Power/Frequency Control in Hydro Generator System

An isochronous governor is a speed governor that keeps the turbine speed (frequency) exactly constant at the set value for every load. "Isochronous" means "same speed". It uses integral action: the gate keeps moving as long as there is a speed error.

Closed loop control system

 w_ref +   error  +------+ dY +-------+ dPm
 ---->(+)-------->| K/s  |--->|turbine|---+
       ^ -        +------+    +-------+   |
       |                                  v
       |                         -dPL -->(+)
       |      +-----------+               |
       +------| 1/(2Hs+D) |<--------------+
        speed +-----------+

Governor law:

$$\Delta Y(s) = -\frac{K}{s}\,\Delta\Omega(s)$$

Operation:

1. Load increases, $P_e > P_m$, so the rotor slows down.
2. The speed error $(\omega_{ref} - \omega)$ is integrated; the integrator output opens the gate.
3. $P_m$ increases; once it exceeds $P_e$ the rotor accelerates back.
4. The integrator stops changing only when the error is zero. So in steady state $\omega = \omega_{ref}$ and $\Delta P_m = \Delta P_L$: zero steady-state frequency error.

 speed
  w0 |-----.         .---------- w0 (restored)
     |      \       /
     |       '.___.'
     +-------------------------> t
     Pm rises until Pm = Pe

Why it is not suitable for units in parallel

• Units in parallel are tied to the same frequency. Each isochronous governor tries to hold the frequency at its own set value.
• The set points can never be exactly equal. If unit 1 is set at 50.01 Hz and unit 2 at 50.00 Hz, unit 1 keeps opening its gate to raise frequency while unit 2 keeps closing, so they fight each other. One unit is driven to full load and the other to zero (or motoring).
• With integral control there is no unique load sharing: the load can be divided in any ratio and drifts. The operating point is not stable.
• Therefore only one unit in a system (or an isolated unit) may run isochronous. Other units use a speed droop characteristic, which gives a definite sharing of load in proportion to $1/R$.

Without a governor, the gate (and so mechanical power $P_m$) stays fixed. When load changes, speed is held back only by two natural effects: the inertia of the rotating masses, which slows the rate of speed change, and the load damping, which reduces the load as frequency falls so that a new balance is reached.

Derivation of the transfer function

Inertia (rotating mass): for small changes, in per unit, the swing equation is

$$2H\frac{d\,\Delta\omega}{dt} = \Delta P_m - \Delta P_e$$

where $H$ = inertia constant (s), $\Delta\omega$ = speed deviation (pu). Taking the Laplace transform,

$$\Delta\Omega(s) = \frac{1}{2Hs}\big[\Delta P_m(s) - \Delta P_e(s)\big]$$

Load damping: a composite load has frequency-insensitive parts (lighting, heating) and frequency-sensitive parts (motors). So

$$\Delta P_e = \Delta P_L + D\,\Delta\omega$$

where $\Delta P_L$ = non-frequency-sensitive load change and $D$ = load damping constant (% change in load per % change in frequency, typically 1–2).

Substituting,

$$\begin{aligned} 2Hs\,\Delta\Omega(s) &= \Delta P_m(s) - \Delta P_L(s) - D\,\Delta\Omega(s) \\ \Delta\Omega(s) &= \frac{\Delta P_m(s) - \Delta P_L(s)}{2Hs + D} \end{aligned}$$

Without a governor $\Delta P_m = 0$, so

$$\frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D} = \frac{-1/D}{1 + sT}, \qquad T = \frac{2H}{D}$$

 dPm=0  +    +--------------+
 ------>(+)->|      1       |----+--> d(omega)
         ^ - |   2Hs + D    |    |
         |   +--------------+    |
       dPL

Response to a small step change in load

For a step $\Delta P_L(s) = \Delta P_L / s$:

$$\Delta\omega(t) = -\frac{\Delta P_L}{D}\left(1 - e^{-t/T}\right), \qquad T = \frac{2H}{D}$$

 d(omega)
   0 |-----.
     |      \
     |       '-.
     |          '--.______  -dPL/D
     +------------------------> t
           T = 2H/D

Role of inertia and load damping

• Inertia (H): At the first instant $P_m$ is unchanged, so the extra load is supplied from the kinetic energy of the rotor. Speed starts to fall at a rate $d\Delta\omega/dt = -\Delta P_L/2H$. Larger $H$ gives a slower fall and a larger time constant $T = 2H/D$. Inertia does not change the final value; it only slows the change.
• Load damping (D): As frequency falls, motor loads take less power ($D\,\Delta\omega$). The fall stops when this reduction equals $\Delta P_L$, giving steady-state deviation $\Delta\omega_{ss} = -\Delta P_L/D$. Larger $D$ gives a smaller final speed error and a faster settling.
• Since $D$ is small (1–2), the steady speed error is large: e.g. $\Delta P_L = 0.01$ pu and $D = 1$ gives a 1 % drop in frequency. So a governor is needed to restore speed; inertia and damping only make the system self-regulating for small changes.

With a speed governor, the turbine-generator responds to a sudden load increase by first giving up stored kinetic energy and then, through governor action, increasing the water flow until mechanical input again equals electrical output.

System model

Inertia (rotating mass): for small changes, in per unit, the swing equation is

$$2H\frac{d\,\Delta\omega}{dt} = \Delta P_m - \Delta P_e$$

where $H$ = inertia constant (s), $\Delta\omega$ = speed deviation (pu). Taking the Laplace transform,

$$\Delta\Omega(s) = \frac{1}{2Hs}\big[\Delta P_m(s) - \Delta P_e(s)\big]$$

Load damping: a composite load has frequency-insensitive parts (lighting, heating) and frequency-sensitive parts (motors). So

$$\Delta P_e = \Delta P_L + D\,\Delta\omega$$

where $\Delta P_L$ = non-frequency-sensitive load change and $D$ = load damping constant (% change in load per % change in frequency, typically 1–2).

