Home › Electrical Engineering Forum › General Discussion › how to find apparent power and reactive power?
 This topic has 6 replies, 3 voices, and was last updated 10 years ago by admin.

AuthorPosts

2011/06/05 at 7:58 pm #10502adminKeymaster
how to find apparent power and reactive power?
is it different for static and rotating devices?
sorry but i want to clear my doubt.
2011/06/29 at 8:19 pm #12238adminKeymasterS = 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 3phase i.e. 1,732
Static or rotary same rules apply.
2011/06/30 at 6:06 am #12240adminKeymastershah 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.
2011/06/30 at 11:02 am #12244sParticipantHi
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
2011/06/30 at 8:48 pm #12246عبده همدParticipantapparent 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 .
2011/07/01 at 11:29 am #12247sParticipantHi
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
2011/07/01 at 4:07 pm #12252adminKeymasternael.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.

AuthorPosts
 You must be logged in to reply to this topic.