Question 16.1: A certain 5-hp three-phase induction motor operates from a 4...
A certain 5-hp three-phase induction motor operates from a 440-V-rms (line-to line) three-phase source and draws a line current of 6.8 A rms at a power factor of 78 percent lagging [i.e., cos(θ) = 0.78] under rated full-load conditions. The full-load speed is 1150 rpm. Under no-load conditions, the speed is 1195 rpm, and the line current is 1.2 A rms at a power factor of 30 percent lagging. Find the power loss and efficiency with full load, the input power with no load, and the speed regulation.
The rated output power is 5 hp. Converting to watts, we have
P_{out} = 5 × 746 = 3730 W
Substituting into Equation 16.1, we find the input power under full load:
P_{in} = \sqrt{3}V_{rms}I_{rms} \cos(θ) (16.1)
= \sqrt{3}(440)(6.8)(0.78) = 4042 W
The power loss is given by
P_{loss} = P_{in} − P_{out} = 4042 − 3730 = 312 W
The full-load efficiency is
η = \frac{P_{out}}{ P_{in}} × 100\% = \frac{3730}{ 4042} × 100% = 92.28%
Under no-load conditions, we have
P_{in} = \sqrt{3}(440)(1.2)(0.30) = 274.4 W
P_{out} = 0
P_{loss} = P_{in} = 274.4 W
and the efficiency is
η = 0%
Speed regulation for the motor is given by Equation 16.7. Substituting values,
we get
speed regulation = \frac{n_{no-load} − n_{full-load}} {n_{full-load}} × 100% (16.7)
= \frac{1195 − 1150}{ 1150 }× 100% = 3.91%