Kaplan Afternoon #14 question

[ekluempers]

Attached is Kaplan Afternoon problem 14 along with the solution.

In the solution it mentions that P/(sin delta) is constant (if P goes up, sin delta goes up).

Then it goes on to say (sin delta 2)=(P1/P2)*(sin delta 1). This looks like P*sin delta is constant (if P goes up, sin delta goes down).

Is it me or should it be (sin delta 2)=(P2/P1)* (sin delta 1)?



Thanks,

Eric

 

[soma]

Attached is Kaplan Afternoon problem 14 along with the solution.

In the solution it mentions that P/(sin delta) is constant (if P goes up, sin delta goes up).

Then it goes on to say (sin delta 2)=(P1/P2)*(sin delta 1). This looks like P*sin delta is constant (if P goes up, sin delta goes down).

Is it me or should it be (sin delta 2)=(P2/P1)* (sin delta 1)?



Thanks,

Eric

You are right, the new load angle would be 28.26. The reactive power supplied will be around 2.8 KVAR.

 

[Silkworm]

Agreed