The full-load current of a 3.3 kVA, 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. Calculate the excitation emf, torque angle, efficiency and shaft output of the motor. Assume the mechanical stray load loss to be 30 kW.
Question 8.41: The full-load current of a 3.3 kVA, star-connected synchrono...
Refer Fig. 8.113.
X_{s}=0.8+j 5.5=5.56 \angle 81.7^{\circ} \Omega\bar{E}_{f}=1.905-5.56 \angle 81.7^{\circ} \times 160 \angle-36.9^{\circ} \times 10^{-3}
=1.42 \angle-26.2^{\circ}
Or E_{f} = 1.42 kV (phase) or 2.46 kV (line)
P_{\text {mech }}(\text { dev })=3 \times 1.42 \times 160 \cos \left(-36.9^{\circ}+26.2^{\circ}\right)= 670 kW
Shaft output = 670 – 30 = 640 kW
Power input =\sqrt{3} \times 3.3 \times 160 \times 0.8=731.5 kW
\eta=640 / 731.5=87.5 \%
Related Answered Questions
(a) Total load = 260 MW; full load each generator ...
\phi=-\cos ^{-1} 0.8=-36.9^{\circ}...
The speed-load characteristics of the two generato...
On equivalent star basis
X_{d}=0.12 / 3=0....
\bar{V}_{t}=6.6 / \sqrt{3} \quad \angle 0^...
Base ( MVA )_...
(a) Refer Fig. 8.15
SCR =\text { of }^{\pr...
The circuit diagram of the system is drawn in Fig....
230 MW at UPF
I_{a}=\frac{230}{\sqrt{3} \t...
I_{a}(f l)=\frac{1000}{\sqrt{3} \times 6.6...