Power Systems Test 5

Source impedance (Z_S) = j5 Ω

Load impedance (Z_L) = (20 + j 15) Ω

The equivalent impedance (Z_eq) = Z_S + Z_L = (20 + j20) Ω

To get a unity power factor, the circuit should be purely resistive.

So, we need to the required series impedance = -j20 Ω

Therefore, the element should be a capacitor i.e. X_C = 20 Ω

\( \Rightarrow \frac{1}{{2\pi fC}} = 20\)

\(Q = \frac{{{V^2}}}{{{X_c}}}\) 

\(\Rightarrow Q\;\alpha \;{V^2}f\) 

\(\Rightarrow \frac{{{Q_1}}}{{{Q_2}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^2} \times \frac{{{f_1}}}{{{f_2}}}\) 

Given that V₂ = 1.05 V₁

And f₂ = 0.9 f₁

\(\Rightarrow {Q_2} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^2} \times \left( {\frac{{{f_2}}}{{{f_1}}}} \right) \times {Q_1}\) 

= (1.05)² × (0.9) × 100

kW rating of alternator (P) = 400 W

Power factor = cos ϕ = 0.5

kVA rating of alternator, \(S = \frac{P}{{\cos \phi }} = \frac{{400}}{{0.5}} = 800\;kVA\)

kW at unity power factor = 800 × 1 = 800 W

Source impedance (Z_S) = j5 Ω

Load impedance (Z_L) = (20 + j 15) Ω

A series compensator is used to raise the load voltage to the input voltage.

To achieve this, the voltage drop across the source impedances should be zero.

So, the impedance of series compensator element = -j5 Ω

Therefore, the element should be a capacitor i.e. X_C = 5 Ω

\( \Rightarrow \frac{1}{{2\pi fC}} = 5\)

Motor input, P = 6 kW

Cosϕ₁ = 0.6 lag

Cosϕ₂ = 0.9 lag

Leading KVAR taken by capacitor bank

= P (tanϕ₁ – cosϕ₂)

= 6 (tan (cas^-1(0.6))) – tan (can¹ (0.9))

= 5.094 kVAR

Rating of capacitors connected in each phase.

\( = \frac{{5.094}}{3} = 1.698\;kVAR\)

Rated Line Voltage = V_LO

Rated frequency = f_o

3 – ϕ VAR rating

Q₀ = ω₀ CV²_L0 = 2πf₀ CV²_L0

At f = 0.95f₀, V_L = 0.90 V_L0

Q = 2πfc V²_L= 2π × (0.95f₀) C × (0.9 V_L0)²

= 0.7695 Q₀

Percentage reduction \(= \frac{{{Q - }{Q_0}}}{{{Q_0}}} \times100\)

\(= \frac{{{Q_0} - 0.7695{Q_0}}}{{{Q_0}}} \times 100\)

Load 1: Lighting load of 20 kW at unity power factor

P₁ = 20 kW, Q₁ = 0

Load 2: Induction motor load of 100 kW at pf of 0.707 lagging

P₂ = 100 kW, cos ϕ₂ = 0.707 lagging

Q₂ = P₂ tan ϕ₂ = 100 kVAR

Q₂ is negative as the power factor is lagging

Load 3: Synchronous motor load of 50 kW at pf 0.9 leading

P₃ = 50 kW, cos ϕ₂ = 0.9 leading

Q₃ = P₃ tan ϕ₃ = 24.21 kVAR

Q₃ is positive as the power factor is leading

Real and reactive power of generator are

P = P₁ + P₂ + P₃ = 20 + 100 + 50 = 170 kW

Q = Q₁ + Q₂ + Q₃ = 0 – 100 + 24.21 = - 75.79 kVAR

Negative sign indicates the lagging power factor.

\(S = \sqrt {{P^2} + {Q^2}} = \sqrt {{{170}^2} + {{75.79}^2}} = 186.13\;kVA\)

Power factor \( = \frac{P}{S} = \frac{{170}}{{186.13}} = 0.913\;lagging\)

Concept:

Series Compensation:

Series compensation is the method of improving the system voltage by connecting a capacitor in series with the transmission line.

The basic idea behind series capacitive compensation is to decrease the overall effective series transmission impedance from the sending end to the receiving end.

\(P = \frac{{{V_r}{V_s}}}{{{X_L}}}\sin \delta \)

P is the power transferred per phase (W)

Vs is the sending-end phase voltage (V)

Vr is the receiving-end phase voltage

X_L is the series inductive reactance of the line

δ is the phase angle between Vs and Vr

If a capacitor having capacitance reactance Xc is connected in series with the line, the reactance of the line is reduced from X_L to (X_L– X_C).

X_eff = (X_L – X_C) = (1 – K) X_L

Where \(K = \frac{{{X_C}}}{{{X_L}}}\) is the degree of series compensation

Now, the power transfer capability is given by \(P = \frac{{{V_r}{V_s}}}{{{X_{eff}}}}\sin \delta \)

\(P = \frac{{{V_r}{V_s}}}{{\left( {1 - K} \right){X_L}}}\sin \delta \)

Application:

Given that, the degree of series compensation (K) = 0.5

\(P = \frac{{{V_r}{V_s}}}{{\left( {1 - 0.5} \right){X_L}}}\sin \delta = P = 2\frac{{{V_r}{V_s}}}{{{X_L}}}\sin \delta \)

Given that P_L = 200 kW

cos ϕ₁ = 0.8, cos ϕ₂ = 0.9

The kVAR need to supply by the synchronous motor to improve the load power factor to 0.9 is

Q = P_L (tan ϕ₂ – tan ϕ₁) = 53.13 kVAR

kW rating of synchronous motor (P) = 80 kW

\(S = \sqrt {{P^2} + {Q^2}} = \sqrt {{{80}^2} + {{53.13}^2}} = 96.04\;kVA\)