Gilgamesh07
Student
- Apr 11, 2022
- 4
Problem:
The propulsion system of a train consists of inverter-fed three-phase induction traction motor. At base speed, the train has the speed 100 km/h. The maximum torque is (1.5Tn) where Tn is the rated torque and the breakdown torque is (1.875 Tn).
The maximum speed of the train is increased to 250 km/h by increasing the rated power of the three-phase inverters (e.g. by changing the base speed) and by using traction motors with a 10% higher breakdown torque. Determine the increase in rated power of the three-phase motor inverters.
Solution:
In the region above base speed, the breakdown torque is inversely proportional to the square of the speed whereas the maximum torque is inversely proportional to the speed.
The new motors have 10% higher breakdown torque => Breakdown torque = 1.10 * 1.875*Tn = 2.0625*Tn Where Tn is rated torque.
Above base speed : (Breakdown torque / maximum torque ) = (2.0625*Tn/w^2)/(1.5*Tn/w) = 1.375/w
For w = 2.5 => (Breakdown torque / maximum torque ) = 0.55 < The requirement of being 25% higher breakdown torque over the whole speed range.
The new rated torque is changed by (Tn' = k Tn)
(Breakdown torque / maximum torque ) = (2.0625*k*Tn/w^2)/(1.5*Tn/w) =1.25
=> (1.375*k/w) = 1.25
For speed w = 2.5 => k = (1.25*2.5)/1.375 = 2.27
The motors must have 127% higher rated power.
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ANY COMMENTS ON THE SOLUTION ABOVE ?
I am thinking if the train has 10 motors this mean an increase of 12.7% of the rated power of each motor.
Shouldn't this be achievable?
The propulsion system of a train consists of inverter-fed three-phase induction traction motor. At base speed, the train has the speed 100 km/h. The maximum torque is (1.5Tn) where Tn is the rated torque and the breakdown torque is (1.875 Tn).
The maximum speed of the train is increased to 250 km/h by increasing the rated power of the three-phase inverters (e.g. by changing the base speed) and by using traction motors with a 10% higher breakdown torque. Determine the increase in rated power of the three-phase motor inverters.
Solution:
In the region above base speed, the breakdown torque is inversely proportional to the square of the speed whereas the maximum torque is inversely proportional to the speed.
The new motors have 10% higher breakdown torque => Breakdown torque = 1.10 * 1.875*Tn = 2.0625*Tn Where Tn is rated torque.
Above base speed : (Breakdown torque / maximum torque ) = (2.0625*Tn/w^2)/(1.5*Tn/w) = 1.375/w
For w = 2.5 => (Breakdown torque / maximum torque ) = 0.55 < The requirement of being 25% higher breakdown torque over the whole speed range.
The new rated torque is changed by (Tn' = k Tn)
(Breakdown torque / maximum torque ) = (2.0625*k*Tn/w^2)/(1.5*Tn/w) =1.25
=> (1.375*k/w) = 1.25
For speed w = 2.5 => k = (1.25*2.5)/1.375 = 2.27
The motors must have 127% higher rated power.
------------------------------------------------------------------------------------------------------------------------------------
ANY COMMENTS ON THE SOLUTION ABOVE ?
I am thinking if the train has 10 motors this mean an increase of 12.7% of the rated power of each motor.
Shouldn't this be achievable?
