--- title: "Speed control by inertia and load damping" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:inertia-damping-speed-control" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/inertia-damping-speed-control site: BCT Engineering --- # Speed control by inertia and load damping Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2076 Asoj, 8 marks; 2079 Bhadra, 5 marks; 2070 Asar, 8 marks; 2078 Bhadra, 5 marks; 2075 Chaitra, 6 marks; 2071 Chaitra, 6 marks (asked 6 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/inertia-damping-speed-control - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question Below figure shows the block diagram of water turbine driving an electric generator without speed governor. Explain how the inertia and load damping constant can control the speed of the system against small change in load. Derive the transfer function showing the effect of inertia and load damping constant. [Figure: turbine coupled by a shaft to a generator, which supplies a composite load] ## Short answer Without a governor, gate position is fixed; inertia (H) determines the initial rate of speed fall, while load damping (D) reduces load as frequency drops, establishing a steady-state deviation of -Delta P_L/D. The transfer function is DeltaOmega(s)/DeltaP_L(s) = -1/(2Hs + D) with time constant T = 2H/D. ## Answer 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 $$ \frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D} = \frac{-1/D}{1 + sT}, \qquad T = \frac{2H}{D} $$ ```text dPm=0 + +--------------+ ------>(+)->| 1 |----+--> d(omega) ^ - | 2Hs + D | | | +--------------+ | dPL ``` #### Response to a small step change in load For a step $\Delta P_L(s) = \Delta P_L / s$: $$ \Delta\omega(t) = -\frac{\Delta P_L}{D}\left(1 - e^{-t/T}\right), \qquad T = \frac{2H}{D} $$ ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation relates speed change to power imbalance (per unit)? A. 2H d(omega)/dt = Delta Pm - Delta Pe B. H d(omega)/dt = Delta Pm x Delta Pe C. 2H Delta omega = Delta Pm + Delta Pe D. d(omega)/dt = D Delta Pe Answer: A. The swing equation says rotor acceleration is proportional to Pm - Pe. (Key point: Swing equation) 2. With no governor (Delta Pm = 0), the transfer function Delta Omega / Delta PL is: A. 1/(2Hs - D) B. -1/(2Hs + D) C. -D/(2Hs) D. 2H/(s + D) Answer: B. Substituting Pe = PL + D omega in the swing equation gives -1/(2Hs + D), with T = 2H/D. (Key point: Transfer function) 3. What is the effect of larger inertia constant H? A. Larger final speed error B. Smaller final speed error C. Slower speed change (larger time constant) but the same final value D. Faster response and larger final change Answer: C. T = 2H/D; the final deviation is -Delta PL/D, independent of H. (Key point: Role of inertia) 4. For Delta PL = 0.01 pu and D = 1, the steady speed deviation is: A. -0.0001 pu B. -0.02 pu C. -0.001 pu D. -0.01 pu (1 % drop) Answer: D. Delta omega_ss = -Delta PL / D = -0.01 pu, so a governor is needed to restore speed. 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