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how to find apparent power and reactive power?
is it different for static and rotating devices?
sorry but i want to clear my doubt.
S = P + jQ
P = 1.732 VI Cos(phi) Real power
Q = 1.732 VI Sin(phi) Reactive power
S = 1.732 VI Apparent power
Use root 3 for 3-phase i.e. 1,732
Static or rotary same rules apply.
shah said:
how to find apparent power and reactive power?
is it different for static and rotating devices?
sorry but i want to clear my doubt.
dear Shah, if you want to make the mathematical calculation then i dont know,
if u want to measure it actually then use FLUKE POWER ANALYSERmodel
43B or 435 your job will be done in minutes.
Hi
Static or rotating device : this is a mechanical difference (even for AC inductive motors which have electrical rotating magnetic flux)
But the general rule is : as long as there is inductance (or capacitor) inside the load, their is an apparent power and reactive power,
Please note that reactive power is the power responsible of creating the magnetic flux (which is 90% of cases apply on rotating machines :)
To know the apparent and reactive power on any machine, I'll use an inductive motor as an example,
On every motor there is a name plate, which must indicates the apperant power ( Pn in W or KW or HP ) , Voltage ( in V ), and the phase shift ( cos phy ), and nominal current ( In in A)
Those values are for the full load,
for lower loads you should measure the current I and compare it to the nominal current In and devide with the calculated ration I/In=A
P =Pn x A, Q=Qn x A
Now for Q reactive power in VAR
Q=P x tg (Phy)
Phy can be known by any scientefic calculator,
Please note that the Harmonics effects are not considered here.
Take care
apparent power is the balance power between real power (p) and reactive power (Q) and has equation that P sqr and Q sqr both under radical and gives you aparent power (s) if it three phase times three when under the radical .
Hi
One mistake at my post
P is the real power in Watt , S is the apperant power in VA
S=Square root (Psqr + Q sqr)
Just replace the word “REAL” instead of “apperant” in my previous post
nael.samman said:
Hi
One mistake at my post
P is the real power in Watt , S is the apperant power in VA
S=Square root (Psqr + Q sqr)
Just replace the word “REAL” instead of “apperant” in my previous post
apparant power= voltagex current (vA or KVA)
REAL POWER= VOLTAGE X CURRENT X POWER FACTOR(COS Phi.) – WATTS /KW
APPARANT POWER IS VECTOR SUM OF REAL POWER(KW) AND REACTIVE POWER KVAR(KVAR).
kva= square root ( (kw)square+(kvar) square )
power factor=kw/kva
IF YOU KNOW ANY TWO OF THESE (KW,KVA .KVAR OR POWER FACTOR) YOU CAN CALCULATE OTHER PARAMETERS.
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