- #1

topcat123

- 78

- 1

FIGURE 3(a) represents a 50 Hz, three-phase, high-voltage, transmission

line. For one phase, the relationships between the sending end voltage

and current and the receiving end voltage are given by the complex

ABCD equations:

Vsp = Vrp(A1 + jA2) + Irp(B1 + jB2)

Isp = Vrp(C1 + jC2) + Irp(D1 + jD2)

where VSP is the sending-end phase voltage, ISP the sending-end phase

current and VRP is the magnitude of the open-circuit receiving end phase

voltage.

(a) Given the parameter values in TABLE A and if the magnitude of the

receiving-end line voltage VRL is measured as 154 kV when feeding

a balanced load of 40 MVA at a power factor of 0.9, calculate the

value of the sending-end phase voltage VSP and sending-end phase

current ISP.

[N.B. VSL = √3 × VSP and the total power in a three-phase load is

given by P = √3VI cos θ.]

(b) Hence or otherwise calculate the sending-end power and thus the

power lost in the cable.

(c) If the line is modeled by the Π-circuit of FIGURE 4(b), see if you

can estimate the primary line coefficients R, L, G and C. The line is

50 km long.

A1= 0.8698

A2= 0.03542

B1= 47.94 Ω

B2= 180.8 Ω

C1= 0 S

C2= 0.001349 S

D1= 0.8698

D2= 0.03542

so

a) P=sqrt(3)VI

I=40000000/(sqrt(3)*154000) = 149.961 A

Vrp=Vrl/sqrt(3) = 88912 V

Using formulas given for ABCD.

Vsp =84524.79 + j30262.16 = 89778.8∠19.70

Isp = 130.44 +j125.252 = 180.83∠43.84

b) p= sqrt(3)VI cos θ = sqrt(3)*154000*149.961*0.9 =36 MW

Vsl = sqrt(3)*89778.8 = 155501.4*149.961*0.9 = 36.3MW

350960.2 W lost in the cable.

c) I don't have a clue?

line. For one phase, the relationships between the sending end voltage

and current and the receiving end voltage are given by the complex

ABCD equations:

Vsp = Vrp(A1 + jA2) + Irp(B1 + jB2)

Isp = Vrp(C1 + jC2) + Irp(D1 + jD2)

where VSP is the sending-end phase voltage, ISP the sending-end phase

current and VRP is the magnitude of the open-circuit receiving end phase

voltage.

(a) Given the parameter values in TABLE A and if the magnitude of the

receiving-end line voltage VRL is measured as 154 kV when feeding

a balanced load of 40 MVA at a power factor of 0.9, calculate the

value of the sending-end phase voltage VSP and sending-end phase

current ISP.

[N.B. VSL = √3 × VSP and the total power in a three-phase load is

given by P = √3VI cos θ.]

(b) Hence or otherwise calculate the sending-end power and thus the

power lost in the cable.

(c) If the line is modeled by the Π-circuit of FIGURE 4(b), see if you

can estimate the primary line coefficients R, L, G and C. The line is

50 km long.

A1= 0.8698

A2= 0.03542

B1= 47.94 Ω

B2= 180.8 Ω

C1= 0 S

C2= 0.001349 S

D1= 0.8698

D2= 0.03542

so

a) P=sqrt(3)VI

I=40000000/(sqrt(3)*154000) = 149.961 A

Vrp=Vrl/sqrt(3) = 88912 V

Using formulas given for ABCD.

Vsp =84524.79 + j30262.16 = 89778.8∠19.70

Isp = 130.44 +j125.252 = 180.83∠43.84

b) p= sqrt(3)VI cos θ = sqrt(3)*154000*149.961*0.9 =36 MW

Vsl = sqrt(3)*89778.8 = 155501.4*149.961*0.9 = 36.3MW

350960.2 W lost in the cable.

c) I don't have a clue?