Substituting,

$$\begin{aligned} 2Hs\,\Delta\Omega(s) &= \Delta P_m(s) - \Delta P_L(s) - D\,\Delta\Omega(s) \\ \Delta\Omega(s) &= \frac{\Delta P_m(s) - \Delta P_L(s)}{2Hs + D} \end{aligned}$$

The governor with speed droop $R$ gives

$$\Delta P_v(s) = \frac{1}{1 + sT_g}\left[\Delta P_{ref}(s) - \frac{1}{R}\Delta\Omega(s)\right]$$

and the hydraulic turbine gives

$$\frac{\Delta P_m(s)}{\Delta P_v(s)} = \frac{1 - T_w s}{1 + 0.5T_w s}$$

 dPref +      +-------+ dPv +-------+ dPm
 ---->(+)---->|   1   |---->|turbine|---+
       ^ -    |1+sTg  |     | Gt(s) |   |
       |      +-------+     +-------+   v
    +-----+                   -dPL --->(+)
    | 1/R |                             |
    +-----+      +-----------+          |
       ^     dw  |     1     |          |
       +---------|  2Hs + D  |<---------+
                 +-----------+

Transient response to a sudden load increase

When a load $\Delta P_L$ is suddenly added:

1. Instant $t = 0^+$: the electrical output $P_e$ rises at once by $\Delta P_L$, because the generator must supply the connected load immediately. The mechanical input $P_m$ cannot change instantly (gate and water need time).
2. Deceleration: the deficit $P_e - P_m$ is supplied from the kinetic energy of the rotor, so speed starts to fall with slope $-\Delta P_L/2H$.
3. Load damping: as speed falls, frequency-sensitive load drops by $D\Delta\omega$, so $P_e$ comes down slightly from its initial jump.
4. Governor action: the speed fall is sensed; through the droop $1/R$ the governor opens the gate. For a hydro turbine, $P_m$ first dips (water inertia, non-minimum phase) and then rises.
5. Recovery: when $P_m$ exceeds $P_e$, the rotor accelerates and the fall in speed is arrested. After some oscillation, $P_m = P_e$ and speed settles at a new, lower value.

Steady state, from the block diagram ($s \to 0$):

$$\Delta\omega_{ss} = \frac{-\Delta P_L}{D + \frac{1}{R}}, \qquad \Delta P_{m,ss} = \Delta P_{e,ss} = \Delta P_L + D\,\Delta\omega_{ss}$$

 power
     |   Pe
 P0+ |   ._____
 dPL |   |     '--.________ Pm = Pe (final)
     |   |         _.-'
     |   |      _.'  Pm
  P0 |---'.   .'
     |     '.'  (hydro: Pm dips first)
     +---------------------------> t
 speed
     |----.
     |     \        .--------
     |      '._  _.'   dw_ss = -dPL/(D+1/R)
     |         ''
     +---------------------------> t

• Electrical power $P_e$: jumps by $\Delta P_L$ at once, then reduces slightly as frequency falls (load damping) and settles at $\Delta P_L + D\Delta\omega_{ss}$.
• Mechanical power $P_m$: starts at the old value (dips first in a hydro unit), rises with the governor and turbine time constants, overshoots slightly, and settles equal to $P_e$.
• Speed: falls, reaches a minimum (frequency nadir), recovers partly and settles at a lower value $\Delta\omega_{ss}$ fixed by the droop. With an isochronous governor ($R \to 0$) speed would return to nominal. Supplementary control (load reference change) is then used to bring the frequency back to 50 Hz.

Conclusion: inertia limits the initial rate of fall, the governor limits the depth of the fall and brings $P_m$ up to match $P_e$, and the droop leaves a small steady frequency error which is removed by changing the load reference.

A hydro generating unit has two main control loops. The P-f loop (automatic load frequency control, ALFC) controls real power and frequency through the turbine gate. The Q-V loop (automatic voltage regulator, AVR) controls reactive power and terminal voltage through the field excitation.

 P-f loop (ALFC)

 f_ref -->(+)--> speed governor --> hydraulic
           ^ -                      amplifier
           |                            |
           |                            v
   frequency sensor               wicket gates
           ^                            |
           |                            v
           +----- speed ------ turbine-generator

 Q-V loop (AVR)

 V_ref -->(+)--> amplifier --> exciter
           ^ -                    |
           |                      v
   PT + rectifier          generator field
           ^                      |
           |                      v
           +------ Vt ----- generator terminals

P-f loop

1. A speed (frequency) sensor measures the shaft speed.
2. The speed governor compares it with the reference and, through the speed droop, gives a command to the hydraulic amplifier (pilot valve and servomotor).
3. The servomotor moves the wicket gates (or needle), changing the water flow and the turbine power $P_m$.
4. Real power balance decides frequency: if $P_m < P_e$ frequency falls. The loop restores balance. Primary control uses droop; secondary control changes the load reference to restore 50 Hz.
5. It is slow (seconds) because of large mechanical and water time constants.

Q-V loop

1. A potential transformer measures terminal voltage; it is rectified and filtered.
2. The comparator finds the error $V_{ref} - V_t$.
3. The amplifier boosts it and drives the exciter, which changes the field current of the generator.
4. Field current changes the generated emf and hence the reactive power output and terminal voltage.
5. It is fast (fraction of a second) because electrical time constants are small. Stabilising feedback is used to avoid oscillations.

