Solved Problems · Solid-State Drives

Topic 2: Starting Methods

DOL · Star-Delta · Rotor Resistance — Induction Motor Drives

Prof. Mithun MondalBITS Pilani HyderabadFree & Open Access

Quick Revision of Key Formulae \& Concepts

Quick Revision — Starting Methods

Star-Delta Starter
Current ratio: $I_{\text{line}}^{Y}/I_{\text{line}}^{\Delta}=1/3$
Torque ratio: $T^{Y}/T^{\Delta}=1/3$
Voltage per winding reduced to $V_L/\sqrt{3}$
Autotransformer — Ratio $a_T < 1$
\[ I_{\text{supply}} = a_T^2\,I_{\text{DOL}} \] \[ T_{\text{start}} = a_T^2\,T_{\text{DOL}} \]
Rotor Resistance Insertion (Slip-ring)
Maximum torque condition: \[ s_{T_{\max}} = \frac{\Rr+R_e}{\sqrt{\Rs^2+(\Xs+\Xr)^2}} \] At $s=1$: $R_e = \sqrt{\Rs^2+(\Xs+\Xr)^2}-\Rr$
for $\Tmax$ at starting.
Series Reactor / Impedance
\[ V_{\text{motor}} = V_s \cdot \frac{Z_{\text{motor}}}{Z_{\text{motor}}+Z_e} \]

Problems: Starting Methods

Problem 3 — Starting Torque \& Maximum Torque

Problem Statement
A 2200 V, 2600 kW, 735 rpm, 50 Hz, 8-pole, 3-phase squirrel-cage, Y-connected motor: $\Rs=0.075\,\Omega$, $\Rr=0.1\,\Omega$, $\Xs=0.45\,\Omega$, $\Xr=0.55\,\Omega$. Find: (a) starting torque as ratio of rated torque, (b) maximum torque as ratio of rated torque, (c) autotransformer ratio required to limit the supply-side starting line current to the rated full-load line current.
Step 1: Rated quantities \[ n_s=\frac{120\times50}{8}=750\;\text{rpm} \] \[ \omegams=\frac{2\pi\times750}{60}=78.54\;\text{rad/s} \] \[ s_{\text{rated}}=\frac{750-735}{750}=0.02 \] \[ V_\phi=\frac{2200}{\sqrt{3}}=1270.1\;\text{V} \] Rated torque: $T_r=\dfrac{2600\times10^3}{2\pi\times735/60}=33{,}778\;\text{N-m}$
Step 2: Starting current (DOL, $s=1$) \[ I_{st}=\frac{1270.1}{\sqrt{(0.075+0.1)^2+(0.45+0.55)^2}} \] \[ =\frac{1270.1}{\sqrt{0.0306+1}}=\frac{1270.1}{1.0152} \] \[ I_{st}=1251\;\text{A} \] Step 3: Starting torque \[ T_{st}=\frac{3\times1251^2\times0.1}{78.54} \] \[ T_{st}=5979\;\text{N-m} \] \[ \frac{T_{st}}{T_r}=\frac{5979}{33778}=\ans{0.177} \]

Problem 3 — Maximum Torque \& Autotransformer

Step 4: Maximum torque \[ \sqrt{\Rs^{2}+(\Xs+\Xr)^{2}}=\sqrt{0.0056+1}=1.0028 \] \[ \Rs+\sqrt{\dots}=0.075+1.0028=1.0778\;\Omega \] \[ \Tmax=\frac{3V_\phi^{2}}{2\omegams(1.0778)}=\frac{3\times1270.1^{2}}{2\times78.54\times1.0778} \] \[ \ans{\Tmax=28{,}588\;\text{N-m}} \] \[ \ans{\frac{T_{\max}}{T_r}=\frac{28588}{33778}=0.846} \] Note: $T_{\max}
Step 5: Rated line current
Assuming $\eta\approx0.93$, $\cos\phi\approx0.88$: \[ I_r=\frac{P_{\text{out}}}{\sqrt{3}\,V_L\,\eta\cos\phi}=\frac{2.6\times10^{6}}{\sqrt{3}\times2200\times0.93\times0.88} \] \[ I_r\approx 834\;\text{A} \] Step 6: Autotransformer ratio
To limit the supply-side line current to rated value $I_r=834\;\text{A}$: \[ a_T^{2}\,I_{\text{st}}^{\text{DOL}} = I_r \] \[ a_T^{2}=\frac{834}{1251}=0.667 \Rightarrow \ans{a_T=0.816} \] Motor current at start: $a_T\cdot I_{\text{st}}^{\text{DOL}}=0.816\times1251=1021\;\text{A}$

