Per-Unit System, Load Flow, Voltage Control and Economic Operation
What to remember
- Per-unit value = actual value ÷ base value. Base MVA is common to the whole system; base kV changes only across transformers. Transformer impedance in per-unit is the same on both sides.
- Load flow solves for voltage magnitude and angle at every bus. Buses are slack, PV (generator) or PQ (load). Newton-Raphson converges fastest; Gauss-Seidel is simplest.
- Economic operation = equal incremental cost. Without losses, all units run at the same incremental cost λ. With losses, each unit also carries a penalty factor.
Per-unit system
In the per-unit (pu) system every quantity is written as a fraction of a chosen base. Two bases are chosen freely (base MVA and base kV). The other two follow from them.
- Base current (three-phase) Ibase = Sbase / (√3 × Vbase), where Sbase is three-phase MVA and Vbase is line-to-line kV.
- Base impedance Zbase = (kVbase)² / MVAbase (ohm).
- Z(pu) = Z(actual) / Zbase. Voltage (pu) = V / Vbase. Power (pu) = S / Sbase.
Change of base. A machine rating gives Z in pu on its own rating. To refer it to a new base:
Z(new) = Z(old) × (MVA new / MVA old) × (kV old / kV new)²
Why use per-unit?
- Transformer impedance has the same pu value whether seen from the high-voltage or low-voltage side, so ideal transformers vanish from the diagram.
- Pu values of machines of the same type lie in a narrow range, so errors are easy to spot.
- Calculations become simple, and the √3 factors are avoided.
Worked example 1. Base 100 MVA, 11 kV. Zbase = 11² / 100 = 1.21 ohm. A reactance of 0.605 ohm is 0.5 pu.
Worked example 2. A generator is rated 50 MVA, 11 kV with X = 0.2 pu. On a 100 MVA, 11 kV base, X = 0.2 × (100/50) = 0.4 pu.
Worked example 3. A transformer of 0.1 pu on 50 MVA, 20 kV is moved to a base of 100 MVA, 10 kV. X = 0.1 × 2 × (20/10)² = 0.8 pu.
Choosing bases in a system with transformers. Pick one MVA base for the whole system. Pick the kV base in one zone, then find the other zones by the transformer turn ratio (line-to-line voltage ratio). Convert each equipment value to this base before drawing the pu impedance diagram.
Load flow (power flow) studies
A load flow study finds the steady-state voltages (magnitude and angle), line flows and losses for a given generation and load pattern. It is used for planning, operation and as the starting point for fault and stability studies.
Bus classification
| Bus type | Specified | Unknown | Notes | ||
|---|---|---|---|---|---|
| Slack (swing, reference) | V | and angle δ | P and Q | One per system; takes up the losses | |
| PV (generator, voltage-controlled) | P and | V | Q and δ | Q limits of the machine apply | |
| PQ (load) | P and Q | V | and δ | Most buses are of this type |
Bus admittance matrix (Ybus). Diagonal element Yii is the sum of all admittances connected to bus i. Off-diagonal element Yij is the negative of the admittance between buses i and j. Ybus is square, symmetric (no phase shifters) and sparse for large systems. The equation is I = Ybus × V.
Methods
| Method | Key feature |
|---|---|
| Gauss-Seidel | Simple; uses Ybus; slow, needs many iterations; iterations grow with bus number |
| Newton-Raphson (NR) | Uses Jacobian matrix; quadratic convergence; few iterations; best for large systems |
| Fast decoupled (FDLF) | Decouples P-δ and Q-V; constant matrices; very fast; works when R/X is small |
In NR the Jacobian is made of partial derivatives of P and Q with respect to δ and |V|. P is strongly coupled to δ and Q to |V|; the fast decoupled method exploits the weak P-|V| and Q-δ coupling, while full NR keeps all four Jacobian blocks. Real power mainly follows the angle difference; reactive power mainly follows the voltage magnitude difference.
Acceleration factor. In Gauss-Seidel, an acceleration factor (about 1.6) speeds convergence.
Outputs. Bus voltages, power flow on each line, losses, and the reactive power the generators must supply.
Voltage control
Voltage must stay within a small band of the rated value for consumers. Reactive power flow is the main reason for voltage change. A lagging load draws reactive power and lowers voltage.
Voltage regulation = (Vno-load − Vfull-load) / Vfull-load × 100%. If the no-load voltage is 10.5 kV and the full-load voltage is 10 kV, regulation is 5%.
