- #1
Lunat1c
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Hi, I got a small problem which I'm not sure I'm solving correctly. A small induction motor has a short circuit current equal to 5 times the full-load current. I need to find the starting torque as a percentage of full load torque if the motor is started by a star-delta starter. It is also known that the motor has a full load slip of 4%.
Torque = \frac{3I_{R}^2R_r}{(s*\omega_s)}
T_{start}= \frac{3(\frac{5}{3} I_{FL})^2R_r}{\omega_s}
Note that in the above equation i divided the current by three because in a star-delta starter the current is supposed to be 1/3 the direct online current.
and T_{FL} = \frac{3I_{FL}^2R_r}{0.04\omega_s}
Then dividing (1) by (2) i should get \frac{T_{start}}{T_{FL}}=(25/9)*0.04=0.11.
That is equivalent to 11.1% of the Full load torque. However my lecturer got the answer as 33.33%, which is probably because he did \frac{i^2}{3} instead of (\frac{i}{3})^2. Is my answer incorrect?
Torque = \frac{3I_{R}^2R_r}{(s*\omega_s)}
T_{start}= \frac{3(\frac{5}{3} I_{FL})^2R_r}{\omega_s}
Note that in the above equation i divided the current by three because in a star-delta starter the current is supposed to be 1/3 the direct online current.
and T_{FL} = \frac{3I_{FL}^2R_r}{0.04\omega_s}
Then dividing (1) by (2) i should get \frac{T_{start}}{T_{FL}}=(25/9)*0.04=0.11.
That is equivalent to 11.1% of the Full load torque. However my lecturer got the answer as 33.33%, which is probably because he did \frac{i^2}{3} instead of (\frac{i}{3})^2. Is my answer incorrect?
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