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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