--- title: "Water turbine transfer function and characteristics" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:water-turbine-tf-characteristics" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/water-turbine-tf-characteristics site: BCT Engineering --- # Water turbine transfer function and characteristics Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2081 Bhadra, 6 marks; 2072 Kartik, 8 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-tf-characteristics - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question Derive the transfer function of water turbine relating gate position and mechanical power output. Explain the special characteristic of hydraulic turbine. ## Short answer The transfer function is (1 - Tw s)/(1 + 0.5 Tw s), with water starting time Tw = L U0/(g H0). It is non-minimum phase: a step gate opening initially reduces power to -2 pu before rising to +1 pu, due to water column inertia. This requires a special governor with transient droop for stability. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After linearising, which relation between per-unit power, flow velocity and gate is obtained? A. dPm = 3dU - 2dG B. dPm = 2dU - 3dG C. dPm = dU + dG D. dPm = 3dG - 2dU Answer: A. With dH = 2(dU - dG), dPm = dH + dU = 3dU - 2dG. (Key point: Linearised relations) 2. What is the water starting time Tw? A. Tw = g H0/(L U0) B. Tw = L U0/(g H0) C. Tw = L H0/(g U0) D. Tw = U0/(g L H0) Answer: B. Tw = L U0/(g H0), the time for head H0 to accelerate the water column to U0. (Key point: Water starting time) 3. What is the final turbine transfer function dPm/dG? A. (1 + Tw s)/(1 - 0.5 Tw s) B. 1/(1 + 0.5 Tw s) C. (1 - Tw s)/(1 + 0.5 Tw s) D. (1 - 0.5 Tw s)/(1 + Tw s) Answer: C. 3/(1+0.5Tw s) - 2 = (1 - Tw s)/(1 + 0.5 Tw s). (Key point: Transfer function) 4. Why is a hydro turbine called non-minimum phase? A. It has a pole at the origin B. It has no inertia C. Its output follows the gate immediately D. Its transfer function has a right-half-plane zero, so power first moves opposite to the gate Answer: D. The zero at s = 1/Tw makes the initial response reverse, caused by water inertia. 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