Ferranti effect. In a long lightly loaded line, the receiving-end voltage rises above the sending-end voltage, because of line charging current flowing through the line inductance.
Methods of voltage control
| Device | Action |
|---|---|
| Shunt capacitor | Supplies lagging VAr; raises voltage; also improves power factor |
| Shunt reactor | Absorbs VAr; lowers voltage on lightly loaded long lines |
| Series capacitor | Cancels part of line reactance; reduces voltage drop |
| Synchronous condenser | Over-excited: supplies VAr; under-excited: absorbs VAr; smooth control |
| On-load tap changer (OLTC) | Changes turns ratio under load; used in transformers at substations |
| Booster transformer | Injects a voltage in series with the line |
| SVC (static VAr compensator) | Thyristor-controlled reactor plus capacitor; fast, stepless control |
| STATCOM | Voltage source converter based; faster than SVC; output current does not fall with voltage |
| Automatic voltage regulator (AVR) | Controls generator excitation |
Key points
- Reactive power from a capacitor is proportional to V². At 0.9 pu voltage a 100 kVAr capacitor gives 81 kVAr.
- Tap changers work in steps. A ±10% range in 1.25% steps gives 8 steps on each side of the nominal tap.
- Tap changing controls voltage but does not generate reactive power. It only shifts the problem to another part of the network.
- Excitation control affects generator terminal voltage and the reactive power it delivers.
- Voltage control and reactive power are linked; real power is controlled by the turbine governor and frequency.
Economic operation of power systems
Fuel cost curve. The cost of a thermal unit is usually written as C = a + bP + cP² (Rs/h), where P is the output in MW.
Incremental cost (IC). IC = dC/dP (Rs/MWh). For C = 0.1P² + 20P, IC = 0.2P + 20.
Economic dispatch without losses. For minimum total cost with a fixed demand, all units that are not at limits operate at equal incremental cost: dC1/dP1 = dC2/dP2 = … = λ. Also ΣP = demand.
Worked example. Two units with IC1 = 0.2P1 + 20 and IC2 = 0.2P2 + 30; demand 200 MW.
Set 0.2P1 + 20 = 0.2(200 − P1) + 30. So 0.4P1 = 50, P1 = 125 MW, P2 = 75 MW, λ = 45 Rs/MWh.
With losses (coordination equation). dC/dP × L = λ, where L = 1 / (1 − ∂PL/∂PG) is the penalty factor. If ∂PL/∂PG = 0.2, penalty factor = 1/0.8 = 1.25. A plant far from the load has a higher penalty factor and is loaded less.
Transmission loss formula (B-coefficients). PL = Σ Pi Bij Pj. The loss coefficients are found from load flow data.
Generator limits. Each unit has Pmin and Pmax. A unit that reaches a limit is held there and the remaining demand is shared by the others at equal λ.
Related concepts
- Unit commitment: deciding which units are on or off for each hour, considering start-up cost, minimum up/down time and reserve. A simple method is the priority list (switch on the cheapest full-load cost unit first).
- Spinning reserve: extra capacity of units already running and synchronised, ready to take sudden load.
- Hydrothermal scheduling: water is limited, so hydro units are scheduled to save fuel in thermal units. Run-of-river plants run as base load.
- Load duration curve: hours for which load equals or exceeds a value; used to plan base, intermediate and peak plants.
- Merit order: low-cost plants (hydro, nuclear, efficient thermal) are loaded first.
- AP-specific: the state's generation is shared among thermal stations, hydro projects such as Srisailam, and solar and wind parks. Dispatch must respect water release rules and grid code limits.
Exam traps
- Base MVA is common to the whole system; base kV changes across a transformer. Do not mix them up.
- Zbase = kV² / MVA, not kV / MVA.
- Transformer pu impedance is the same on both sides; actual ohm values are not.
- Slack bus has |V| and δ specified; PV bus has P and |V| specified. P and |V| at PV and P and Q at PQ are easy to swap.
- Newton-Raphson has quadratic convergence; Gauss-Seidel has linear convergence.
- Capacitor VAr varies as V², not as V.
- An OLTC changes voltage, not the generated reactive power.
- Equal incremental cost is correct only when losses are ignored and limits are not hit.
- Penalty factor greater than 1 means the plant contributes more to losses.
- Ferranti effect needs a light load or open-ended long line; it is not a heavy-load effect.
- Shunt reactor lowers voltage; shunt capacitor raises it.
- Unit commitment decides on/off status; economic dispatch decides the output of running units.