Problem 4 — Star-Delta Starting

Problem Statement
A 400 V, 50 Hz, 4-pole, 15 kW, Δ-connected induction motor has per-phase parameters $\Rs=0.5\,\Omega$, $\Rr=0.4\,\Omega$, $\Xs=\Xr=1.5\,\Omega$ (referred to stator). Compare DOL and star-delta starting: compute the line current and starting torque for each.
Note: Star-delta starting requires the motor to be rated for Δ operation at 400 V; the windings are temporarily re-connected in Y during start.
DOL in Δ ($s=1$)
Phase voltage $V_\phi = V_L = 400\;\text{V}$ \[ |Z_{\text{st}}|=\sqrt{(0.5+0.4)^{2}+3^{2}}=3.132\;\Omega \] \[ I_\phi^{\Delta}=\frac{400}{3.132}=127.7\;\text{A} \] \[ \ans{I_L^{\Delta}=\sqrt{3}\,I_\phi^{\Delta}=221.2\;\text{A}} \] Starting torque ($\omegams=157.08$ rad/s): \[ T_{st}^{\Delta}=\frac{3\times127.7^{2}\times0.4}{157.08} \] \[ \ans{T_{st}^{\Delta}=124.6\;\text{N-m}} \]
Star (Y) Start
Each winding sees $V_\phi = 400/\sqrt{3}=231\;\text{V}$ \[ I_\phi^{Y}=\frac{231}{3.132}=73.7\;\text{A} \] \[ \ans{I_L^{Y}=I_\phi^{Y}=73.7\;\text{A}} \] Starting torque: \[ T_{st}^{Y}=\frac{3\times73.7^{2}\times0.4}{157.08} \] \[ \ans{T_{st}^{Y}=41.5\;\text{N-m}} \]
Comparison: The line current ratio $I_L^{Y}/I_L^{\Delta}=73.7/221.2=\tfrac{1}{3}$ and the torque ratio $T^{Y}/T^{\Delta}=41.5/124.6=\tfrac{1}{3}$ — the classical star-delta reduction to one-third.

Problem 5 — Rotor Resistance for Maximum Starting Torque

Problem Statement
A 400 V, 6-pole, 50 Hz, star-connected slip-ring induction motor has: $\Rs=1\,\Omega$, $\Rr=1\,\Omega$, $\Xs=\Xr=2\,\Omega$. Find the external rotor resistance required to obtain maximum torque at starting ($s=1$). The stator-to-rotor turns ratio is $a=2.5$.
Step 1: Condition for $T_{\max}$ at $s=1$ \[ \Rr+R_e' = \sqrt{\Rs^2+(\Xs+\Xr)^2} \] \[ 1+R_e' = \sqrt{1+16} = 4.123\;\Omega \] \[ R_e' = 3.123\;\Omega\;\text{(referred to stator)} \]
Step 2: Actual rotor resistance \[ R_e = \frac{R_e'}{a^2} = \frac{3.123}{2.5^2} = \frac{3.123}{6.25} \] \[ \ans{R_e = 0.5\;\Omega\;\text{per phase}} \]
Step 3: Starting torque with $R_e'$ \[ n_s=\frac{120\times50}{6}=1000\;\text{rpm} \] \[ \omegams=104.72\;\text{rad/s},\quad V_\phi=231\;\text{V} \] At $s=1$ with $R_e'$ inserted: \[ |Z|=\sqrt{(1+4.123)^{2}+16}=\sqrt{42.24}=6.50\;\Omega \] \[ I_{st}=\frac{231}{6.50}=35.5\;\text{A} \] \[ T_{st}=\frac{3\times35.5^{2}\times 4.123}{104.72} \] \[ \ans{T_{st}=149.1\;\text{N-m} = T_{\max}} \]