Module 9. DC Machines. Version 2 EE IIT, Kharagpur


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1 Module 9 DC Mchines Version EE IIT, Khrgpur
2 esson 40 osses, Efficiency nd Testing of D.C. Mchines Version EE IIT, Khrgpur
3 Contents 40 osses, efficiency nd testing of D.C. mchines (esson40) Gols of the lesson Introduction Mjor losses Swinburne s Test Hopkinson s test rocedure oding the mchines Clcultion of efficiency Condition for mximum efficiency Mximum efficiency for motor mode Mximum efficiency for Generl mode Tick the correct nswer Solve the following 14 Version EE IIT, Khrgpur
4 Chpter 40 osses, efficiency nd testing of D.C. mchines (esson40) 40.1 Gols of the lesson 1. To know wht re the vrious power losses which tke plce in d.c mchine.. To understnd the fctors on which vrious losses depend upon. 3. To know how to clculte efficiency. 4. To know how to estimte nd predict efficiency of d.c shunt motor by doing simple tests. 40. Introduction In the previous sections we hve lernt bout the principle of opertion of d.c. genertors nd motors, (strting nd speed control of d.c motor). Motors convert electricl power (input power) into mechnicl power (output power) while genertors convert mechnicl power (input power) into electricl power (output power). Whole of the input power cn not be converted into the output power in prcticl mchine due to vrious losses tht tke plce within the mchine. Efficiency η being the rtio of output power to input power, is lwys less thn 1 (or 100 %). Designer of course will try to mke η s lrge s possible. Order of efficiency of rotting d.c mchine is bout 80 % to 85 %. It is therefore importnt to identify the losses which mke efficiency poor. In this lesson we shll first identify the losses nd then try to estimte them to get n ide of efficiency of given d.c mchine Mjor losses Tke the cse of loded d.c motor. There will be copper losses ( Irnd I frf VI f) in rmture nd field circuit. The rmture copper loss is vrible nd depends upon degree of loding of the mchine. For shunt mchine, the field copper loss will be constnt if field resistnce is not vried. Recll tht rotor body is mde of iron with slots in which rmture conductors re plced. Therefore when rmture rottes in presence of field produced by sttor field coil, eddy current nd hysteresis losses re bound to occur on the rotor body mde of iron. The sum of eddy current nd hysteresis losses is clled the core loss or iron loss. To reduce core loss, circulr vrnished nd slotted lmintions or stmping re used to fbricte the rmture. The vlue of the core loss will depend on the strength of the field nd the rmture speed. Aprt from these there will be power loss due to friction occurring t the bering & shft nd ir friction (windge loss) due to rottion of the rmture. To summrise following mjor losses occur in d.c mchine. Version EE IIT, Khrgpur
5 1. Field copper loss: It is power loss in the field circuit nd equl to I f Rf VI f. During the course of loding if field circuit resistnce is not vried, field copper loss remins constnt.. Armture copper loss: It is power loss in the rmture circuit nd equl to IR. Since the vlue of rmture current is decided by the lod, rmture copper loss becomes function of time. 3. Core loss: It is the sum of eddy current nd hysteresis loss nd occurs minly in the rotor iron prts of rmture. With constnt field current nd if speed does not vry much with loding, core loss my be ssumed to be constnt. 4. Mechnicl loss: It is the sum of bering friction loss nd the windge loss (friction loss due to rmture rottion in ir). For prcticlly constnt speed opertion, this loss too, my be ssumed to be constnt. Aprt from the mjor losses s enumerted bove there my be smll mount loss clled stry loss occur in mchine. Stry losses re difficult to ccount. ower flow digrm of d.c motor is shown in figure A portion of the input power is consumed by the field circuit s field copper loss. The remining power is the power which goes to the rmture; portion of which is lost s core loss in the rmture core nd rmture copper loss. Remining power is the gross mechnicl power developed of which portion will be lost s friction nd remining power will be the net mechnicl power developed. Obviously efficiency of the motor will be given by: η net mech in Similr power flow digrm of d.c genertor cn be drwn to show vrious losses nd input, output power (figure 40.). Version EE IIT, Khrgpur
