Power Systems Test 10

Concept:

The value of the resistance to be used, across the contact space is given by

\(R = \frac{1}{2}\sqrt {\frac{L}{C}} \)

Where,

L is inductance

C is capacitance

Calculation:

Given that inductance (L) = 5 H

Capacitance (C) = 0.02 μF

= 0.02 × 10^-6 F

Rated symmetrical braking current \(= \frac{{rated\;making\;current}}{{2.55}} = \frac{{90}}{{2.55}} = 35.29\;kA\)

Rated secondary current of current transformer = 5 A

Pickup current(I_pickup) = Current setting(A) \(= \frac{{current\;setting\left( \% \right)}}{{100}} \times {I_{relay}}\)

Current setting(A) = 1.5 × 5 = 7.5 A

Plug-setting multiplier\( = \frac{{{\rm{Fault\;current}}}}{{{\rm{C.T. ratio\times current\;setting(A)}}}}\)

\(= \frac{{5000}}{{\frac{{400}}{{5}}\;\times7.5}} = 8.33\)

Concept:

When low inductive current is being interrupted and the arc quenching force of the circuit breaker is more than necessary to interrupt a low magnitude of current, the current will be interrupted before its natural zero instant.

In this situation, the energy stored in the magnetic field appears in the form of high voltage across the stray capacitance which will cause restriking of the arc.

The energy stored in the magnetic field is \(\frac{1}{2}L{i^2}\) where i is the instantaneous value of the current which is interrupted.

This will appear in the form of electrostatic energy equal to \(\frac{1}{2}C{v^2}\).

Where V is called the prospective value of the voltage.

As these two energies are equal, they can be related as

\(\frac{1}{2}L{i^2} = \frac{1}{2}C{v^2}\)

\( \Rightarrow v = i\;\sqrt {\frac{L}{C}}\)

Calculation:

Given that, short circuit current (I_f) = 8000 A.

Current chopping occurs at 2.5% of peak value & current.

\(i = \frac{{2.5}}{{100}} \times 8000 \times \sqrt 2 = 282.84\;A\)

Inductance (L) = 30 mH = 30 × 10^-3 H

Stray capacitance (C) = 100 × 10^-12 F

Prospective voltage, \(V = i\sqrt {\frac{L}{C}} \)

\(= 282.84\;\sqrt {\frac{{30 \times {{10}^{ - 3}}}}{{100 \times {{10}^{ - 12}}}}} \)

= 4898.98 kV

Concept:

\(e = {V_{ar}}\left[ {1 - \cos \left( {\frac{t}{{\sqrt {LC} }}} \right)} \right]\)

V_ar = k₁ k₂ k₃ E_m

k₁ = sin ϕ

k₂ takes into account armature reaction effect.

k₃ is phase factor = 1 for both neutral and fault grounded

= 1.5 for any one of the two not grounded.

E_m is the peak value of voltage.

Frequency of oscillations.

\({f_n} = \frac{1}{{2\pi \sqrt {LC} }}\)

Average RRRV = (Maximum restriking voltage)/(Time to reach maximum restriking voltage)

Maximum restriking voltage = 2 V_ar

Time to reach maximum restriking voltage, \({t_m} = \pi \sqrt {LC}\)

Calculation:

Power factor = cos ϕ = 0.4

⇒ ϕ = 66.42°

k₁ = sin ϕ = 0.916

Recovery voltage is 0.9 times full line value

⇒ k₂ = 0.9

As both neutral and faults are grounded,

k₃ = 1

\({E_m} = \frac{{132}}{{\sqrt 3 }} \times \sqrt 2 \;V\)

Maximum restriking voltage

= 2 V_ar

\(= 2 \times 0.916 \times 0.9 \times 1 \times \frac{{132}}{{\sqrt 3 }} \times \sqrt 2 \times {10^3}\)

= 177.7 kV

Frequency of oscillations, (f_n) = 16 kHz

\(\Rightarrow \frac{1}{{2\pi \sqrt {LC} }} = 16 \times {10^3}\)

\(\Rightarrow \pi \sqrt {LC} = 3.125 \times {10^{ - 5}}\) sec = 31.25 μ sec

Average \(RRRV = \frac{{177.7}}{{31.25}} = 5.6864\) kV/μ sec

Concept:

The frequency of damped oscillations is given by

\(f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}} - \frac{1}{{4{C^2}{R^2}}}}\)

Calculation:

Given that, inductive reactance (X­_L) = 5Ω

⇒ 2πfL = 5

⇒ 2π × 50 × L = 5

⇒ L = 15.91 mH

Capacitance (C) = 0.02 μF

Resistance (R) = 500 Ω

\(f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{15.91\; \times \;{{10}^{ - 3}}\; \times\; 0.02\; \times \;{{10}^{ - 6}}}} - \frac{1}{{4\; \times\; {{\left( {0.02\; \times \;{{10}^{ - 6}}} \right)}^2}\; \times\; {{\left( {500} \right)}^2}}}} \)

= 4.03 kHz  

Let %x be the stator winding which is unprotected.

Emf induced in this unprotected winding \( = \frac{{13 \times {{10}^3}}}{{\sqrt 3 }} \times \frac{x}{{100}} = \frac{{130}}{{\sqrt 3 }}x\;V\)

Fault current, \({I_f} = \frac{{\frac{{130}}{{\sqrt 3 }}x}}{4} = 18.76x\;A\)

The minimum current which will operate the relay

\( = \frac{{1500}}{4} \times 0.8 = 300\) 

Now, 18.76x = 300

Percentage of winding to be protected = 85%

Percentage of unprotected winding = 15%

Voltage (V) = 11 kV

MVA rating = 100 MVA

Full load current \(\left( {{I_L}} \right) = \frac{{MVA}}{{\sqrt 3 \times kV}}\)

\(= \frac{{100 \times {{10}^6}}}{{\sqrt 3 \times 11 \times {{10}^3}}}\)

= 5248.63 A

Relay operates when there is 20% out of balance current.

Minimum fault current (I_f) = 0.2 × 5248.63

= 1049.73 A

Voltage across unprotected winding

\( = \frac{{0.15 \times 11 \times {{10}^3}}}{{\sqrt 3 }}\)

= 952.63 V

Let R_n is the resistance to be placed in the neutral to ground connection.

\(\Rightarrow 1049.73 = \frac{{952.63}}{{{R_n}}}\)