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: $a_T\cdot I_{\text{st}}^{\text{DOL}}=0.816\times1251=1021\;\text{A}$
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): \[ T_{st}^{\Delta}=\frac{3\times127.7^{2}\times0.4}{157.08} \] \[ \ans{T_{st}^{\Delta}=124.6\;\text{N-m}} \]
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: \[ T_{st}^{Y}=\frac{3\times73.7^{2}\times0.4}{157.08} \] \[ \ans{T_{st}^{Y}=41.5\;\text{N-m}} \]
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:
\[
|Z|=\sqrt{(1+4.123)^{2}+16}=\sqrt{42.24}=6.50\;\Omega
\]
\[
I_{st}=\frac{231}{6.50}=35.5\;\text{A}
\]
\[
T_{st}=\frac{3\times35.5^{2}\times 4.123}{104.72}
\]
\[
\ans{T_{st}=149.1\;\text{N-m} = T_{\max}}
\]