Interaction

Real power depends mainly on rotor angle and frequency, while reactive power depends mainly on voltage magnitude. Also, the AVR loop is much faster than the P-f loop. So the two loops are weakly coupled and are analysed and designed separately (decoupled control).

When load rises, both units move down their droop lines to a common lower frequency; each picks up load in proportion to its stiffness $P_r/R$.

Method: with droop $R_i$ (pu, on its own rating $P_{ri}$) and rated frequency $f_0$, the droop line of each unit gives

$$\Delta P_i = -\frac{P_{ri}}{R_i\, f_0}\,\Delta f = -K_i\,\Delta f$$

where $K_i = P_{ri}/(R_i f_0)$ is the stiffness of the unit in MW/Hz. For the units together, $\Delta P_L = -\Delta f\,\sum K_i$.

Data: $P_{r1} = 250$ MW, $R_1 = 0.20$; $P_{r2} = 500$ MW, $R_2 = 0.10$; $f_0 = 50$ Hz. Initial loading: 125 MW and 250 MW. $\Delta P_L = +60$ MW.

Stiffness of each unit:

$$\begin{aligned} K_1 &= \frac{250}{0.20 \times 50} = 25\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.10 \times 50} = 100\ \text{MW/Hz} \\ K_1 + K_2 &= 125\ \text{MW/Hz} \end{aligned}$$

i) New frequency

$$\begin{aligned} \Delta f &= -\frac{\Delta P_L}{K_1 + K_2} = -\frac{60}{125} = -0.48\ \text{Hz} \\ f_{new} &= 50 - 0.48 = 49.52\ \text{Hz} \end{aligned}$$

ii) New loading of each unit

$$\begin{aligned} \Delta P_1 &= K_1 \times 0.48 = 25 \times 0.48 = 12\ \text{MW} \\ \Delta P_2 &= K_2 \times 0.48 = 100 \times 0.48 = 48\ \text{MW} \end{aligned}$$

Check: $12 + 48 = 60$ MW.

Answer: New frequency = 49.52 Hz; unit 1 carries 137 MW and unit 2 carries 298 MW. The 500 MW unit, with the smaller droop (stiffer), takes 80 % of the increase.

Transfer function without governor

Without a governor, the turbine gate is fixed, so $\Delta P_m = 0$ and the speed is decided only by rotor inertia and load damping.

Inertia (rotating mass): for small changes, in per unit, the swing equation is

$$2H\frac{d\,\Delta\omega}{dt} = \Delta P_m - \Delta P_e$$

where $H$ = inertia constant (s), $\Delta\omega$ = speed deviation (pu). Taking the Laplace transform,

$$\Delta\Omega(s) = \frac{1}{2Hs}\big[\Delta P_m(s) - \Delta P_e(s)\big]$$

Load damping: a composite load has frequency-insensitive parts (lighting, heating) and frequency-sensitive parts (motors). So

$$\Delta P_e = \Delta P_L + D\,\Delta\omega$$

where $\Delta P_L$ = non-frequency-sensitive load change and $D$ = load damping constant (% change in load per % change in frequency, typically 1–2).

Substituting,

$$\begin{aligned} 2Hs\,\Delta\Omega(s) &= \Delta P_m(s) - \Delta P_L(s) - D\,\Delta\Omega(s) \\ \Delta\Omega(s) &= \frac{\Delta P_m(s) - \Delta P_L(s)}{2Hs + D} \end{aligned}$$

With $\Delta P_m = 0$:

$$\frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D}$$

 dPm=0  +    +--------------+
 ------>(+)->|      1       |----+--> d(omega)
         ^ - |   2Hs + D    |    |
         |   +--------------+    |
       dPL

For a step load $\Delta P_L$:

$$\Delta\omega(t) = -\frac{\Delta P_L}{D}\left(1 - e^{-tD/2H}\right)$$

The speed falls exponentially with time constant $2H/D$ and settles at $-\Delta P_L/D$, which is a large error because $D$ is small.

Incorporating an isochronous governor

An isochronous (constant speed) governor uses integral control. It measures speed, compares with the reference, and integrates the error to move the gate:

$$\Delta Y(s) = -\frac{K}{s}\,\Delta\Omega(s)$$

The gate keeps moving as long as there is any speed error, so speed returns exactly to the set value.

                          dPL
                           | -
 +-------+  dPm            v      +-----------+
 |turbine|--------------->(+)---->| 1/(2Hs+D) |--+--> dw
 | Gt(s) |                 +      +-----------+  |
 +-------+                                       |
     ^ dY                                        |
 +-------+    -                                  |
 |  K/s  |<---(+)<-------------------------------+
 +-------+     ^ +
               | dw_ref = 0

Here $G_t(s)$ is the turbine transfer function (for hydro, $\frac{1 - T_w s}{1 + 0.5T_w s}$).

Working: a load rise makes speed fall; the integrator output grows, gates open, $P_m$ rises above $P_e$, the rotor accelerates and speed comes back to the reference. In steady state $\Delta\omega = 0$ and $\Delta P_m = \Delta P_L$.

The isochronous governor works well for a single unit supplying an isolated load. It cannot be used on several units in parallel, because each would try to fix the frequency at its own set point and they would fight for the load; for that, speed droop is added.

The water turbine transfer function relates a small change in gate position to the change in mechanical power. Because the water column in the penstock has inertia, the turbine is a non-minimum phase system.

Derivation

Assumptions: lossless turbine, incompressible water, rigid penstock, and small changes about an operating point.

Velocity of water in the penstock: $U = K_u\,G\sqrt{H}$, where $G$ = gate opening and $H$ = head at the gate.

