--- title: "Power angle characteristics of salient pole machine" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:salient-power-angle-derivation" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-power-angle-derivation site: BCT Engineering --- # Power angle characteristics of salient pole machine Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2080 Bhadra, 5+2 marks - Video: [Derivation of Power equation for salient pole machine](https://www.youtube.com/watch?v=SwYdSXYxByQ) — Crash Course in Electrical Engineering - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-power-angle-derivation - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question Derive the power angle characteristics of salient pole synchronous machine and also derive condition for maximum power. ## Short answer For salient-pole machine, P = (Ef V/Xd) sinδ + (V²/2)(1/Xq–1/Xd) sin2δ per phase. Maximum power occurs at δ_m < 90°, found from dP/dδ=0, giving cosδ_m = [–a+√(a²+8b²)]/(4b) where a=EfV/Xd, b=V²(1/Xq–1/Xd). ## Answer The power angle characteristic gives the power of a salient pole machine as a function of the load angle $\delta$. Because the air gap is not uniform, the armature current is split into a direct-axis part $I_d$ (sees $X_d$) and a quadrature-axis part $I_q$ (sees $X_q$), with $X_d > X_q$ (two-reaction theory). #### Derivation (generator, $R_a$ neglected) ```text E (q-axis) /| jXq Iq / | / | jXd Id / | V /delta| -----+------+------> d-axis I lags V by phi; psi = delta + phi ``` Take $E$ along the q-axis and $V$ lagging $E$ by $\delta$. From the phasor diagram, resolving $V$ along the two axes: $$ \begin{aligned} V\cos\delta &= E - X_d I_d \ \Rightarrow\ I_d = \frac{E - V\cos\delta}{X_d} \\ V\sin\delta &= X_q I_q \ \Rightarrow\ I_q = \frac{V\sin\delta}{X_q} \end{aligned} $$ Power per phase is $P = VI\cos\phi$. Projecting $I_d$ and $I_q$ on $V$ (the angle between $V$ and the q-axis is $\delta$): $$ \begin{aligned} P &= V\left(I_q\cos\delta + I_d\sin\delta\right) \\ &= V\left[\frac{V\sin\delta}{X_q}\cos\delta + \frac{E - V\cos\delta}{X_d}\sin\delta\right] \\ &= \frac{EV}{X_d}\sin\delta + V^2\sin\delta\cos\delta\left(\frac{1}{X_q} - \frac{1}{X_d}\right) \end{aligned} $$ Using $\sin\delta\cos\delta = \tfrac12\sin2\delta$: $$ P = \frac{EV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin 2\delta \quad \text{(per phase)} $$ For three phases multiply by 3. The same expression holds for a motor (with $\delta$ the angle by which $E$ lags $V$). - First term: **excitation power**, depends on field current. - Second term: **reluctance power**, due to saliency; it exists even when $E = 0$ and vanishes for a cylindrical rotor ($X_d = X_q$). ```text P | .--. resultant | .' '. | .' exc. '. | / .-.reluct.\ |/.-' '-. \ +-----------+----+---- delta 0 ~60-75 90 180 deg ``` #### Condition for maximum power $$ \frac{dP}{d\delta} = \frac{EV}{X_d}\cos\delta + V^2\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\cos2\delta = 0 $$ Let $a = \dfrac{EV}{X_d}$ and $b = V^2\left(\dfrac{1}{X_q}-\dfrac{1}{X_d}\right)$. With $\cos2\delta = 2\cos^2\delta - 1$: $$ 2b\cos^2\delta + a\cos\delta - b = 0 \ \Rightarrow\ \cos\delta_{m} = \frac{-a + \sqrt{a^2 + 8b^2}}{4b} $$ So maximum power occurs at $\delta_m < 90^\circ$ (unlike the cylindrical machine, where $P_{max}$ is at $90^\circ$). The salient pole machine is therefore "stiffer": it gives more synchronising power for a given $\delta$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is armature current split into Id and Iq in a salient-pole machine? A. Air gap is non-uniform, so the d-axis and q-axis see different reactances Xd and Xq B. The machine has two sets of windings C. Ra is large D. Two supplies feed it Answer: A. Non-uniform air gap gives different reluctance along the two axes, with Xd > Xq. (Key point: Two-reaction theory) 2. The derived per-phase power of a salient-pole machine is: A. (EV/Xd) sin(delta) only B. (EV/Xd) sin(delta) + (V^2/2)(1/Xq - 1/Xd) sin(2 delta) C. (EV/Xq) cos(delta) D. (V^2/Xd) sin(2 delta) only Answer: B. Substituting Id = (E - V cos delta)/Xd and Iq = V sin delta/Xq into P = V(Iq cos delta + Id sin delta) gives these two terms. (Key point: Power expression) 3. What is the second term (reluctance power)? A. Power depending on field current B. Power due to saliency; it exists even at E = 0 and vanishes if Xd = Xq C. Power lost in the damper D. Power of the prime mover Answer: B. It depends on (1/Xq - 1/Xd), so it exists without excitation and is zero for a cylindrical rotor. (Key point: Reluctance power) 4. Compared to a cylindrical machine, where does maximum power occur in a salient-pole machine? A. Exactly 90 degrees B. At delta less than 90 degrees C. At 180 degrees D. At 0 degrees Answer: B. Setting dP/d(delta) = 0 gives cos(delta_m) positive, so delta_m < 90 degrees, and the machine gives more synchronising power at a given delta. 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