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

Question 1 of 6: Armature Reaction and Full-Load Generated Voltage of a Round-Rotor Synchronous Generator

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

Question 1: Armature Reaction and Full-Load Generated Voltage of a Round-Rotor Synchronous Generator (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.

(a) Armature reaction; synchronous versus magnetizing reactance

When a synchronous machine is loaded, the three-phase armature currents themselves set up a rotating magnetomotive force. For a balanced set of currents in a distributed winding this armature mmf wave rotates at exactly synchronous speed, so it stands still relative to the rotor field and simply adds vectorially to the main field mmf. Armature reaction is this modification of the resultant air-gap flux by the armature current. Its consequence is that the air-gap flux under load is no longer the open-circuit flux: a lagging (inductive) armature current produces an armature mmf that is largely in opposition to the field mmf and therefore demagnetising, while a leading current produces an mmf that partly aids the field and is magnetising.

The modelling step that makes this tractable is the observation that, in a linear (unsaturated) magnetic circuit, the flux produced by armature reaction is proportional to the armature current and the voltage it induces lags that flux by 90 electrical degrees. An effect proportional to current and in quadrature with it is exactly what a reactance does, so armature reaction can be replaced by a fictitious magnetizing reactance (also called the armature-reaction reactance) $X_m$ in series with the winding. This is legitimate for a round-rotor (cylindrical, non-salient) machine because the air gap is uniform: the reluctance seen by the armature mmf is the same whatever the rotor position, so one number suffices. A salient-pole machine has different direct- and quadrature-axis air gaps and needs two reactances, $X_d$ and $X_q$.

The synchronous reactance is the total per-phase reactance seen at the terminals. It is the sum of the magnetizing reactance and the armature leakage reactance $X_l$ — slot, tooth-tip and end-winding flux that links only the stator conductors and never crosses the air gap:

So the distinction is one of physical origin and of scope. The magnetizing reactance accounts only for the mutual, air-gap flux as altered by armature reaction; the synchronous reactance lumps that together with the purely local leakage flux into the single number that appears in the round-rotor equivalent circuit. In practice $X_m \gg X_l$, so $X_s$ is dominated by armature reaction, and the terminal behaviour of the machine follows the one-line-per-phase model used in parts (b) to (d).

(b), (c), (d) Full-load generated voltage per phase

Given. A three-phase, Y-connected, two-pole, 60 Hz turbine generator with the ratings and per-phase constants below.

Find. The magnitude (and phase) of the internally generated voltage per phase, $E$, at full-load current for unity power factor, 0.8 lagging and 0.8 leading.

Approach. Convert the ratings to a per-phase voltage and a full-load current, take $V$ as the phase reference, write the current phasor at the angle set by the power factor, and evaluate $E = V + I(R_a + jX_s)$ once per power factor.

1. Reduce the wye rating to one phase. For a wye connection the phase voltage is the line voltage divided by $\sqrt{3}$:$$V = \frac{V_{LL}}{\sqrt{3}} = \frac{13\,800}{\sqrt{3}} = 7967.4\ \text{V per phase}$$This is the terminal voltage the machine must hold; it is the reference phasor, $V = 7967.4\angle 0^\circ$ V.
2. Find the full-load armature current. The kVA rating fixes the current magnitude independently of power factor:$$I = \frac{S}{\sqrt{3}\,V_{LL}} = \frac{9.375\times 10^{6}}{\sqrt{3}\times 13\,800} = 392.15\ \text{A}$$The same 392.15 A flows in all three cases — only its angle changes.
3. Assemble the synchronous impedance. Combining the two given per-phase constants,$$Z_s = R_a + jX_s = 0.064 + j1.79\ \Omega = 1.7911\angle 87.95^\circ\ \Omega$$The impedance angle is within two degrees of 90°, which is why synchronous-machine problems are so nearly reactive.
4. Unity power factor — part (b). With $I = 392.15\angle 0^\circ$ A the drop is $IZ_s = 25.10 + j701.95$ V, so$$E = 7967.4 + 25.10 + j701.95 = 7992.5 + j702.0\ \text{V}$$Taking the magnitude and angle,$$\boxed{E_{\text{upf}} = 8023.3\ \text{V per phase} \;\angle\; 5.02^\circ}$$The resistive drop adds directly to $V$ while the reactive drop stands in quadrature, so it raises the magnitude only through the Pythagorean term — a 0.70 % rise in this case.
5. 0.8 power factor lagging — part (c). A lagging power factor means the current lags the terminal voltage, so $I = 392.15\angle -36.87^\circ = 313.72 - j235.29$ A. Multiplying out,$$IZ_s = (313.72 - j235.29)(0.064 + j1.79) = 441.25 + j546.50\ \text{V}$$and therefore $E = 8408.7 + j546.50$ V, giving$$\boxed{E_{0.8\,\text{lag}} = 8426.4\ \text{V per phase} \;\angle\; 3.72^\circ}$$Notice what happened: rotating $I$ backwards by 36.87° rotated the reactive drop $jIX_s$ into near-alignment with $V$, so almost the whole 702 V now adds arithmetically. This is the demagnetising armature reaction of part (a) showing up as a larger required excitation.
6. 0.8 power factor leading — part (d). Now the current leads, $I = 392.15\angle +36.87^\circ = 313.72 + j235.29$ A, and the same multiplication gives$$IZ_s = -401.09 + j576.62\ \text{V}$$The real part has changed sign, so $E = 7566.3 + j576.62$ V and$$\boxed{E_{0.8\,\text{lead}} = 7588.3\ \text{V per phase} \;\angle\; 4.36^\circ}$$The leading current magnetises the machine, so less excitation is needed than at no load — indeed $E$ is now below the terminal voltage of 7967 V.

Collecting the three results, and converting each to its equivalent line-to-line value for comparison with the 13 800 V nameplate:

The ordering $E_{\text{lead}} < E_{\text{upf}} < E_{\text{lag}}$ is the physical signature of armature reaction and is worth using as a sanity check on any answer of this type.