Turbine power: $P_m = K_p\,H\,U$.

Linearising and normalising by initial values ($U_0, G_0, H_0, P_{m0}$), with bar for per-unit change:

$$\begin{aligned} \Delta\bar U &= \tfrac{1}{2}\Delta\bar H + \Delta\bar G \\ \Delta\bar P_m &= \Delta\bar H + \Delta\bar U \end{aligned}$$

Eliminating $\Delta\bar H = 2(\Delta\bar U - \Delta\bar G)$:

$$\Delta\bar P_m = 3\Delta\bar U - 2\Delta\bar G$$

Acceleration of the water column (length $L$, area $A$): the net force from the head change accelerates the water mass,

$$\rho L A\,\frac{d\Delta U}{dt} = -A\rho g\,\Delta H \;\Rightarrow\; T_w\,\frac{d\Delta\bar U}{dt} = -\Delta\bar H$$

where the water starting time is

$$T_w = \frac{L\,U_0}{g\,H_0}$$

In Laplace form: $T_w s\,\Delta\bar U = -2(\Delta\bar U - \Delta\bar G)$, so

$$\frac{\Delta\bar U}{\Delta\bar G} = \frac{1}{1 + \tfrac{1}{2}T_w s}$$

Substituting in $\Delta\bar P_m = 3\Delta\bar U - 2\Delta\bar G$:

$$\frac{\Delta\bar P_m(s)}{\Delta\bar G(s)} = \frac{3}{1 + 0.5T_w s} - 2 = \frac{1 - T_w s}{1 + 0.5T_w s}$$

Special characteristics of a hydraulic turbine

1. Non-minimum phase response: the transfer function has a zero in the right-half plane at $s = 1/T_w$. Power initially changes opposite to the gate movement. For a unit step in gate opening, power first falls to $-2$ pu and then rises to $+1$ pu with time constant $T_w/2$:

$$\Delta\bar P_m(t) = 1 - 3e^{-2t/T_w}$$

2. Water inertia decides the response. $T_w$ is typically 0.5–4 s and changes with load, because $U_0$ and $H_0$ change.
3. Governor needs transient droop: a fast governor with normal droop would be unstable. So hydro governors have a large temporary droop with a reset time (or slow integral action), making hydro units slower to respond than steam units.
4. Water hammer: quick gate closing causes pressure rise; gate speed must be limited.
5. Hydro turbines can start and pick up load quickly from standstill, so they are good for peaking and frequency control once the above is allowed for.

Both units share the load increase in proportion to $P_r/R$, and the frequency falls to a common new value.

Method: with droop $R_i$ (pu, on its own rating $P_{ri}$) and rated frequency $f_0$, the droop line of each unit gives

$$\Delta P_i = -\frac{P_{ri}}{R_i\, f_0}\,\Delta f = -K_i\,\Delta f$$

where $K_i = P_{ri}/(R_i f_0)$ is the stiffness of the unit in MW/Hz. For the units together, $\Delta P_L = -\Delta f\,\sum K_i$.

Data: Unit 1: 600 MVA, $R_1 = 0.04$; Unit 2: 500 MVA, $R_2 = 0.06$; $f_0 = 60$ Hz. Initial: $P_1 = 500$ MW, $P_2 = 400$ MW. $\Delta P_L = 90$ MW. (MVA ratings are used as the MW base, i.e. unity power factor rating.)

Stiffness:

$$\begin{aligned} K_1 &= \frac{600}{0.04 \times 60} = 250\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.06 \times 60} = 138.89\ \text{MW/Hz} \\ K_1 + K_2 &= 388.89\ \text{MW/Hz} \end{aligned}$$

New frequency:

$$\begin{aligned} \Delta f &= -\frac{90}{388.89} = -0.2314\ \text{Hz} \\ f_{new} &= 60 - 0.2314 = 59.769\ \text{Hz} \end{aligned}$$

(In per unit, $\Delta f = -0.003857$ pu.)

New generation:

$$\begin{aligned} \Delta P_1 &= 250 \times 0.2314 = 57.86\ \text{MW} \\ \Delta P_2 &= 138.89 \times 0.2314 = 32.14\ \text{MW} \end{aligned}$$

Check: $57.86 + 32.14 = 90$ MW.

Answer: New frequency ≈ 59.77 Hz (drop of 0.231 Hz); unit 1 generates 557.86 MW and unit 2 generates 432.14 MW.

The water turbine transfer function relates a small change in gate opening to the change in mechanical power output. Due to the inertia of water in the penstock, power first moves in the wrong direction when the gate is moved.

Derivation of the transfer function

Assumptions: lossless turbine, incompressible water, rigid penstock, and small changes about an operating point.

Velocity of water in the penstock: $U = K_u\,G\sqrt{H}$, where $G$ = gate opening and $H$ = head at the gate.

Turbine power: $P_m = K_p\,H\,U$.

