In the per-unit (pu) notation, any electrical quantity may be expressed in pu as the ratio of the actual quantity and the chosen base value for the quantity expressed either in decimal form or percent. Four basic quantities must be considered, namely, power Sb, voltage Vb, current Ib, and impedance Zb.
In single phase systems, the relationships among these quantities are:
Sb = VbIb
Vb = IbZb
With only two equations relating the four basic quantities, it is necessary to specify two base values. The power and voltage bases are usually chosen equal to the rated value and the other two are computed from the above equations as follows.
The specified power base is applicable to all parts of the power system. The voltage base varies across a transformer and so do the current base and impedance base. The pu electrical quantities are calculated as follows:
Su =
It can be seen by inspection of any power system diagram that:
a. Several voltage levels exist in a system
b. It is common practice to refer to plant MVA in terms of per unit or percentage values c. Transmission line and cable constants are given in ohms/km.
Before any system calculation can take place, the system parameters must be referred to ‘base quantities’ and represented as a unified system of impedances in the either ohmic, percentage, or per unit values.
The base quantities are power and voltage. Normally, they are given in terms of the three-phase power in MVA and the line voltage in kV. The base impedance resulting from the above base quantities is:
and, provided the system is balanced, the base impedance may be calculated using either single-phase or three-phase quantities.
The per unit or percentage value of any impedance in the system is the ratio of actual to base impedance values.
Hence:
Simple transposition of the above formulae will refer the ohmic value of impedance to the per-unit or percentage values and base quantities.
Having chosen base quantities of suitable magnitude all system impedances may be converted to those base quantities by using the equations given below:
where suffix b1 denotes the value to the original base and suffix b2 denotes the value to the new base
The choice of impedance notation depends upon the complexity of the system, plant impedance notation and the nature of the system calculations envisaged.
If the system is relatively simple and contains mainly transmission line data, given in ohms, then the ohmic method can be adopted with advantage. However, the per-unit method of impedance notation is the most common for general system studies since:
1. Impedances are the same referred to either side of a transformer if the ratio of base voltages on the two sides of a transformer is equal to the transformer turns ratio
2. Confusion caused by introduction of powers of 100 in percentage calculations is avoided
3. By a suitable choice of bases, the magnitudes of the data and results are kept within a predictable range, and hence errors in data and computations are easier to spot.
Most power system studies are carried out using software in per unit quantities. Irrespective of the method of calculation, the choice of base voltage, and unifying system impedances to this base, should be approached with caution, as shown in the following example (Fig. 3.6).
Wrong selection of base voltage
11.8kV 132kV 11kV
Right selection
11.8kV 141kV
132
141
x 11 = 11.7kV Fig 3.6 Selection of base voltagesWORKED EXAMPLE
A 500KVA, 6.6kv alternator is ∆ connected. Calculate the current in each phase of the alternator when it delivers its full output to a balanced 3-φ load at a power factor of 0.8 lagging.
Solution.
Power supplied by the alternator P =
3 V
LI
Lcos ϕ
A balanced 3-φ load is connected to 2.2 kV feeder, the load draws a line current of 60A at a P.f of 0.9 lagging. Calculate
VL = 2200V IL = 60A
P.F = 0.9 lag
⇒
φ = -cos-1(0.9) = -25.80 P =3
VLIL cosφ = 205.8KW Φ =3
VLIL sinφ = 99.7KVAR S =3
VLIL = 228KVA WORKED EXAMPLEA 3-φ 1.5 KW, unity power factor; rating unit, and a 5 horse power induction motor with a full load efficiency of 80%
and P.f of 0.85 are served from the same 3Q, 3 wire 208v system. Find the magnitude of the line current for rated O/P from the 5HP motor. Note that 1 HP = 746W.
Solution
Motor Output = 5x746=3730w Motor Input =
η
3730
=4662 W 8
. 3730 = 0
The motor is a balanced 3-φ load so that P =
3
VLIL cos φ4662 =
3
x 208 x IL x 0.85 IL =15 . 25 A
85 . 0 208 3
4662 =
×
×
Ф = cos-1 (0.85) = 31.70
Hence, the motor line current is IL = 15.25
∠
-31.70A For the heater P =3
VLIL1500 =
3
x 208 x ILILH =
4 . 16 0
0A 208
3
1500 = ∠
×
cosφ = 1Therefore, total line current is the phasor sum of motor and the later line current.
IL = 15.25
∠
-31.70 + 416∠
00IL = 18.9
∠
-25.10A WORKED EXAMPLEA ∆- connected load consists of three identical impedance of 8+j6w each and supplied from a three phase 200V source.
Calculate (a) the power factor (b) phase current and line current (c) the real, reactive and apparent power taken by the load.
Solution
Z ∆ = 8 + j6 =10
∠
36.90 Ω VL = Vph. = 200∠
00 (a) pf = cos36.90 = 0.8 lagging(b) Iph = 0
0
9 . 36 10
0 200
∠
= ∠
A ph
Z
V
Iph = 20∠
-36.90IL =
3
Iph∠
-300 = 34.6∠
-66.90A(c) P = 3VphIph cosφ
= 3 x 20 x 200 x 0.8 = 9600W
QL = 3Vph x Iph sin = 3 x 20 x 200 x sin(36.90) QL = 7200VAR
SL = 3VphIph = 3 x 200 x 20 = 7200 kVA
PL =
3
VLIL cosφ = 96 kW Q =3
VLIL sinφ = 7.2 kVAR S =3
VLIL = 12kVA If the impedance is Y- connected we have Zy = 8 + j6Ω VL = 200∠
0o VWORKED EXAMPLE
A generator supplied a load through a feeder whose impedance is Zf = 1+j202. the load impedance ZL = 8+j602 the voltage across the load is 120V. find the real power and reactive power supplied by the generator using per unit representation. Take the load voltage as the reference phasor and choose Sb = 1500VA, Vb = 120V
Solution
The real voltage for generator is 1.204