--- title: "Positive synchronizing power coefficient" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:synchronizing-coefficient-positive" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/synchronizing-coefficient-positive site: BCT Engineering --- # Positive synchronizing power coefficient Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2083 Baishakh (new course), 5 marks - Video: [SM32 Synchronizing power Coefficient](https://www.youtube.com/watch?v=ERoR6UzgRz4) — Lectures in Electrical Engineering - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/synchronizing-coefficient-positive - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question Justify with appropriate expression that synchronizing power coefficient must be positive for a stable equilibrium power system operating condition. ## Short answer For small deviation delta = delta0 + Delta delta, linearising P_e gives M d^2(Delta delta)/dt^2 + P_s Delta delta = 0. If P_s > 0, roots are imaginary and the rotor oscillates stably; if P_s < 0, one root is positive and the angle grows exponentially, causing loss of synchronism. Thus P_s = P_max cos delta0 must be positive, so delta0 < 90 degrees. ## Answer **Synchronizing power coefficient** $P_s$ is the rate of change of electrical power output with rotor angle at the operating point: $$ P_s = \left.\frac{\partial P_e}{\partial\delta}\right|_{\delta_0} = \frac{EV}{X}\cos\delta_0 = P_{max}\cos\delta_0\ \ \text{(pu power/rad)} $$ It gives the change in power that tends to pull the rotor back to synchronism for a small change in rotor angle: $\Delta P_e = P_s\,\Delta\delta$. It is also called the stiffness of the coupling between the machine and the system. #### Small-signal analysis Let the machine operate at $\delta_0$ with $P_m = P_{max}\sin\delta_0$, and let the angle change by a small amount: $\delta = \delta_0 + \Delta\delta$. Then $$ P_e = P_{max}\sin(\delta_0+\Delta\delta) \approx P_{max}\sin\delta_0 + (P_{max}\cos\delta_0)\Delta\delta = P_m + P_s\Delta\delta $$ Substituting in the swing equation $M\dfrac{d^2\delta}{dt^2} = P_m - P_e$ (with $M = H/\pi f$): $$ M\frac{d^2\Delta\delta}{dt^2} + P_s\,\Delta\delta = 0 $$ The characteristic equation is $Ms^2 + P_s = 0$, so $$ s = \pm\sqrt{-\frac{P_s}{M}} $$ - **$P_s > 0$:** roots are imaginary, $s = \pm j\omega_n$ with $\omega_n = \sqrt{P_s/M}$. $\Delta\delta$ oscillates with constant amplitude (and decays once damping is included). The rotor is pulled back: the operating point is **stable**. - **$P_s < 0$:** one root is real and positive, $\Delta\delta$ grows exponentially; the machine falls out of step: **unstable**. - **$P_s = 0$** ($\delta_0 = 90^\circ$): the steady state stability limit. The frequency of natural oscillation is $$ f_n = \frac{1}{2\pi}\sqrt{\frac{P_s}{M}} = \frac{1}{2\pi}\sqrt{\frac{\pi f P_s}{H}}\ \text{Hz} $$ #### Physical reasoning ```text Pe ^ .-''-. | Pm .' Ps<0 '. |-----a'---------b.--- | Ps>0 \ +-+---+-----+------+--> d 0 d0 90 180 ``` At point $a$ ($\delta_0 < 90^\circ$, $P_s > 0$) a small increase in $\delta$ raises $P_e$ above $P_m$, decelerating the rotor back. At point $b$ ($\delta > 90^\circ$, $P_s < 0$) the same increase lowers $P_e$, the rotor accelerates further and synchronism is lost. Hence a **positive synchronizing power coefficient is necessary for a stable equilibrium**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The linearised swing equation for small change is: A. M d2(delta)/dt2 + Ps delta = 0 B. M d2(delta)/dt2 - Ps delta = Pm C. M d(delta)/dt = Ps D. M d2(delta)/dt2 = Pm Answer: A. Since Pe = Pm + Ps delta, M d2(delta)/dt2 = Pm - Pe gives M d2(delta)/dt2 + Ps delta = 0. (Key point: Linearised swing equation) 2. If Ps > 0 the roots s = +/- sqrt(-Ps/M) are: A. Real and positive B. Real and negative C. Both zero D. Imaginary, giving oscillation about the equilibrium: stable Answer: D. Imaginary roots mean bounded oscillation with natural frequency sqrt(Ps/M). (Key point: Roots for Ps > 0) 3. If Ps < 0, what happens to delta(delta)? A. It oscillates forever B. It decays C. It grows exponentially (a positive real root) and the machine loses synchronism D. It stays constant Answer: C. A positive real root gives exponential growth of the angle deviation. 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