Power Systems Test 11

Shunt capacitors are used with individual equipment to improve power factor. For fixed power loss and voltage, as power factor increases, current decreases. Therefore, I²R loss decrease. The power losses are reduced and the efficiency is increased. The smaller voltage drop in the line results in good voltage regulation. Thus, shunt capacitors regulate the voltage and reactive power flows at the points where they installed.

Rated line voltage = V_LO

Rated frequency = f_o

3 - ϕ VAR rating

\(\begin{array}{l}{Q_o} = {\omega _o}CV_{Lo}^2 = 2\pi {f_o}CV_{Lo}^2\\At\ f = 0.98\ {f_o},{V_L} = 0.95{V_{Lo}}\\Q = 2\pi fcV_L^2 = 2\pi \times \left( {0.98{f_o}} \right)C \times {\left( {0.95{V_{L0}}} \right)^2}\end{array}\)

= 0.885 Q_O

\(Percentage\ reduction = \frac{{Q - {Q_o}}}{{{Q_o}}} \times 100\)

= -11.5%

Speed regulation (R) = 0.05

Change in frequency, Δf = f – f₀ = 59.7 – 60 = -0.3 Hz

\({\rm{\Delta }}f = - \frac{{0.3}}{{60}} = - 0.005\;pu\)

ΔP_m = ΔP_L

\( = - \frac{{{\rm{\Delta }}f}}{R} = 0.1\;pu\)

ΔP_m = 0.1 × 250 = 25 MW

\({\rm{\Delta }}f = \frac{{\% change\;in\;f}}{{{P_{fl}}}} \times {\rm{\Delta }}P\)

ΔP₁ + ΔP₂ = 500 MW

⇒ ΔP₂ = 500 – ΔP₁

Δf₁ = Δf₂

\( \Rightarrow \frac{{0.06}}{{250}} \times {\rm{\Delta }}{P_1} = \frac{{0.064}}{{400}} \times {\rm{\Delta }}{P_2}\)

\( \Rightarrow \frac{{0.06}}{{250}} \times {\rm{\Delta }}{P_1} = \frac{{0.064}}{{400}} \times \left( {500 - {\rm{\Delta }}{P_1}} \right)\)

⇒ ΔP₁ = 200 MW, ΔP₂ = 300 MW

Let, Load on generator 1 = x MW

Load on generator 2 = (450 – x) MW

Reduction in frequency = Δf

Now,

\(\frac{{{\rm{\Delta }}f}}{x} = \frac{{0.04 \times 60}}{{250}}\)      ----(i)

\(\frac{{{\rm{\Delta }}f}}{{\left( {450 - x} \right)}} = \frac{{0.06 \times 60}}{{400}}\)       ---(ii)

From equations (i) & (ii), we get

\(\frac{{450 - x}}{x} = \frac{{0.04 \times 60}}{{250}} \times \frac{{400}}{{0.06 \times 60}} = 1.066\)

∴ x = 217.81 MW (load on generator 1)

450 – x = 232.18 MW (load on generator 2)

and Δf = 2.09Hz

∴ System frequency = (60 – 2.09) = 57.909 Hz.

Load current, \({I_2} = \frac{{25 \times {{10}^6}}}{{\sqrt 3 \times 33 \times {{10}^3} \times 0.8}} = 546.8\ A\)

\(\begin{array}{l}{I_P} = {I_2}\cos {\phi _2} = 437.4\ A\\{I_q} = {I_2}\sin {\phi _2} = 328.1\ A\end{array}\)

R = 5 Ω; x = 20 Ω

Sending end voltage, V₁ = receiving end voltage V₂

\(= \frac{{33 \times {{10}^3}}}{{\sqrt 3 }} = 19053\)

Let I_m be the current taken by the synchronous condenser

\(\begin{array}{l}V_1^2 = {\left[ {{V_2} + {I_P}R - \left( {{I_m} - {I_q}} \right)X} \right]^2} + {\left[ {{I_P}X + \left( {{I_m} - {I_q}} \right)R} \right]^2}\\ \Rightarrow {\left( {19053} \right)^2} = {\left[ {19053 + \left( {437.4} \right)\left( 5 \right) - \left( {{I_m} - 328.1} \right)\left( {20} \right)} \right]^2} + {\left[ {\left( {437.4} \right)\left( {20} \right) + \left( {{I_m} - 328.1} \right)\left( {20} \right)} \right]^2} + {\left[ {\left( {437.4} \right)\left( {20} \right) + \left( {{I_m} - 328.1} \right)\left( 5 \right)} \right]^2}\end{array}\)

⇒ I_m = 579.5 A

Capacity of synchronous condenser \(= \frac{{3{V_2}{I_m}}}{{{{10}^6}}}MVAR\)

= 33.12 MVAR