One-liners
- 1. Per-unit value = actual value / base value; it has no unit.
- 2. Ibase = Sbase / (√3 Vbase) for a three-phase system.
- 3. Zbase = (kV)² / MVA in ohm.
- 4. Z(new pu) = Z(old pu) × (MVA new / MVA old) × (kV old / kV new)².
- 5. The slack bus supplies the system loss and provides the angle reference.
- 6. A load bus is a PQ bus; a generator bus is a PV bus.
- 7. Ybus diagonal element is the sum of connected admittances.
- 8. Newton-Raphson needs the Jacobian matrix and converges quadratically.
- 9. Fast decoupled load flow uses weak P-V and Q-δ coupling.
- 10. Shunt capacitors supply lagging reactive power and raise voltage.
- 11. Synchronous condenser is an over- or under-excited motor running with no mechanical load.
- 12. Economic dispatch: equal incremental cost; with losses, equal ICs multiplied by penalty factors.
Practice questions
What is the base impedance, in ohm, for a base of 33 kV and 100 MVA?
- 3.3
- 10.89
- 108.9
- 0.33
Answer
B. 10.89
Zbase = kV²/MVA = 1089/100 = 10.89 ohm.
What is the base current for a three-phase base of 100 MVA and 11 kV (line-to-line)?
- 1.75 kA
- 5.25 kA
- 9.09 kA
- 3.03 kA
Answer
B. 5.25 kA
Ibase = 100/(√3 × 11) = 5.25 kA.
A generator of 50 MVA, 11 kV has a reactance of 0.2 pu. What is its reactance on a base of 100 MVA, 11 kV?
- 0.4 pu
- 0.2 pu
- 0.8 pu
- 0.1 pu
Answer
A. 0.4 pu
X = 0.2 × (100/50) = 0.4 pu.
A transformer reactance is 0.1 pu on 50 MVA, 20 kV. What is it on a base of 100 MVA, 10 kV?
- 0.2 pu
- 0.4 pu
- 0.8 pu
- 0.05 pu
Answer
C. 0.8 pu
X = 0.1 × (100/50) × (20/10)² = 0.8 pu.
On a base of 100 MVA and 10 kV, an actual impedance of 5 ohm is equal to:
- 0.05 pu
- 50 pu
- 0.5 pu
- 5 pu
Answer
D. 5 pu
Zbase = 10²/100 = 1 ohm, so Z = 5/1 = 5 pu.
A bus voltage is 0.95 pu on a 220 kV base. The actual line-to-line voltage is:
- 190 kV
- 231 kV
- 220.95 kV
- 209 kV
Answer
D. 209 kV
V = 0.95 × 220 = 209 kV.
A system has 10 buses with one slack bus and three generator (PV) buses. How many load (PQ) buses are there?
- 7
- 5
- 6
- 3
Answer
C. 6
PQ buses = 10 − 1 − 3 = 6.
How many elements does the bus admittance matrix of a 4-bus system have?
- 4
- 16
- 8
- 12
Answer
B. 16
Ybus is a 4 × 4 matrix, so it has 16 elements.
A transformer has no-load secondary voltage of 10.5 kV and full-load secondary voltage of 10 kV. Its voltage regulation is:
- 5%
- 0.5%
- 4.76%
- 10%
Answer
A. 5%
Regulation = (10.5 − 10)/10 × 100 = 5%.
A transformer has taps of ±10% in steps of 1.25%. How many steps are there on each side of the nominal tap?
- 4
- 10
- 16
- 8
Answer
D. 8
10 ÷ 1.25 = 8 steps.
A shunt capacitor rated 100 kVAr at 1 pu voltage is operated at 0.9 pu voltage. The reactive power supplied is:
- 100 kVAr
- 110 kVAr
- 81 kVAr
- 90 kVAr
Answer
C. 81 kVAr
Q varies as V²: 100 × 0.81 = 81 kVAr.
Two units have IC1 = 0.2P1 + 20 and IC2 = 0.2P2 + 30 (Rs/MWh). For a total demand of 200 MW with no losses, the economic output of unit 1 is:
- 100 MW
- 125 MW
- 75 MW
- 150 MW
Answer
B. 125 MW
Equal IC: 0.2P1 + 20 = 0.2(200 − P1) + 30 gives P1 = 125 MW.
