UNIT-IV

SYSTEM ANALYSIS
Syllabus:
Voltage drop and power – loss calculations: Derivation for voltage drop and power 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 feeder which are not necessarily in three phase, 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 Two Wire Laterals with Ungrounded Neutral:

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.

............................ (1)
Where the subscripts 1 and 3 refer to the single phase and three phase circuits,
respectively than equation (1) can be written as
 .........(2)

Where 𝑉𝑠 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 – phase lateral. The VD in the three phase lateral can be expressed as

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

And in the single – phase lateral as

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

Where 𝐾𝑅 and 𝐾𝑋 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

𝐾𝑅 = 2.0
𝐾𝑋 = 2.0 when underground cable is used

𝐾𝑋 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

𝑉𝐷1∅
= 2√3 ..................................... (8)
𝑉𝐷3∅

i.e

𝑉𝐷1∅ = 2√3𝑉𝐷3∅ ............................. (9)

Therefore 𝑉𝐷1∅ ungrounded lateral is approximately 3.46 times larger than the one in the
equivalent 3 lateral. Since base voltages for the 1- and 3∅ laterals are

𝑉𝐵(1∅) = √3𝑉𝑆,𝐿−𝑁 ............................... (10)

And

𝑉𝐵(3∅) = 𝑉𝑆,𝐿−𝑁 ......................... (11)

In per unit, it can be expressed as, from equations (8)
𝑉𝐷𝑃𝑈,1∅
= 2......................... (12)
𝑉𝐷𝑃𝑈,3∅

Which means that the pu VD single phase ungrounded lateral is two times larger than the one in
the equivalent three – 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 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 – phase lateral can be expressed as

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

and in the single phase lateral as

 ..................... (5)
Where 𝐾𝑅 = 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, 𝐾𝑅 >2.0 when a reduced capacity
neutral is used, and 𝐾𝑋 2.0 when a overhead line is used. Therefore, if 𝐾𝑅 = 2.0 and 𝐾𝑋 =
2.0 , equation (5) can be 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 laterals with multigrounded 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
𝐼𝑎 = current in phase conductor
𝐼𝑛 = current in neutral wire
𝐼𝑑 = current in ground conductor

The return current in the neutral wire is

𝐼𝑛 = 𝜉1 𝐼𝑎
Where
𝜉1 = 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 𝐾𝑅 is less than 2.0 and constant 𝐾𝑋 is more or less equal to 2.0 because of
conflictingly large 𝐷𝑚 (i.e., mutual geometric mean distance or geometric mean radius, GMR) of
the Carson’s equivalent ground(neutral) conductor.

Therefore

𝑉𝐷𝑝𝑢,1∅ = 𝜉2 × 𝑉𝐷𝑝𝑢,3∅

Where

𝜉2 = 3.8 to 4.2

And

𝑃𝐿𝑆,1∅ = 𝜉3 × 𝑃𝐿𝑆,3∅

Where
 𝜉3 = 3.5 to 3.7

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

And

Where

𝐾𝑅 < 2.0 and 𝐾𝑋 < 2.0

Two- phase plus neutral (open – 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 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 (𝑍𝑎 ) is zero.
The VD in each phase is

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

Where

𝐾𝑅 = 1.0 and 𝐾𝑋 = 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 (𝑍𝑛 ) is larger than
zero.

.......................... (11)

Therefore in this case some unbalanced voltage are inherent.

If the neutral is multi grounded and 𝑍𝑛 > 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 loss is

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

Where

𝐾𝑅 = 3.0 when a full capacity neutral is used and 𝐾𝑅 > 3.0 when a reduced capacity
neutral is used. Therefore, if 𝐾𝑅 = 3.0

................... (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: