Question

Q1. Design a three-phase balanced power system where the load is connected with the power supply through a resistive transmission line. Your design needs to meet the following criteria: 1. Phase sequence for the power supply is ACB. You can take any phase or line as reference. 2. For the power supply, the line and phase voltages are related by a factor of V3. Please do not ignore the phase shift between the line and phase voltage or current. 3. Please assume the supply voltage (either phase or line) as XYZ and neutral as N with X ranges from 1 to 7, Y ranges from 0 to 5, and Z ranges from 0 to 9 while N as 0 or 5 (if necessary). 4. The loads need to be configured in such a way that the line-to-line voltage and phase voltage do not become same. Please do not ignore the phase shift between the line and phase voltage or current. 5. Please consider the loads in such a way that the current lags the voltage and the magnitude of the load current (this is the phase current for the load) does not exceeds a threshold T (which ranges from 0.1 A to 1 A) while the angle is within the limit of 1° to 89° for the reference phase. Please justify your design. You can make any reasonable assumption to complete your design.

Answer 1

(yet) point A so rotation 1) The phase Sequence Demanded is "cs". Tragen? I will treat supply voltage as phase voltager with A as reference for my Nollage selection -> Note? ka'e need to design baboad power systemand transmission lines should boovide Resistive drop. Boetically transmission line cloop are less compared to load. TRANS- mission lines ideally should provide No drop in vdioge and neto cause No power loss during transmission blence we shocell shoco choose it in order few ohms. fco this essay Let me choose each line has tx liºne resistance of 0.5.12 coming to criteria 1 mom mm > PLEASE NOTE THE Sequence taken is. ACB with A as reference. of phesor. In the direction of w rotation 100E 100'e "A appears then " appears at ref point the "B" appears at reference point 84 As we have three phases for balance we need to equally shift the phasors for complete 366°clectrical = for balance a phases symmetrical angle is 3 => 120 E phase shift is thus maintained. VA phase = MM Lo reference is choo sen 09 Ve phase = f (-120 = 11 l-120 No phase = lvl | +120 These are decided by shifting reference phase je VA of VA is in accordance with W rotation "+angle is assigned else " angle". 120€ -40 atret

Cphase) VA NOTE? Bilanced Voltages are alunys equal in magnitude and phone displaced by angle 120° for 3 phase system. Page @ ve cphases Line to line voltage is difference *. în Voltages between transmission dines. VAB = u rase o prose equally divides 120° 7120 VBC = cphase) base ve phase) Cphase) Line AVCA a veultant Notes WC a Mare phase Vresultant equally civicles 60'e Vresultant i found from cosine daw of resultant vectors - NOTE!- IVAl - 118) = V for balanced System) Vresultant = V (vo)2 + (No)? + DgVA VẠCauckB) (AB) Er 12 + v2 +2vcos (60°) ? tween VA anel Vo is (AB) Vresultant - W 202 +v? + 2V%? → 73027 → VE V VAB = VE V w is magnitice of ra phase. 3. z vresultant - VAB NOW! 800 VAB = 11B vl L-300 NAB lags Vaphase by 30° angle

. VAB = va v |-30° +6° = V8v 1-30 -120 Vervov | -30°+120° -) These were shifted as per 120° as explained in Page 1 = 13'v 1-80 Page 3 = V3'v 1 - 150° = 13'v [ +90° ACB Phase Sequence with - criteria 3 :- leb are supposed to assume Voltages as XY 7 and Delitzat "N" it needed. Please note in previoces pages "v" was left as magnitele. Now we need to choose lui as per given criterias . for X,Y,Z VA - Vx = V LOO V=5 (choose) Vc - Vy - VL-1200 v = 5 (choose) VB = Vz. = V [+120. V = 5 choose) for easy calculations iam choosing IV = 5 which falls in criteria given for X, Y, Z as well. . (N = o volts will be choosen of needed. Vx = 5loº Yy = 5 (-1200 vz = 5 +120° wye) (wye) Criteriene. =for VAD VA phase we can choose any configuration - of source and load of same nature score - Ford - Asource = load 3) I am choosing source -d load for easy onderstand 'ings and ☆ Easy calculatióne. > Vline + Va cobase) these fore souse - lead configuration

Current. + VI 1+ VE 145 criteria 5 - Page 4 choosing a lag load. 7 = Quib C AIRE LAGGING NATURE) Z = a-lb Case of lending current) 1. nature). → 7=A+iB provide Il-tan (B) => lagging current bence are called lagging LOADS Example :: LETS Assume Ve I loo 7 = atib be = sti 110 077=17 from choms law = I2109 - 10 - 2.446 I= 0.707 +45° There fore I lage v by 45° a ns > Vref Ilo Hence B for load to be Lag 7 = AtiB is choosen. 3) LOAD Current should not exceed 1 Amp or should fall uncles 0.1 to 1 Amp - » UNDER ANY SITUATION Line vee Voltage magnitude is 13 V phase bence lets assume VB Vphase is applied. to load "z" and limit current to o.1 to lamp -45 I. range. VaB - (V3)(5) 203) 5VBLo volts 7 = (A + 9B) ® Lets Keep A - B =A 7 = A (1+i) e for easy aralysis I. Yoo , 15 (5) . Re: < 1amp as per given statement.

fom explaining Randomly How to one can set II = 0.5 but set a constraint on Current. [Pages] I: V5 (5) A (15) A => (VB)(5) <A V2 => 6.1244 <A co A must be greater than 6:1244 wese in worst cases even if VAB line voltage appears at load the Is through load would be in range of on to lamp. Let me choose TA = 10] er 7 = A+A = 10+109 Fload = 10 +10°) frnally the correlate to x = 5400 6 x = 5200 IN E5 L-120 - []z=10+10 0 . 5 Nz=51112014 loi 10+10° = 7 10+10 0.5 >> As load is to choosen balan ced and symmetrical Zx = Zy = Z, is choosen as 10+10 i , =) for balanced System Vectoal has to be Zero" (This is verkied) - in next page)

[Page 6 >x -n. - VERIFICATION OF DESIGN- Aply KVL in Apply KVL N + x "N" is neutral Supply VN = VATTB t te VN = T at Ty the Vue 5L0 +5)-120° + 5)+120 & Tn zo on Applying KvL 5 Lo° = (I) (0.5) + ſ (10+10) کہا نے phase I Tx line drop supply a load drop per phase. 5.*= (0.5 + 10 + 9 10 ) 5 LOT = Í (10.5 + 110 ) . Iz 5 100.5+109) 14:5|443.60 I = 0.3448 | -4360 1 amp that O.1 to of oo (I falls in range is 0.3448 Amps. NLOAD = (0.3448 1-43-60)( 10+101) a clovz) (0.34489 |_43-60745 VLOAD = 3.9341 / +104 are symmetrical and balance we will VY LOAD , VZLOAD as follocus - As loads es . LOAD ] Z LOAD

vxl = 1 Vyl=1zl [Page 7 vY = X = 2,4341 | 1+4 -420 - 3.4541 -118-4 vz 3.9341 , 1.4 +420" = 8.9341 [134.4° Tx + Fy + F z = Vn Vo = CON ADDITIAN) thus entire system is Balancediere - - - - - - - - - - - - - - - - -