| sas gd Ww | a C re NNT | THESIS EDN RB LI ABET VEEL ne vee A TAL see ST BUA MU AU Tt R. P. KELLEY R. W. SHEEHAN SDs - enemnmmerargnent tte . Copy l. THERMS “Comparison of Electrical and liechanical liethods for Determinins the Ffficiency of Motore, A Thesis Submitted to The Faculty of MICHIGAN AGRICULTURAL COLLEGE, By ; ' : - Norsk te . Ne qT eb eo K.P Kelley. RW Sheehan, ~ = Candidates for the Degree of 4 Bachelor of Science, June I9I7. THESIS vop: | PREFACE, The object of the tests on the Induction Motor was to obtain all of the data possible for the comparison of the different methods of determining the efficiency of this type of Alternating Current epparatus;and to become familiar with the different methods used commercially for testing the efficiency of motors, Data was taken for the different tests and curves plotted from the calculated results.These curves furnish an easy and accurate method of comparing the relative merits of the different methods, A study was made of the different kinds of brakes used for the Brake Test and the kind shown in our sketch was finally selected .We are indebted to Prof. Polson for advice on this subject. The main reason for selecting the type of brake used was that the length of the brake .azm is constant at all conditions of load, RPK. R.W.S,. * 401892 BIBLIOGIAPHY, Experimental Electrical Engineering------ Karapetoff 970= 584 p61 = 564, Standard Hand Book;---------------------- Sect .7 II5 = 126 199 = 209, Sheldon ,Mason and Hausmen---<------------ 23I = 233, Penders American Hand Booke-------------- I000 = I005. Anerican Correspondence Schools-<-------<- Volume 4, 360 = 364, Steinmetz A.C, Phenomena----------------- 306 = 310, LIST OF APPARATUS USED. LIOTOR: G.E.Induction Motor # 183030, Type: KT=-65A= I200 = Form C, 60 cycle,5 H.P., 1200 R.P.M, I23.2 Amperes, 220 Volts. VATTIFTFRS: , Weston A.C, Wattmeters # 207 and # 208. sodel 310, Max, capacity 20 amperes. High range in volts 300, low 150, High range in watts 3000, low 750. Weston Portable A.C. Wattmeter lax, capacity 25 amperes, High range in volts 300, low I50. High range in watts 6000, low 3000, Model 16, # 4995. CURRENT TRANSFORMERS: Weston Switchboard Current Trensfoxrmers,. Model 236,Type 1,# 330, # 325. Amperes 25 = 5, Line voltage 2200,Frequency 25 = 125, Watts 5. AITLETERS : Veston A.C, Ammeters,. Model 155, i 8892, Range 5 amperes. Model 155, # 8889, Range 5 amperes. Model 155, # I6615, Range 25 amperes. Model 155, # 16451, Range 50 amperes. VOLT! TTER: Weston A.C, Voltmeter. liodel 156, # 4143, CRAKE: Rope Brake as shown in blue print. SCALFS : Platform scales,Range on arm I0 7, Max, range 100 STEED COUNTER, BRAKE TEST, The object of @ brake test on any motor is to obtain the data for calculating the E.P, output and input of the machine, To do this suitable electrical measuring instruments are placed in the line and their readings ,multiplied by some constant,will the power input into the machine, The output is measured by means of the brake and from these two sets of readings the efficiency is determined, The set up was made as shown in the wiring diagram consisting of two wattmeters with the current coils connected in the two outside leads and the potentialcoils connected across the middle lead and the line in which the current coil was connected, Ammeters were placed in the two outside lines and a voltmeter across the lines, The voltage was held at the name plate value of the machine and simultaneous readings of all instruments were teken, The scale weight and the speed of the motor were recorded, Using the formula: H.P, Output= 2 YT Wil1n Where W= Weight at end of the arm, 1 i} Leneth of the brake arm, 3 Hl R.P.M. of the motor the H.P, output was calculated for each point of the test end the efficiency of the motor determined from the formula.: Efficiency = H.P, Output meme mma eee x IOO. H.P. Input BRAKE TEST # I, DATA, Vt .