--- title: "Water turbine response to step gate change" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:water-turbine-step-gate" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/water-turbine-step-gate site: BCT Engineering --- # Water turbine response to step gate change Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2078 Bhadra, 6 marks; 2072 Chaitra, 6 marks (asked 2 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/water-turbine-step-gate - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question Derive the transfer function of a water turbine and explain the response against unit step change in gate opening. ## Short answer Transfer function: (1 - T_w s)/(1 + 0.5 T_w s). A unit step gate change causes mechanical power to instantly drop to -2 pu due to water inertia, then rise to +1 pu with time constant T_w/2. ## Answer 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: $$ \Delta\bar P_m(t) = 1 - 3e^{-2t/T_w} $$ ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a unit step in gate opening, what is the immediate change in turbine power? A. -2 pu B. +1 pu C. -1 pu D. +2 pu Answer: A. Initial value theorem gives -Tw/(0.5Tw) = -2. (Key point: Initial value) 2. What is the final value of dPm for a unit step in gate? A. +2 B. +1 C. -2 D. +3 Answer: B. Setting s to 0 in (1 - Tw s)/(1 + 0.5Tw s) gives 1. (Key point: Final value) 3. What is the time response and its time constant? A. 1 - e^(-t/Tw), time constant Tw B. 1 + 3e^(-2t/Tw), time constant Tw/2 C. 1 - 3e^(-2t/Tw), time constant Tw/2 D. 1 - 3e^(-t/Tw), time constant Tw Answer: C. Partial fractions give dPm(t) = 1 - 3e^(-2t/Tw), time constant Tw/2. (Key point: Time response) 4. Why does power first fall when the gate is opened suddenly? A. The generator loses excitation B. Water hammer raises the head C. The governor is too slow D. Flow cannot rise at once, so the head at the turbine drops Answer: D. Inertia of the water column keeps flow nearly unchanged, so the larger opening lowers the pressure and power. (Key point: Physical reason) ## Related in Power Plant Equipment - [Water turbine transfer function and characteristics](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-turbine-tf-characteristics) — Power Plant Equipment, chapter 2, asked 2× - [Speed response without governor and isochronous governor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aspeed-response-isochronous-block) — Power Plant Equipment, chapter 2, asked 2× - [Transient response of turbine-generator with governor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agovernor-transient-response) — Power Plant Equipment, chapter 2, asked 3× - [Excitation system with stabilizing transformer: transfer function](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aexcitation-stabilizing-transformer-model) — Power Plant Equipment, chapter 3, asked 13× - [Function of excitation system and its transfer function](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aexcitation-function-tf) — Power Plant Equipment, chapter 3, asked 2× - [Control of water delivery to turbine](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-delivery-control) — Power Plant Equipment, chapter 1 ## Related in other subjects - [Unit step response of 4/[s(s+2)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Aunit-step-4-s-s2) — Control Systems, chapter 3 - [Underdamped second order unit step response derivation](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asecond-order-step-derivation) — Control Systems, chapter 3 - [Unit step response for h(t) = exp(-at)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aunit-step-response-exp) — Data Communication, chapter 2 - [Impulse response, transfer function and distortionless transmission](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adistortionless-transfer-function) — Communication Systems, chapter 1