Question

Question #4 A 60-Hz, three-phase, transmission line is 210 km long. It has a total series impedance of (0.06 +0.053) 2/km and a shunt admittance of j5.6 x 10-6 S/km. The supply provides 200MW (P) at 275kV (line voltage i.e. Vs), with 0.9 power-factor lagging (i.e. pfs). Using the nominal-T approximation, (i) ABCD parameters (ii) Receiving end voltage and current (iii) Receiving end power and power factor; (iv) Line voltage regulation (v) line efficiency (vi) Propagation constant (vii) Attenuation constant

Answer 1

Given:-

Z=(0.06+$\large i$0.053)/KM *210KM

=(12.6+$\large i$11.13)$\large ohm$

Y=$\large i$5.6*10-6  S/KM *210

=$\large i$1.176*10-3

1) In T-approximation:

ABCD parameters:

A= 1+YZ/2

B=Z(1+YZ/4)

C=Y

D=1+YZ/2

Now;

put value of Y&Z and find ABCD parameters

A=(1+((12.6+$\large i$11.13)*( $\large i$ 1.176*10-3 ))/2)

=(1+$\large i$0.0074-0.0065)

=0.9935+$\large i$0.0074

=0.9935$\large \angle$0.42

B=(12.6+$\large i$11.13)(1+(($\large i$1.176*10-3) *(12.6+$\large i$11.13))/4)

=(12.6+$\large i$11.13)(1+$\large i$0.0035-0.00325)

=(12.6+$\large i$11.13)(0.99675+$\large i$0.0035)

=12.52+$\large i$11.13

C=$\large i$1.176*10-3

D=(1+((12.6+$\large i$11.13)*( $\large i$ 1.176*10-3 ))/2)

=(1+$\large i$0.0074-0.0065)

=0.9935+$\large i$0.0074

2.)We know

Vs =AVR+BIR   ........(1)

IS =CVR+DIR ...........(2)

Where:

Vs =Sending end Voltage

VR =Receiving end voltage

IR =Receiving end current

Given :

supply power=200MW

Supply voltage=275KV((Line)

=275/1.732

=158.77 kv(phase)

pf=0.9(lagging)

suppl power =$\large \sqrt{}$3 *Vs*Is *cos(pf)

200 *106 = $\large \sqrt{}$ 3* 275*Is* 0.9* 103

Is = 466.54amp (Line)

= 269.36 amp (phase)

Now from equation (1)&(2)

  Vs =AVR+BIR  

158.77*103=(0.9935+$\large i$0.0074)VR+(12.52+$\large i$11.13)IR ................(4)

IS =CVR+DIR

  269.36  =($\large i$1.176*10-3)VR +(0.9935+$\large i$0.0074) IR ................(5)

From (4) & (5)

VR=158.853$\large \angle$-1.75KV

  IR =289.33$\large \angle$35.29

3.)Receiving end power(PR) =3 *VR*IR*cos(pf)

Receiving end Power factor= Cos(Angle between  VR& IR )

=Cos(37.01)

=0.79 (Leading)

Now Power= 3 * 158.853$\large \angle$-1.75 *289.33$\large \angle$35.29 *0.79

=108.92MW

4.) Voltage regulation:

=(VS/A-VR )VR * 100 Percent

= (158.77/0.993$\large \angle$0.4 -158.853$\large \angle$-1.75)/158.853$\large \angle$-1.75

=2.4 percent