Document e7k716J9exnov8rkanb65ynQp

value o! load. Reducing (he *ped, by /n;rea<iagtbe rotor Teaittance, results In increased f*R toss In the rotor. When this occurs, the efficiency decreases in direct proportion to the Increased slip and rotor resistance. It is also possible that the conveyor load i too light lor the motor, keeping In mind load U necessary for effective speed reduction. As this point It fre quently overlooked when sllprlng mo tors are used it is suggested that a watt, meter be connected to the motor end the true load readings obtained. Study of tbe above curves shows the speed regulation of o stipring motor poor as the motor increase* it* speed with a reduction of load, or viee versa, fra A Batcher, Ptney River, Virginia, JiiV does not state what speed be wants st full load. If he changes the variable resistor from 1.04 to 0.52 ohms per phase, speed will vary according to curve 0- When carrying full load the speed would be 600 rpm. On the other hand, i( resistances are doubled to 2.08 per phase, we get curve C end the motor won't start until iho load Is re duced 50%. At half speed tbe load is reduced to 25% (point C< on curve C). UtV sayt the speed is BOO rpm at full load. This would correspond to curve D and by interpolation it is evident that the resistance per phase is less then that used in curve 8. Curve E is drawn to show the motor speed with rings short eireuited. Following is proof of the formula: tions under actual opersi;ni . ,, On curve below, marked iff' torque-speed relations si ^ celeroied. Only nine steps Assuming the motor Cso the load on the second notch Formula and Proof Shown Tue formula to dstesmine secondary resistance in e slipring motor is: ,, . ffpX249 nP . IfiOXWfl , . --twt5-- " 1-* Where R equals external secondary re sistance in ohms per phase causing motor to exert 100% torque at sero speed; Hp is the full-load rating (ISO); f equals full-load secondary current (190 ftmpl. If three 1.04-ohm resistors are con PH 8? ~5iT With sufficient resistance in the motor secondary to barely stall the motor un. der full load, a total of 150 hp is being dissipated in beat. In other words, tbe current squared in'each leg multiplied by 3, since there are 3 rings, is 3PR. Divide this amount by 746 for 746 wotts = I hp. Hence: 3/7? -+- 746 = 1*8 -f249 s hp dissipoted, which in this cose -- 150 hp^--RFE Space Taps Evenly, Adjust With Load \4od `aoo *Tlod * t&F Cn ttovtns iMU-towi two*** swafto conftM loro*. corroisanf >o roton cewrowr rofcsts. torque required by tbe load is the load value, tbe motor will accelers point A on curve Mo. 2 eemupoa to 260 rpm. On further moveiuet third notch, the rotor speed shUl No. 9 curve and steadies awsy si ( B, or $20 rpm. This Is repeated ss trailer is moved until rotor Is shall cuited at the rings. J A Cooldan* G Dion, Ontario. Values op rotor current and rotor volt age between rings st necessary to find tbe required resistance. The following formula Is used to find the ring voltage: hp X 746 rmnrsrK Torque Curves Analyzed IT (IB TAKES MEASUneMEKTS |J isi cated, JHV can plot curves like tho nected in Y, the motor will not start under full load. This includes internal rotor resistance which we can neglect for practical purposes. Under this con dition, all the rotor energy is being dis sipated in the resistor. - to gal a clear picture of what the speed-load conditions ore, refer to the above curve. Point Ai shows motor speed as zero at 100% toad. If the load la re duced by halt, speed is 600 rpm as in dicated by point At- Whm i s voltage between rings, f, = rotor current, tip ss horsepower, K s 1 -- % slip. The slip is 4.15% with rings short circuited at full load. Hence: 'raxrsSxTSB"358' Resistance required lor a given start ing torque is found as follows: Where T ss torque required on 6rst step of controller, expressed for 200% full-load torque as 1.00, or 150% fit as 1.50, and so on. Assuming 100% fit: 1 ) *!** 1gW*tRUM| WW.I vjsj'k t--n l**`N --t1 1 v\1 \ sN VF ss -K \\ JJ AA v 3 " rVaxigbxToo ~ 1.09 ohms per leg la a Y-connectioo. To obtain even acceleration, I would spaee the taps on the grids equal resis tance values apart end make correc shown. Typical readings end estadttions are given as an example for < 3-phase 440-v 60-cyele induction mow*-. (Continued on page 138) 114 <?<5| POWER Apdl f***. acts when danger threatens Costly damage to turbines can result from-- Back flow of water Into extraction lines..* Accidental change of steam pressure. . Back flow of re-evaporated condensate In heater An A Bleeder Line Protecting Valve.. Acts when danger threatens... Closes to prevent back flow, . Guards egofnst damage. A Bleeder Line Protecting Valves are available... With air operating cylinders *,, ^ With oil operating cylinders.. With screw spindles and with plain covers. Bulletin 8-K describes these valves In detail. Write for a copy--today. SCHUTTE & KOERTING COMPANY '?* April 1,48 7KctMufac&t>U*f m I ISO THOMPSON SftCIf . PWtt ADCIPHM 22, PA. mifpturn itir rturmtcsinmiir * Irwins * atastat no mm rein Da nuns umui niuitun uo fisi ntiutQts caafn tm urns * tttn mnu m ifuuni sui rein i 'ii i.v : f'.l