# ReP.M. Volts. Amps Amps Watts Watts. AB A B AB BC 0 T2I0 220 5.05 5.25 620 = 360 2 1199 220 5.15 5.30 680 = 310 4 II85 220 5.15 5.30 740 = 220 6 II80 220 5.35 5.45 840 = 180 8 II75 220 5.40 5.50 866 = I20 TO II70 220 5.50 2+ 29 920 - 80 15 II70 220 5.75 5.85 1060 + 40 20 II70 220 6.15 +33 I200 + I60 25 II70 220 6.65 6.85 1400 + 320 30 II60 220 6.85 25 1490 + 450 by II50 220 g-65 .00 1680 + 600 0 1150 220 50 8.90 1870 + 760 45 II55 220 8.90 9.30 2020 + 880 0 II55 220 9.75 10.15 2220 1020 55 II55 220 10.35 10.50 2390 1290 60 II50 220 II.50 0 3=6.: 12,25 2640 1380 65 II40 220 I2.90 13.00 2800 1540 Raudiue of pulley plus I/2 thickness of rope = .344 ft. Calculations for Brake Test, HP, Seacnn----- x R.P.M. x load in pounds = ,06000656xKPMx¥ = ,0000656 x 1140 x 65 = 4,8 5 BRAKE TEST # I, (continueé) RESULTS , Wt. #. Total Watts H.P, H.P, Efficiency, Input. Input. Output. “3 0 260 . 348 0.0 0.0 ; 329 a PS 6 660 "BBs “262 52.6 8 46 I.00 615 61.5 IO 40 I,120 766 63-0 15 IIO0O I.475 I,150 f .0 20 1360 I ,82 I. 530 4.0 25 1720 2.30 1.91 83.2 30 1940 2.60 2.26 87.0 35 2280 3.05 2.64 86 .0 40 2630 322 3.02 85.5 45 2900 3.89 3.41 87 6 50 3240 4.34 3.78 87.0 55 3580 4,80 4.16 86.5 60 4020 5-37 4038 84 .0 65 ~ 4340 5 ol 4.85 83.5 wry te ee BRAKE TEST # 2, The efficiency curves from Brake Test i} I ,the Circle Diagram and the Loss method,while checkirg within 3 % at the full load point were over 5 % different at the quarter and half load points.Not being quite sure of our results from the first set of readings, we ran a second brake test on the same motor end under the same conditions, DATA, Wt.#. R.P.if, Volts Amps. Amps. Watts Watts. AB A C AB BC 0 I200 220 4.9 4.9 560 =- 360 5 II95 220 4.95 4.95 680 - 190 I0 1193 220 5.1 $I 820 = 30 15 II30 220 5.55 55 960 + 120 20 II80 220 6.1 6.0 II60 +280 25 1270 220 6.7 6.7 1320 + 380 30 II65 220 g-3 g-3 1440 495 35 II60 220 .0 .0 1640 630 40 II55 220 8.7 8.75 1820 730 45 II50 220 9.5 9.6 2040 980 50 II45 220 10.2 10.3 2200 1080 60 II40 220 I2.0 12.0 2560 1390 0 II30 220 13.5 13.3 2600 1640 0 II20 220 16.0 15, 2400 1900 RESULTS, Wt .#, Total Watts H.P, H.P, Efficiency Input, Input, Output. fo 5 490 656 +331 59.8 IO 730 I.060 781 3.6 I5 1080 I.45 I.I6 0.0 20 1440 1.9 1.55 80.5 25 I700 2.2 1.92 84.0 30 1935 2.59 2.28 88 I 35 2270 3.04 2.67 87.7 40 2610 3.50 3.02 86 .4 45 3020 4,05 3.39 83,8 50 3280 4.40 3-78 85.2 60 3950 +3 4.4 Shee 70 4240 6.69 5.09 88, 80 4300 7.1 5.89 83.0 CIRCLE DIAGRAI, Actual load tests on Induction “iotors are avoided whenever it is possible because they involve a consider- able expenditure of energy and require special apvliances if accurate results are to be obtained, In machines beyond a certain size the waste of power and the arrangements for load make load tests practically impossible, Therefore methods have been developed for predetermining the perform ance of the machine under actual load conditions from a few simple readings taken at no load. One of these methods is by means of the circle diagram, We will first explain the theory and the construction of the circle diagram and then give the proof of its accuracy. t I, > i I, a. ° Let OF in the figure represent the impressed voltage in one of the phases and OI the current in the same phase ata certain load. The angle # between the two them cor- responds to the nower factor of the motor at this load, Theory and experiment both show that, when the load is varied, the current changes in such a way that the locus of its vector is revresented by a semicircle I IK o The current at