22-Elec-B8 Power Electronics and Drives · December 2013

Question 2 of 6: Synchronous-Machine Reactances, Capability Curves, and Salient-Pole Power Transfer

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Synchronous-Machine Reactances, Capability Curves, and Salient-Pole Power Transfer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — leakage reactance versus synchronous reactance

Leakage reactance $X_l$ accounts for that part of the armature flux which links the stator winding but never crosses the air gap to link the field winding. It is made up of slot leakage, tooth-tip leakage, end-winding leakage and the differential (belt and zigzag) leakage associated with the harmonic content of the winding distribution. Because these flux paths are largely in air, $X_l$ is small, essentially independent of saturation, and typically of the order of 0.1 to 0.2 per unit.

Synchronous reactance $X_s$ is the whole effective series reactance that appears in the steady-state per-phase equivalent circuit, $E = V + I(R_a + jX_s)$. It is the sum of the leakage reactance and the magnetizing (armature-reaction) reactance, $X_s = X_l + X_m$, where $X_m$ represents the flux that does cross the air gap and therefore interacts with the field. Since the air-gap flux path is the dominant one, $X_m$ is much the larger term and $X_s$ commonly lies between 1.0 and 2.0 per unit — roughly an order of magnitude above $X_l$. Two practical distinctions follow. First, $X_s$ is sensitive to saturation because it contains an iron-path component, whereas $X_l$ is not, which is why the saturated and unsaturated synchronous reactances differ while $X_l$ does not. Second, it is $X_l$ and not $X_s$ that limits the fault current in the first instants after a short circuit, because armature reaction takes time to establish itself; that is the physical origin of the subtransient and transient reactances lying between $X_l$ and $X_s$.

Part (b) — reactive capability curves and coolant pressure

A reactive capability curve is the locus, drawn on the $P$–$Q$ plane at rated terminal voltage, that bounds the region in which a synchronous generator may be operated continuously. Three separate physical limits form the boundary. The armature-current (stator heating) limit is a circle of radius equal to the rated apparent power centred on the origin, because $S = \sqrt{P^{2}+Q^{2}} = 3V_tI_a$ and the armature copper loss depends only on $I_a$. The field-current (rotor heating) limit is a second circle, of radius $3V_tE_f/X_s$ centred at $Q = -3V_t^{2}/X_s$ on the reactive axis; it binds in the overexcited (lagging, VAR-exporting) region, where the field must be strong. The underexcited limit closes the leading side of the diagram and arises from end-region core heating — with a weak field, the end leakage flux drives the stator end iron into saturation and heats it — together with the steady-state stability limit, since operating too far underexcited erodes the synchronizing torque. In practice a further vertical limit is imposed by the prime mover’s rated output.

Because two of the three limits are thermal, they move whenever the machine’s ability to reject heat changes. Large turbo-generators are hydrogen cooled, and the manufacturer publishes a family of capability curves — typically at 30, 45 and 60 psig hydrogen. Raising the coolant pressure raises the gas density and hence the mass flow and heat-transfer coefficient, so the same temperature rise is reached at a higher current. Both the armature-current circle and the field-current circle expand outwards, the machine’s continuous MVA rating rises, and a wider band of reactive output becomes available at any given active power. Conversely, a loss of hydrogen pressure forces the operator onto a smaller curve and the unit must be derated immediately. Hydrogen is chosen for this duty because its density can be traded against its very high thermal conductivity and low windage loss, so a pressure increase buys cooling capacity at modest extra friction.

Parts (c) and (d) — power transfer to the infinite bus

Given.

Find. (c) the active and reactive power delivered to the infinite bus at a power angle of $30^\circ$, and (d) the torque angle and reactive power when the active power is 0.85 pu.

Approach. Add the link reactance to both axis reactances — the tie line is magnetically symmetric, so it loads the $d$- and $q$-axes equally — then apply the two-term salient-pole power-angle equations, solving the cubic-like $P(\delta)$ relation numerically in part (d).

