Transcript Line Diagrams And The Per Unit System

Line Diagrams And The Per Unit System
Line Diagrams
• In power engineering, one-line diagram or single-line diagram
is a simplified notation for representing a three-phase power
system. Electrical elements such as circuit breaker,
transformer, capacitors, bus bars, and conductors are shown by
standized schematic symbols. Instead of representing each of
three phases with a separate line or terminal, only one
conductor is represented. It is a form of block diagram
graphically depicting the paths for power flow between entities
of the system. Elements on the diagram do not represent the
physical size or location of the equipment. An important point
to note is that the diagram often represents the application, i.e.
if load flow is being depicted then protective devices may not
be shown
Region - 3
2
Region - 1
Region 2
2
1
T2
L1
T2
1
2
L1 = impedance
R1 = 5Ω
X1 = 20Ω
G1 Rating;
30 MVA
13.8KV
R = 0.1 pu
X = 1.0 pu
T1 Rating;
35 MVA
13.2/115KV
R = 0.01 pu
X = 0.10 pu
T2 Rating;
30 MVA
120/12.5KV
R = 0.01 pu
X = 0.08pu
M1 Rating;
20 MVA
12.5KV
R = 0.1 pu
X = 1.1 pu
M2 Rating;
10 MVA
12.5KV
R = 0.1 pu
X = 1.1 pu
The Per-Unit System
• In the analysis of power network, instead of using actual values of
quantities it is usual to express them as fractions of reference
quantities, such as rated or full load values. These fractions are
called per unit ‘p.u.’ the ‘p.u.’ value of any quantity is defined as :
𝐴𝑛𝑦 𝑣𝑎𝑙𝑢𝑒 (𝑖𝑛 𝑎𝑛𝑦 𝑢𝑛𝑖𝑡)
𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑝. 𝑢 =
𝐵𝑎𝑠𝑒 𝑜𝑟 𝑅𝑒𝑓𝑟𝑒𝑒𝑛𝑐𝑒 𝑉𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑢𝑛𝑖𝑡
The ‘p.u.’ value can also be expressed as a percentage of the base or
reference value. The per-unit system is widely used in the power
system industry to express values of voltages, currents, powers, and
impedances of various power equipment. It is mainly used for
transformers and AC machines.
The advantages of using the ‘p.u.’
i.
The apparatus considered may vary widely in size, losses and volt
drops will also vary considerably. For apparatus of the same general
type the ‘p.u.’ voltage drops and losses are in the same order regardless
of size.
ii.
By choice of the appropriate voltage bases the solution of networks
containing several transformer is facilitated.
iii. When values are expressed in ‘p.u.’ the comparison of electrical
quantities with their “normal” values is straightforward.
iv. Calculations are simplified because quantities expressed as per-unit are
the same regardless of the voltage level.
Base Apparent Power ‘ SB’
The base apparent power, or base volt-amperes, represents the following
equation :
𝑆𝐵 = 3 × 𝑉𝐵 × 𝐼 𝐵
where, ‘VB’ = Line Voltage in a 3 phase System
‘IB’ = Line/Phase Current in a 3 phase System
Using the above equation for base apparent power it is possible to derive the
base current and base voltage;
𝐼𝐵 =
𝑆𝐵
3×𝑉𝐵
Once the base current is defined then the base impedance is given
by;
• 𝑍𝐵 =
𝑉𝐵
3
𝐼𝐵
• 𝑍𝐵 =
• 𝑍𝐵 =
• 𝑍𝐵 =
𝑉𝐵
3
𝑆𝐵
3×𝑉𝐵
𝑉𝐵
3
𝑆𝐵
3×𝑉𝐵
𝑉𝐵2
𝑆𝐵
Hence, the per-unit impedance of an electrical item of equipment in the
power system is given by:
𝑍𝑝𝑢 =
𝑧Ω
𝑍𝐵
𝑍𝑝𝑢 =
𝑍Ω
𝑉𝐵2
𝑆𝐵
𝑍𝑝𝑢 =
𝑍Ω ×𝑆𝐵
𝑉𝐵2
In load flow analysis, where the base apparent power may be different to
that of the equipment rating, it is necessary to convert the per-unit
impedance referenced to the ‘old base’ to the ‘new base’ per unit
impedance, using the base apparent power selected for the whole power
system. This conversion is achieved using the equation:
𝑍𝑝. 𝑢. 𝑛𝑒𝑤 = 𝑍𝑝. 𝑢. 𝑔𝑖𝑣𝑒𝑛
𝑉𝑔𝑖𝑣𝑒𝑛
𝑉𝑛𝑒𝑤
2
𝑆𝑛𝑒𝑤
𝑆𝑔𝑖𝑣𝑒𝑛
In some instances the per-unit impedance referenced to the ‘old base’
Is given as a percentage. When this is the case it is simple to convert the
given percentage to its per-unit equivalent using ‘Zpu = Z%/100%.
Transformer PU
In an ideal transformer the following equation apply :
|𝑉1|
𝑉2 =
& 𝐼2 = 𝑛|𝐼1|
𝑛
Where n is the turn’s ration of the transformer 𝑛 =
𝑁1
𝑁2
Similarly
|𝑉𝑏1|
𝑉𝑏2 =
𝑛
|𝑆𝑏|
|𝑆𝑏|
𝐼𝑏2 =
= 𝑛.
= 𝑛|𝐼𝑏1|
|𝑉𝑏2|
|𝑉𝑏1|
Dividing the voltage relation of the ideal transformer by Vb2 and
substituting Vb2 = Vb1/n results in :
|𝑉2|
|𝑉1|
|𝑉1|
=
=
|𝑉𝑏2| 𝑛|𝑉𝑏2| |𝑉𝑏1|
Similarly it can be shown that dividing the current relationship of the
ideal transformer by the base current Ib2 and substituting Ib2 = nIb1
results in :
|𝐼2|
|𝐼1|
|𝐼1|
= 𝑛.
=
|𝐼𝑏2|
|𝐼𝑏2|
|𝐼𝑏1|
Exercise 1 :
A certain power element in a series circuit as impedance of 4+ J7Ω, if the
output of the element is 5..773KV at a current of 150 at -36.9 degrees,
determine the input voltage using conventional circuit theory, by phase
or diagram and by using p.u. method. (assume Sbase = 106).
Answer Vs.p.u = 1.19 + j1.19/V
Region 3
Region 1
Region 2
2
2
1
G1
T2
L1
T2
1
L1 impedance
R=5Ώ
X = 20Ώ
T1 Rating;
35 MVA
13.2/115KV
R = 0.01 pu
X = 0.10 pu
T2 Rating;
30 MVA
120/12.5KV
R = 0.01 pu
X = 0.08pu
2
M1 Rating;
20 MVA
12.5KV
R = 0.1 pu
X = 1.1 pu
M2 Rating;
10 MVA
12.5KV
R = 0.1 pu
X = 1.1 pu