Question

6. Short Questions (10 x 5 = 50) a. Why 39 AC system is used in power transmission, why not 10, 20, 49, 50, 60 or any other combination of AC systems? b. Mention four factors to control power flow to the load bus. Why two of these are controllable and the other two are not? c. Suppose you need to transmit 300 MW of power between locations A and B, which are 150 km apart. What level of transmission voltage would you recommend and why? d. Draw a circle diagram and show and explain briefly the different factors that limit maximum power that can be transferred to a load. e. Explain the concept of decomposing an unbalanced system voltage into three sets: positive, negative and zero sequence.

Answer 1

for part (e) we decompose into sequence components to solve the unbalanced faults in power system thats known as fortseque theorem which represents into sum of 3 balanced phasors

Basically any number of phase possible but only 3% supply is transmitted usually because of many reasons as follows I since even number phases like 2,916 they have to be evenly distributed ใก space they will have equal angle in them and as such pair of phases in such system will always be in opposite direction to each other this 2 1 R losses means for neutral return wire į but in 3d power system none of the phases are in exact phase opposition and hence current through neutral is lower than previous case Further 51719. all the odd ones are possible but there isn't any significant increase in the current so want to increase the cost by increasing phase ₂ we don't cscanned with CamScanner

b 3-phase is already ideal in the sum of power output from rom all the three phases is constant means motors are all being driven with constnart torque without AC frequency Components. L 3-phase electricity can easily iged for generate rotating magnetic field the 4 Factors to control the power flow to load bus Real power (p) Reactive power (0) Voltage at bus (u) power angle (8) only two at a time are controllable because applying power balance equation at each node (bus) we got only two equations per node remaining two are 'evalerated using loved CS$canned with CamScanner

L also since the power flow equations are non-linear as they involve sinusoidal sources the system of such type equations can not be solved in one step С. can be For 300mW power to be transferred at for 150km long line a double cirwit 220kV AC line used because higher the the voltage level lower the transmission losses and above the 220kV the voltage af receiving end need to be step down so that is difficult above 22070 medium length line 22030 is prefer so for CScanned with CamScanner

Assume a medium line for power fronsmission Vrloo NsLS A 1 > Is C D 7 I r So Ps tj Og Sr - Prtjar By solving ABCT besed V3, I3 IAI/V21 CR-4)- //</a (6+8) (3 ། 1B1 28 JAV21 8in18-d) 131 transfer power Nellur Sin (8+8) BI for marm dps 8 O I us! I VR! 3in18+8) - 1B 1 18+8=.*) CSScanned with CamScanner

hence Pmax. Talli?! Cos (8-4) + 1 vs 1 IVA/ | BI ། 3 ། BO maximum power transfer depends on Vsi impedance angle (B), power angle of A - ALQ° and Bl MUAR MW Ng VR B positive, negative and zero sequence Components are parts of unbalanced System these are defined es CSscanned with CamScanner

ปคม VR2 ATRO - tyo Woo NB2 Yi Vy2 Noi (zero) (-ne (the any phase faults voltage is made of ale three components. And for unbalanced usually we decompose the into sequence network theet n phasors can be represented as the Sum of n balanced Components ( Fortsquels theorem ) CS$canned with CamScanner