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 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 to limit starting current to 2$\times$ rated.
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
\[
\Tmax=\frac{3}{2\times78.54}\cdot\frac{(1270.1)^2}{0.075+\sqrt{0.075^2+1^2}}
\]
\[
=\frac{3\times1612154}{2\times78.54\times1.0028}
\]
\[
\Tmax=30{,}690\;\text{N-m}
\]
\[
\frac{T_{\max}}{T_r}=\frac{30690}{33778}=\ans{0.908}
\]
Note: Motor marginally overloaded at breakdown. External starting resistance recommended for this machine.
Step 5: Autotransformer ratio
Rated current: $I_r = T_r\,\omegams/V_\phi$
$\approx 33778\times78.54/(3\times1270)=695\;\text{A}$
Limit line current: $I_{\text{line}}\leq 2\times695=1390\;\text{A}$ \[ a_T^2 = \frac{I_{\text{line,DOL}}}{I_{st}}=\frac{1390/I_{st}\cdot I_{st}}{I_{st}} \] \[ a_T^2 = \frac{1390}{1251}\Rightarrow a_T=\sqrt{\frac{2\times695}{1251}} \] \[ \ans{a_T = 0.627} \] Motor starting current $=I_{st}\times a_T^2=1251\times0.393=491\;\text{A}$
Line current $=491/0.627=783\;\text{A}$
Rated current: $I_r = T_r\,\omegams/V_\phi$
$\approx 33778\times78.54/(3\times1270)=695\;\text{A}$
Limit line current: $I_{\text{line}}\leq 2\times695=1390\;\text{A}$ \[ a_T^2 = \frac{I_{\text{line,DOL}}}{I_{st}}=\frac{1390/I_{st}\cdot I_{st}}{I_{st}} \] \[ a_T^2 = \frac{1390}{1251}\Rightarrow a_T=\sqrt{\frac{2\times695}{1251}} \] \[ \ans{a_T = 0.627} \] Motor starting current $=I_{st}\times a_T^2=1251\times0.393=491\;\text{A}$
Line current $=491/0.627=783\;\text{A}$
Problem 4 — Star-Delta Starting
Problem Statement
A 400 V, 50 Hz, 4-pole, Y-connected, 15 kW induction motor has the following parameters:
$\Rs=0.5\,\Omega$, $\Rr=0.4\,\Omega$, $\Xs=\Xr=1.5\,\Omega$.
Compare DOL vs.\ star-delta starting: find starting current and torque for each method.
DOL Starting ($s=1$, Y-connected)
\[
V_\phi = 400/\sqrt{3}=231\;\text{V}
\]
\[
I_{st}^{\text{DOL}}=\frac{231}{\sqrt{0.9^2+3^2}}=\frac{231}{3.132}=73.8\;\text{A}
\]
\[
T_{st}^{\text{DOL}}=\frac{3\times73.8^2\times0.4}{(2\pi\times1500/60)}=\frac{6554}{157.1}
\]
\[
T_{st}^{\text{DOL}}=41.7\;\text{N-m}
\]
Star-Delta Starting
In $\Delta$ mode: $V_\phi=400\;\text{V}$ (phase)
At start, each winding at $400/\sqrt{3}=231\;\text{V}$ \[ I_{st}^{Y}=I_{st}^{\Delta}/3=73.8\;\text{A (line)} \] \[ \Rightarrow I_{\text{per-phase}}^{Y}=73.8/\sqrt{3}=42.6\;\text{A} \] \[ T_{st}^{Y}=\frac{1}{3}T_{st}^{\Delta}=\frac{41.7}{3}=\ans{13.9\;\text{N-m}} \] Line current $=\ans{73.8\;\text{A}}$ (same as DOL per-phase but 1/3 of DOL line)
In $\Delta$ mode: $V_\phi=400\;\text{V}$ (phase)
At start, each winding at $400/\sqrt{3}=231\;\text{V}$ \[ I_{st}^{Y}=I_{st}^{\Delta}/3=73.8\;\text{A (line)} \] \[ \Rightarrow I_{\text{per-phase}}^{Y}=73.8/\sqrt{3}=42.6\;\text{A} \] \[ T_{st}^{Y}=\frac{1}{3}T_{st}^{\Delta}=\frac{41.7}{3}=\ans{13.9\;\text{N-m}} \] Line current $=\ans{73.8\;\text{A}}$ (same as DOL per-phase but 1/3 of DOL line)
Comparison: Y-$\Delta$ reduces line current and torque to 1/3 of DOL values.
DOL line current $=\sqrt{3}\times73.8=127.8\;\text{A}$; Y-$\Delta$ line current $=73.8\;\text{A}$.
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 $R_e'$
\[
n_s=\frac{120\times50}{6}=1000\;\text{rpm}
\]
\[
\omegams=104.72\;\text{rad/s}
\]
\[
I_{st}=\frac{231}{\sqrt{(1+4.123)^2+16}}=\frac{231}{6.31}=36.6\;\text{A}
\]
\[
T_{st}=\frac{3\times36.6^2\times4.123/1}{104.72}
\]
\[
\ans{T_{st}=154.1\;\text{N-m (maximum possible)}}
\]