--- title: "Transient response of turbine-generator with governor" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:governor-transient-response" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/governor-transient-response site: BCT Engineering --- # Transient response of turbine-generator with governor Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2070 Chaitra, 10 marks; 2081 Baisakh, 5 marks; 2074 Asoj, 8 marks (asked 3 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/governor-transient-response - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question The figure below shows a turbine-generator coupled system with speed governor. Explain the transient responses of electrical power output (Pe), mechanical power input (Pm) and speed due to sudden increase in electrical load by an amount of ΔPL. [Figure: water enters the turbine through a valve controlled by the governor; the turbine (Pm) is coupled to the generator, which delivers Pe to the load] ## Short answer Upon a sudden load increase, electrical power jumps instantly by ΔPL. Speed drops as the deficit is supplied from rotor inertia, then a governor opens the gate; for a hydro turbine, mechanical power initially dips before rising. Speed recovers to a lower steady-state value determined by droop and load damping. ## Answer 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 $$ \frac{\Delta P_m(s)}{\Delta P_v(s)} = \frac{1 - T_w s}{1 + 0.5T_w s} $$ ```text 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 ($s \to 0$): $$ \Delta\omega_{ss} = \frac{-\Delta P_L}{D + \frac{1}{R}}, \qquad \Delta P_{m,ss} = \Delta P_{e,ss} = \Delta P_L + D\,\Delta\omega_{ss} $$ ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At the instant of a sudden load increase, what happens to Pe and Pm? A. Both rise instantly B. Pe rises at once by Delta PL while Pm cannot change instantly C. Pm rises first, Pe later D. Neither changes until the governor acts Answer: B. The generator must supply load immediately, while gate and water take time to respond. (Key point: Instant response of Pe) 2. What supplies the power deficit immediately after the load increase? A. The reservoir directly B. The exciter C. The kinetic energy of the rotor, so speed falls at a rate -Delta PL / 2H D. The surge tank only Answer: C. The rotor slows down, giving up stored kinetic energy. (Key point: Initial speed fall) 3. How does Pm behave in a hydro unit after the governor opens the gate? A. It rises monotonically B. It first dips (non-minimum phase due to water inertia) and then rises C. It falls permanently D. It stays constant Answer: B. The transfer function (1 - Tw s)/(1 + 0.5 Tw s) gives an initial reverse response. (Key point: Hydro turbine behaviour) 4. What is the steady-state speed deviation with a droop governor? A. Zero B. -Delta PL / D C. -Delta PL / (D + 1/R) D. -Delta PL x R Answer: C. Both load damping and governor droop contribute; an isochronous governor would give zero, and load reference restores 50 Hz. 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