A problem for drawing circle diagram for 3-phase induction motor

Given a 3- phase induction motor, 400V, Y connected, No, of poles =2, frequency =50HzNo load test:VNL = 400V INL=20A PNL=4500W

Blocked rotor testVBR=100V IBR=40A PBR=2000W

R2= 0.5 R1

(i) Draw the circle diagram after choosing a suitable scale(ii) Draw the output power line and torque line(iii) Find the maximum power factor then at this power factor find the current,

slip, efficiency, rotor copper loss, and torque(iv) Find the maximum Torque and at this power find the current, slip, efficiency,

power factor, stator copper loss and toque

Solution

The blocked rotor values should be corrected to 400VVBR =400V IBR = 40*400/100 =160A PBR= 2000*4*4=32000WThe power scale shall be taken as 4500W = 1 cmHence 32000W shall be 32000/4500 = 7.1 cmShort circuit power factor = 32000 / √3/160 /400 = 0.288 hence θ =73.2°Horizontal distance of the short circuit point = 7.1 * tan θ = 23.5 cmPF at no load = 4500 / √3/20 /400 = 0.324 hence θ =71°Horizontal distance of the noload point = 1 * tan θ = 2.9 cmThe distance ON =3.08cm = 20A Hence the scale of the current = 20/3.08 =6.5A/cmThe distance OP = 24.65 cm = 160ADraw ON at 3.08 cm and an angle θ =73.2°Draw OP at 24.65 cm and an angle θ =71°Draw PN and find the mid distance point MDraw MC perpendicular to PN. It intersects the horizontal line from N at CDraw a semicircle with center C and radius CN or CPDraw a vertical line from P. It intersects with horizontal line from N at DDivide PD into PR and RD so that PR/RD = R2/ R1= 0.5Draw the torque line RNThe power line is NPDraw maximum torque line tangent to the circle at T and parallel to NRDraw maximum power factor line tangent to the circle at F from originOF is the current at maximum power factor = 8.5 cm = 55.25AMaximum power factor = 6.75/8.5 = 0.794Locate F1, F2 and F3Slip = F1-F2/F-F2 = 0.25/5.25 = 0.047

Efficiency = F-F1/F-F3 = 5/6.7 = 74.6%Rotor cupper loss = F1-F2 = 0.25cm = 0.25*4500=1125WTorque = Airgap power F-F2 divided by synchronous speed = 5.25 * 4500/(3000*2π/60) =75N.m.At maximum torque the slip is T1-T2/T-T2 = 0.75/8 = 0.09375Efficiency = T-T1/T-T4 = 7.2/10.5 = 68.5%Stator Cupper Loss = T2-T3 = 1.55cm =1.55*4500 = 6975WTorque = T-T2/ωs = 8* 4500/(3000*2π/60) = 114N.m.