Power Systems: Generation, Transmission, and Distribution

Posted on Feb 22, 2025 in Technology

UNIT 1

Power Systems: Generation, Transmission, Distribution

Generation (1-30kV), Transmission (115-800kV), Distribution (2.4-69kV), Consumers (120/240V).
Generated power = Consumed power.
Base demand: Needed all the time (100%).
Intermediate demand: Constant for a long time.
Peak demand: Changing all the time.

Hydro Power (1)

P = 9.8 * q * h

  • P (kW)
  • q (flow rate)
  • h (water head)

V = A * d

Time (t) = V / q

Hydro Power (2)

  • Base power: 100MW
  • Peak power: 60MW

Thermal Power (1)

η = (1 – T2 / T1)

  • η = efficiency
  • T1 = temperature of entering steam [K]
  • T2 = temperature of leaving steam [K]
  • [K] = T[°C] + 273

Thermal Power (2)

  1. Boiler
  2. Drum
  3. Turbines
  4. Turbines
  5. Turbines
  6. Condenser
  7. Re-heater
  8. Pump
  9. Burner
  10. Fan (in)

Plant: Pin = 30MW, Pout = 12MW, efficiency: η = 12/30 = 40%

Turbine: Pin = 27MW, Pout = 12MW, efficiency: η = 12/27 = 44.4%

Nuclear Power (1)

Nuclear is the biggest portion (35% Generation), (61% Consumed).
Nuclear reaction → heat → steam → turbine → generator → electricity.

Nuclear Power (2)

E = mc2

  • E = energy [J]
  • m = mass loss [kg]
  • c = speed of light [3 × 108 m/s]

CANDU Reactor (Canada Deuterium Uranium).

Wind Power (2)

Energy possessed by a moving mass:
Formula: Ek = ½mv2

  • Ek = Kinetic Energy [J]
  • m = mass [kg]
  • v = velocity (speed) [m/s]

Moving air:

  • Cross section area A [m2]
  • Moving speed v [m/s]
  • After the time: t [s]
  • Volume: V = A * v * t [m3]
  • Mass: m = 1.2 * V = 1.2 * A * v * t [kg]
  • Energy: E = ½mv2 = ½ * (1.2 * A * v * t) * v2 = 0.6 * A * t * v3 [J]
  • Power: P = E / t = 0.6 * A * v3 [W]
  • vt = 2πr * n/60

Wind Power (3)

Cb = P * t / E

Where:

  • Cb = battery capacity (AH)
  • P = power of charging or discharging (W)
  • E = battery voltage (V)
  • t = time of charging or discharging (h)

Solar Power (1)

Break-even time: Investment / Income per year

Solar Power (2)

Pin = G * A

  • G: power density [W/m2]
  • A: area [m2]

Po = V * I

  • V: voltage [V]
  • I: current [A]

η = Po / Pin

UNIT 2

Transmission

  1. Low voltage (LV): <600V
  2. Medium Voltage (MV): 2.4-69kV
  3. High Voltage (HV): 115-230kV
  4. Extra High Voltage (EHV): 345-800kV

Transmitted Power: P3 = 3ELILcosθ

Insulators:
Electrical Cables? For LV wiring, not for HV transmission!
ACSR Cable: Aluminum Cable Steel Reinforced.
Insulators support and anchor the conductors to insulate the conductors from ground.
Types:

  • Pin-type
  • Suspension-type
  • Polymer Disc

Lightning:
Isurge = Esurge / (Esurge / Isurge)

e * Esys,ph = Esys / 3

HV BIL:
BIL: Basic Impulse Insulation Level (The peak value of applied DC impulse before the insulator breaks down).
Voltage rating: The RMS value of applied 60Hz sine wave voltage before the insulator breaks down.

  • RMS = 0.707 * Peak
  • Peak = 1.414 * RMS

Equivalent Circuit

Per phase:

  • Resistance (r)
  • L Reactance (xL)
  • C Reactance (xC)
  • Line to neutral

In series, cause losses I2R.
In series, absorb reactive power.
In parallel, deliver reactive power.

Typical Reactance Value per km:

Type of linexL [Ω/km]xC [Ωkm]
Aerial line0.5300,000
Underground cable0.13,000

Lumped resistance: R = r * l

  • R = lumped resistance [Ω]
  • r = resistance per km [Ω/km]
  • l = cable length [km]

Lumped L reactance: XL = xL * l

  • XL = lumped L reactance [Ω]
  • xL = L reactance per km [Ω/km]

At both terminals: 2XC,Lumped

XC: Xc = xc / l

Simplified Equivalent Circuit

Power Flow:

  • P = active power to load [W]
  • PJ = I2 * R, losses [W]
  • QL = I2 * XL, absorbed reactive power [Var]
  • QC = Er2 / XC, delivered reactive power [Var]

Power Limit

  • Es: voltage source
  • X: L reactance only, determined by length
  • R: load resistance
  • P: active power transmitted to load
  • ER: voltage delivered to load

Active Power Calculation:

  • Load R changes → active P changes
  • Total Impedance: Z = √(X2 + R2)
  • Line current: I = Es / Z
  • Load voltage: ER = IR
  • Active Power: P = ER * I

Power limitation of a transmission line:
When R = X, ER = 0.707 * Es Pmax = Es2 / 2X

Compensated Line

When: Adjust Xc, so that: QC = ½QL

Active Power:

  • Active Power Transmitted: P = (Es * ER / X) * sinδ
  • P = active power transmitted [MW]
  • Es, ER = line to neutral voltages [kV]
  • X = L reactance [Ω]
  • δ = phase angle between two voltages
  • Period: T = 1 / f
  • Time delay: ΔT = θ / 360 * T
  • P3 = 3P

Line Voltage:

  • Uncompensated line: Pmax = E2 / 2X
  • Compensated line: Pmax = E2 / X
  • L reactance: X = xl
  • Phase voltage: E = √(cxlPmax)
  • Line voltage: EL = √3E = √3√(cxlPmax) = k√(Pl)

Where:

  • c = 1 for compensated line, c = 2 for uncompensated line.
  • EL = k√(Pl)
  • E = line voltage [kV]
  • P = power to be transmitted [kW]
  • l = length of transmission line [km]
  • k = coefficient, 0.1 for uncompensated, 0.06 for compensated
  • I = P3 / √3EL