00:01
Hi there, so for this problem, we are told that at 600 and 600 balls in three -phase, a start -connected synchronous motors draws a fuel load current of 80 ampers.
00:19
And at 0 .8, leaden, now the armature resistance is 2 .2 oms, and the synchronous reactance is 22 oms per phase.
00:31
If the straight loses of the machines are 3 ,200 whaps, we need to determine the following.
00:46
So the first question for part a is the induced emph.
00:54
Now, first, what we need to calculate is the value that we are given.
01:14
And the potential is going to be equal to the value that we are given, 6 ,600 volts divided by the square root of the value for the face.
01:28
So this will give us a value of 3 ,810 .6 volts.
01:38
So this is the face value.
01:40
Now the current that we are given for this case is 80 ampers.
01:46
The cosine of v, the cosine of the phase is 0 .8.
01:51
The sign of the phase is equal to 0 .6.
01:57
The resistance that we are given for the armature resistance, we're going to call that capital r, is equal to 2 .2 oms.
02:09
And the synchronous reactance that we're going to call just simply as x is equal to 22.
02:17
Now, for the leading power factor, we are going to have the following expression.
02:25
So that will be just simply the potential difference minus the current times the resistance, d times the cosine of the phase, this plus the current times the reactance, the sign of the face.
02:42
And this minus the imaginary part that is going to be equal to um.
02:53
Yes the current times this cosine of fee that plus the current times the resistance times the sign of the phase so if we substitute all of the values that we are given we will obtain that this is just simply this is this factor is just 4 ,725 .8 minus the imaginary part that is equal to 1 ,513 .6.
03:33
So we can also write this as 4 ,962 and in a direction that is minus 17 .76 degrees.
03:48
And this involves.
03:50
Now, with that said, the induced line nph, so the induce line, mf let's call that e is a value of 8 ,000 well it's going to be the square root of two sorry the square root of three the square root of three that times the magnitude that we obtained from before which is 4 ,962 so the value that we obtained from this is 8 ,594 volts so that's the solution for part a of this problem.
04:33
Now for part b, we are asked about the output power.
04:40
Now, first we need to calculate the power input.
04:45
So the power input is equal to the square root of three times the potential that we are given initially...