Hi,
Can some one explain the relationship between Voltage, Frequency. Is
frequency is affecting the electrical consumption.

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having
a relationship between V or I..

Can some one clarify this.

Regards,
Sridhar

What do you mean Cos(Pi)?

Frequency just describes the cycles per second, or Hertz, that the
Voltage is. Here in the US its 60 Hertz. It is constant. The Pi in the
formula is also a constant.

"Hi,
Can some one explain the relationship between Voltage, Frequency. Is
frequency is affecting the electrical consumption.


According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having
a relationship between V or I..

Can some one clarify this. "


The power formula you have is for a three phase AC load. For a pure
resistance load, eg a simple load like a heater, the voltage and
current are always in line with each other. If you drew graphs of the
two, and placed one over the other, they would line up perfectly.

That is not true for a load that has inductance or capacitance, eg a
motor. In that case, the inductance of the motor will cause a phase
shift between the voltage and the current. If you place the graphs
together, you will see that while the frequency is exactly the same,
one curve is shifted slightly relative to the other.

Since power is the product of voltage and current, the instantaneous
power is still V*I, but the average power is affected by the amount of
shift between the voltage and current curves. That's where the Cos()
function comes in. The angle used in the cosine function is the angle
between the voltage and current in the load. For a pure resistive
load, the voltage and current would be in perfect alignment and the
angle would be zero, giving cos(0)=1. As you add inductance or
capacitance, the angle will become non-zero, resulting in a reduction
in power. Another way of looking at this intuitively is that as the
the voltage and current go out of alignment, when multiplying
instantaneous voltage and power along the two curves, since they no
longe line up, the power will obviously be reduced.

Just to further clarify, regarding frequency, the OP is correct.
Frequency of an AC load does not affect power

"Frequency just describes the cycles per second, or Hertz, that the
Voltage is. Here in the US its 60 Hertz. It is constant. The Pi in the
formula is also a constant. "


The argument of the cosine function is the angle between voltage and
current in an AC load. It varies depending on the inductance and/or
capacitance of a particular load. The symbol used for the angle is
normally theta, not Pi as the OP stated.

wrote in message
oups.com...
Just to further clarify, regarding frequency, the OP is correct.
Frequency of an AC load does not affect power

Are you sure about that? Frequency affects induction, and if induction
affects affects power, then frequency affects power.

"Are you sure about that? Frequency affects induction, and if
induction
affects affects power, then frequency affects power"

Well, you've got me there. That was an incorrect statement. I had the
equation for 3 Phase power that the OP gave in mind and the fact that
freq is not part of it. But of course you are right, the frequency has
a big effect on any load with inductance or capacitance. And that
effect gets into the OP's power equation by virture of the fact that
the phase angle between voltage and current contained in the equation
is itself a function of the frequency.

Thanks for correcting that!

dean wrote:

What do you mean Cos(Pi)?

-1?

Nick

Exactly! Why not just say -1 then? LOL

"Exactly! Why not just say -1 then? LOL "

It's an obvious mistake in an equation which is otherwise correct.
Instead of Pi, the correct variable usually used is theta.

wrote in message
oups.com...
"Exactly! Why not just say -1 then? LOL "

All of this reminds me of the guy who was taking Electricity 101.
He was trying to remember the equation for power for a test.
His buddy said "Just remember 'twinkle, twinkle little star. Power is equal
to I(squared) R.'"
When the test results came back our student had not done very well.
"I got mixed up" he said. "All I could remember was ' Shining in the sky so
high, Power is equal to R(squared) I' "

Sorry about straying a little off topic. The devil made me do it.
Charlie

wrote in message
oups.com...
Hi,
Can some one explain the relationship between Voltage, Frequency. Is
frequency is affecting the electrical consumption.

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having
a relationship between V or I..

Can some one clarify this.

Regards,
Sridhar


Some good answers here...Might I just add that in a resistive circuit, as
posted, voltage and current are in phase so there is 0 phase angle between
voltage and current and thus no losses other than the resistance. In a
capacitive or inductive circuit there is reactance. The formula for
capacative reactance is 1 divided by 2pi*f*c and for inductive reactance it
is 2pi *f*l. As you can see, as the frequency goes up so does the reactance
of the circuit and thus the overall power is reduced. With typical AC power
the 60 cycles are constant so the only variable is the capacitance or
inductance of the circuit which will cause the current to lead or lag the
voltage and leave you with less power. Hope that helps........Ross

wrote:


The power formula you have is for a three phase AC load. For a pure
resistance load, eg a simple load like a heater, the voltage and
current are always in line with each other. If you drew graphs of the
two, and placed one over the other, they would line up perfectly.




