How to calculate the Operating region for Directional over current relay ?
Relay characterstic Angle (RCA)
The Relay characterstic value is used to define the region to operate the relay in the
specified direction.
Example :-
Let RCA = 45 ͦ
In normal condition angles between Current & Voltages
IR IY IB
0 ͦ 120 ͦ 240 ͦ
VR VY VB
0 ͦ 120 ͦ 240 ͦ
VRY VYB VBR
120 ͦ 360 ͦ 240 ͦ
If any fualt occure in any phase the voltage will be drop and heavy current will flow.
So to see the direction of the current flow relay will take other helth phase voltages.
Operating region = ( Phase displaced + RCA) to ( Phase displaced + RCA +180 )
Let's take R phase grounded and relay will take VYB voltage value is referance
VYB displaced 0 or 360 ͦ from IR
Operating region = ( 0 + 45 ͦ to 45 ͦ + 180 ͦ)
= 45 ͦ to 225 ͦ
Let's take Y phase grounded and relay will take VBR voltage value is referance
VYB displaced 240 ͦ from IR
= (240 ͦ + 45 ͦ ) = 285 ͦ and
= 285 ͦ + 180 ͦ = 465
= 465 - 360 = 105 ͦ
Operating regien = 105 ͦ to 285 ͦ
Let's take B phase grounded and relay will take VBR voltage value is referance
VRY displaced 120 ͦ from IB
= (120 ͦ + 45 ͦ ) = 165 ͦ and
= 165 ͦ + 180 ͦ = 345 ͦ
Operating regien = 165 ͦ to 345 ͦ
Thursday, August 30, 2012
Wednesday, August 29, 2012
CT TEST IN LIVE STATION
1 . Open the Isolator 1 & 2
2 . Close the Isolator 3
3 . Close the CB
4 . Close Line earth switch
5 . Close Maintenance earth switch 1 & 2
6 . Now the Induction voltage is zero in dead portion
7. Remove the earth link in the Maintenance earth switch
8. Connect the primary leads on that.
9. Start the testing
10. There is no leakage current in earth because primary current enter in to Main earth switch -1 and return at Main earth switch -2.
Thursday, August 23, 2012
SIMPLE Q & A
1. Why DC supply used as control supply in all protection and control panels?
DC supply is storable and if any problem in HV side AC voltage then Alarm and protection is easy by using DC supply.
2. Why CT & VT wires are bigger in size comparatively other control cables?
The I²R loss is more when small Sq.mm cables used.
Resistance is inversely proportional to cross section area of the wire or cable.
When area increase Resistance decrees.
R ∞ 1/a
Example :
VT secondary voltage = 115v
Burden of the VT secondary circuit = 300 VA
2.5 Sq.mm cable RL = 2 Ω
Current flow = 300 / 115 = 2.6 Amps
Loss per meter = I² RL = 35 watts
Voltage drop = I X RL = 2.6 * 2= 5.2 volts
3. What is Inrush current in Transformer or motor?
During commissioning time the residual magnetic flux in the
Transformer and motor coil is almost zero. There is no back emf for opposing
the main emf in the coil. So the magnetic current very high during starting
stages.
During charging time the inrush current value is less when
HV primary voltage frequency cross the zero value and comparatively high during
frequency cross its peck value.
Inrush current contain 3rd and 5th harmonics and it is detected by protection relays and block the tripping.
4. Why most of the Power Transformer primary side connection arrangements Star connection?
Per Phase voltage value of star connection = V/ √3
But per phase voltage in Delta connection is same as Phase
to Phase voltage.
The insulation requirement of transformer is depending open the voltage value. If voltage reduces then insulation requirement reduced and size & cost of transformer reduced.
DC supply is storable and if any problem in HV side AC voltage then Alarm and protection is easy by using DC supply.
2. Why CT & VT wires are bigger in size comparatively other control cables?
The I²R loss is more when small Sq.mm cables used.
Resistance is inversely proportional to cross section area of the wire or cable.
When area increase Resistance decrees.
R ∞ 1/a
Example :
VT secondary voltage = 115v
Burden of the VT secondary circuit = 300 VA
2.5 Sq.mm cable RL = 2 Ω
Current flow = 300 / 115 = 2.6 Amps
Loss per meter = I² RL = 35 watts
Voltage drop = I X RL = 2.6 * 2= 5.2 volts
3. What is Inrush current in Transformer or motor?
During commissioning time the residual magnetic flux in the
Transformer and motor coil is almost zero. There is no back emf for opposing
the main emf in the coil. So the magnetic current very high during starting
stages.
During charging time the inrush current value is less when
HV primary voltage frequency cross the zero value and comparatively high during
frequency cross its peck value.
Inrush current contain 3rd and 5th harmonics and it is detected by protection relays and block the tripping.
