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Default Frequency and Voltage

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

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dean
 
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Default Frequency and Voltage

What do you mean Cos(Pi)?

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Mikepier
 
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Default Frequency and Voltage

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.

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Default Frequency and Voltage

"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.

  #5   Report Post  
 
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Default Frequency and Voltage

"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.



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Default Frequency and Voltage

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

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Toller
 
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Default Frequency and Voltage


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.


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Default Frequency and Voltage

"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!

  #9   Report Post  
Calvin Henry-Cotnam
 
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Default Frequency and Voltage

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
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Default Frequency and Voltage

"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.

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Default Frequency and Voltage

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



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Default Frequency and Voltage

dean wrote:

What do you mean Cos(Pi)?


-1?

Nick

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dean
 
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Default Frequency and Voltage

Exactly! Why not just say -1 then? LOL

  #18   Report Post  
 
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Default Frequency and Voltage

"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.

  #19   Report Post  
Charlie Bress
 
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Default Frequency and Voltage


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




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Stormin Mormon
 
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Default Frequency and Voltage

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




  #21   Report Post  
Ross Mac
 
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Default Frequency and Voltage


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


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