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Loads currents are exactly the same as in above example (delta connected reactive loads). Next , analyze currents to terminals of connected capacitors Id = { Sxn /Vxn}* = { -jQ/Vp/_0°}* = { (Q/Vp) /_-90°}* = { (Q/Vp) /_90°} Ie = { Syn /Vyn}* = { -jQ/Vp/_-120°}* = { (Q/Vp) /_(-90° + 120°) } * = { (Q/Vp) /_30°}*= { (Q/Vp) /_-30°} If = { Szn /Vzn}* = { -jQ/Vp/_-240°}* = { (Q/Vp) /_(-90° + 240°)}* = { (Q/Vp) /_150°}*= { (Q/Vp) /_-150°}
Lastly, analyzing utility supply curents of phase R, Y and B. Ia = Id + Ig = { (Q/Vp) /_90°} + | Ig | /_ - cos -1 pfi Ib = Ie+ Ih = { (Q/Vp) /_-30°} + | Ih | /_ { (- cos -1 pfi ) - 120° } Ic = If+ Ii = { (Q/Vp) /_-150°} + | Ii | /_ { (- cos -1 pfi ) - 240° } Let's find value of capacitance, C :
Xc = Vp / i = 1/(wC)
Typical ExampleSimilarly, the resultant currents at load side and supply side are identical to that calculated in the case of delta connected reactive load A 3-phase 415V supply delivers a balanced 3-phase load of 300kW , power factor lagging 0.75 is to be corrected to 0.87 Load current of line R, Ig = | Ig | /_ - cos -1 pfi = 100kW /(240V x 0.76) /_- cos - 0.75 =555.55 /_-41.4° A
From Table 4D1A of 16th Edition IEE Wiring Regulation (BS 7671), 4 x 1core mm 185 sq PVC Cu carry 296A Reactive power (single phase), Q=100kW x [ ( tan cos -1 0.75 ) - (tan cos -1 0.87) ] = 31.5kVar
Capacitor line current, Id = (Q/Vp) /_90° = 31,500/240 /_90°A = 131.25 /_90° Supply current of line R, Ia = Ig + Id = 555.55 /_-41.4° + 131.25 /_90° = 416.7 - j236.1 =478.9 /_ -29.5°
From Table 4D1A of 16th Edition IEE Wiring Regulation (BS 7671), 4 x 1core 120 mm sq PVC Cu carry 296A Notice that besides PF is improved from 0.75 to 0.87, smaller sized cables of 2 x 150mm per phase is required to carry supply current when cap bank is installed as compared to bigger cable of 2 x 185 mm sq at load side. POINTS DISCUSSED:
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