Power Systems Test 9

Inertia constant, H = (stored kinetic energy)/(rating of machine)

Given that, f = 50 Hz, P = 4, V = 11 kV

S = 100 MVA

H = 5 MJ/MVA

Stored KE = H × S

= 5 × 100 = 500 MJ

Reactive power compensation, X is applied at midpoint hence it becomes \(\frac{X}{2},\) and hence the power transfer limit is \(\frac{{{V_s}{V_m}}}{{\frac{X}{2}}}.\)

Where, V_s =sending end voltage

V_m = maintained voltage at other end of the line

Given that, V_s = 1 pu, V_m = 0.98, X = 0.2

Steady-state power transfer limit \(= \frac{{{V_s}{V_m}}}{{X/2}}\)

\(= \frac{{1\; \times \;0.98}}{{\frac{{0.2}}{2}}} = 9.8\;pu\) 

When two synchronous machines do not swing together, \({H_{eq}} = \frac{{{H_1}{H_2}}}{{{H_1} + {H_2}}}\)

Also, acceleration angle α is \(\frac{{{d^2}\delta }}{{d{t^2}}}\)

Given that, H₁ = 6 MJ/MVA, H₂ = 4 MJ/MVA, f = 50 Hz

\({H_{eq}} = \frac{{{H_1}{H_2}}}{{{H_1}\; + \;{H_2}}} = \frac{{6\; \times \;4}}{{6\; + \;4}} = 2.4\) 

\(M = \frac{{{H_{eq}}}}{{180\; \times \;f}} = \frac{{2.4}}{{180\; \times \;50}} = \frac{1}{{3750}}\) 

\(\alpha = \frac{{{d^2}\delta }}{{d{t^2}}} = \frac{{{P_a}}}{M} = \frac{{1 - 0.6}}{{\frac{1}{{3750}}}}\) 

⇒ α = 1500 elec-deg/sec²

The critical clearing angle is given by,

\({\delta _c} = {\cos ^{ - 1}}\left[ {\frac{{{P_s}\left( {{\delta _{max}} - {\delta _o}} \right) + {P_{m3}}\cos {\delta _{max}} - {P_{m2}}{\delta _o}}}{{{P_{m3}} - {P_{m2}}}}} \right]\) 

\({\delta _0} = {\sin ^{ - 1}}\left( {\frac{{{P_s}}}{{{P_{m1}}}}} \right)\)

\({\delta _{max}} = \pi - {\sin ^{ - 1}}\left( {\frac{{{P_s}}}{{{P_{m3}}}}} \right)\)

Where P_m1 is the maximum power transfer at the pre-fault condition

P_m2 is the maximum power transfer at the during-fault condition

P_m3 is the maximum power transfer at the post-fault condition 

Calculation:

Given that, P_max1 = 1.75 pu, P_max2 = 0.4 pu, P_max3 = 1.25 pu

P_s = 1.0 pu

Now, P_s = P_m1 sin δ_o

\(\Rightarrow \sin {\delta _o} = \frac{1}{{1.75}}\) 

⇒ δ_o = 0.608 radians = 34.85°

\({\delta _{max}} = \pi - {\sin ^{ - 1}}\left( {\frac{{{P_s}}}{{{P_{m3}}}}} \right)\) 

\(= \pi - {\sin ^{ - 1}}\left( {\frac{1}{{1.25}}} \right)\) 

⇒ δ_max = 2.214 radians = 126.86°

Critical clearing angle, δ_c

\({\delta _c} = {\cos ^{ - 1}}\left[ {\frac{{{P_s}\left( {{\delta _{max}} - {\delta _o}} \right) + {P_{m3}}\cos {\delta _{max}} - {P_{m2}}\cos {\delta _o}}}{{{P_{m3}} - {P_{m2}}}}} \right]\) 

\(= {\cos ^{ - 1}}\left[ {\frac{{1.0\left( {2.214 - 0.608} \right) + 1.25\cos \left( {126.86} \right) - 0.4 \times \cos \left( {34.85} \right)}}{{1.25 - 0.4}}} \right]\) 

⇒ δ_c = 51.6°

A round rotor generator with internal voltage E₁ = 2.0 p.u. and X = 1.1 p.u.

A round rotor synchronous motor with internal voltage E₂ = 1.3 p.u. and X = 1.2 p.u.

Power generated by generator = 0.5 pu

\({P_e} = \frac{{EV}}{X}\sin \delta = \frac{{{E_1}{E_2}}}{{{X_{eq}}}}\sin \left( {{\delta _1} - {\delta _2}} \right)\)

\( \Rightarrow 0.5 = \frac{{2 \times 1.3}}{{1.1 + 0.5 + 1.2}}\sin \left( {{\delta _1} - {\delta _2}} \right)\)

⇒ δ₁ – δ₂ = 32.58°

Concept:

New frequency \({{f}_{n}}={{f}_{i}}\sqrt{\frac{H.S\pm \left( \text{ }\!\!\Delta\!\!\text{ }{{P}_{D}} \right){{T}_{d}}}{H.S}}\)

Where ΔP_D = Load changes

T_d = time delay

H = inertia constant

S = generator rating

f_i = initial frequency

+ is used when there is decrease in load

-is used when there is increase in load

Calculation:

Given H = 5

T_d = 0.4 sec

ΔP_D = 100 - 50 = 50

S = 100

Maximum steady state power that can be transmitted over the line

\({P_{max}} = \frac{{\left| {{V_s}} \right|\left| {{V_R}} \right|}}{{\left| z \right|}} - \frac{{{{\left| {{V_R}} \right|}^2}}}{{\left| z \right|}}cos\theta\)

\(\theta = {\tan ^{ - 1}}\left( {\frac{2}{2}} \right) = 45^\circ\)

\({P_{max}} = \frac{{{{\left( {220 \times {{10}^3}} \right)}^2}}}{{2\sqrt 2 }} - \frac{{{{\left( {220 \times {{10}^3}} \right)}^2}}}{{2\sqrt 2 }} \times \frac{1}{{\sqrt 2 }}\)

P_s = 0.5,  P_m1 = 1

\({\delta _0} = {\sin ^{ - 1}}\left( {\frac{{{P_s}}}{{{P_{m1}}}}} \right) = 30^\circ\)

δ_c = 75°

\({t_c} = \sqrt {\frac{{2 \times \frac{{SH}}{{\pi f}}\left( {{\delta _c} - {\delta _o}} \right)}}{{{P_s}}}}\)

\(= \sqrt {\frac{{\frac{{2 \times 1.0 \times 6}}{{\pi \times 50}} \times \left( {75 - 30} \right) \times \frac{\pi }{{180}}}}{{0.5}}}\)

= 0.346 sec