Linearising and normalising by initial values ($U_0, G_0, H_0, P_{m0}$), with bar for per-unit change:

$$\begin{aligned} \Delta\bar U &= \tfrac{1}{2}\Delta\bar H + \Delta\bar G \\ \Delta\bar P_m &= \Delta\bar H + \Delta\bar U \end{aligned}$$

Eliminating $\Delta\bar H = 2(\Delta\bar U - \Delta\bar G)$:

$$\Delta\bar P_m = 3\Delta\bar U - 2\Delta\bar G$$

Acceleration of the water column (length $L$, area $A$): the net force from the head change accelerates the water mass,

$$\rho L A\,\frac{d\Delta U}{dt} = -A\rho g\,\Delta H \;\Rightarrow\; T_w\,\frac{d\Delta\bar U}{dt} = -\Delta\bar H$$

where the water starting time is

$$T_w = \frac{L\,U_0}{g\,H_0}$$

In Laplace form: $T_w s\,\Delta\bar U = -2(\Delta\bar U - \Delta\bar G)$, so

$$\frac{\Delta\bar U}{\Delta\bar G} = \frac{1}{1 + \tfrac{1}{2}T_w s}$$

Substituting in $\Delta\bar P_m = 3\Delta\bar U - 2\Delta\bar G$:

$$\frac{\Delta\bar P_m(s)}{\Delta\bar G(s)} = \frac{3}{1 + 0.5T_w s} - 2 = \frac{1 - T_w s}{1 + 0.5T_w s}$$

Response to a unit step change in gate opening

For a unit step $\Delta\bar G(s) = 1/s$:

$$\Delta\bar P_m(s) = \frac{1 - T_w s}{s(1 + 0.5T_w s)}$$

Initial value ($s \to \infty$): $\Delta\bar P_m(0^+) = \dfrac{-T_w}{0.5T_w} = -2$

Final value ($s \to 0$): $\Delta\bar P_m(\infty) = 1$

Time response:

$$\Delta\bar P_m(t) = 1 - 3e^{-2t/T_w}$$

 dPm
  1 |                 ____________
    |             _.-'
  0 |----.     _-'
    |    |   .'
    |    |  /   time constant Tw/2
 -2 |    |_/
    +----+-------------------------> t
       gate step at t = 0

Explanation: when the gate opens suddenly, the flow cannot increase at once because the water column has inertia. The larger opening with the same flow lowers the pressure (head) at the turbine, so power first drops to $-2$ times the gate change. As the water accelerates, flow and power rise, and power settles at $+1$ times the gate change with time constant $T_w/2$. Similarly, closing the gate first gives a power rise (with pressure rise, i.e. water hammer).

This behaviour (zero in the right-half plane) is why hydro governors are given a large transient droop and a slower response.

Speed droop is the drop in speed (frequency) of a governed unit as its load goes from no load to full load, expressed in per unit of rated speed. It is defined by

$$R = \frac{\Delta f / f_0}{\Delta P / P_r} \quad \text{(pu)}$$

A 4 % droop means the frequency falls by 4 % (2 Hz at 50 Hz) from no load to full load. Droop gives every governor a definite speed–load line, so parallel units share a load change in a fixed, stable ratio. Without droop (isochronous governors), parallel units fight each other and the load division is undefined.

Block diagram (two units on a common bus)

   +-->[1/R1]--+
   |           v -
   |  dPref1->(+)->[Gov1]->[Turb1]--dPm1--+
   |                                      v
   |                        -dPL ------->(S)--+
   |                                      ^   |
   |  dPref2->(+)->[Gov2]->[Turb2]--dPm2--+   |
   |           ^ -                            v
   +-->[1/R2]--+                    [1/(2Hs+D)]
   |                                          |
   +--------------- d(omega) -----------------+

S = summing point: $\Delta P_{m1} + \Delta P_{m2} - \Delta P_L$.

Both units see the same frequency deviation $\Delta f$ because they are synchronised.

Derivation

In steady state, the governor of unit $i$ obeys its droop line:

$$\Delta P_i = \Delta P_{ref,i} - \frac{1}{R_i}\frac{\Delta f}{f_0}P_{ri}$$

With fixed references ($\Delta P_{ref,i} = 0$):

$$\Delta P_1 = -\frac{P_{r1}}{R_1}\frac{\Delta f}{f_0}, \qquad \Delta P_2 = -\frac{P_{r2}}{R_2}\frac{\Delta f}{f_0}$$

The total change must supply the load change:

$$\begin{aligned} \Delta P_L &= \Delta P_1 + \Delta P_2 = -\frac{\Delta f}{f_0}\left(\frac{P_{r1}}{R_1} + \frac{P_{r2}}{R_2}\right) \\ \frac{\Delta f}{f_0} &= \frac{-\Delta P_L}{\frac{P_{r1}}{R_1} + \frac{P_{r2}}{R_2}} \end{aligned}$$

and the ratio of sharing is

$$\frac{\Delta P_1}{\Delta P_2} = \frac{P_{r1}/R_1}{P_{r2}/R_2}$$

If $R_1 = R_2$ (same per-unit droop), $\Delta P_1/\Delta P_2 = P_{r1}/P_{r2}$, i.e. load is shared in proportion to ratings.

 f                     f
 f0|\ unit 1           |\  unit 2
   | \                 |  \
 f'|--*-----common f'--|---*
   |   \               |    \
   +----+---> P1       +-----+---> P2
       P1'                  P2'

Importance

• Gives a unique, stable division of load among parallel units.
• A smaller droop makes a unit "stiffer", so it takes a bigger share of any change.
• Equal per-unit droop on all units makes each carry load proportional to its rating, avoiding overload of small units.
• The frequency drop needed is small (typically 4–5 % droop), and the load reference setting is then used to bring frequency back to nominal.

A turbine-generator with a speed governor is a closed loop: the governor senses speed and adjusts the gate so that mechanical input matches electrical load.

