Power System Protection
Tutorial
Instrument Transformer
Q1:
A CO-8 relay with a current tap setting of 5 A is used with
150:5 CT. The CT secondary current I’ is the input to the relay
operating coil. The CO-8 relay burden is shown in Table for
various relay input currents.
Relay input current I’ (A) 5 8 10 13 15
Relay burden ZB (Ω) 0.5 0.8 1.0 1.3 1.5
• (i) Evaluate the primary current I and CT error for the given
relay input current in the table.
• (ii) Plot relay input current I’ versus primary current I for the
values of I’ in (i) .
Instrument Transformer
Excitation curve for a multi-ratio bushing CT
with a C100 ANSI accuracy classification.
CT Ratio Secondary resistance, Ω
50:5 0.061
100:5 0.082
150:5 0.104
200:5 0.125
250:5 0.146
300:5 0.168
400:5 0.211
450:5 0.230
500:5 0.242
600:5 0.296
• I’ = 5A
• E’
• Ie
•I
• CT error..
• Repeat the steps..
I’_5=5; Zb_5=0.5; N= 150/5
Based on Table,
Secondary resistance for CT ratio 150:5, Z=0.104
E’_5=(Z’+Zb_5) x I’_5=(0.104+0.5)x5=3.02V
From the curve we get,
Ie_5= 0.13
Ip_5=(n)(I’_5+Ie_5)=(150/5) x (5 + 0.13) = 153.9 A
CT Error = Ie/(I’+Ie) x 100
CTerr_5= [Ie_5/( I’_5+Ie_5)] x 100 = [0.13/(5+0.13)]x 100= 2.534%
-----------------
I’_5=8; Zb_5=0.8; N= 150/5
Based on Table,
Secondary resistance for CT ratio 150:5, Z=0.104
E’_8=(Z’+Zb_8) x I’_8=(0.104+0.8)x8=7.232V
From the curve we get,
Ie_8= 0.25
Ip_8=(n)(I’_8+Ie_8)=(150/5) x (8 + 0.25) = 247.5 A
CT Error = Ie/(I’+Ie) x 100
CTerr_8= [Ie_8/( I’_8+Ie_8)] x 100 = [0.25/(8+0.25)]x 100=3.03%
I’_10=10; Zb_10=1.0; N= 150/5
Based on Table,
Secondary resistance for CT ratio 150:5, Z=0.104
E’_10=(Z’+Zb_10) x I’_10=(0.104+1.0)x10=11.04V
From the curve we get,
Ie_10= 0.3
Ip_10=(n)(I’_10+Ie_10)=(150/5) x (10 + 0.3) = 309 A
CT Error = Ie/(I’+Ie) x 100
CTerr_10= [Ie_10/( I’_10+Ie_10)] x 100 = [0.3/(10+0.3)]x 100=2.91 %
------------------------------------------------------------------------------------------------------- I’_15=15; Zb_15=1.5; N= 150/5
I’_13=13; Zb_5=1.3; N= 150/5 Based on Table,
Based on Table, Secondary resistance for CT ratio 150:5, Z=0.104
E’_15=(Z’+Zb_15) x I’_15=(0.104+1.5)x15=24.06V
Secondary resistance for CT ratio 150:5, Z=0.104
E’_13=(Z’+Zb_13) x I’_13=(0.104+1.3)x13=18.252V From the curve we get,
From the curve we get, Ie_15= 0.55
Ie_13= 0.5
Ip_13=(n)(I’_13+Ie_13)=(150/5) x (13 + 0.43) = 405A Ip_15=(n)(I’_15+Ie_15)=(150/5) x (15 + 0.55) = 466.5 A
CT Error = Ie/(I’+Ie) x 100
CTerr_13= [Ie_13/( I’_13+Ie_13)] x 100 = [0.5/(13+0.5)]x 100=3.7% CT Error = Ie/(I’+Ie) x 100
------------------------------------------------------------------------------------------------------- CTerr_10= [Ie_15/( I’_15+Ie_15)] x 100 = [0.55/(15+0.55)]x
100=3.537 %
16
14
12 I’ Ip
0 0
5 153.9
10
8 247.5
8
I'
10 310.5
6
13 402.9
4
15 466.5
0
153.9 247.5 310.5 402.9 466.5
Ip
Instrument Transformer
Voltage transformer (minggu depan)
• Given the open-delta PT connection shown in Figure 10.38, both
VTs having a voltage rating of 240 kV : 120 V, the voltages are
specified as VAB = 230∠ 0º, VBC = 230∠ -120ºand VCA = 230 ∠
120ºkV. Determine Vab, Vbc, and Vca.
Instrument Transformer(minggu depan)
The primary conductor in Figure is one phase of three phase transmission line operating at 345 kV,
600 MVA, 0.95 power factor lagging. The CT ratio is 1200:5 and the VT ratio is 3000:1. Determine the
CT secondary current I’ and the VT secondary voltage V’. Assume zero CT error.
Power Factor
• When connected to a 120 V (rms), 50 Hz power line, a load absorbs 5 kW at a lagging power factor
of 0.7. Find the value of capacitance necessary to rise the pf to 0.90.
Three phase
• A three-phase motor can be regarded as a balanced Y-load. A three-
phase motor draws 6.6 kW when the line voltage is 220 V and the line
current is 20 A. Determine the power factor of motor.
• Tips.. Example in lecture note..
Overcurrent relay
• The input current to a CO-8 relay is 10 A. Determine the relay
operating time for the following current tap settings (TS) and time dial
settings (TDS):
• (a) TS=1.0,TDS=1/2;
• (b) TS=2.0, TDS=1.5;
• (c) TS=2.0, TDS=7;
• (d) TS=3.0, TDS=7;
• (e) TS=12.0, TDS=1.