1. Part (c) — refer both axis reactances to the infinite bus. The link is a plain reactance, identical in both axes, so it simply adds: $$X_d = X_{d,\text{m}} + X_e = 0.9 + 0.2 = 1.10 \ \text{pu}, \qquad X_q = X_{q,\text{m}} + X_e = 0.65 + 0.2 = 0.85 \ \text{pu}.$$ From here on $\delta$ is the angle of the excitation voltage $E$ (which lies on the $q$-axis) ahead of the infinite-bus voltage $V$, and all powers computed are those crossing into the bus, because the intervening reactances are lossless.
2. Write the salient-pole power-angle equation. A salient machine develops power by two distinct mechanisms — the field-excited term and the reluctance term that exists because $X_d \neq X_q$: $$P = \frac{EV}{X_d}\sin\delta + \frac{V^{2}(X_d - X_q)}{2X_dX_q}\sin 2\delta .$$ The second term peaks at $\delta = 45^\circ$ and is what shifts the maximum of $P(\delta)$ below $90^\circ$, as the figure shows.
3. Evaluate the two terms at $\delta = 30^\circ$. The excitation term is $$\frac{EV}{X_d}\sin\delta = \frac{1.3 \times 1.0}{1.10}\sin 30^\circ = 1.18182 \times 0.5 = 0.5909 \ \text{pu},$$ and the reluctance term is $$\frac{V^{2}(X_d-X_q)}{2X_dX_q}\sin 2\delta = \frac{1.0 \times 0.25}{2(1.10)(0.85)}\sin 60^\circ = 0.13369 \times 0.86603 = 0.1158 \ \text{pu}.$$ Adding them, $$\boxed{\;P = 0.5909 + 0.1158 = 0.7067 \ \text{pu}\;}$$ Note that saliency contributes about 16 percent of the total at this angle — not a term to be dropped.
4. Evaluate the reactive power at the same angle. The companion expression resolves the current into its two axis components before recombining them: $$Q = \frac{EV}{X_d}\cos\delta - V^{2}\left(\frac{\cos^{2}\delta}{X_d} + \frac{\sin^{2}\delta}{X_q}\right).$$ Substituting, $$Q = 1.18182(0.86603) - \left(\frac{0.75}{1.10} + \frac{0.25}{0.85}\right) = 1.02348 - (0.68182 + 0.29412),$$ $$\boxed{\;Q = 1.02348 - 0.97594 = 0.0475 \ \text{pu}\;}$$ The machine is therefore only marginally overexcited: it supplies $S = \sqrt{0.7067^{2}+0.0475^{2}} = 0.708$ pu at a power factor of 0.998 lagging.
5. Part (d) — solve $P(\delta) = 0.85$ pu. The relation is transcendental, so it is solved by iteration on $$1.18182\sin\delta + 0.13369\sin 2\delta = 0.85 .$$ Trial values bracket the root quickly: $\delta = 37^\circ$ gives 0.8398 pu and $\delta = 38^\circ$ gives 0.8573 pu, so the root lies between them. Interpolating and refining, $$\boxed{\;\delta = 37.58^\circ\;}$$ and back-substitution returns $1.18182(0.60964) + 0.13369(0.96471) = 0.7205 + 0.1290 = 0.8495 \approx 0.85$ pu, confirming the angle. Note that the machine is comfortably inside the stability limit, since $P(\delta)$ does not peak until about $72^\circ$.
6. Evaluate the reactive power at the new angle. Using the same $Q$ expression with $\cos 37.58^\circ = 0.79268$ and $\sin 37.58^\circ = 0.60964$: $$Q = 1.18182(0.79268) - \left(\frac{0.62834}{1.10} + \frac{0.37166}{0.85}\right) = 0.93676 - (0.57122 + 0.43724),$$ $$\boxed{\;Q = 0.93676 - 1.00846 = -0.0720 \ \text{pu}\;}$$
7. Interpret the sign. The negative result means the machine now absorbs 0.072 pu of reactive power from the bus rather than supplying it: at 0.85 pu of active loading, an excitation of only 1.3 pu leaves the machine underexcited, and it operates at a leading power factor of $0.85/\sqrt{0.85^{2}+0.072^{2}} = 0.996$. Raising the excitation would be required to return it to VAR export, and part (b) is the reason one cannot simply do so without checking the capability curve.