Answer to Question #203134 in Electrical Engineering for Anusha

Question #203134

A 3 phase, 400 V, 50 Hz, 4 pole, star connected SCIM has the following per phase constants

in ohms referred to stator:

r1= 0.15, x1=0.45, x2=0.45, Xm= 28.5

Rotational losses= 400W.

Compute stator current, rotor speed, output torque and efficiency when the motor is operated at

rated voltage and frequency at 4% slip.


1
Expert's answer
2021-06-07T05:46:24-0400

I2=vpn(Rs+RrS)+j(Xis+Xir)=230.940(0.15+0.120.04)+j(0.45+0.45)=70.49415.950I'_2=\frac{v_{pn}}{(R_s+\frac{R_r}{S})+ j(X_{is}+X_{ir})}=\frac{230.94 \angle0}{(0.15+\frac{0.12}{0.04})+ j(0.45+0.45)} =70.494 \angle -15.95^0

Im=VpmjXm=230.940j28.5A=8.103900I_m=\frac{V_{pm}}{jX_m}=\frac{230.94 \angle0}{j28.5} A = 8.103 \angle -90^0

a. State current, Is=Im+Is=73.13722.060I_s=I_m+I'_s=73.137 \angle -22.06^0

b. Synchronuos speed Ns=120fP=120504=1500rpmN_s=\frac{120f}{P}=\frac{120*50}{4}=1500 rpm

Nr=Ns(1s)=1500(10.04)=1440rpmN_r=N_s(1-s)=1500(1-0.04)=1440 rpm

c. Air gap power, =3I22Rrs    Pg=3(70.414)2(0.120.04)=44.7246kW=3 I_2'^2* \frac{R_r}{s} \implies P_g=3(70.414)^2*(\frac{0.12}{0.04})=44.7246 kW

Rotor power developed =(1-s)Pg =(1-0.04)(44.7246)kW=42.936 kW

Rotational loss =400 W = 0.4 kW

Shaft power = 42.936 -0.4 =42.536 kW

Output torque, Tsh=602πNsShaftpoweroutput=602π144042.5361000=282.07NmT_{sh}=\frac{60}{2 \pi N_s}* Shaft power output=\frac{60}{2 \pi *1440}*42.536*1000=282.07 Nm

d. Total power output = 42.536 kW

Total power input = 3vpmIscosϕs=3(230.94)(73.137)cos22.06W=46.961kW3v_{pm}I_s cos\phi_s=3(230.94)(73.137) cos 22.06 W= 46.961 kW

Efficiency=42.53646.961100=90.577%Efficiency = \frac{42.536}{46.961}*100=90.577 \%


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog