GATE Electrical Engineering (EE) 2020 Solved Paper

Question : 54 of 65

Marks: +1, -0

Bus 1 with voltage magnitude

{V}_{1}=1.1 {p} . {u} .

is sending reactive power

{Q}_{12}

towards bus 2 with voltage magnitude

V_{2}=1

p.u. through a lossless transmission line of reactance

X

. Keeping the voltage at bus 2 fixed at 1 p.u., magnitude of voltage at bus 1 is changed, so that the reactive power

Q_{12}

sent from bus 1 is increased by

20 \% .

Real power flow through the line under both the conditions is zero. The new value of the voltage magnitude,

V_{1}

, in p.u. (rounded off to 2 decimal places), at bus 1 is____________.

Solution:

{Q}_{12}=\;{| {V}_{1}|^{2}}/{ {x}}-\;{| {V}_{1}|| {V}_{2}|}/{ {X}} \cos δ

For a highly stable system,

δ

is very small,

\cos δ ∼eq 1

{Q}_{12}=\;{| {V}_{1}|^{2}}/{ {x}}-\;{| {V}_{1}|| {V}_{2}|}/{ {X}}

{Q}_{12}=\;{| {V}_{1}|}/{ {x}}-(| {V}_{1}|-| {V}_{2}|)

.....(1)
Let, for

{Q}_{12} → 102 {Q}_{12}

(

20 \%

increase)

{V}_{1} → {V}_{1}^{'}

Then,

\;{1.2 {Q}_{12}}/{ {Q}_{12}}=\;{| {V}_{1}^{'}|(| {V}_{1}^{'}|| {V}_{2}^{'}|)}/{ {X}} × \;{ {X}}/{| {V}_{1}|(| {V}_{1}|-| {V}_{2}|)}

1.2=\;{| {V}_{1}^{'}|(| {V}_{1}^{'}|-1)}/{1.1(1.1-1)}

|V^{'}{ }_{1}|^{2}-|V^{'}{ }_{1}|-0.132=4

∼eq 1.12

(Rounded off to two decimal places)

Go to Question:

Prev Question

Next Question