3. Loads are connected across a 240V, 60Hz AC mains as...

3. Loads are connected across a 240V, 60Hz AC mains as follows: Load 1 is an inductive load of 2kVA at a power factor of 0.88. Load 2 is a capacitive load of 1.5kVAr at a reactive factor of 0.25. Load 3 is a 1.25hp, 0.89 power factor an efficiency of 0.90. Determine the total power, reactive power and apparent power. Determine also the system power factor. 4. If in number 4, we wish to make the power factor equal to 1.0, what kVAr rating of reactor needed to achieve this new power factor.

Transcript

00:02 Hello, in the question we have given loads are connected across 240v 60hz ac main as follows.

00:07 So we have the load 1 which is inductive load and the unit is given that is 2 kv is given, load 2 is capacitive load and load 3 is power that is 1 .2 horse power.

00:22 So we will see how to solve it.

00:24 So these two triangles we will require.

00:26 So this is the power triangle.

00:29 So i have specified what these are.

00:31 So this is the real power.

00:33 So this is the apparent power and this is the reactive power along with the units.

00:38 So the units are given.

00:39 So it is fair enough to mention with the units for inductive load.

00:44 So these two formulas are also important for inductive load.

00:47 We have this and for the capacitive load we have s is equal to p minus j times qc.

00:52 So for load 1, so see load 1 is inductive.

00:56 So s is given.

00:58 It is kva.

00:59 So kva means s is given that is apparent power is given.

01:03 So if i use the trigonometry over here, so if i use cos phi, so cos phi is adjacent upon hypotenuse.

01:10 So cos phi will be p by s.

01:12 So from here p will be s cos phi.

01:17 So cos phi is given that is 0 .88.

01:20 So p1 will be s cos of phi that is p1 i will get it as 1 .76 kw.

01:27 Now q similarly if i use the trigonometry, i will get s sin phi.

01:32 So phi from here is cos inverse of 0 .88.

01:36 So that we have plugged over here.

01:37 So q1 we will get it as 0 .94 kvar.

01:41 So that is the reactive power unit.

01:43 So s1 will be equal to this.

01:46 We have written the formula over here.

01:48 So that is p plus j times ql.

01:51 So that is what we have done over here.

01:53 And load 2, for load 2 q is given because they have mentioned the unit is 1 .5 kvar.

02:00 So that is the reactive power unit.

02:03 So kvar means q is given.

02:06 So we will be using this triangle now because this is the capacitive power formula and this is the sorry capacitive load and this is inductive load formula triangle.

02:17 So from that from that triangle if we see so p2 will be q tan of q divided by tan phi.

02:23 So 1 .5 tan of cos inverse of 0 .25 because cos phi is 0 .25.

02:31 So from here p2 if we calculate, we will get it as 0 .387 kw and s2 is nothing but so q is already given...

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