/\.\ \ /:/\:\ \
/:/Z/\:\ \ /:/Z/\:\ \
/:/Z/__\:\ \____/:/Z/ \:\ \
/:/Z/____\:\ \___\/Z/ \:\ \
\:\ \ZZZZZ\:\ \ZZ/\ \ \:\ \
\:\ \ \:\ \ \:\ \ \:\ \
\:\ \ \:\ \_\;\_\_____\;\ \
\:\ \ \:\_________________\
\:\ \ /:/ZZZZZZZZZZZZZZZZZ/
\:\ \ /:/Z/ \:\ \ /:/Z/
\:\ \/:/Z/ \:\ \/:/Z/
\:\/:/Z/________\;\/:/Z/
\::/Z/_______itz__\/Z/
\/ZZZZZZZZZZZZZZZZZ/

On Mon, 05 Dec 2011 13:57:32 -0000, PeterC wrote:

On Mon, 05 Dec 2011 09:36:30 -0000, Lieutenant Scott wrote:

On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:

On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.

What is it a ZX spectrum?

Dual-core Brisbane, 1900MHz - not cutting-edge but fast enough for now.

TWO cores? Ugh.

--
http://petersparrots.com
http://petersphotos.com

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On Mon, 05 Dec 2011 12:43:26 -0000, whisky-dave wrote:

On Dec 5, 12:27 pm, NT wrote:
On Dec 5, 9:36 am, "Lieutenant Scott" wrote:

On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:
On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

We really need to design a 12vDC ring standard for all the electronics..

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.

Good grief! Mine's 80W inc. the 19" monitor.

What is it a ZX spectrum?

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.


yes the power supplies are rated at 350W liike a car is rated as it's
maxium MPH 120
that doesn;t mean that it's the average speed or even the likeliy
speed.
My G4 PSU is rated at 338W or there about, typical wattage measured by
one of those
Maplin tpyoe ddevices was 130-140W.

My PSU is rated at 1kW. It gets 350 watts on the maplin type device.

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On Mon, 05 Dec 2011 12:29:29 -0000, The Natural Philosopher wrote:

NT wrote:
On Dec 5, 9:36 am, "Lieutenant Scott" wrote:
On Mon, 05 Dec 2011 08:34:41 -0000, PeterC wrote:
On Mon, 05 Dec 2011 01:03:16 -0000, Lieutenant Scott wrote:

12vDC for computers? This thing uses 350 watts! That would be a big cable with lots of voltage drop.
Good grief! Mine's 80W inc. the 19" monitor.
What is it a ZX spectrum?

Theres a lot of misinformation about concerning pc power consumption,
fuelled by the 'bigger number is better' philosophy of sales of PSUs.
Its en exceptional system that uses 350w.



My server I hope is sub 10W..

What?!!??


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On Mon, 05 Dec 2011 10:10:50 -0000, Mike Clarke wrote:

On 05/12/2011 01:01, Lieutenant Scott wrote:
Or is it lagging by 270 degrees? That makes sense. Voltage had to
happen before current surely?

No, it's leading. The voltage has to "happen" at source (the power
station) but the current leads the voltage measured at the load (your
appliance).

Imagine this circuit (you'll need a fixed width font):

A o-------------/\/\/\/\/\/\/\-------------
10 Ohm |
Resistor |
--------- C
Capacitor
--------- D
|
|
B o----------------------------------------

Consider the voltage measured across the capacitor between points C and
D when 10 volts are applied to the input terminals A and B. Initially
both plates of the capacitor are at 0V so there's a potential difference
of 10V across the resistor and a current of 1A will flow through the
resistor into the capacitor. As the current flows it will charge up the
capacitor increasing the voltage across C-D. As the voltage of C rises
the voltage across the resistor falls and the current flowing into the
capacitor falls until we reach the stage when the capacitor is fully
charged with 10V across C-D and no current is flowing.

So the current flowing into the capacitor leads the voltage measured
across its terminals.

With an AC supply a similar thing happens in reverse for the negative
half cycle and you see a current waveform which leads the voltage at the
capacitor by 90 degrees.

You might find it easier to imagine the capacitative load delaying the
voltage waveform seen at your end of the wire.

Ah. Understood.


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On 05/12/2011 09:36, Lieutenant Scott wrote:

What is it a ZX spectrum?

An early home computer.

http://en.wikipedia.org/wiki/ZX_Spectrum

--
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Tim Streater wrote:

In article op.v50uzzdxytk5n5@i7-940, "Lieutenant Scott"
wrote:

On Mon, 05 Dec 2011 13:14:29 -0000, Tim Streater
wrote:

In article op.v5zrvqp5ytk5n5@i7-940, "Lieutenant Scott"
wrote:

On Sun, 04 Dec 2011 21:16:36 -0000, The Natural Philosopher
wrote:

We really need to design a 12vDC ring standard for all the
electronics..

12vDC for computers? This thing uses 350 watts!

Why?

A decent processor uses 130 watts. A decent graphics card uses 75, 150,
or 225 watts.

More twaddle from the Loo-tenant. My Mini uses no more than 110 for
everything. And usually, considerably less than that (fan almost never
goes on).


Dell R610 1U rack mount server - currently 184W assuming 0.95 PF.

That's nearly idle - VMWare ESXi 4.1, one Windows Server 2008 VM running
vCenter.

But not bad for a 12 core, 96GB RAM server. Be happy to report on its
consumption with 30 VMs on it in about 2 weeks...

--
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On Mon, 05 Dec 2011 16:15:35 +0000, Tim Streater wrote:

More twaddle from the Loo-tenant.

Oh, is he still here?
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