EARTHING CALCULATION FOR TOUCH AND STEP VOLTAGEFull description
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In this guide, a design engineer works on a converter drive design and determines the voltage change coefficient during the drive start, compares it to permissible levels and suggests solutions to ...
In this guide, a design engineer works on a converter drive design and determines the voltage change coefficient during the drive start, compares it to permissible levels and suggests solutions to ...
In this guide, a design engineer works on a converter drive design and determines the voltage change coefficient during the drive start, compares it to permissible levels and suggests solutions to ...Full description
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Knee exercises from a Chikung perspective, article extracted from prestigiuos Qi magazine by Master Michael Tse
dislokasi sendi lutut. genu. pembagian anatomi dan posisional.Full description
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Knee point calculation for 765kV CT's CT Ratio :3000-2000-1000-500/lA CT Class :PS Secondary Winding Resistance: · 15 ohm at 3000 tap · 10 ohm at 2000 tap . 5 ohm at 1000 tap . 2.5 ohm at 500 tap Length of line assumed =25 kms.
Where IF= Max. secondary fault current at zone-l reachpoint. X/R = Primary system reactance/resistance ratio System short circuit Fault Calculations System Impedance,Zs =Vs/ IF 765xl03
IF = 40 kA(Fault Level)
- ,J3x40x 103 =11.042 ohms
-.
Line parameters are Rr. = 1.9513 X 10-6 Pu/kM/Ckt
XL= 4.475 X 10-5pu/kM/ckt
]
On a 100 MVA ,765 kV Base
XL _ 4.475xlO-s RL -1.9513x10-6
= 22.93 =23
e=tan-I
4.475XlO-S
( 1.953
x 1O-{i
)
= 87.5°
Page 1 of 4
Knee point calculation for 765kV CT's Therefore ('
Xs = Zs sin 87.5° = 11.042 X 0.999 = 11.031 ohms Rs = Zs cos 87.5° = 11.042 X 0.0436 = 0.481 ohms Zs = 0.481 + j11.031 Line Impedance: Rr. = 1.9513 X 10-6 pulkM/ckt.: XL= 4.475 X 10-5pu/kM/ckt :
On a 100 MVA ,765 kV Base
Therefore Base Impedance =765 X 765/100 = 5852.25 ohms Rr. = 1.9513 X 10-6X 5852.25 = 0.01142 ohmslkM XL= 4.475 X 10-5X 5852.25 = 0.2619 ohmslkM For a Line length of25 kMs (zone-1 reach=20 kMs) Rr.' = 0.01142 X 20 = 0.2284 ohms XL'= 0.2619 X 20 = 5.238 ohms
Therefore Total Fault Impedance upto zone-1 reach point R= Rr.'+ Rs = 0.2284+0.481 = 0.7902 ohms X=Xs+ XL' = 11.031+ 5.238 = 16.269 ohms Z = 0.7094 + j 16.269 Z = 16.284 ohms Therefore Fault current at zone-1 reach
= IF' 765x103
- J3 x16.284 = 27123.12 A
IF'= 27.12 kA (fault level at zone-1 reach point)
Page 2 of 4
Knee point calculation for 765kV CT's Lead Burden: Cable Length= 800 m Lead resistance = 8.71 ohms/kM ( x 2.5 mm2 copper cable is used) Therefore 2R L
= 2x 800 x 8.71 = 6.968 2 1000
= 7.0ohms CT secondaryresistance, Rcr = 15.00 at 3000/1 A tap Knee point voltage requirement VK~0.6IF( 1
+;)Rcr +RL)
Relay burden is negligible .Therefore Rrelay=0. 27120 VK~ 0.6x-(1 +23)(15 +7.0) 3000 For 3000/1 A tap VK~ 2863.87Vat30001lTap Selected knee point voltage for 3000/1 A tap =6000 V(plese see note below) RCf ~15.00hm
Imag < 20mAatVK
"For 2000 A tap,
27120 VK(2000)=0.6x-x(I+23)(10+7.0) 2000
= 3319.5 V Selected knee point voltage for 2000/1 A tap = 4000V(same as that for 400 kV CTs) Note: 1. Since the requirement at 2000/1 A tap is 2V per turn,the selected knee point voltage for 3000/1 A tap is 6000V.This is in line with the knee point voltage requirement being specified for 400 kV CT's . 2. It is intended to use only 3000/1 A or 000/1 A taps for 765 kV line protection.
Page 3 of4
Knee point calculation for 765kV CT's 3. 1000/1 A and 500/1 A taps shall be used for Generator circuits (Protection Overall Differential,Overhead Line Differential). CT requirement for busbar protection IF
= 40 kA
VK~2IF(RCT+RL) For 3000 A tap Rcr = 15.0ohms Rr.= 7.0 ohms 40000 VK~2x-x(15.0+7.0) 3000
Selected VK at 3000/1 A tap = 3000V For 2000 A tap Rcr = 10.0 ohms Rr.= 7.0 ohms 40000 VK~2x-x(IO.0+7.0) 2000
Selected VK at 2000/1 A tap = 2000V VK~ 680V For 1000 A tap Rcr = 5.0 ohms Rr.= 7.0 ohms 40000 VK~2x-x(5.0+7.0) 1000
Selected VK at 1000/1 A tap = 1000V Note: 1. 500/1 A tap shall not be used for bus-bar protection. 2. 500/1 A tap shall be used for Generator circuit protection(Overall Differential, Overhead Line Differential)
