Question

The full load current of a 3.3 kV, star connected synchronous motor is 160 A at 0.8 p.f. lagging. The resistance and synchronous reactance of motor are 0.8 $\Omega$ and $5.5\Omega$ per phase respectively. The mechanical stray load loss to be 30 kW. The efficiency of the motor at full load is ____%.

• A. 87.46

Answer 1

The correct option is A 87.46
Consider the following circuit

$\text{Full load current}=160\angle -{36.86}^{\circ }A$

$\text{Synchronous impedance,}{Z}_s=(0.8+j5.5)\Omega$

$=5.56\angle {81.724}^{\circ }\Omega$

$\text{From circuit diagram we can write,}$
$\overrightarrow{E_f}=1.905\times {10}^3\angle 0^{\circ }-5.56\angle {81.724}^{\circ }\times 160\angle -{36.86}^{\circ }$

$E_f=1.42\angle -{26.22}^{\circ }kV$

$\text{Now}$ $P_{mech\left(dev\right)}=3\times 1.42\times 160cos(-{36.86}^{\circ }+{26.22}^{\circ })$ $=669.88kW$

$\text{shaft output}=669.88-30=639.88kW$

$\text{Power input =}\sqrt{3}\times 3.3\times 160\times 0.8=731.62kW$

${\eta }_{fullload}=\frac{Output}{Input}\times 100=\frac{639.88}{731.62}\times 100=87.46\%$