By changing the motor’s connection from delta to star, by how much will the actual power decrease?
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Dear sea friends: There are two wiring methods for electric motors, one is star connection and the other is delta connection. If the wiring configuration of a 75KW motor is changed from delta to star, by how much will the actual power decrease?
This post was last edited by zhaohh3211 on 2011-8-29 at 12:49. I never thought about it before, but now I really want to know. I’m sharing my own thoughts in the hope of sparking further discussion; I look forward to answers from those who are more knowledgeable. My immediate thought is that, after the connection is changed, the voltage drop across each phase of the coil becomes 1/1.732 of the original value, while the impedance of the coil remains unchanged; accordingly, the current also decreases to 1/1.732 of its original value. So the power would be 1/3 of the original value? It seems like it’s dropped too much, so I’m not sure.
Reply 1# Wu Ningyi Tang: P∝U2; p1/p = (220/380)2 = 0.335 (times)
The formula for calculating motor power is P=3*U_phase*I_phase. With U_phase remaining constant at 220V, the value of I_phase in a delta connection is 1.73 times that in a star connection. Therefore, I believe that the power in a delta connection is 1.73 times the power in a star connection. 1/1073=0.58, so the power decreases by 42% when changing from a triangular configuration to a star configuration, (1-0.58=0.42).
The original power has been reduced by 1/3!
It’s indeed a rather complicated issue~ When the input voltage and power remain constant, if the number of turns in the star-connected winding is low, the actual load voltage per phase winding is 220V, and the wire diameter has to be larger； The delta-connected winding has 1.732 times as many turns as the star-connected winding; the load voltage for each phase winding is 380V, and the wire diameter is relatively smaller. Therefore, when a motor that was originally connected in a star configuration is connected to the power supply in a delta configuration, the voltage across its windings becomes √3 times that in the star configuration. Since power generally increases in proportion to the square of the voltage, the power will increase to 3 times its original value.
The power should remain unchanged, right? For example, star-delta starting only reduces the starting current; it seems that the power doesn’t change at all
The power drops by 2/3 when connected in Y. Only 1/3 of the rated power.
I don’t quite understand it; I’ll take my time. You can’t expect to achieve everything at once!
This post was last edited by zhaohh3211 on 2011-8-29 at 12:55. Generally, motors with a power rating of 4 kW or less use a star connection, while in practical applications, a reduced-voltage starting method is used only for motors of 75 kW.
This post was last edited by zhaohh3211 on 2011-3-8 at 20:33. By connecting the triangle configuration to a star configuration, the operating current decreases by about one-third. Based on this principle, it seems that the power should also decrease by about one-third. However, there is one factor that cannot be ignored when discussing this issue: the power factor. Will the power factor change as a result of the change in winding connection during prolonged operation? What is the approximate rate of change?