(2.) (20 points) A three-phase balanced transmission line is...

(2.) (20 points) A three-phase balanced transmission line is 400 km long. The voltage at the sending end is 220kV line-to-line. The line parameters are known to be r = 0.125 ?/km xL = 0.400 ?/km y = 2.8 x 10^-6 ?^-1/km Answer the following questions. (a.) (10 points)The sending-end current and the magnitude of the receiving-end line-to-line voltage need to be computed when there is NO LOAD at the receiving end of the line as shown in Figure 1 (b.) (10 points) If the maximum allowable no-load voltage is 235kV, then what must the line length be for the no-load voltage not to exceed this value of voltage? Assume that Vs remains at 220 kV line-to-line. Make the following approximation to find the length l: A = cosh(?ZY) ? 1 + ZY/2 (1.0) HINT: What is A equal to at NO LOAD? It is a number greater than ZERO and less than ONE. Use this value for solving equation (1.0),

Transcript

00:01 So here in this question we are considering a three phase balanced transmission line which length is given that is equal to 400 kilometer long.

00:09 The voltage is sending at the end that is equal to 220 kilo volt.

00:13 Line parameters are given where radius is 0 .125 ohm per kilometer.

00:19 We are having the value of xl that is equal to 0 .400 ohm per kilometer and we are having the value of y that is equal to 2 .8 multiplied by 10 raised to the power minus 6 ohm inverse per kilometer.

00:31 So we have to find out the value of this leading core current when the no load is present here.

00:36 So we are having the value of r that is equal to this.

00:41 So this from here is equal to 0 .125 multiplied by the 400 that is equal to 50 ohm and this from here is equal to this is multiplied by the 400 that become equal to 160 ohm.

00:53 Again this is connected to the 400 that become equal to 1 .12 multiplied by the 10 raised to the power minus 3 ohm inverse.

01:03 So this is the value.

01:04 So in the first part we are asked about the sending current and the magnitude of the remaining voltage.

01:09 So abc parameters represented here.

01:11 So we can say is equal to a of vr plus b of ir and the value of is is equal to c of vr plus d of ir.

01:20 So if we make the matrix in this case that is abcd.

01:24 So this value from here is equal to 1 plus yz divided by 2z y of 1 plus y of z that is divided by 4 and 1 plus y of z that is divided by 2.

01:36 So the value of become equal to r plus j of xl that become equal to 50 plus j of 160 ohm and the value of y in this case become equal to 1 .12 multiplied by the 10 raised to the power minus 3 ohm inverse.

01:51 So the value of a is equal to 1 plus y of z that is divided by 2.

01:55 Plugging into the values we get the value of a that from here is equal to 1 .02 to the angle of 4 .98.

02:01 So this is the value of a.

02:02 When we are considering a no load on ir.

02:07 So this is equal to 0.

02:09 So vs is equal to a multiplied by the vr.

02:11 So mod of vr is equal to v of s which is divided by a that is equal to 220 which is divided by 1 .03 to the angle of 4 .98.

02:21 So the value of vr mod in this case become equal to 213 .59 kilowatt.

02:26 So this is the value of vr...

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