no load is represented by the vector OI, The current is small and the phase angle EOI,is large, the power factor being low, The no load current may ve cénside ered as consisting of a wattless component Om, which pro- duces the magnetic flux, and a power component m0,which overcomes lron and friction loss in the motor, As the load increases, the point I,moves along the circle, the value of the current increasing as well as the power factor.When the motor becomes overloaded the power factor decreases again, due to a more pronounced influence of the magnetic leakage, When the load is such that the motor stops, the c Current it takes at standstill is represented by a certain vector OI, . This current is called the short circuited current or the current with the armature locked, Thus if the circle has been determined for the motor, the power factor Cos @ may be measured from the diagram for any given value of the current input OI, The proof of this can be verified by experiment with an accurate load test by plotting the vectors of current to the observed value of the phase angle @, It will be found that the load points lie on a semécircle. The circle may be constructed knowing only one point on it, if in addition, the vnosition of the point Ig is known, Ags such a point , the point Lis usually selected, the vector Ol, representing the current with the armature loeked, No brake is really required in either case and the readings are both simple and accurate, OI beins the vrimary current, its component Id in phase with the voltage OK renresents the workings comnonent of the current;therefore Id multiplied by the voltage, gives true watts input per phase, The circle gives watts input and the value of the power factor for any given value of the primary current, Output, torque, slip, efficiency, im fact all the “sta lesessary for plotting the performahce curves may be pre= determined from the circle diagram, PROOF OF THE CIRCLE DIAGRAM, Instead of considering the rotor revolving and deliver- ing mechanical power, it may be assumed as locked and loaded electrically on noneinductive external resistances.The same currents can be produced in the primary windings, by suit- ably varying the resistences, as if the rotor were revolving and were loaded mechanically, / D L ‘ve LY Cc K Ww 5 = x By sstopping the motor the secondary induced emf is in- creased n,+ (n,- n,) times , where n,is the synchronous wpeed of the motor, and n,is it s actual speed, The revolv- ins flux cuts the secondary conductors at a rate proportional to ( no Nn, ) when the rotor is revolving ; with the armature locked the speed at which the conductors are cut by the flux is equal to n,. The frequency of the secondary currents being thus greatly increased, the reactance of the secondary Ls thereby also increased the same number of times. By add- inc such an external resistance that the rotor resistance is is incresed n,+ (n - n,) times, the same current and the same electrical relations are obtained in the secéndary as with the rotor revolving; consequently the primary current will have the same vakue and the same power factor, Vary- ing the exterha} resistance is equivalent to varying the byake load with the armature revolving. In this way the Induction motor is reduced to a stationary transformer with an abnormal leakage and with a secondary load of non-induct= ive resistances , varying in value from 0 to infinity. Such a transformer can be replaced by an equivalent res- istance and an equivalent reactance and the problem is then reduced to mereky this;A non=inductive load is connected to a constant potential line, with some resistance and some in- ductance in series with it. The locus of the current vector