Question

The full load current of a 3.3 kV star connected synchronous motor is 150 A at 0.6 pf lagging. The resistance and synchronous reactance of motor are 0.6 $\Omega$ and 5.0 $\Omega$ per phase respectively. The mechanical stray load losses are 35 kW. The efficiency of motor at full load is____.

• A. 69.74%
• B. 81.42%
• C. 63.92%
• D. 88.71%

Answer 1

The correct option is C 63.92%

Consider the following circuit:



Full load current $=150\angle -{53.13}^o$
Synchronous impedance, $Z_s=(0.6+j5.0)\Omega$
$=5.035\angle {83.157}^o\Omega$

For circuit diagram we can write
${\overrightarrow{E}}_f=1.905\times {10}^3\angle 0^o-5.035\angle {83.15}^o\times 150\angle -{53.13}^o$
${\overrightarrow{E}}_f=1.306\angle -16.80$ kV

Now, $P_{mech\left(dev\right)}=3\times 1.306\times 150cos(-{53.13}^o-16.80)$
$=473.46kW$

Shaft output $=473.46kW-35kW=438.46$ kW

Power input $=\sqrt{3}\times 3.3\times 150\times 0.8=685.0kW$

Efficiency, ${\eta }_{fullload}=\frac{Output}{Input}\times 100\%=\frac{438.46}{685.9}\times 100\%$
$=63.92\%$