Power System - Section 2
Complex Power, Load Characteristics and Per-Unit System
This section introduces three important concepts used extensively in power-system analysis: complex power, load characteristics, and the per-unit system. Complex power helps us understand and calculate real power, reactive power, apparent power, and power factor in AC circuits. Load characteristics describe how electrical loads behave as voltage, frequency, and time vary. The per-unit system simplifies power-system calculations by expressing electrical quantities as fractions of selected base values. These concepts are fundamental for analysing transformers, transmission systems, generators, and distribution networks.
Q1. What is Complex Power?
A1. In an AC circuit, voltage and current may have a phase difference. Therefore, power cannot always be represented by a single real number.
The total or complex power is represented as: S = P + jQ
where:
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S = complex power in VA
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P = active or real power in W = V.I.cosϕ (for 1ph) or √3.VL.IL.cosϕ (for 3ph)
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Q = reactive power in VAR = V.I.sinϕ (for 1ph) or √3.VL.IL.sinϕ (for 3ph)
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j = √(-1)
Apparent Power
The magnitude of complex power is called apparent power. |S| = V.I (for 1ph) or √3.VL.IL (for 3ph)
Unit: VA or kVA or MVA etc.
Complex Power Using Phasors
Complex power can also be calculated using: S = V.I*
where is the complex conjugate of current.
If: V = V∠θv and I = I∠θi, then S = V.I∠(θv - θi).
Power Triangle
The power triangle represents the relationship between P, Q, and S.
Power factor, cosϕ = P/S.
Therefore: Power factor = Real Power / Apparent Power.
A lagging power factor normally indicates an inductive load, while a leading power factor normally indicates a capacitive load.
Q2. What is Load Characteristics in power system?
A2. An electrical load does not necessarily consume constant power. Its behaviour can change with voltage, frequency, time, and operating conditions.
Load characteristics describe the relationship between load power and different operating variables.
a. Constant-Resistance Load
For a resistance: P = V²/R,
Therefore, P ∝ V²
b. Constant-Current Load
For constant current: P = V.I
Since I is constant: P ∝ V
Thus, power changes approximately linearly with voltage.
c. Constant-Power Load
For a constant-power load: V.I = P = constant
Therefore, if voltage decreases, current increases approximately according to: I = P/V.
Examples can include controlled power-electronic loads.
Load Variation with Voltage
A general load model may be represented approximately by: P = P0 * (V / V0)ⁿ
where:
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P0 = power at reference voltage
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V0 = reference voltage
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V = actual voltage
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n = load characteristic exponent
Different values of represent different types of load behaviour.
Load Curve
A load curve shows how electrical demand varies with time.
For example, the demand of a power system may be:
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Low during early morning
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Increasing during daytime
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High during peak hours
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Decreasing during late night
Important quantities obtained from a load curve include:
Maximum Demand : The highest demand recorded during a specified period.
Average Load : Pavg = Energy Consumed / Time
Load Factor : LF = Pavg / Max. Demand = Energy Consumed / (Time * Max. Demand)
A higher load factor generally means the available generating capacity is being used more uniformly.
Q3. What is Per-Unit System?
A3. The per-unit (p.u.) system expresses an electrical quantity as a fraction of a selected base value.
Per unit value = Actual value / Base value
Therefore,
Actual value = Per unit value * Base value
The per-unit system is widely used in power-system analysis because it simplifies calculations involving transformers and different voltage levels.
Base Quantities
Usually two base quantities are independently selected:
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Base apparent power SB
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Base voltage VB
The other base quantities can then be calculated.
Base Current
For a three-phase system: IB = SB / √3*VB
Base Impedance
For a three-phase system: ZB = VB² / SB
where is VB line-to-line voltage and SB is three-phase apparent power.
Per-unit Voltage: Vpu = V / VB
Per-unit Current: Ipu = I / IB
Per-unit Impedance: Zpu = Z / ZB
Q4. What are Advantages of the Per-Unit System?
A4.The per-unit system provides several useful advantages:
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Transformer turns ratios can be handled more easily.
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Calculations become simpler.
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Quantities from different voltage levels can be compared conveniently.
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Numerical values of system impedances are usually small and convenient.
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Errors in calculations can be easier to identify.
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It is widely used in power-flow, fault, and stability studies.
Q5. How to change base in the Per-Unit System?
A5. Sometimes an impedance is given on one base but needs to be converted to another base.
The general conversion is:
If the base voltage is unchanged:
Practice MCQ
1. Complex power is represented by::
A. P - Q
B. P + jQ
C. P/Q
D. P*Q
2. The unit of reactive power is:
A. W
B. VA
C. VAR
D. Hz
3. Apparent power in a single-phase AC circuit is:
A. V.I
B. V.I.cosϕ
C. V.I.sinϕ
D. V/I
4. The power factor of an inductive load is generally:
A. Leading
B. Lagging
C. Unity only
D. Zero only
5. For a resistance load, power is proportional to:
A. V
B. 1/V
C. V²
D. 1/V²
6. Load factor is defined as:
A. Maximum load / Average load
B. Average load / Maximum demand
C. Energy / Voltage
D. Current / Voltage
7. The per-unit value of a quantity is:
A. Actual value × Base value
B. Base value / Actual value
C. Actual value / Base value
D. Actual value + Base value
8. For a three-phase system, base current is:
A. SB / VB
B. SB / 3*VB
C. SB / √3*VB
D. VB / SB
9. The three-phase base impedance is:
A. SB / VB²
B. VB² / SB
C. VB / SB
D. SB * VB
10. The main purpose of using the per-unit system in power-system analysis is to:
A. Increase system voltage
B. Eliminate reactive power
C. Simplify calculations and comparison of system quantities
D. Increase power factor automatically
Answer: 1. B, 2. C, 3. A, 4. B, 5. C, 6. B, 7. C, 8. C, 9. B, 10. C.
Practice Problems
Problem 1 — Complex Power
A single-phase load operates at 230V, draws 10A current, and has a power factor of 0.8 lagging. Find:
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Apparent power
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Real power
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Reactive power
Problem 2 — Three-Phase Power
A balanced three-phase load is supplied 400V at and draws 20A current at 0.8 power factor lagging. Find the active power.
Problem 3 — Load Factor
A generating station supplies 7200 kWh of energy in 24 hours. Its maximum demand is 500 kW. Find the load factor.
Problem 4 — Per-Unit Current
A three-phase system has a base apparent power of 10 MVA and a base voltage of 11 kV. Find the base current.
Problem 5 — Per-Unit Impedance
A three-phase system has a base power of 100 MVA and base voltage of 132 kV. Find the base impedance.