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Synchronous-motor power equation

For all but very small synchronous machines the armature resistance is negligible compared with the synchronous reactance. Under this assumption the general terminal-equation of the synchronous motor can be simplified.

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The equivalent circuit and the phasor diagram used for machine analysis are shown below. These representations are the standard tools for studying how a synchronous motor responds to changes of mechanical load and to changes of field excitation.

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Multiplying the phasor relation by the terminal voltage V_T and rearranging gives the expression for the active power input per phase.

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With the armature winding resistance assumed negligible, the active power input per phase is equal to the electromagnetic power developed per phase. Therefore, the electromagnetic power developed per phase can be written as

(65)

or equivalently

(66)

For a three-phase machine the developed electromagnetic power becomes

(67)

or

(68)

The relation in Eqn. 66 is the synchronous-machine power equation for a cylindrical-rotor machine. It expresses the developed electromagnetic power in terms of the excitation E_f (counter emf), the terminal voltage V_T, the synchronous reactance X_s, and the power angle δ.

Assuming the supply voltage V_T and the supply frequency are constant, Eqns. 65 and 66 give useful proportionalities for machine behaviour:

(69)

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Figure 54: Equivalent-circuit of a synchronous motor, assuming armature resistance is negligible

Figure 55: Phasor diagram model for a synchronous motor, assuming armature resistance is negligible

Effect of changes in load on armature current, power angle, and power factor of a synchronous motor

The influence of mechanical (shaft) load on the armature current, the power angle and the power factor is conveniently seen from the phasor diagram shown in the figure below. The supply voltage, supply frequency and the field excitation are assumed constant while load changes.

The initial steady-state condition is shown using heavier lines. Doubling the shaft load produces the lighter phasors that represent the new steady state. From Eqns. 69 and 70, doubling the shaft load doubles both I_a cos φ_i and E_f sin δ. When redraw­ing the phasor diagram for the new steady state the reactance drop j I_a X_s must remain perpendicular to the current phasor I_a.

With excitation unchanged, increasing the shaft load causes the tip of the E_f phasor to move along a circular arc and the phase of E_f increases as load increases. At the same time the angle φ_i (power-factor angle between I_a and V_T) decreases, therefore the power factor improves (becomes less lagging). As the shaft load is increased the rotor angle δ increases: the rotor lags the rotating stator field by a larger angle, and stator current magnitude increases.

Figure 56: Phasor diagram showing effect of changes in shaft load on armature current, power angle and power factor of a synchronous motor

The average speed of the machine remains essentially constant at synchronous speed except during transients while the rotor moves to a new steady position. A further increase of load produces a limit: when an increase in δ no longer produces an increase of developed torque the rotor loses synchronism. For the ideal cylindrical-rotor model the maximum developed torque (pull-out torque) occurs at about δ = 90°, as indicated by Eqn. 68. Practically a machine is not operated near δ = 90° because armature current would be many times rated at such high torque.

Effect of changes in field excitation on synchronous motor performance

Changing the field excitation changes the counter emf magnitude E_f and therefore alters δ so as to keep the developed torque equal to the shaft load (at steady state). Increasing excitation (E_f) strengthens the rotor field and tends to reduce the power angle δ for a given shaft load.

If the shaft load is constant, from Eqn. 69 we have

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Graphically, for constant shaft load the tip of the E_f phasor moves along a straight line parallel to the V_T phasor. Similarly, from Eqn. 69, for constant shaft load

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so that the tip of the I_a phasor moves along a line perpendicular to V_T.

Figure 57: Phasor diagram showing effect of changes in field excitation on armature current, power angle and power factor of a synchronous motor

Increasing excitation from a low value to a high value in the phasor diagram causes the stator current phase to shift from lagging to leading with respect to V_T. The excitation that gives unity power factor is called normal excitation. Excitation greater than normal is over-excitation, and excitation less than normal is under-excitation. When over-excited |E_f| > |V_T| and the machine supplies reactive power to the system; such a synchronous motor operated for reactive support is often called a synchronous condenser.

V curves

Plots of armature current magnitude |I_a| versus field current I_f (or equivalently versus excitation voltage scaled appropriately) at a fixed shaft load are called V curves. These curves illustrate how armature current and power factor vary with field excitation for different loads. For higher shaft loads the minimum armature current (unity power factor point) occurs at higher excitation.

The left-most locus point of the V curves corresponds to the stability limit (δ = -90°). If excitation is reduced below that limit for a given load the machine will lose synchronism.

The V curves may be obtained experimentally by varying the field current at constant shaft load and recording the armature current, or they may be plotted using the phasor relations. The mathematical expression for the V curves, obtained assuming R_a negligible, is

Eqn. 74 is derived from the phasor diagram under the assumption of negligible armature resistance. If the quantity under the square root sign becomes negative the expression is not physically real; this corresponds to instability when the developed torque is less than the shaft load plus friction and windage losses.

The family of V curves shown here was computed using the data for a representative three-phase 10 hp synchronous motor with V_ph = 230 V and X_s = 1.2 Ω/phase.

Figure 58: Family of representative V curves for a synchronous motor

Synchronous-motor losses and efficiency

The flow of power through a synchronous motor from stator terminals to shaft output and the various loss components are commonly represented by a power-flow diagram.

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• P_scl = stator (armature) copper loss
• P_fcl = field copper loss
• P_core = core (iron) loss
• P_f,w = friction and windage loss
• P_stray = stray load loss

Except during transients while field energy is being stored or released, the electrical energy supplied to the field winding appears as I²R losses in the field. The overall (mechanical) efficiency is given by

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Manufacturers usually specify overall efficiency at rated load and a few other conditions only; this gives the total losses for those conditions. Separating the total losses into the components listed above requires detailed tests (such as open-circuit and short-circuit tests, blocked-rotor tests, and additional measurements) that are typically performed in a laboratory.

A practical approximation of the developed mechanical power is obtained by subtracting the known copper losses (armature and field) from the input electrical power. The shaft power is then found by subtracting mechanical losses (friction and windage, stray losses, core loss share if appropriate) from the developed mechanical power.

Figure 59: Power-flow diagram for a synchronous motor