For a unit with C = 0.02P² + 10P + 100 (Rs/h), the incremental cost at P = 50 MW is:
- 12 Rs/MWh
- 10 Rs/MWh
- 13 Rs/MWh
- 11 Rs/MWh
Answer
A. 12 Rs/MWh
IC = 0.04P + 10 = 0.04 × 50 + 10 = 12.
If ∂PL/∂PG = 0.2 for a plant, its penalty factor is:
- 0.8
- 0.2
- 1.25
- 1.2
Answer
C. 1.25
PF = 1/(1 − 0.2) = 1.25.
In the per-unit system, a quantity is expressed as:
- Actual value multiplied by base value
- Actual value minus base value
- Base value divided by actual value
- Actual value divided by base value
Answer
D. Actual value divided by base value
Per-unit = actual/base; it is dimensionless.
The base impedance in ohm is given by:
- (kV base)² / MVA base
- kV base / MVA base
- MVA base / (kV base)²
- (MVA base)² / kV base
Answer
A. (kV base)² / MVA base
Zbase = kV²/MVA.
The per-unit impedance of a transformer, when the bases follow the voltage ratio, is:
- Higher on the low-voltage side
- The same on the primary and secondary sides
- Higher on the high-voltage side
- Zero on the secondary side
Answer
B. The same on the primary and secondary sides
Per-unit impedance is unchanged by the turns ratio.
In a load flow study, which bus has its voltage magnitude and angle specified?
- Any load bus
- Slack bus
- PV bus
- PQ bus
Answer
B. Slack bus
The slack bus is the reference; |V| and δ are fixed, and P and Q are found.
At a PV bus, which quantities are specified?
- Real power and voltage magnitude
- Real and reactive power
- Reactive power and angle
- Voltage magnitude and angle
Answer
A. Real power and voltage magnitude
A generator bus has P and |V| fixed; Q and δ are computed.
At a load (PQ) bus, the unknown quantities are:
- Real power and angle
- Reactive power and magnitude
- Voltage magnitude and angle
- Real and reactive power
Answer
C. Voltage magnitude and angle
P and Q are given; |V| and δ are found.
The diagonal element Yii of the bus admittance matrix equals:
- The sum of all admittances connected to bus i
- The negative of the admittance between buses i and j
- The reciprocal of the bus impedance
- The shunt admittance at bus i only
Answer
A. The sum of all admittances connected to bus i
Yii sums every admittance terminating at bus i; Yij is the negative of the admittance between i and j.
Which load flow method has quadratic convergence?
- Gauss-Seidel with acceleration factor
- Direct substitution
- Gauss-Seidel
- Newton-Raphson
Answer
D. Newton-Raphson
Newton-Raphson converges quadratically near the solution.
Fast decoupled load flow is based on the weak coupling between:
- P and Q
- P and angle, and Q and voltage magnitude
- Voltage and current
- P and voltage magnitude, and Q and angle
Answer
D. P and voltage magnitude, and Q and angle
P is mainly linked to angle and Q to magnitude; the cross couplings are weak and neglected.
The Ferranti effect refers to:
- Skin effect in conductors
- A rise in receiving-end voltage in a long, lightly loaded line
- A fall in receiving-end voltage in a heavily loaded line
- Loss of power by corona
Answer
B. A rise in receiving-end voltage in a long, lightly loaded line
Line charging current through the inductance raises the receiving-end voltage.
A shunt reactor is used mainly to:
- Improve stability by adding real power
- Reduce fault level to zero
- Lower voltage on lightly loaded long lines
- Raise voltage at heavy load
Answer
C. Lower voltage on lightly loaded long lines
A shunt reactor absorbs reactive power and so lowers voltage.
A synchronous condenser absorbs reactive power when it is:
- Over-excited
- Delivering real power
- Under-excited
- Running at no field
Answer
C. Under-excited
An under-excited machine takes lagging VAr (absorbs reactive power); over-excited supplies it.
Which of the following is a thyristor-based fast reactive power compensator?
- Booster transformer
- Static VAr compensator
- Buchholz relay
- Peterson coil
Answer
B. Static VAr compensator
An SVC uses a thyristor-controlled reactor with capacitors.
Voltage drop in a transmission line is mainly governed by:
- The flow of reactive power
- Frequency
- Corona loss
- The flow of real power
Answer
A. The flow of reactive power
Because X is much larger than R, the drop depends mostly on Q.
The condition for economic load dispatch without losses is that:
- All units share equal load
- All units run at full load
- All units have equal efficiency
- All units operate at equal incremental cost
Answer
D. All units operate at equal incremental cost
Total cost is minimum when dC/dP is the same for all units not at limits.
