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A three phase squirrel cage induction motor has maximum torque equal to twice the full load torque. The ratio of motor starting torque to its full load torque, if it is started by start-delta starter given slip at maximum torque is 0.2 is –
Correct answer is between '0.2,0.3'. Can you explain this answer?
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A three phase squirrel cage induction motor has maximum torque equal t...
When using, Direct on line starter,
Starting torque of star delta starter is one third of that obtained by DOL starter.
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A three phase squirrel cage induction motor has maximum torque equal t...
Given:
Maximum torque (Tmax) = 2 x Full load torque (TFL)
Slip at maximum torque (smax) = 0.2
We know that:
Tmax = (3V^2 / 2ω) x ((smax / R2) / (1 + (smax / R2)^2))
Where V = line voltage, ω = angular frequency, R2 = rotor resistance
Let the full load torque (TFL) be T
The starting torque (Tst) of the motor when started by star-delta starter can be calculated as:
Tst = T / (1 - sst)
Where sst is the slip at starting
In a star-delta starter, the starting current is reduced to 1/3rd of the full load current. Therefore, the starting torque is reduced by a factor of (1/3)^2 = 1/9 compared to the full load torque. Hence,
Tst = T / 9
We need to find the ratio of starting torque to full load torque, i.e., Tst / T
Substituting the values of Tmax and smax in the equation for Tmax, we get:
2TFL = (3V^2 / 2ω) x ((0.2 / R2) / (1 + (0.2 / R2)^2))
Simplifying the above equation, we get:
R2 = 0.4 / sqrt(3TFL)
Substituting the value of R2 in the equation for Tmax, we get:
2TFL = (3V^2 / 2ω) x 0.2 / (0.04 + 0.6TFL)
Simplifying the above equation, we get:
TFL^2 - 5TFL + 10 = 0
Solving the above quadratic equation, we get:
TFL = 2.56 or 1.94
Since TFL cannot be negative, we take TFL = 2.56 N-m
Substituting the value of TFL in the equation for R2, we get:
R2 = 0.4 / sqrt(3 x 2.56) = 0.162 Ω
Now, substituting the values of T and sst in the equation for Tst, we get:
Tst = T / (1 - sst) = TFL / (1 - 0.2 / 3) = 2.56 / 0.9333 = 2.743 N-m
Therefore, the ratio of starting torque to full load torque is:
Tst / TFL = 2.743 / 2.56 = 1.072
Hence, the required ratio is 1.072.
Maximum torque (Tmax) = 2 x Full load torque (TFL)
Slip at maximum torque (smax) = 0.2
We know that:
Tmax = (3V^2 / 2ω) x ((smax / R2) / (1 + (smax / R2)^2))
Where V = line voltage, ω = angular frequency, R2 = rotor resistance
Let the full load torque (TFL) be T
The starting torque (Tst) of the motor when started by star-delta starter can be calculated as:
Tst = T / (1 - sst)
Where sst is the slip at starting
In a star-delta starter, the starting current is reduced to 1/3rd of the full load current. Therefore, the starting torque is reduced by a factor of (1/3)^2 = 1/9 compared to the full load torque. Hence,
Tst = T / 9
We need to find the ratio of starting torque to full load torque, i.e., Tst / T
Substituting the values of Tmax and smax in the equation for Tmax, we get:
2TFL = (3V^2 / 2ω) x ((0.2 / R2) / (1 + (0.2 / R2)^2))
Simplifying the above equation, we get:
R2 = 0.4 / sqrt(3TFL)
Substituting the value of R2 in the equation for Tmax, we get:
2TFL = (3V^2 / 2ω) x 0.2 / (0.04 + 0.6TFL)
Simplifying the above equation, we get:
TFL^2 - 5TFL + 10 = 0
Solving the above quadratic equation, we get:
TFL = 2.56 or 1.94
Since TFL cannot be negative, we take TFL = 2.56 N-m
Substituting the value of TFL in the equation for R2, we get:
R2 = 0.4 / sqrt(3 x 2.56) = 0.162 Ω
Now, substituting the values of T and sst in the equation for Tst, we get:
Tst = T / (1 - sst) = TFL / (1 - 0.2 / 3) = 2.56 / 0.9333 = 2.743 N-m
Therefore, the ratio of starting torque to full load torque is:
Tst / TFL = 2.743 / 2.56 = 1.072
Hence, the required ratio is 1.072.
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A three phase squirrel cage induction motor has maximum torque equal to twice the full load torque. The ratio of motor starting torque to its full load torque, if it is started by start-delta starter given slip at maximum torque is 0.2 is –Correct answer is between '0.2,0.3'. Can you explain this answer?
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