Question 3.1: An induction machine with the following nameplate data is in......
An induction machine with the following nameplate data is initially operating under its rated condition in steady state, supplying its rated torque. Calculate initial values of flux linkages and the load torque. Assume that the phase-a voltage has its positive peak at time t = 0.
Nameplate Data
Power: 3 HP/2.4 kW
Voltage: 460 V (L-L, rms)
Frequency: 60 Hz
Phases: 3
Full-Load Current: 4 A
Full-Load Speed: 1750 rpm
Full-Load Efficiency: 88.5%
Power Factor: 80.0%
Number of Poles: 4
Per-Phase Motor Circuit Parameters
R_s = 1.77 Ω
R_r = 1.34 Ω
X_{\ell s} = 5.25 Ω (at 60 Hz)
X_{\ell r} = 4.57 Ω (at 60 Hz)
X_m = 139.0 Ω (at 60 Hz)
Full-Load Slip = 1.72%
The iron losses are specified as 78 W and the mechanical (friction and windage) losses are specified as 24 W. The inertia of the machine is given. Assuming that the reflected load inertia is approximately the same as the motor inertia, the total equivalent inertia of the system is J_{eq} = 0.025 kg · m².
A MATLAB file EX3_1.m on accompanying website is based on the following steps:
Step 1 Calculate by phasor analysis \bar{V}_a, \bar{I}_a, and \bar{I}_A\left(=-\bar{I}_{r a}^{\prime}\right), given that the phase-a voltage has a positive peak at time t = 0.
Step 2 Calculate the current space vectors \vec{i}_s^a and \vec{i}_r^a at time t = 0 from the phasors for phase-a.
Step 3 In the dq analysis, choose \omega_d=\omega_{\text{syn}} and \theta_{da}(0)=0. Calculate i_{s \alpha}, i_{s \beta}, i_{r \alpha}, and i_{r \beta} from the space vectors in step 2 , using equations similar to Eq. (3-64) and Eq. (3-65).
i_{s d}(0)=\sqrt{\frac{2}{3}} \times \text { projection of } \vec{i}_s(0) \text { on } d \text {-axis }=\sqrt{\frac{2}{3}} \underbrace{\left\lgroup\frac{3}{2} \hat{I}\right\rgroup}_{\hat{I}_s} \cos \left(\theta_i\right)\enspace \text{(3-64)}
i_{s q}(0)=\sqrt{\frac{2}{3}} \times \text { projection of } \vec{i}_s(0) \text { on } q \text {-axis }=\sqrt{\frac{2}{3}} \underbrace{\left\lgroup\frac{3}{2} \hat{I}\right\rgroup}_{\hat{I}_s} \sin \left(\theta_i\right)\text{. (3-65)}
Step 4 Calculate flux linkages of dq windings using Eq. (3-61).
\left[\begin{array}{c}\lambda_{s d} \\\lambda_{s q} \\\lambda_{r d} \\\lambda_{r q}\end{array}\right]=\underbrace{\left[\begin{array}{cccc}L_s & 0 & L_m & 0 \\0 & L_s & 0 & L_m \\L_m & 0 & L_r & 0 \\0 & L_m & 0 & L_r\end{array}\right]}_{[M]}\left[\begin{array}{l}i_{s d} \\i_{s q} \\i_{r d} \\i_{r q}\end{array}\right].\hspace{30 pt} \text{(3-61)}
Step 5 Calculate torque T_L(0), which equals T_{e m} in steady state, from Eq. (3-46) or Eq. (3-47).
T_{e m}=\frac{p}{2}\left(\lambda_{r q} i_{r d}-\lambda_{r d} i_{r q}\right).\hspace{30 pt} \text{(3-46)}
T_{e m}=\frac{p}{2} L_m\left(i_{s q} i_{r d}-i_{s d} i_{r q}\right).\hspace{30 pt} \text{(3-47)}
The results from EX3_1.m are listed below:
\begin{aligned}& \lambda_{s d}(0)=0.0174 \text { Wb-turns } \\& \lambda_{r d}(0)=-0.1237 \text { Wb-turns } \\& \lambda_{s q}(0)=-1.1951 \text { Wb-turns } \\& \lambda_{r q}(0)=-1.1363 \text { Wb-turns } \\& T_{e m}(0)=T_L(0)=12.644\text{ Nm.}\end{aligned}