The formula works just as well for single phase power.
Your description is quite correct for any number of phases.

Bill Gill

wrote:


The power formula you have is for a three phase AC load. For a pure
resistance load, eg a simple load like a heater, the voltage and
current are always in line with each other. If you drew graphs of the
two, and placed one over the other, they would line up perfectly.




The formula works just as well for single phase power.
Your description is quite correct for any number of phases.

Bill Gill

wrote:


The power formula you have is for a three phase AC load. For a pure
resistance load, eg a simple load like a heater, the voltage and
current are always in line with each other. If you drew graphs of the
two, and placed one over the other, they would line up perfectly.




The formula works just as well for single phase power.
Your description is quite correct for any number of phases.

Bill Gill

wrote:


The power formula you have is for a three phase AC load. For a pure
resistance load, eg a simple load like a heater, the voltage and
current are always in line with each other. If you drew graphs of the
two, and placed one over the other, they would line up perfectly.




The formula works just as well for single phase power.
Your description is quite correct for any number of phases.

Bill Gill

"The formula works just as well for single phase power.
Your description is quite correct for any number of phases. "


P= Root (3) V . I. Cos (Pi).

It does? The cube root arises because it's a three phase circuit.

Toller ) said...


wrote in message
roups.com...
Just to further clarify, regarding frequency, the OP is correct.
Frequency of an AC load does not affect power

Are you sure about that? Frequency affects induction, and if induction
affects affects power, then frequency affects power.

Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never
really have a purely inductive or capacitive load!).

Since the impedance will be affected by frequency, the total current
draw will be changed as a result of a frequency change.

The cosine of the angle between current and voltage is known as the
"power factor". In a purely resistive load, it is 1. In a purely inductive
or capacitive load, it is zero. Typical loads lie inbetween, but it is
best to keep it as close to 1 as possible.

Industrial customers of electric utilities must maintain as high a power
factor as possible. They usually have banks of capacitors that can be
automatically switched on and off as needed as they will have heavy
inductive loads from motors. If they maintain too low a power factor,
they will be charged for kVA-hours instead of kW-hours.

For instance, if you maintained a 0.5 power factor, then a 100 A load
at 120 V would be 100 x 120 x 0.5 = 6000 watts. An hour of this would
be 6 kWh, but it is 12 kVAh.

Why be charged for more energy than you actually used? Simply because you
are a burden on the system. Even though your load was only 6000 watts, you
drew double the current than was really necessary for that amount of
power. Therefore, the infrastructure needed to deliver that power had to
have twice the capacity than was really necessary.

Low power factor loads tend to be inductive, so a bank of capacitors
can cancel it out. Inductors cause current to lag behind the voltage, while
capacitors cause current to lead voltage. The two cancel each other out.
In the example above, since capacitors store and release current, they
supply the "extra" current needed for the inductive load, so the only
current draw on the supply is for the current actually needed to provide
power. If perfectly matched, the 6000 watt load would draw 50 amps from
the supply while the other 50 amps would be current between the inductive
load and the capacitors.

As an interesting side note: your home probably has a leading power factor
most of the time. The wiring in your home actually acts as a capacitor.
When I was a student, I worked on weekends as a watchman at a factory.
There was a power factor meter where the bank of capacitors was. When the
factory was shut down and the only load was lighting, the power factor
was usually 0.7-0.8 leading. As machines were started up and the inductance
of the load increased, the PF would rise to 1 then start dropping on the
lagging side. I believe a bank of capacitors would be switched in when it
dropped to 0.7. In addition to the kWh meter, there was a kVAh meter.

--
Calvin Henry-Cotnam
"Never ascribe to malice what can equally be explained by incompetence."
- Napoleon
-------------------------------------------------------------------------
NOTE: if replying by email, remove "remove." and ".invalid"

"Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never

really have a purely inductive or capacitive load!). "

That's true for a purely inductive or capacitve load, but isn't true in
the general case. Consider a load consisting of a resistor in series
with an inductor. Connect a voltage source with a frequency of zero
(DC) and the inductor offers no impedance and the load is purely
resistive. It looks just like a resistor and the power dissipated is
V**2/R.