The penalty factor of a plant is given by:
- 1 / (1 + ∂PL/∂PG)
- ∂PL/∂PG
- 1 − ∂PL/∂PG
- 1 / (1 − ∂PL/∂PG)
Answer
D. 1 / (1 − ∂PL/∂PG)
Penalty factor L = 1/(1 − ∂PL/∂PG).
Unit commitment is the process of deciding:
- Which generating units are on or off at each hour
- The exact output of each running unit in real time
- The tap setting of transformers
- The bus voltages
Answer
A. Which generating units are on or off at each hour
Unit commitment is the on/off schedule; dispatch decides outputs.
Spinning reserve is:
- Water stored in a dam
- Spare capacity in running synchronised units
- Capacity of cold-standby units
- Reserve of coal at a station
Answer
B. Spare capacity in running synchronised units
Spinning reserve is available immediately from units already on the bus.
Statements on the per-unit system: 1. The base MVA is the same for the whole system. 2. The base kV is the same on both sides of a transformer.
- 1 only
- 2 only
- Both 1 and 2
- Neither 1 nor 2
Answer
A. 1 only
Base kV changes across a transformer in the ratio of its voltages; base MVA is common.
Statements on bus types: 1. The slack bus has real and reactive power specified. 2. A PQ bus has voltage magnitude as an unknown.
- 1 only
- Both 1 and 2
- Neither 1 nor 2
- 2 only
Answer
D. 2 only
The slack bus has |V| and δ specified, so statement 1 is wrong; a PQ bus has |V| unknown.
Statements on Newton-Raphson load flow: 1. It has quadratic convergence. 2. It uses a Jacobian matrix.
- 1 only
- 2 only
- Both 1 and 2
- Neither 1 nor 2
Answer
C. Both 1 and 2
Both are true; it needs few iterations, each costing more computation.
Statements on voltage control: 1. A shunt capacitor raises voltage. 2. A shunt reactor raises voltage.
- 2 only
- 1 only
- Both 1 and 2
- Neither 1 nor 2
Answer
B. 1 only
A shunt reactor absorbs reactive power and lowers voltage.
Statements on economic dispatch: 1. Without losses, units are loaded at equal incremental cost. 2. With losses, the penalty factor is included.
- Both 1 and 2
- 1 only
- 2 only
- Neither 1 nor 2
Answer
A. Both 1 and 2
Both are correct; the coordination equation is IC × penalty factor = λ.
Statements on the bus admittance matrix: 1. It is symmetric when there are no phase-shifting transformers. 2. It is dense for large practical systems.
- 2 only
- Both 1 and 2
- Neither 1 nor 2
- 1 only
Answer
D. 1 only
Ybus is sparse for large systems, since most buses connect to only a few others.
Statements on tap-changing transformers: 1. An OLTC changes the voltage ratio while on load. 2. An OLTC generates reactive power.
- 2 only
- Both 1 and 2
- 1 only
- Neither 1 nor 2
Answer
C. 1 only
An OLTC redistributes voltage but does not generate reactive power.
Statements on dispatch and commitment: 1. Incremental cost is dC/dP. 2. Unit commitment decides the output of each running unit minute by minute.
- 2 only
- 1 only
- Both 1 and 2
- Neither 1 nor 2
Answer
B. 1 only
Unit commitment decides on/off status; economic dispatch sets the output.
Which pair is correctly matched?
- Shunt reactor – raises voltage at heavy load
- PV bus – P and Q specified
- Ferranti effect – voltage rise in a lightly loaded long line
- Slack bus – P and Q specified
Answer
C. Ferranti effect – voltage rise in a lightly loaded long line
The other pairs state wrong specified quantities or effects.
Statements on voltage regulation and reactive power: 1. Capacitor reactive power varies as the square of voltage. 2. A lagging load lowers the receiving-end voltage.
- 1 only
- Both 1 and 2
- 2 only
- Neither 1 nor 2
Answer
B. Both 1 and 2
Q = V²/Xc and a lagging load draws reactive power, increasing the voltage drop.
Statements on hydro and thermal scheduling: 1. Run-of-river plants are usually run as base load. 2. Water use in hydro plants has no limit, so fuel saving is not a consideration.
- 1 only
- 2 only
- Both 1 and 2
- Neither 1 nor 2
Answer
A. 1 only
Water is limited, so hydrothermal scheduling aims to save fuel in thermal plants.