Tuesday, May 6, 2014
Thursday, April 17, 2014
Control Cable Required For 33/11KV Sub-Station
Control Cable Required For 33/11KV Sub-Station
10 Core Cable:
Used for Wiring from Control Room To Breaker.
a. Within Breaker (4 X 6M) = 24 M
b. LV Breaker to Control Room (1 X 25M) = 25 M
c. Feeder Breakers to Control Room (3 X 40M) = 120 M
(35m+37m+40m) arrox 40m
d. One Length for Group Control (1 X 40M) = 40 M
Total 210 M (250 M Approx).
4 Core Cable:
a. CTs to Breaker (3 X 10M) = 30 M
b. 1 LV Breaker + 3 Feeder Breakers (4 X 30M) = 120 M
c. LV to PTR (1 X 10M) = 10 M
d. LV To Control Room(Voltmeter) (1 X 25M) = 25 M
e. PT Supply - AC, PT Section Box to Breakers (1 X 60M) = 60 M
(10m+15m+20m+15m)
Total 200 M (250 M Approx).
2 Core Cable:
a. Chargers to Breaker (4 X 20M) = 80 M
b. Station DTR Junction Box to Each Charge
and Charger to Breaker (5m+5m+10m+10m) = 30 M
Total 110 M.
All above Calculations were done by assuming that the control room will be constructed on 33 KV Side.
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| 11 KV Polymer Disc and MP |
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| 11 KV Guy Insulator |
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| 11 KV Pin Insulator (Polymer Type) |
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| Stay Set Bow and I Bolt |
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| NSF Sub Station |
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| NSF Sub Station |
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| Stay Rod |
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| DTR Structure Mounting Channel & H X Arm |
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| DTR Structure Mounting Angle |
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| DTR Structure Mounting Angle |
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| LT 3 Ph X Arm |
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| LT 3 Ph X Arm |
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| Stay Set Base Plat |
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| Stay Wire |
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| 11 KV Top Cleat |
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| 11 KV Top Cleat |
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| 11 KV X Arm |
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| 11 KV X Arm |
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| 11 KV X Arm |
Sunday, April 6, 2014
Calculate % Voltage Regulation of Distribution Line - Method-2 (Load Base)
Method-2 (Load Base)
% Voltage Regulation =(I x (RcosǾ+XsinǾ)x Length ) / No of Cond.per Phase xV (P-N))x100
Voltage drop at Load A
- Load Current at Point A (I) = KW / 1.732xVoltxP.F
- Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.
- Distance from source= 1.500 Km.
- Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No
- Voltage Drop at Point A = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
- Voltage Drop at Point A =((98x(0.272×0.8+0×0.6)x1.5) / 1×6351) = 0.52%
- % Voltage Regulation at Point A =0.52 %
Voltage drop at Load B
- Load Current at Point B (I) = KW / 1.732xVoltxP.F
- Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
- Distance from source= 1500+1800=3.3Km.
- Required No of conductor / Phase =118 / 205 =0.57 Amp =1 No
- Voltage Drop at Point B = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
- Voltage Drop at Point B =((118x(0.272×0.8+0×0.6)x3.3)/1×6351) = 1.36%
- % Voltage Regulation at Point A =1.36 %
Voltage drop at Load C
- Load Current at Point C (I) = KW / 1.732xVoltxP.F
- Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131Amp.
- Distance from source= 1500+1800+2000=5.3Km.
- Required No of conductor / Phase =131/205 =0.64 Amp =1 No
- Voltage Drop at Point C = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100
- Voltage Drop at Point C =((131x(0.272×0.8+0×0.6)x5.3)/1×6351) = 2.44%
- % Voltage Regulation at Point A =2.44 %
Here Trail end Point % Voltage Regulation is 2.44% which is in permissible limit.
Calculate % Voltage Regulation of Distribution Line - Method-1 (Distance Base)
Calculate % Voltage Regulation of Distribution Line
Calculate Voltage drop and % Voltage Regulation at Trail end of following 11 KV Distribution system:
- System have ACSR DOG Conductor (AI 6/4.72, GI7/1.57)
- Current Capacity of ACSR Conductor = 205Amp,
- Resistance = 0.2792Ω and Reactance = 0 Ω,
Permissible limit of % Voltage Regulation at Trail end is 5%.
Method-1 (Distance Base)
Voltage Drop = ( (√3x(RCosΦ+XSinΦ)x I ) / (No of Conductor/Phase x1000))x Length of Line
Voltage drop at Load A
- Load Current at Point A (I) = KW / 1.732xVoltxP.F
- Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.
- Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No
- Voltage Drop at Point A = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
- Voltage Drop at Point A =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x1500) = 57 Volt
- Receiving end Voltage at Point A = Sending end Volt-Voltage Drop= (1100-57) = 10943 Volt.
- % Voltage Regulation at Point A = ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
- % Voltage Regulation at Point A = ((11000-10943) / 10943 )x100 = 0.52%
- % Voltage Regulation at Point A =0.52 %
Voltage drop at Load B
- Load Current at Point B (I) = KW / 1.732xVoltxP.F
- Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.
- Distance from source= 1500+1800=3300 Meter.
- Voltage Drop at Point B = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
- Voltage Drop at Point B =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x3300) = 266 Volt
- Receiving end Voltage at Point B = Sending end Volt-Voltage Drop= (1100-266) = 10734 Volt.
- % Voltage Regulation at Point B= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
- % Voltage Regulation at Point B= ((11000-10734) / 10734 )x100 = 2.48%
- % Voltage Regulation at Point B =2.48 %
Voltage drop at Load C
- Load Current at Point C (I) = KW / 1.732xVoltxP.F
- Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131 Amp
- Distance from source= 1500+1800+2000=5300 Meter.
- Voltage Drop at Point C = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line
- Voltage Drop at Point C =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x5300) = 269 Volt
- Receiving end Voltage at Point C = Sending end Volt-Voltage Drop= (1100-269) = 10731 Volt.
- % Voltage Regulation at Point C= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
- % Voltage Regulation at Point C= ((11000-10731) / 10731 )x100 = 2.51%
- % Voltage Regulation at Point C =2.51 %
Here Trail end Point % Voltage Regulation is 2.51% which is in permissible limit.
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