Now change the frequency to the other extreme, making it virtually
infinite. Now the inductor effectively blocks current and the power
dissipated (in the resistor) is close to zero.

wrote in message
oups.com...
"Are you sure about that? Frequency affects induction, and if
induction
affects affects power, then frequency affects power"

Induction is NOT 'affected' by frequency!

For example the frequency of the AC mains supply is, in North America = 60
hertz) and elsewhere is often = 50 hertz.

Maybe what is being referred to is the affect of 'reactance' in an AC
circuit?
Reactance (as compared to pure DC resistance usually designated 'R') is the
affect on current/voltage by the presence of inductors or capacitors in a
circuit.

In an inductive circuit the current flowing will 'lag' behind the applied
voltage due inductive reactance often designated Xl. In a capacitive circuit
the current flowing will 'lead' the applied voltage due to capacitive
reactance often designated Xc.

Barring anything else frequency will not 'affect' induction. Current and
voltage will be 'affected' (if we must use that affected word) by the
frequency of the voltage applied; this is how, for example, low pass and
high pass filters work!

Also in this discussion thread suppose the frequency is zero? i.e. DC
(Direct Current). Inductive reactance would be nil. Capacitive reactance
would be infinite (after initial charging). The affect of 'inductive
reactance' on the current flowing would be nil.

"Are you sure about that? Frequency affects induction, and if
induction
affects affects power, then frequency affects power"


"Induction is NOT 'affected' by frequency! "


"Induction" most certainly is affected by frequency. The basic
definition of induction is the inducing of a voltage in a conductor by
a changing magnetic field. With zero frequncy, you have zero induced
current. Start increasing the frequency of the magnetic field, eg by
moving a magnet back and forth, or allowing a nearby coil's magnetic
field to vary, and there will be an increasing induced current in the
conductor.

I think what your referring to is that the "inductance" of say a
particular coil is fixed and determined strictly from the physical
properties of the coil and it's measured in henrys and not dependent on
frequency. I think what Toller meant was fairly clear from the
context of the discussion concerning power. And that is that in the
general case, the amount of power consumed by a given piece of
equipment depends on the frequency of the source it is being driven
with. That's because induction is frequency dependent and the
impedance of a load containing inductance and/or capacitance will
change as a function of frequency.

wrote:

P= Root (3) V . I. Cos (Pi).

It does? The cube root arises because it's a three phase circuit.

I believe you will find that's the square root of 3.

Nick

"I believe you will find that's the square root of 3. "

And I believe you are right!

None. Voltage is electrical intensity. Frequency is the inverse of the
duration of time necessary to cycle.

--

Christopher A. Young
Do good work.
It's longer in the short run
but shorter in the long run.
..
..


wrote in message
oups.com...
Hi,
Can some one explain the relationship between Voltage, Frequency. Is
frequency is affecting the electrical consumption.

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having
a relationship between V or I..

Can some one clarify this.

Regards,
Sridhar

"Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never

really have a purely inductive or capacitive load!). "

Frequency does not affect power ONLY in a load that is purely inductive
or capacitive, because there is no power dissipation there. However,
in the general case, frequency does affect power consumption. Consider
a load consisting of a 5 ohm resistor and an inductor in series driven
by a 10volt source of variable frequency. At 0 freq, ie DC, the
series load looks just like a resistor and the power dissipated is
20watts. At inifinite frequency, zero current flows through the
resistor, and the power dissipated is 0 watts.

wrote in message
oups.com...
"Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never

really have a purely inductive or capacitive load!). "

Frequency does not affect power ONLY in a load that is purely inductive
or capacitive, because there is no power dissipation there. However,
in the general case, frequency does affect power consumption. Consider
a load consisting of a 5 ohm resistor and an inductor in series driven
by a 10volt source of variable frequency. At 0 freq, ie DC, the
series load looks just like a resistor and the power dissipated is
20watts. At inifinite frequency, zero current flows through the
resistor, and the power dissipated is 0 watts.


I have been wondering why this thread did not get moved to:

alt.engineering.electrical