syafruddin hasan. power flow diagram figure motor energy balance flow diagram
TRANSCRIPT
Syafruddin Hasan
Motorinputin
stator
Statorcopperandironlosses
Rotorinput
Rotorcopperlosse
MechPower
developedor
GrossTorque
Windageand
Frictionlosses
RotoroutputorBHP
Power Flow DiagramPower Flow Diagram
Input powerPsup
Stator Copper loss3 Ista
2 Rsta
Rotor Copper loss3 Irot
2 Rrot
Stator Iron loss3 Vsta
2 / Rc
Ventilation andfriction losses
Output powerPout
Air gappower Pag
Developed powerPdv = 3 Irot
2 Rrot (1-s)/s
Air gap
Figure Motor energy balance flow diagram.
Induction MotorsInduction MotorsFigure 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.
*stasupsup IVS 3Re)Re(Psup
Induction MotorsInduction Motors
The 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
s
sR trot
13P _
2
dv rot_tI
mlossdvout PPP
Power and Torque in an IMPower and Torque in an IM
From Fig.7-12Input 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
Induction MotorsInduction MotorsThe motor efficiency:
Motor torque (shaft load torque):
sup
out
P
P
m
outP
T
Equivalent CircuitEquivalent Circuit
Stator Rotor
Xrot_m = rot Lrot Rrot
Irot
Vrot = s Vrot_s
Rsta Irot_t
VstaVsup
IstaXmRc
Xsta = sy Lsta
Figure Single-phase equivalent circuit of a three-phase induction motor.
Induction MotorsInduction Motors
Stator Rotor
Xsta Xrot Rrot/s
IrotVrot_s
Rsta Irot_t
VstaVsup
IstaXmRc
Figure Modified equivalent circuit of a three-phase induction
motor.
The rotor impedance is transferred to the stator side. This eliminates the transformer
Induction MotorsInduction Motors
Figure Simplified equivalent circuit of a three-phase induction motor.
Stator Rotor
Xsta Xrot_t Rrot_t/s
Irot_tVsta
Rsta
Vsup
IstaXmRc
Air gap
Induction MotorsInduction 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
trottrot
trot Rs
sR
s
R__
_ 1
trotR
s
s_
1
Blocked-Rotor testBlocked-Rotor test (cont…)(cont…)
Vblocked_ln
Vblocked
3 Vblocked_ln 21.939 V
Pblocked_A
Pblocked
3 Pblocked_A 160 W
Re
Pblocked_A
Iblocked2
Rrot_t Re Rsta
The stator resistance was measured directly
IM Design ClassesIM Design Classes
NEMA & 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
Starting Induction MotorsStarting Induction Motors
Problems : - High Starting current - Low starting torque
Determining of Istart
Read the rated voltage, hp and code letter from name plate.
Starting apparent power :Sstart = (rated hp)(code letter factor)
Istart = Sstart / √3 VT
Example 7-7
Q U I Z Q U I Z 04-10-200504-10-2005
1. Why the circuit equivalent of an induction motor can be approached by equivalent circuit of a transformer and draw it
2. 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 this motorb). Determine the input powerc). Calculate the efficiency
Speed Control of IMSpeed Control of IM ns = 120 f / p
by changing the electrical frequency (f)
by changing the number of poles (p)(1) The methode of consequent poles(2) Multiple stator windings
by changing the line voltage:n proportional to V2
by changing the rotor resistance Solid-state IM drives
Motor ProtectionMotor ProtectionOver Current ProtectionOver Load Protection
IM RatingsIM Ratings1. Output power2. Voltage3. Current4. Power Factor5. Speed6. Nominal efficiency7. NEMA design Class8. Starting Code9. Service Factor (SF)