Overview: Induction Motors. Review Questions. Why the Rotor Moves: Motor Speed

Transcription

1 Overview: nduction Motor Motor operation & Slip Speed-torque relationhip Equivalent circuit model Tranformer Motor efficiency Starting induction motor Smith College, EGR 35 ovember 5, 04 Review Quetion ) Why doe a motor rotor rotate? ) How/Why doe a motor tart? 3) Once tarted, why doe the motor reach a teady-tate and not accelerate forever? 4) What define the teady-tate peed (and torque) of the motor? 5) Why are thee motor called induction motor? Why the Rotor Move: Motor Speed f the tator i energized with a 3ϕ voltage, a rotating flux at ynchronou peed n i induced in the airgap The relative peed, Δn, i the difference between the peed of the rotor (winding) and the airgap flux Δn n n f the rotor i tationary, then Δn n 3 4

2 Motor Speed con t Thu, the rotating field induce a voltage, e, in the rotor winding e electro-motive force (emf) With the rotor winding horted together, current flow, producing a torque on the winding (& the rotor).. So the rotor pin. What i the teady tate peed of the Faraday & Lorentz for Rotating Motion e Bl B lδ v e f ( φ, Δn ) F B l i T f ( φ,i ) Linear Motion Angular Motion induction motor? (<, >, n?) 5 6 i a Δ n φ c i b φ a φ b i c n n f (T load ) Apply 3-phae to Stator to Stator Field rotate in airgap Voltage induced in rotor circuit e B l Δ n Torque i developed in rotor wdg f B l i e i z What if motor peed i n? What happen to the induced voltage and the developed torque? Apply 3-phae to Stator Δn 0 Voltage induced in rotor circuit e B l Δn 0 Torque i developed in rotor wdg f B l i 0 Motor low down e i z 0 and the motor doe what? Motor rotate 7 8

3 What i the motor peed with load connected? Load Apply require 3-phae certain to Stator Torque Motor develop equivalent force f B l i That force induce a current Current induce a voltage e i z nduced voltage induce motor rotation à peed e B l Δn 9 Δn n Motor Slip () Baic meaure of motor performance and attached load Per unit value of the relative peed Δω ω n n n ω ω ω n i in rev/min (rpm) ω i in rad/ec 0 0 max Small Slip, typical operating range Maximum Torque Large Slip T tarting T max Torque n n S Δ n n n f (T ) load Example A -pole, 60 Hz induction motor operate at a lip of 0.0. Compute it 60 f f rotor peed. (Recall n 0 rev / min ) Solution f 60 n rpm p n n S n pp ( ) ( ) rpm p 3

4 Determine Synchronou Speed Compute n for, 4, 6, 8 and 0 pole machine, operating at 60Hz. umber of Pole Synchronou Speed (rpm) Motor Speed Example The teady-tate peed of a 60Hz induction motor i 735 rpm. What i the lip of thi motor? At teady-tate, lip i jut below the ynchronou peed, o 735 i jut below 800 Thi motor ha 4 pole S ( )/ % kV-750kV Ditribution Tranformer Tranformer 5 kv- 5kV Tranmiion Tranformer Service Tranformer 08V- 46V 5 6 4

5 Service Tranformer Service Tranformer 7 Service Tranformer 9 8 Ditribution Tranformer 0 5

6 Ditribution Tranformer Tranmiion Tranformer Low power Tranformer Why do we need tranformer? 3 Step up ncreae voltage of generator output Generator output ~ 000 V Tranmit high power at low current (up to 765kV) Reduce cot of tranmiion ytem, becaue Reduce tranmiion loe ( R loe) Step down to adjut voltage to a uable level Create electrical iolation 4 6

7 nput Power (P in ) Stator Loe: Copper loe (P cu ) Core loe (P iron ) Airgap Power (P g ) Objective: Determine Motor Loe & Efficiency Baic Tranformer Model ron Core Rotor Copper Loe (P cu ) Developed Power (P d ) Rotational Loe (P rotational ) Output Power (P out ) 5 6 e dφ dt i i e e e dφ dt V Tranformer Equivalent Circuit What doe each circuit element repreent? ' R X R X R c X c c E E V load Primary Secondary 7 E E V V ' 8 7

8 Squirrel Cage M nduction Motor Equivalent Circuit (at Stand Still) R X X R c r V R c X c E E E E 9 30 nduction Motor Equivalent Circuit (Stator Circuit) R X R R X c V R c X c E E R ' ( ) X c V R c X c E Equivalent to tranformer econdary winding 3 Equivalent to tranformer primary winding & core Equivalent to tranformer load 33 8

9 nput Power (P in ) Loe & Efficiency Pin 3 V coθ R X X R m E E R ' V Rm Xm ( ) Stator Loe: Copper loe (P cu ) Core loe (P iron ) Rotor Copper Loe (P cu ) Airgap Power (P g ) Developed Power (P d ) ' Pcu 3 R ' R P g 3 ( ) Td ω V Piron 3 Rm cu ' ' ( ) R Pg P 3 ' ' R ( ) ( ) P ( T ω P d 3 g ) d Rotational Loe (P rotational ) Output Power (P out ) P rotational P out T ω Example Solution A 50 hp, 60 Hz, three-phae induction motor operate at full load at a peed of 764rpm. v.34hp kw The rotational loe of the motor are 950 W, the tator copper loe are.6 kw and the iron loe are. kw. Compute the motor efficiency. 50 P out.34 f n 0 p 37.3 kw ß hp to kw converion pole machine ha n 3600 rev/min 4 pole machine ha n 800 rev/min 6 pole machine ha n 00 rev/min 8 pole machine ha n 900 rev/min n 764. Since the lip i very mall, n 800 rev/min n n n

10 P P + P + in g cu P iron kw P iron. P d P cu.6 P g 39 kw P cu P g η P P out in.89 or 89% P P + d out P rotational kw Starting nduction Motor..5 Problem: Low tarting torque (relative to full-load T) High tarting current (relative to full-load ) Why are thee problem? 0 Small Slip Maximum Torque max P rotational P out kw 39 Large Slip T t T max Torque 40 Summary Sytem f coupling generator to end-ue (motor) Tranformer Operation Equivalent circuit nduction Motor Slip, and teady-tate motor peed Speed-torque relationhip Equivalent circuit Starting induction motor 4 0