when the load changes from zero to infinity is to be determineé This can be solved by the use of the above figure. CD represents the vector of the line voltage, which is constant; CI is the current at a certain load, CF, in phase with the current ,represents the resistance drop. FD, perpen- dicular to the current vector represents the reactive drop; CD the geometrical sum of CF and FD is constant. As the latter two vectors are perpendiculer to each other, the voint F moves on a sémicircle having CD for its diameter. The equivalent reactance being constant under all conditioms of load, the vector FD represents the current to a certain scale and ié@ proportional to it, Thus when the load varies, the extremity of the vector of the clrrent moves on a semicircle, This semicircle CIK, has its diameter perpendicular to the true position of the current x vector,This proof applies to the induction motor, since it has been demonstrated that the circuits in the two cases are electrically equivalent. The only difference is that in addition to the load current CI, the inductiom motor takes a magnetizing current Om and a small power component mC for overcoming iron loss and friction, This brings the total pri-«- mary current to the value OI and this figure is identical with that of the circle diagram of the induction motor, Volts, AB 250 235 220 203 190 175 T60 I45 120 Rated Voltage, 220 : + > MOUW WW FS Buy Or e WO P~YD HINO ONO MP AHN AO O RESULTS FROM EXCITATION TEST, CIRCLE DIAGRAM TEST, DATA, EXCITATION TEST, Amps, C NWWW & Puy ® WO PY HN ONONC WIHT U1 O OO Watts Input. Watts AB, 00 I0 600 510 430 360 330 250 190 259 Watts BC, 60 £30 0 380 210 180 160 I20 70 Amos, > : oeoeeoeod®@ WO PY BNW Ow], SP MUILN NOW OO MUWWIWWY f BNO Total Watts, 40 360 — 240 230 220 180 170 130 120 Amps. Input. Volts, AB, 48 Q aC 62.5 80 102.5 122.0 I40 I60 T80 200 250 220 The Amps, CIRCLE DIA GRAM TEST, SHORT CIRCUIT TEST. DATA, Amps, Watts Watts Amps Total C, AB, BC, Av, Watts. I2.0 70 5 I2, 75 16.0 40 20 16.0 2 0 18.1 890 25 18.05 915 or. 1300 45 21.5 1345 25. I eRe 85 25.6 I9I5 32.8 200 32.7 3260 40.0 4360 370 39.7 $730 6180 8210 10410 13000 14300 15775 last six values of the total watts were calculated from tne power component, Phase i:ean BC Mean Ch liean RESISTAUCH OF STATOR WINDINGS BY DROP OF POTENTIAT, METHOD, Amperes, Volts Drop Resistance, ohms, 14 r3 93 I2 I2 I.00 L7 I7 1.00 _—_ 99 7 7 I .00 II II I,00 17.5 17,0 “2f T5 14.5 J 99 £7. 17.0 ° 1h°6 rho 1°06 9.5 9.5 I.00 7.0 7.0 1.00 99 Sennen, a PeinaRnanUeee Re CS wace-ve( am fife: Ghar pOtertstics sa cook Eat seek 2B dS ar oz ee OS, ane aaa EPCOT ror (Va te¥ 3 . sdk mace 20 M-on* 30.5 1720 6.75 100.1 2.89 1408.1 40.7 1940 7.05 104.9 3.82 1683.3 64.4 2280 p83 II6.5 4.55 1991.2 89.0 2630 .70 129.0 4.55 2289.2 104.0 2900 9.10 133.0 4.30 Ba" 109.5 3240 9.95 148.0 4.30 80.2 124.0 3580 10.42 153 +0 4.30 3213, 2 138.0 4020 II .87 178.0 4.55 3631.2 165.0 4340 12.50 186.0 5.36 3942.2 212.0 Watts Output. Efficiency. 0.0 0.0 wi 22.0 abet f 43.5 360. i 54.6 (442.0 59.4 530.4 63.0 778.9 70.6 1024.7 7569 1367.4 £9-2 1619.0 3.3 1862.7 81.7 2185.0 83.2 2444 0 84.5 2776 .0 85.5 3075 86.0 3466 86.2 3730 86.0 Resistance of one phase = .495 ohms, Primary no load copper loss I°R x 3/2 520° x .99 x 3/2 = 40.2 Watts Irom and friction loss = No load input = no load copper loss, 252 = 40,2 = 211.8 Watts, i SECISE SHES baseSasars cantsas . . - ; : . ~ oSssenncs a: - — — | 1: seis | / ; : i t Beuetadeccasn sean8 ee meet ; / : : : ; ' ; PSSsSeegs } a a a : : ! vas Banoane & ost } - . ' / : pire Persea: Pow ak conees| ; = / Sasa st | Maa 50 ees ree, : ms peoee ' ; ; _ . an ae ++ +4-- t J } : on io : : ied! 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