Model and block diagram

Rotor and load: $\Delta\Omega(s) = \dfrac{\Delta P_m - \Delta P_L}{2Hs + D}$

Governor with droop: $\Delta P_v(s) = \dfrac{1}{1+sT_g}\left[\Delta P_{ref} - \dfrac{\Delta\Omega}{R}\right]$

Turbine: $\Delta P_m = G_t(s)\,\Delta P_v$ with $G_t(s) = \dfrac{1 - T_w s}{1+0.5T_w s}$ for hydro.

 dPref +      +-------+ dPv +-------+ dPm
 ---->(+)---->|   1   |---->|turbine|---+
       ^ -    |1+sTg  |     | Gt(s) |   |
       |      +-------+     +-------+   v
    +-----+                   -dPL --->(+)
    | 1/R |                             |
    +-----+      +-----------+          |
       ^     dw  |     1     |          |
       +---------|  2Hs + D  |<---------+
                 +-----------+

Steady state behaviour

With $\Delta P_{ref} = 0$ and a step load $\Delta P_L$, the closed loop gives

$$\frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D + \dfrac{G_t(s)}{R(1+sT_g)}}$$

Using the final value theorem ($G_t(0) = 1$):

$$\Delta\omega_{ss} = \frac{-\Delta P_L}{D + \frac{1}{R}} = \frac{-\Delta P_L}{\beta}$$

where $\beta = D + 1/R$ is the area frequency response characteristic. Since $1/R \gg D$ (e.g. $R = 0.05$ gives $1/R = 20$), the governor reduces the frequency error about 20 times compared with no governor.

• Speed-power relation: $\Delta P_m = \Delta P_{ref} - \Delta\omega/R$. This is the droop line: as load rises, speed falls in proportion.
• Units in parallel share a load change in proportion to $1/R$ (i.e. to rating, if all have the same per-unit droop).
• Changing $\Delta P_{ref}$ shifts the droop line up or down; this restores frequency to nominal (secondary control).

Transient behaviour

When a load $\Delta P_L$ is suddenly added:

1. Instant $t = 0^+$: the electrical output $P_e$ rises at once by $\Delta P_L$, because the generator must supply the connected load immediately. The mechanical input $P_m$ cannot change instantly (gate and water need time).
2. Deceleration: the deficit $P_e - P_m$ is supplied from the kinetic energy of the rotor, so speed starts to fall with slope $-\Delta P_L/2H$.
3. Load damping: as speed falls, frequency-sensitive load drops by $D\Delta\omega$, so $P_e$ comes down slightly from its initial jump.
4. Governor action: the speed fall is sensed; through the droop $1/R$ the governor opens the gate. For a hydro turbine, $P_m$ first dips (water inertia, non-minimum phase) and then rises.
5. Recovery: when $P_m$ exceeds $P_e$, the rotor accelerates and the fall in speed is arrested. After some oscillation, $P_m = P_e$ and speed settles at a new, lower value.

Steady state, from the block diagram ($s \to 0$):

$$\Delta\omega_{ss} = \frac{-\Delta P_L}{D + \frac{1}{R}}, \qquad \Delta P_{m,ss} = \Delta P_{e,ss} = \Delta P_L + D\,\Delta\omega_{ss}$$

 power
     |   Pe
 P0+ |   ._____
 dPL |   |     '--.________ Pm = Pe (final)
     |   |         _.-'
     |   |      _.'  Pm
  P0 |---'.   .'
     |     '.'  (hydro: Pm dips first)
     +---------------------------> t
 speed
     |----.
     |     \        .--------
     |      '._  _.'   dw_ss = -dPL/(D+1/R)
     |         ''
     +---------------------------> t

• Electrical power $P_e$: jumps by $\Delta P_L$ at once, then reduces slightly as frequency falls (load damping) and settles at $\Delta P_L + D\Delta\omega_{ss}$.
• Mechanical power $P_m$: starts at the old value (dips first in a hydro unit), rises with the governor and turbine time constants, overshoots slightly, and settles equal to $P_e$.
• Speed: falls, reaches a minimum (frequency nadir), recovers partly and settles at a lower value $\Delta\omega_{ss}$ fixed by the droop. With an isochronous governor ($R \to 0$) speed would return to nominal. Supplementary control (load reference change) is then used to bring the frequency back to 50 Hz.

Both generators move down their droop lines to the common frequency of 49.5 Hz; the load increase is the sum of what each picks up.

Method: with droop $R_i$ (pu, on its own rating $P_{ri}$) and rated frequency $f_0$, the droop line of each unit gives

$$\Delta P_i = -\frac{P_{ri}}{R_i\, f_0}\,\Delta f = -K_i\,\Delta f$$

where $K_i = P_{ri}/(R_i f_0)$ is the stiffness of the unit in MW/Hz. For the units together, $\Delta P_L = -\Delta f\,\sum K_i$.

Data: $P_{r1} = 250$ MW, $R_1 = 0.04$; $P_{r2} = 500$ MW, $R_2 = 0.02$; $f_0 = 50$ Hz. Initial loads: 125 MW and 250 MW (half load). $\Delta f = 49.5 - 50 = -0.5$ Hz.

Stiffness:

$$\begin{aligned} K_1 &= \frac{250}{0.04 \times 50} = 125\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.02 \times 50} = 500\ \text{MW/Hz} \end{aligned}$$

Increase in load on each generator:

$$\begin{aligned} \Delta P_1 &= 125 \times 0.5 = 62.5\ \text{MW} \\ \Delta P_2 &= 500 \times 0.5 = 250\ \text{MW} \end{aligned}$$

Increase in system load:

$$\Delta P_L = 62.5 + 250 = 312.5\ \text{MW}$$

Answer: The system load increased by 312.5 MW. G1 shares 62.5 MW (now 187.5 MW) and G2 shares 250 MW (now 500 MW, i.e. fully loaded). Any further load increase would overload G2.

