Per Unit RepresentationLoad Flow Analysis

Power System StabilityPower Factor Improvement

Ashfaq Hussain

8.1 Per Unit Representation• Sometimes it is more convenient to represent the value of current,

voltage, impedance, power in per unit(pu) value rather than in Amps, Volts, Ohms, kW.

• It is unit less.• It is the ratio of same dimension• Per unit value is the ratio of an actual value to a reference

value(Base value)

• Apu= (Aactual/Abase)

• Some times per unit quantities is mentioned as per cent quantity by multiplying a factor of 100.

pu=actual/base• Apu= (Aactual/Abase)

• Ipu= (Iactual/Ibase)

• Vpu= (Vactual/Vbase)

• Spu= (Sactual/Sbase)

• Ppu= (Pactual/Sbase)

• Qpu= (Qactual/Sbase)

• Zpu= (Zactual/Zbase)

• Rpu= (Ractual/Zbase)

• Xpu= (Xactual/Zbase)

pu=actual/base• If any two of the I, V, Z, S is known, remaining two could be found

out.

• Z= V/I

• Zbase= (Vbase/Ibase)

• Zb= (Vb/Ib)

• Zb= (V2b/Sb)

• Zpu= (Vpu/Ipu)

• Spu= VpuI*pu

Example 8.1• A 5 KVA 400/200 V, 50Hz, single phase transformer has the

primary and secondary leakage reactance each of 2.5 ohm. Determine the total reactance in per unit.

• Sb= 5000 VA

• Primary Base Voltage Vb1 = 400 V

• Secondary Base Voltage Vb2 = 200 V

• X1e=X1+a2X2

• a=N1/N2=400/200=2

• X1e=X1+a2X2=2.5 + 2.5 * 22 =12.5 ohm

Example 8.1• A 5 KVA 400/200 V, 50Hz, single phase transformer has the

primary and secondary leakage reactance each of 2.5 ohm. Determine the total reactance in per unit.

• Xpu= (Xactual/Zbase)

• Zb1= (V2b1/Sb)=4002/5000=32 ohm

• Xpu= (Xactual/Zbase)= 12.5/32 = 0.390625 pu

• X1pu = X1/Zb1

• X2pu = X2/Zb2

Sudden TestA 7 KVA 1000/250 V, 50Hz, single phase transformer has the primary and secondary leakage reactance each of 5 ohm. Determine the total reactance in per unit.

Advantages of per unit representation1. ordinary parameters vary considerably with variation of physical

size, terminal voltage and power rating etc. while per unit parameters are independent of these quantities over a wide range of the same type of apparatus. In other words, the per unit impedance values for the apparatus of like ratings lie within a narrow range.

2. It provide more meaningful information.

3. The chance of confusion between line and phase values in a three-phase balanced system is reduced. A per unit phase quantity has the same numerical value as the corresponding per unit line quantity regardless of the three-phase connection whether star or delta.

4. Impedances of machines are specified by the manufacturer in terms of per unit values.

Advantages of per unit representation5. The per unit impedance referred to either side of a single-phase

transformer is the same.

6. The per unit impedance referred to either side of a three -phase transformer is the same regardless of the connection whether they are ∆-∆, Y-Y or ∆-Y.

7. The computation effort in power system is very much reduced with thee use of per unit quantities.

8. Usually, the per unit quantities being of the order of unity or less can easily be handled with a digital computer. Manual calculation are also simplified. Per unit quantities simply theoretical deduction and give them more generalizes forms.

14.1 Load Flow Analysis1. Load flow analysis is the determination of current, voltage, active

and reactive volt-amperes at various points in a power system operating under normal steady-state or static conditions.

2. This is made to plan the best operation and control of the existing system as well as to plan the future expansion to keep pace with the load growth.

3. Information from load flow studies serve to minimize the system losses and to provide a check on the system stability.