4. Why most of the Power Transformer primary side connection arrangements Star connection?
Per Phase voltage value of star connection = V/ √3
But per phase voltage in Delta connection is same as Phase
to Phase voltage.
The insulation requirement of transformer is depending open the voltage value. If voltage reduces then insulation requirement reduced and size & cost of transformer reduced.
Thursday, August 16, 2012
TRANSFORMER PROTECTION OVER CURRENT
CONSIDER YNyn0 TRANSFORMER
HV side Over current protection
1.Directional over current HV side (67HV)
2.Directional over current HV side (Neutral current) (67NHV)
3.Non directional Time delayed over current HV Neutral side (51GHV)
2.Directional over current HV side (Neutral current) (67NHV)
3.Non directional Time delayed over current HV Neutral side (51GHV)
LV side Over current protection
(Incase Transformer to LV side Bus tie normally close ( NC ) )
1. Directional over current LVside (67LV)
2. Directional over current LVside (Neutral current) (67NLV)
3. Non directional Time delayed over current LV Neutral side (51GLV)
LV side Over current protection
(Incase Transformer to LV side Bus tie normally open ( NO ) )
1. Non directional over current LV side (50/51 LV)
2. Non directional over current LV side (Neutral current) (50/51 NLV)
3. Non directional Time delayed over current LV Neutral side (51GLV)
TV side Over current protection
1. Non directional Time delayed over current LV side (51TW HS &LS)
Why 67HV instant of 50/51HV on HV side?
The Transformer HV side protection should be directional over current because it should operate only for fault in the transformer HV side. If fault on HV bus then the Bas bar protection should operate.
For the above fault Bus bar protection will operate because 67HV relay is directional over current with curve setting but bus bar protection setting is instantaneous.
FAULT IN TRANSFORMER HV SIDE
Now 67HV transformer protection will operate and BB protection will not operate because BB protection CT will cover this fault.
In this cause first HV side relay will trip and open the HV side CB and full fault current will flow through the LV side and LV side relay will trip and open the LV side breaker.
FAULT IN TRANSFORMER LV SIDE
FAULT IN TRANSFORMER LV SIDE
LV SIDE BUS TIE NORMALLY OPEN CONDITION
CHANGE THE ANY ONE TAP POSITION OF TRANSFORMER AND MAKE THE CIRCULATING CURRENT TO TEST THE DIRECTIONAL OVER CURRENT FUNCTION
What is the meaning of 67NHV?
What is the meaning of 67NHV?
This is Negative Sequence Overcurrent Protection (NPS).
This scheme worked based on residual current produced during earth fault and phase to phase faults. In normal the vectorial sum of all three phase current at neutral point should be zero.
But any earth or phase to phase fault accure a residual current generate due to unbalance.So a leakage current flow through the CT neutral path.This current sence by the NPS element and initiat the trip or alarm.
Negative Phase sequence overcurrent elements give more sensitive to resistive P-T-P faults, where phase over current elements may not be operate.
In certain application residual current may not be detected by an earth fault relay due to the system configuration. Example :- an earth fault relay applied on the Delta Side of a Delta - Star transformer is unable to detect earth faults on on the star side. However negative sequence current will be present on the both sides of the transformer for any fault condition , irrrespective of the transformer configuration. Therefore, a negative phase sequence overcurrent element may be employed to provide time-delayed protection for any any uncleared asymmentrical faults.
Tuesday, August 14, 2012
% IMPEDANCE OF A TRANSFORMER
The value mention on name plate for % Z is the percentage of voltage dropped by the
transformer impedance.
Z=R + j(XL+XC)
Z - Impedance
R - Transformer coil resistance ( Fixed Value)
XL - 2∏fL
XC - 1/2∏fC
The impedance of the Transformer is depending upon voltage applied.
Example:
33 / 11 KV
87.5 /262.5 A 6.81 %Z
The value of voltage drop in nominal voltage = 33000X6.81/100 = 2247 V
The value of impedance on full load condition = 2247.3/87.5 = 25.82 Ω
Conform this % impedance by site testing
1.Short the secondary side terminals
2. Apply the 380 V 3 phase in the HV side.
3. Before that calculate the primary current for applied voltage
If applied voltage 2247.3 v to HV side then full loade current 87.5 A will flow on primary side ( Secondary shorted).
Calculate the drop voltage if we are apply 380v
X /380 = 87.5 / primary current
This current can measure in primary side = 14.8A.
X /380 = 87.5 /14.8
X = 2246.62 Volts ( This is the total voltge drop for transformer total impedance)
Then we have to calculate this value how mach % in the Rated primary voltage
33000 * x / 1000=2246.62
x=6.8
This value is % Z of the transformer the error not more than
± 0.1 %
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