Both generators start from the same no-load frequency (51 Hz) and move down their droop lines until their outputs add up to the load. Rated frequency is taken as $f_0 = 50$ Hz.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Stiffness:

$$\begin{aligned} K_1 &= \frac{600}{0.03 \times 50} = 400\ \text{MW/Hz} \\ K_2 &= \frac{300}{0.06 \times 50} = 100\ \text{MW/Hz} \\ K_1 + K_2 &= 500\ \text{MW/Hz} \end{aligned}$$

(Full-load frequency drop: G1 $= 0.03 \times 50 = 1.5$ Hz, G2 $= 0.06 \times 50 = 3$ Hz.)

Sharing of 600 MW

$$\begin{aligned} \Delta f &= \frac{600}{500} = 1.2\ \text{Hz}, \quad f = 51 - 1.2 = 49.8\ \text{Hz} \\ P_1 &= 400 \times 1.2 = 480\ \text{MW} \\ P_2 &= 100 \times 1.2 = 120\ \text{MW} \end{aligned}$$

Load increased by 200 MW (total 800 MW)

$$\begin{aligned} \Delta f &= \frac{800}{500} = 1.6\ \text{Hz}, \quad f = 51 - 1.6 = 49.4\ \text{Hz} \\ P_1 &= 400 \times 1.6 = 640\ \text{MW} \\ P_2 &= 100 \times 1.6 = 160\ \text{MW} \end{aligned}$$

(Equivalently, the extra 200 MW is shared as $400 \times 0.4 = 160$ MW and $100 \times 0.4 = 40$ MW.)

Answer: At 600 MW: 49.8 Hz, G1 = 480 MW, G2 = 120 MW. At 800 MW: 49.4 Hz, G1 = 640 MW, G2 = 160 MW.

Note: 640 MW is above G1's 600 MW rating. In practice G1's governor would stop at full gate (600 MW) and G2 would carry the remaining 200 MW, so the frequency would settle on G2's line at $51 - 200/100 = 49$ Hz. Load reference settings should be changed to avoid this overload.

Both units drop along their droop lines to a common new frequency, each taking a share of the extra 150 MW in proportion to $P_r/R$.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Data: G1: 600 MW, $R_1 = 0.07$, initial 300 MW; G2: 300 MW, $R_2 = 0.08$, initial 150 MW; $f_0 = 50$ Hz; $\Delta P_L = 150$ MW.

Stiffness:

$$\begin{aligned} K_1 &= \frac{600}{0.07 \times 50} = 171.43\ \text{MW/Hz} \\ K_2 &= \frac{300}{0.08 \times 50} = 75\ \text{MW/Hz} \\ K_1 + K_2 &= 246.43\ \text{MW/Hz} \end{aligned}$$

a) New frequency

$$\begin{aligned} \Delta f &= -\frac{150}{246.43} = -0.6087\ \text{Hz} \\ f_{new} &= 50 - 0.6087 = 49.39\ \text{Hz} \end{aligned}$$

b) Power supplied by each generator

$$\begin{aligned} \Delta P_1 &= 171.43 \times 0.6087 = 104.35\ \text{MW} \\ \Delta P_2 &= 75 \times 0.6087 = 45.65\ \text{MW} \end{aligned}$$

Answer: New frequency ≈ 49.39 Hz; G1 supplies 404.35 MW and G2 supplies 195.65 MW.

A hydraulic turbine responds to gate movement in an unusual way because of the inertia of the water column in the penstock.

Special characteristics

1. Non-minimum phase behaviour: the turbine transfer function

$$\frac{\Delta P_m(s)}{\Delta G(s)} = \frac{1 - T_w s}{1 + 0.5T_w s}$$

has a zero in the right half of the s-plane ($s = +1/T_w$). So the power first changes in the opposite direction to the gate movement. 2. Initial reverse power: when the gate is suddenly opened, the flow cannot rise at once because of water inertia, so the head (pressure) at the turbine falls and power first drops to −2 times the gate change. It then rises to +1 times the gate change with time constant $T_w/2$: $\Delta P_m(t) = [1 - 3e^{-2t/T_w}]\Delta G$. Closing the gate first gives a power (and pressure) rise. 3. Water inertia governs the response: the response speed is set by the water starting time $T_w$, which is not constant; it rises with load because $U_0$ increases. 4. Need for transient droop: a governor with only permanent droop would be unstable. Hydro governors use a large temporary (transient) droop with a reset time, so they respond slowly at first and are slower than steam unit governors. 5. Water hammer and gate rate limits: gate movements must be slow enough to keep pressure rise in the penstock safe.

 dPm
  1 |                ____________
    |            _.-'
  0 |---.     _-'
    |   |   .'
 -2 |   |_/        (unit step in gate)
    +---+----------------------> t

Water starting time

Water starting time $T_w$ is the time needed for the head $H_0$ to accelerate the water in the penstock from standstill to the velocity $U_0$:

$$T_w = \frac{L\,U_0}{g\,H_0}$$

where $L$ = length of penstock (m), $U_0$ = water velocity (m/s), $H_0$ = head at the gate (m), $g = 9.81$ m/s². It varies with load and is typically 0.5–4 s at full load. A large $T_w$ means a stronger reverse-power effect and harder speed control. A surge tank close to the power house reduces $L$ and so reduces $T_w$.

When load falls, both units move up their droop lines; the frequency rises and each unit sheds load in proportion to $P_r/R$.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Data: G1: 500 MW, $R_1 = 0.04$; G2: 250 MW, $R_2 = 0.02$; both fully loaded at 50 Hz (total 750 MW). Load reduced: $\Delta P_L = -100$ MW. Rated frequency 50 Hz.

Stiffness:

$$\begin{aligned} K_1 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\ K_2 &= \frac{250}{0.02 \times 50} = 250\ \text{MW/Hz} \end{aligned}$$

Frequency deviation:

$$\Delta f = -\frac{\Delta P_L}{K_1 + K_2} = -\frac{-100}{500} = +0.2\ \text{Hz}$$

New frequency $= 50.2$ Hz.

