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A 400V, 3-phase, star-connected synchronous motor has armature current of 200A at effective resistance of 0.04 OHMS. The short-circuit load loss at half-full load is _________.
1000 W
2100 W
2000 W
1200 W
1200 W
The short-circuit load loss in a 3-phase machine is defined by the copper loss occurring in the armature windings. For a 3-phase star-connected system, the total power loss is given by PlossтАЛ=3├ЧI2├ЧR.
The short-circuit load loss in a 3-phase machine is defined by the copper loss occurring in the armature windings. For a 3-phase star-connected system, the total power loss is given by PlossтАЛ=3├ЧI2├ЧR.
Similar to water flowing through pipes; if you halve the flow rate, the friction-related energy lost to heat reduces drastically because losses scale with the square of the flow rate.
I squared R is the star for power loss.
P=3├ЧI2├ЧR тАФ formula for total power loss in a 3-phase star connection.
PhalfтАЛ=3├Ч(2IтАЛ)2├ЧR тАФ formula for power loss at half-load condition.
The total armature copper loss is proportional to the square of the armature current I and the effective resistance per phase R. At half-full load, the current IhalfтАЛ=21тАЛ├ЧIfullтАЛ. Therefore, the power loss becomes PhalfтАЛ=3├Ч(2IfullтАЛтАЛ)2├ЧR, which simplifies to 41тАЛ of the full-load copper loss.
Copper losses in synchronous machines are variable losses.
Variable losses depend on the square of the load current.
The resistance R given is the effective resistance per phase.
Allows calculation of efficiency at various loading conditions.
Does not account for iron or stray losses.
Efficiency calculation of synchronous motors.
Thermal design and cooling requirements for electrical machines.
Given: Full load current I=200┬аA, Resistance R=0.04┬а╬й.
Full load loss = 3├Ч2002├Ч0.04=3├Ч40000├Ч0.04=4800┬аW.
Half load loss = 44800тАЛ=1200┬аW.
Options like 2000W or 2100W are incorrect because they fail to account for the I2 relationship or the 3-phase multiplier.
D is correct тАФ The half-full load loss is calculated as one-fourth of the total full-load copper loss, resulting in 1200┬аW.
Always remember that copper loss is a 'variable loss' that follows the square law (I2R). Whenever load changes, remember the square of the ratio of the load current.
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