3 Figure Motor energy balance flow diagram. Power Flow DiagramFigure Motor energy balance flow diagram.
4 Induction Motors Figure shows the energy balance in a motor. The supply power is:The power transferred through the air gap by the magnetic coupling is the input power (Psup) minus the stator copper loss and the magnetizing (stator iron) loss.The electrically developed power (Pdv) is the difference between the air gap power (Pag) and rotor copper loss.
5 Induction MotorsThe electrically developed power can be computed from the power dissipated in the second term of rotor resistance:The subtraction of the mechanical ventilation and friction losses (Pmloss) from the developed power gives the mechanical output power
6 Power and Torque in an IM From Fig.7-12 Input current : I1 = VΦ / Zeq Pcu Stator: Pscl = 3 I12R1 Core Losses : Pcore = 3 E12 Gc Air-gap power : PAG = Pin – Pscl – Pcore = 3I22R2/s Pcu Rotor: PRcl = 3 IR2RR = 3 I22R2 Note PAG : Pconv : PRCL = 1 : (1-s) : S Example 7-3
7 Induction Motors The motor efficiency: Motor torque (shaft load torque):
8 Equivalent CircuitFigure Single-phase equivalent circuit of a three- phase induction motor.
9 Figure Modified equivalent circuit of a three-phase induction motor. Induction MotorsFigure Modified equivalent circuit of a three-phase induction motor.The rotor impedance is transferred to the stator side. This eliminates the transformer
10 Induction MotorsFigure Simplified equivalent circuit of a three-phase induction motor.
11 Induction MotorsThe last modification of the equivalent circuit is the separation of the rotor resistance into two parts:The obtained resistance represents the outgoing mechanical power
12 Blocked-Rotor test (cont…) The stator resistance was measured directly
13 IM Design ClassesNEMA & IEC : Class A: normal Tst, normal Ist, normal S Class B: normal Tst, low Ist, low S Class C: high Tst, low Ist, low S Class D: high Tst, low Ist, high S Class E: low Tst, normal Ist, low S Class F: low Tst, low Ist, normal S
14 Starting Induction Motors Problems : - High Starting current- Low starting torqueDetermining of IstartRead the rated voltage, hp and code letter from name plate.Starting apparent power :Sstart = (rated hp)(code letter factor)Istart = Sstart / √3 VTExample 7-7
15 Q U I Z1. Why the circuit equivalent of an induction motor can be approached by equivalent circuit of a transformer and draw it2. The output of a 3 phase induction motor is 9 kW. Rotor copper losses is 0.5 kW.Motor runs at 5 % of slip. Stator loss is 0.75 kWa). Calculate the mechanical power that converter by thismotorb). Determine the input powerc). Calculate the efficiency
16 Speed Control of IM ns = 120 f / p by changing the electrical frequency (f)by changing the number of poles (p)The methode of consequent polesMultiple stator windingsby changing the line voltage:n proportional to V2by changing the rotor resistanceSolid-state IM drives
17 Motor ProtectionOver Current ProtectionOver Load Protection
18 IM Ratings Output power Voltage Current Power Factor Speed Nominal efficiencyNEMA design ClassStarting CodeService Factor (SF)