Load reduction on each unit:

$$\Delta P_1 = -250 \times 0.2 = -50\ \text{MW}, \qquad \Delta P_2 = -250 \times 0.2 = -50\ \text{MW}$$

Answer: Frequency rises by 0.2 Hz (to 50.2 Hz). G1 now supplies 450 MW and G2 200 MW; both have equal stiffness, so they shed the load equally. The 11 kV rating does not affect the result.

The droop line of each unit passes through its no-load point (51 Hz, 0 MW) and its present operating point (50 Hz, given MW). Extending the line to rated output gives the droop.

Frequency drop from no load to present load: $\Delta f = 51 - 50 = 1$ Hz.

Stiffness of each unit:

$$K_1 = \frac{600}{1} = 600\ \text{MW/Hz}, \qquad K_2 = \frac{200}{1} = 200\ \text{MW/Hz}$$

Droop on own ratings

Frequency drop at rated output: G1: $800/600 = 1.333$ Hz; G2: $300/200 = 1.5$ Hz.

$$\begin{aligned} R_1 &= \frac{1.333/50}{1} = 0.02667 = 2.67\ \% \\ R_2 &= \frac{1.5/50}{1} = 0.03 = 3\ \% \end{aligned}$$

Droop on the common base of 1000 MW

$$R_{new} = R_{own} \times \frac{S_{base}}{S_{rated}}$$ $$\begin{aligned} R_1 &= 0.02667 \times \frac{1000}{800} = 0.0333 = 3.33\ \% \\ R_2 &= 0.03 \times \frac{1000}{300} = 0.10 = 10\ \% \end{aligned}$$

Check: $\Delta P_i = (\Delta f/f_0)\,S_{base}/R_i$: G1 $= 0.02 \times 1000/0.0333 = 600$ MW ✓, G2 $= 0.02 \times 1000/0.1 = 200$ MW ✓.

Load sharing at 50.2 Hz

Frequency drop from no load $= 51 - 50.2 = 0.8$ Hz.

$$\begin{aligned} P_1 &= 600 \times 0.8 = 480\ \text{MW} \\ P_2 &= 200 \times 0.8 = 160\ \text{MW} \end{aligned}$$

Total load $= 640$ MW, i.e. the load has decreased by 160 MW (G1 sheds 120 MW, G2 sheds 40 MW).

Answer: $R_1 = 2.67\ \%$ and $R_2 = 3\ \%$ on own ratings (3.33 % and 10 % on 1000 MW base). At 50.2 Hz, G1 = 480 MW, G2 = 160 MW (total 640 MW).

When frequency rises, both units move up their droop lines and reduce output in proportion to $P_r/R$. Rated frequency is taken as $f_0 = 50$ Hz.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Data: G1: 1000 MW, 2 %; G2: 500 MW, 4 %; fully loaded at 49.5 Hz; frequency rises to 50 Hz, so $\Delta f = +0.5$ Hz.

$$\begin{aligned} K_1 &= \frac{1000}{0.02 \times 50} = 1000\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\ \Delta P_1 &= -1000 \times 0.5 = -500\ \text{MW} \\ \Delta P_2 &= -250 \times 0.5 = -125\ \text{MW} \end{aligned}$$

Answer: At 50 Hz, G1 supplies 500 MW and G2 supplies 375 MW; the system load is 875 MW (a reduction of 625 MW).

All three units are fully loaded at 49 Hz. As frequency rises to 50 Hz, each moves up its droop line and sheds load equal to its stiffness times the rise ($f_0 = 50$ Hz).

$$K_i = \frac{P_{ri}}{R_i f_0}, \qquad \Delta P_i = -K_i\,\Delta f, \qquad \Delta f = +1\ \text{Hz}$$

Answer: At 50 Hz, G1 = 0 MW (it reaches its no-load point, $49 + 1 = 50$ Hz), G2 = 125 MW, G3 = 66.67 MW; total system load ≈ 191.67 MW.

A hydraulic turbine behaves differently from a steam turbine because of the inertia of water in the penstock; its power output does not follow the gate opening directly.

1. Non-minimum phase behaviour: the turbine transfer function

$$\frac{\Delta P_m(s)}{\Delta G(s)} = \frac{1 - T_w s}{1 + 0.5T_w s}$$

has a zero in the right half of the s-plane ($s = +1/T_w$). So the power first changes in the opposite direction to the gate movement. 2. Initial reverse power: when the gate is suddenly opened, the flow cannot rise at once because of water inertia, so the head (pressure) at the turbine falls and power first drops to −2 times the gate change. It then rises to +1 times the gate change with time constant $T_w/2$: $\Delta P_m(t) = [1 - 3e^{-2t/T_w}]\Delta G$. Closing the gate first gives a power (and pressure) rise. 3. Water inertia governs the response: the response speed is set by the water starting time $T_w$, which is not constant; it rises with load because $U_0$ increases. 4. Need for transient droop: a governor with only permanent droop would be unstable. Hydro governors use a large temporary (transient) droop with a reset time, so they respond slowly at first and are slower than steam unit governors. 5. Water hammer and gate rate limits: gate movements must be slow enough to keep pressure rise in the penstock safe.

 dPm
  1 |                ____________
    |            _.-'
  0 |---.     _-'
    |   |   .'
 -2 |   |_/        (unit step in gate)
    +---+----------------------> t

Here $T_w = L U_0/(g H_0)$ is the water starting time, the time needed by the head to accelerate the water to its rated velocity (0.5–4 s).