14.2 Load Flow ProblemThe variables:

1. Magnitude of the voltage |Vi|

2. Phase angle of the voltage, δi

3. Active Power, Pi

4. Reactive Volt-amperes, Qi

The Buses:

5. Swing Bus or reference bus or slack bus

6. Generator bus or voltage control bus or PV bus

7. Load bus or PQ bus.

14.3 Bus Admittance Matrix

I1= (V1-V2)/z12 +(V1-V3)/z31

I2= (V2-V1)/z21+(V2-V3)/z32

-I3= (V3-V1)/z31+(V3-V2)/z32

14.2 Bus Admittance Matrix

I1= (V1-V2)/z12 +(V1-V3)/z31

I1= V1/z12 -V2/z12+ V1/z31 - V3/z31

I1= V1/z12 +V1/z31 -V2/z12 - V3/z31

I1= V1y12 +V1y31 -V2y12 - V3y31

I1= (V1-V2)/z12 +(V1-V3)/z31 I1= V1(y12 +y31) -V2y12 - V3y31

I2= (V2-V1)/z21+(V2-V3)/z32 I2= V2(y12 +y23) –V1y12 - V3y23

-I3= (V3-V1)/z31+(V3-V2)/z23 -I3= V3(y31 +y32) –V1y31 – V2y23

14.2 Bus Admittance MatrixI1= (V1-V2)/z12 +(V1-V3)/z31 I1= V1(y12 +y31) -V2y12 - V3y31

I2= (V2-V1)/z21+(V2-V3)/z32 I2=–V1y12 + V2(y12 +y23) - V3y23

-I3= (V3-V1)/z31+(V3-V2)/z23 -I3= –V1y31 – V2y23+ V3(y31 +y32)

I1= (y12 +y31) -y12 - y31

I2= –y12 + (y12 +y23) - y23

-I3= –y31 –y23 +(y31 +y32)

8.2 Bus Admittance Matrix

8.2 Bus Admittance Matrix

Stability• Ability of a system to reach a normal or stable condition after

being disturbed.• Steady state

o Gradual and small disturbance

• Transient o Sudden and large distrubance

Steady State Satbility• Static stability- Automatic and inherent• Dynamic stability- Artificial

Stability LimitStability Limit: the maximum power that can be transferred in a network between sources and load without loss of synchronism.• Steady State Stability Limit (Gradual increase or decrease of

small load )• Transient Stability Limit (Sudden or large disturbance)• Rapidity if disturbance is responsible for loss of synchronism.• Transient Stability Limit is lower than Steady State Stability

Limit

Infinite BUS• A bus that has constant voltage and constant frequency

regardless the load.

Power Angle Curve• Pe= Pe max sin δ; is the power angle

Swing curves δ versus time in seconds is called the swing curve.

Ashfaq Husain

Electrical Power Systems

Capter 26

26.1 Introduction• P is real power, S is Apparent power, V is the line voltage, I is

the line current and cosφ is the power factor.

• P=Scosφ• cosφ = P/S=P/VI• I= P/(Vcosφ)

• Current is inversely proportional to the power factor.• Decrease of power factor will increase the amount of current

26.2 Disadvantages of a low power factor

1. Uneconomical

2. Higher energy loss

3. Poor voltage regulation

26.3 Causes of low power factor

Inductive load make the current lagged behind the voltage. Thus a lagging power factor.

1. Motors: 3φ induction motor operates at 0.8 lagging at full load and 0.2 to 0.3 at light load. 1φ operates at 0.6 lagging/

2. Due to magnetizing current, power factor is low at light load.

3. Arc lamps, electric discharge lamps, industrial heating furnaces, welding equipment operate at low lagging power factor.

26.5 Power factor improvement by static capacitor

• Figure 26.1 and 26.2

26.6 Capacitor rating calculations

• Equation 26.1 to 26.10• Example 26.1 to 26.4

26.8 Advantages and limitations of static capacitor1. Robust

2. Easy to install

3. Occupy little space

4. No need special foundation

5. Practically loss free

6. Easy maintenance

7. Manufactures in very small size

8. Efficient and trouble free solution

9. Over compensation of light load unless special arrangement for automatic switching of the capacitor are provided

26.9 Location of the capacitor

• Connecting separate capacitor for each load.

• Alternative: Installing capacitor bank for a group of load.

• Larger motors needs individual corrections.