Question 6.2: A three-phase Y-connected 220-V (line-to-line) 7.5-kW 60-Hz ......

Let the impedance Z_f (Fig. 6.11 a) represent the per phase impedance presented to the stator by the magnetizing reactance and the rotor. Thus, from Fig. 6.11 a

Z_f = R_f  +  jX_f = \left(\frac{R_2}{s}  –  jX_2\right)  \text{in parallel with}   j X_m

Substitution of numerical values gives, for s = 0.02,

R_f  +  jX_f = 5.41  +  j3.11  Ω

The stator input impedance can now be calculated as

Z_{in}= R_1  +  jX_1  +  Z_f = 5.70  +  j3.61 = 6.75  ∠32.3°  Ω

The line-to-neutral terminal voltage is equal to

V_1= \frac{220}{\sqrt{3}} = 127  V

and hence the stator current can be calculated as

\hat{I}_1 =\frac{V_1}{Z_{in}}= \frac{127}{6.75  ∠32.3°}= 18.8  ∠-32.3°  A

The stator current is thus 18.8 A and the power factor is equal to cos (-32.3°) = 0.845 lagging.

The synchronous speed can be found from Eq. 4.41

n_s=\left( \frac{120}{poles}\right)f_e=\left(\frac{120}{6}\right) 60 = 1200  r/min

or from Eq. 6.26

ω_s=\frac{4πf_e}{poles} = \left( \frac{2}{poles}\right)ω_e  \quad \quad \quad (6.26)

ω_s=\frac{4πf_e}{poles} = 125.7  rad/sec

The rotor speed is

n = (1 – s)n_s = (0.98)1200 = 1176  r/min

or

ω_m = (1 – s)ω_s = (0.98)125.7 = 123.2  rad/sec

From Eq. 6.17,

P_{gap}=n_{ph}I_2^2 \left( \frac{R_2}{s}\right)

Note however that because the only resistance included in Z_f  is  R_2/s, the power dissipated in Z_f is equal to the power dissipated in R_2/s and hence we can write

P_{gap} = n_{ph}I_1^2R_f = 3(18.8)^2(5.41) = 5740  W

We can now calculate P_{mech} from Eq. 6.21 and the shaft output power from Eq. 6.27. Thus

P_{mech}= n_{ph} I_2^2 R_2 \left( \frac{1-s}{s}\right)  \quad \quad (6.21)

P_{shaft}= P_{mech}  –  P_{rot} \quad \quad \quad (6.27)

\begin{aligned}P_{shaft}&= P_{mech}  –  P_{rot} = (1 -s)P_{gap}  –  P_{rot}\\ &= (0.98)5740  –  403 = 5220  W\end{aligned}

and the shaft output torque can be found from Eq. 6.28 as

T_{shaft}=\frac{P_{shaft}}{ω_m}= T_{mech}  –  T_{rot}             (6.28)

T_{shaft}=\frac{P_{shaft}}{ω_m}=\frac{5220}{123.2} = 42.4  N. m

The efficiency is calculated as the ratio of shaft output power to stator input power. The input power is given by

\begin{aligned}P_{in} & = n_{ph}Re[\hat{V}_1 \hat{I}_1^*] = 3Re[127(18.8  ∠32.3°)] \\ & = 3 × 127 × 18.8  \cos (32.2°) = 6060  W \end{aligned}

Thus the efficiency η is equal to

η=\frac{P_{shaft}}{P_{in}}=\frac{5220}{6060} = 0.861 = 86.1\%

The complete performance characteristics of the motor can be determined by repeating these calculations for other assumed values of slip.