Question 5.11: The 2000-hp, 2300-V synchronous motor of Example 5.7 is assu......
The 2000-hp, 2300-V synchronous motor of Example 5.7 is assumed to have a synchronous reactance X_s = 1.95 Ω/phase. In actual fact, it is a salient-pole machine with reactances X_d = 1.95 Ω/phase and X_q = 1.40 Ω/phase. Neglecting all losses, compute the maximum mechanical power in kilowatts which this motor can deliver if it is supplied with electric power from an infinite bus (Fig. 5.28a) at rated voltage and frequency and if its field excitation is held constant at that value which would result in unity-power-factor operation at rated load. The shaft load is assumed to be increased gradually so that transient swings are negligible and the steady- state power limit applies. Also, compute the value of δ corresponding to this maximum power operation.
The first step is to compute the synchronous motor excitation at rated voltage, full load, and unity power factor. As in Example 5.7, the full-load terminal voltage and current are 1330 V line-to-neutral and 374 A/phase, respectively. The phasor diagram for the specified full-load conditions is shown in Fig. 5.28b. The only essential difference between this phasor diagram and the generator phasor diagram of Fig. 5.25 is that \hat{I}_a in Fig. 5.28 represents motor input current; i.e., we have switched to the motor reference direction for \hat{I}_a. Thus, switching the sign of the current to account for the choice of the motor reference direction and neglecting the effects of armature resistance, Eq. 5.59 becomes
\hat{E}_{af}=\hat{V}_a + R_a \hat{I}_a + j X_d \hat{I}_d + jX_q\hat{I}_q (5.59)
\hat{E}_{af}=\hat{V}_a – j \hat{I}_dX_d – j\hat{I}_qX_q
As in Fig. 5.28b, the quadrature axis can now be located by the phasor
\hat{E}^′= \hat{V}_a – j\hat{I}_aX_q = 1330 – j374(1.40)= 1429 e^{-21.5°}
That is, δ = -21.5°, with \hat{E}_{af} lagging \hat{V}_a. The magnitude of \hat{I}_d is
I_d = I_a \sin |δ| = 374 \sin (21.5°) = 137 A
With reference to the phasor element labeled a^′c in Fig. 5.28b, the magnitude of \hat{E}_{af} can be found by adding the length a^′c = I_d(X_d – X_q) numerically to the magnitude of \hat{E}^′; thus
E_{af} = E^′ + I_d(X_d – X_q) = 1429 + 137(0.55) = 1504 V \text{line-to-neutral}
(Alternatively, E_{af} could have been determined as \hat{E}_{af} = \hat{V}_a – j \hat{I}_dX_d – j \hat{I}_qX_q.) From Eq. 5.65 the power-angle characteristic for this motor is
P = \frac{E_{af}V_{EQ}}{X_{dT}} \sin δ + \frac{V_{EQ}^2 (X_{dT} – X_{qT})}{2X_{dT} X_{qT}} \sin 2δ (6.65)
\begin{aligned}P & = \frac{E_{af}V_{EQ}}{X_d} \sin |δ| + V_{EQ}^2\frac{X_d – X_q}{2X_d X_q} \sin (2|δ|)\\& = 1030 \sin |δ| + 178 \sin(2|δ|) \quad kW/\text{phase}\end{aligned}
Note that we have used |δ| in this equation. That is because Eq. 5.65 as written applies to a generator and calculates the electrical power output from the generator. For our motor, δ is negative and direct use of Eq. 5.65 will give a value of power P < 0 which is of course correct for motor operation. Since we know that this is a motor and that we are calculating electric power into the motor terminals, we ignore the sign issue here entirely and calculate the motor power directly as a positive number.
The maximum motor input power occurs when dP / dδ = 0
\frac{dP}{dδ}= 1030 \cos δ + 356 \cos 2δ
Setting this equal to zero and using the trigonometric identity
\cos 2α = 2 \cos ^2 α – 1
permit us to solve for the angle δ at which the maximum power occurs
δ = 73.2°
Therefore the maximum power is
P_{max}= 1080 kW/\text{phase} = 3240 kW,\text{three-phase}
We can compare this value with P_{max} = 3090 kW found in part (a) of Example 5.7, where the effects of salient poles were neglected. We see that the error caused by neglecting saliency is slightly less than five percent in this case.
