Eds Unit-4

 

UNIT-IV SYSTEM ANALYSIS Syllabus:

Voltage drop and power –  power –  loss  loss calculations: Derivation for voltage drop and power p ower loss in lines, manual methods of solution for radial networks, three phase balanced primary lines. Course Objectives:  

To study the Voltage drop and power loss   To Compare non 3 phase systems with balanced 3 phase systems

Learning outcomes:

Students will be  

Able to Calculate Voltage drop & power loss   Able to Analyze various non 3 phase lines. .

Non three phase primary lines:

Usually there are many laterals on a primary p rimary feeder which are not necessarily in three ph phase, ase, for example, single phase which cause the voltage drop (VD) and power loss due to load current not only in the phase conductor but also in the return path.

Fig. 1 various laterals types

Single –  phase  phase Two Wire Laterals with Ungrounded Neutra Neutral: l: Assume that an overloaded single phase lateral is to be changed to an equivalent three phase three wire and balanced lateral, holding the load constant. As the power input to the lateral is the same as before.

Where Whe re the subscr subscript iptss 1

............................ (1) and 3 refer refer to to the the sing single le phase phase and and thre threee phase phase cir circui cuits, ts,

respectively than equation (1) can be written as

 

  .........(2)

Where   is the line to neutral voltage, therefore from equation (2) ..................... (3) This means that the current in the single phase lateral is 1.73 times larger than the one in the equivalent three –  three –   phase phase lateral. The VD in the three phase lateral can be expressed as .............. (4) And in the single –  single –  phase  phase lateral as ........... (5) Where   and   are conversion constants of R and X and are used to convert them from their three phase values to the equivalent single phase values. Where

  = 2.0   = 2.0

when underground cable is used

   2.0

When overhead line is used

Therefore, equation 5 can be rewritten as ................ (6) Or substituting equation (3) into equation (6), we get ............... (7) By dividing equation (7) by equation (3), we get ∅  = 2√3 ..................................... (8) ∅

i.e

1∅ = 2√3∅  ............................. (9) Therefore 1∅ ungrounded lateral is approximately 3.46 times larger than the one in the equival equi valent ent 3 latera lateral. l. Since Since base base voltag voltages es for the 1-

and 3∅ laterals are

(1∅) = √3,− ............................... (10) And

(∅) = ,− ......................... (11)

 

In per unit, it can be expressed as, from equations (8) ,∅  = 2......................... (12) ,∅

Which means that the pu VD single phase ungrounded lateral is two times larger than the one in the equivalent three –  three –  phase  phase lateral. The power losses due to the load current in the conductors of the single phase lateral and the equivalent three phase lateral are

................. (13) And

............... (14) Respectively, substituting equation (3) into equation (13)

...............(15) By dividing equation (15) by equation (14), we get

........................... (16) Therefore the power loss due to the load currents in the conductors of the 1 ∅ laterals is ‘2’ times larger than the one in the equivalent 3∅ lateral. Conclusion:  By changing a 1- ∅ lateral to an equivalent 3-∅ lateral both the per unit VD and the  power loss due to copper losses in the primary lines are approximately halved.

Single phase Two  –  wire  wire uni grounded laterals: In general, this system is presently not used due to the following disadvantages. There is no earth current in this system. It can be compared with a three phase four wire balanced lateral in the following manner. As the power input the lateral is the same as before, ............................ (1)

Or ................... (2) From which ......................... (3) The VD in the three –  three  –  phase  phase lateral can be expressed as ............................ (4) and in the single phase lateral as

 

..................... (5)   Where   = 2.0 when full  –   capacity neutral is used, that is, if the wire size used for the neutral conductor is the same as the size of the phase wire,  >2.0 when a reduced capacity neutral is used, and

  

2.0 when a overhead line is used. Therefore, if

  = 2.0 and   =

2.0 , equation (5) can be b e written as ........................ (6) Or substituting equation (3) into equation (6),

