A small squirrel-cage induction motor has a starting current of six times the full-load current and a full-load slip of 0.05. Find in pu of full-load values, the current (line) and starting torque with the following methods of starting ((a) to (d)).
(a) direct switching.
(b) stator-resistance starting with motor current limited to 2 pu,
(c) autotransformer starting with motor current limited to 2 pu, and
(d) star-delta starting.
(e) what autotransformer ratio would give 1 pu starting torque?
(a) Direct switching
I_{s}=6 puT_{s}=(6)^{2} \times 0.05=1.8
(b) Stator-resistance starting
I_{s}=2 pu (limited to)
T_{s}=(2)^{2} \times 0.05=0.2 pu (Eq. (9.60)
\frac{T_{s}}{T_{f l}}=\left(\frac{I_{s}}{I_{f l}}\right)^{2} s_{f l} (9.60)
(c) Autotransformer starting
x = 2/6 = 1/3
I_{s}(\text { motor })=2 puI_{s}(\text { line })=\frac{1}{3} \times 2 pu =0.67 pu
T_{s}=(2)^{2} \times 0.05=0.2 pu (Eq. (9.60))
(d) Star-delta starting
I_{s}=\frac{1}{3} \times 6=2 pu (Eq. (9.68)
\frac{I_{s}(\text { star })}{I_{s}(\text { delta })}=\frac{1}{3} (9.68)
T_{s}=\frac{1}{3} \times(6)^{2} \times 0.05=0.6 pu (Eq. (9.69)
\frac{T_{s}(\operatorname{star})}{T_{f l}}=\frac{1}{3}\left(\frac{I_{S C}}{I_{f l}}\right)^{2} s_{f l} (9.69)
(e) Autotransformer starting
T_{s}=x^{2}(6)^{2} \times 0.05=1.0 pu (Eq. (9.64))
\frac{T_{s}}{T_{f l}}=x^{2}\left(\frac{I_{S C}}{I_{f l}}\right)^{2} s_{f l} (9.64)
x=0.745(\approx 75 \% \text { tap })