Question

The full-load current of a 3.3 kV star connected synchronous motor is 160 A at 0.8 pf lagging. The resistance and synchronous reactance of the motor are $0.8\Omega$ and $5.5\Omega$ per phase respectively. The mechanical stray load losses are 30 kW. The efficiency of the motor is

• A. 87.5%
• B. 93.7%
• C. 83.4%
• D. 81.43%

Answer 1

The correct option is A 87.5%



$\overrightarrow{Z_s}=(0.8+j5.5)\Omega$

$=\left(5.56\angle {81.72}^o\right)\Omega$

$\overrightarrow{E_f}=1.905-5.56\angle {81.7}^{\circ }\times 0.16\angle -{26.9}^{\circ }$

$=1.42\angle -{26.2}^o$

or, $E_f=1.42kV\text{(phase 0 or 2.46 kV (line))}$

$P_{mech\left(dev\right)}=3\times 1.42\times 160cos(-{36.9}^{\circ }+{26.2}^{\circ })$ $=670kW$

$\text{Shaft output}=670-30=640kW$

$\text{Power input}=\sqrt{3}\times 3.3\times 160\times 0.8$ $=731.5kW$

$\text{Efficiency,}\eta =\frac{640}{731.5}=87.5\%$