............................. (7) Dividing equation (7) by equation (4), we get

............................. (8) or

......................... (9) Which means that the VD in the single phase two wire unigrounded lateral with full capacity neutral is 6 times larger than the one in the equivalent three phase four wire balanced lateral. The power losses due to the load currents in the conductor of the single phase two wire unigrounded lateral with full capacity neutral and the equivalent three phase four wire balanced lateral are

.......................... (10) and

............................... (11) Respectively. Substituting equation (3) into (10) .................................... (12) And dividing equation (12) by equation (11), we get

............................... (13) Therefore, the power loss due to load currents in the conductors of the single phase two wire unigrounded lateral with full capacity neutral is 6 times larger than the one in the equivalent three phase four wire lateral.

 

Single phase Two  –  wire  wire laterals with multigrounded common common neutrals:  Let the neutral wire is connected in parallel with the ground wire at various places through ground electrodes in order to reduce the current in the neutral. Assume   = current in phase conductor   = current in neutral wire   = current in ground conductor The return current in the neutral wire is

  = 1     Where

1  = 0.25 to 0.33 And it is constant independent of the size of the neutral conductor

Fig.2 single phase two wire lateral with multigrounded common neutral

In figure 2 the constant   is less than 2.0 and constant   is more or less equal to 2.0 because of conflictingly large   (i.e., mutual geometric mean distance or geometric mean radius, GMR) of the Carson’s equivalent ground(neutral) conductor.  conductor.   Therefore

,1∅  = 2  × ,∅  Where

  = 3.8 to 4.2 2

And

,1∅ =   × ,∅  Where

 

 

 = 3.5 to 3.7

The pu VD’s VD’s and the power losses due to load currents can be approximated as

And

Where

  < 2.0 and   < 2.0

Two- phase plus neutral (open  –  wye)  wye) laterals: The neutral conductor can be unigrounded or multigrounded, but because of disadvantages the unigrounded neutral is generally not used. If the neutral is unigriunded, all neutral current conductor itself. Theoretically, it can be expressed that ............................ (1) Where

........................ (2) ............................... (3) It is correct for equal load division between two phases. Assuming equal load division among  phases, the two  –   phase plus neutral lateral can be compared with an equivalent three  –   phase lateral, holding the kilovoltampere load conatant. Therefore

................... (4) Or

................... (5)

 

  Figure3. An open  –  Wye  Wye connected lateral

From which

............................. (6) The voltage drop analysis can be performed depending on whether the neutral is unigrounded or multigrounded. If the neutral is unigrounded and the neutral conductor impedance (  ) is zero. The VD in each phase is

................................ (7) Where

  = 1.0 and   = 1.0 Therefore

..................... (8) Substituting equation (6) into equation (8), we get

............................ (9. a) We have

.............................. (9. b) Dividing equation (9.a) by equation (9.b) side by side,

 

............................. (10) However, if the neutral is unigrounded and the neutral conductor impedance (  ) is larger than zero.

.......................... (11) Therefore in this case some unbalanced voltage are inherent. If the neutral is multi grounded and   > 0, then the pu VD in each phase is ......................... (12)

When a full capacity neutral is used and

......................... (13) When a reduced capacity neutral (i.e., when the neutral conductor employed is one or two sizes smaller than the phase conductors) is used. The power loss analysis also depends on whether the neutral is unigrounded or multigrounded. If the neutral is unigrounded. The power po wer loss is

........................ (14) Where

  = 3.0 when a full capacity neutral is used and neutral is used. Therefore, if   = 3.0

  > 3.0 when a reduced capacity

................... (15) Or

............................... (16) On the other hand, if the neutral is multigrounded,

................................ (17)

 

Therefore the approximate value of this ratio is

................................ (18) Which means that the power loss due to load currents in the conductors of two phase three wire lateral with multigrounded neutral is approximately 1.64 times larger than the one in the equivalent three phase lateral.

 

Calculation of load power factor for which voltage drop is maximum: