Re: "reactive power"

----------------------------
"jclause" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>
> A question please.
>
> Suppose we have 120 volts across a motor
> Current = 5 amps at phase = 33.6 degrees lagging
> Apparent power S = EI = 600 watts
> Real power P = EI cos 33.6 = 499.75 watts
> Reactive VA Q = EI sin 33.6 = 332.03 watts
> Power factor = P/EI = 0.833
>
> So the motor is dissipating 499.75 watts, and creating what is
> sometimes referred to as "reactive power" of 332.03 vars.
> It is said the utility company must supply the "reactive
> power" as well as the real power used by the motor, or
> perhaps better put - supply the current to circulate the
> "reactive power" in the lines between the source and the
> motor.
>
> The question then:
> Is this "reactive power" dissipated in the resistance of the
> lines between the source and the motor?
> If so, then it is not real power as seen by the source?
>
-----------
Daestrom has it right.

The fact that we have inductance (as in a motor) or capacitance means that
there is energy being stored and then returned to the source at a different
part of the cycle. [Draw a voltage wave and a current wave with the current
out of phase with the voltage. Plot the product at each instant of time.
This gives the instantaneous power. The result will be a constant term plus
a sinusoidal term. The constant term is the real (average)power and the
sinusoidal component has a zero average. The "reactive power" is a measure
of the magnitude of the 0 average component which shuttles from source to
load and back] This shuttling of energy has nothing to do with the net real
power (which is generally considered as the average over a cycle) delivered
as its average is 0 over a cycle.

Reactive "power" is NOT dissipated in resistance and is NOT real (average
as that is what we are dealing with in general power seen by the source.

However, there is an associated component of current (e.g. at 0.8 pf, the
current will be 1.25 times the current at 1.0pf for the same power) which
increases circuit loss and (if inductive) voltage drops in the circuit. In
that way, the reactive component does increase real power dissipation in
line resistance but it is not correct to say that the reactive "power" is
dissipated in the line resistance. The utility generator sees the effects of
reactive as well as real power but the turbine driving the generator sees
only the real component and in a motor, only the real component (less real
losses) is converted to mechanical energy.
--

Don Kelly @shawcross.ca
remove the X to answer

You are welcome and from what you said to Daestrom, you have it right.
--

Don Kelly @shawcross.ca
remove the X to answer
----------------------------
"jclause" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
>
>
>
>
> In article <_C9Re.48407$Hk.7212@pd7tw1no>, (E-Mail Removed) says...
>>
>>-----------
>>Daestrom has it right.
>>
>>The fact that we have inductance (as in a motor) or capacitance means that
>>there is energy being stored and then returned to the source at a
>>different
>>part of the cycle. [Draw a voltage wave and a current wave with the
>>current
>>out of phase with the voltage. Plot the product at each instant of time.
>>This gives the instantaneous power. The result will be a constant term
>>plus
>>a sinusoidal term. The constant term is the real (average)power and the
>>sinusoidal component has a zero average. The "reactive power" is a measure
>>of the magnitude of the 0 average component which shuttles from source to
>>load and back] This shuttling of energy has nothing to do with the net
>>real
>>power (which is generally considered as the average over a cycle)
>>delivered
>>as its average is 0 over a cycle.
>>
>>Reactive "power" is NOT dissipated in resistance and is NOT real
>>(average
>>as that is what we are dealing with in general power seen by the source.
>>
>> However, there is an associated component of current (e.g. at 0.8 pf, the
>>current will be 1.25 times the current at 1.0pf for the same power) which
>>increases circuit loss and (if inductive) voltage drops in the circuit. In
>>that way, the reactive component does increase real power dissipation in
>>line resistance but it is not correct to say that the reactive "power" is
>>dissipated in the line resistance. The utility generator sees the effects
>>of
>>reactive as well as real power but the turbine driving the generator sees
>>only the real component and in a motor, only the real component (less real
>>losses) is converted to mechanical energy.
>>--
>>
>>Don Kelly @shawcross.ca
>>remove the X to answer
>
>
> Thanks Don. You're right in there with the good technical stuff
> as usual.
>
> JC
>
>