6 It is importnt to note tht the nme plte kw (or hp) rting of d.c mchine lwys corresponds to the net output t rted condition for both genertor nd motor Swinburne s Test For d.c shunt motor chnge of speed from no lod to full lod is quite smll. Therefore, mechnicl loss cn be ssumed to remin sme from no lod to full lod. Also if field current is held constnt during loding, the core loss too cn be ssumed to remin sme. In this test, the motor is run t rted speed under no lod condition t rted voltge. The current drwn from the supply I 0 nd the field current I f re recorded (figure 40.3). Now we note tht: Input power to the motor, in VI 0 Cu loss in the field circuit fl VI f ower input to the rmture, VI 0  VI f V(I 0 I f ) VI 0 Cu loss in the rmture circuit I0r Gross power developed by rmture VI I r 0 0 ( ) V I r I E I b Version EE IIT, Khrgpur
7 Since the motor is operting under no lod condition, net mechnicl output power is zero. Hence the gross power developed by the rmture must supply the core loss nd friction & windge losses of the motor. Therefore, ( ) core + friction V I0r I0 Eb0I 0 Since, both core nd friction for shunt motor remins prcticlly constnt from no lod to full lod, the sum of these losses is clled constnt rottionl loss i.e., constnt rottionl loss, rot core + friction In the Swinburne's test, the constnt rottionl loss comprising of core nd friction loss is estimted from the bove eqution. After knowing the vlue of rot from the Swinburne's test, we cn firly estimte the efficiency of the motor t ny loding condition. et the motor be loded such tht new current drwn from the supply is I nd the new rmture current is I s shown in figure To estimte the efficiency of the loded motor we proceed s follows: Input power to the motor, in VI Cu loss in the field circuit fl VI f ower input to the rmture, VI  VI f V(I I f ) VI Cu loss in the rmture circuit Gross power developed by rmture Ir VI I r ( ) V I r I EI b Net mechnicl output power, net mech EI b rot efficiency of the loded motor, η EI b rot VI net mech The estimted vlue of rot obtined from Swinburne s test cn lso be used to estimte the efficiency of the shunt mchine operting s genertor. In figure 40.5 is shown to deliver in Version EE IIT, Khrgpur
8 lod current I to lod resistor R. In this cse output power being known, it is esier to dd the losses to estimte the input mechnicl power. Output power of the genertor, out VI Cu loss in the field circuit fl VI f Output power of the rmture, VI + VI f VI Mechnicl input power, in mech VI + Ir + ro t Efficiency of the genertor, η VI in mech VI VI + I r + t ro The biggest dvntge of Swinburne's test is tht the shunt mchine is to be run s motor under no lod condition requiring little power to be drwn from the supply; bsed on the no lod reding, efficiency cn be predicted for ny lod current. However, this test is not sufficient if we wnt to know more bout its performnce (effect of rmture rection, temperture rise, commuttion etc.) when it is ctully loded. Obviously the solution is to lod the mchine by connecting mechnicl lod directly on the shft for motor or by connecting loding rheostt cross the terminls for genertor opertion. This lthough sounds simple but difficult to implement in the lbortory for high rting mchines (sy bove 0 kw), Thus the lbortory must hve proper supply to deliver such lrge power corresponding to the rting of the mchine. Secondly, one should hve lods to bsorb this power Hopkinson s test This s n elegnt method of testing d.c mchines. Here it will be shown tht while power drwn from the supply only corresponds to no lod losses of the mchines, the rmture physiclly crries ny mount of current (which cn be controlled with ese). Such scenrio cn be creted using two similr mechniclly coupled shunt mchines. Electriclly these two mchines re eventully connected in prllel nd controlled in such wy tht one mchine cts s genertor nd the other s motor. In other words two similr mchines re required to crry out this testing which is not bd proposition for mnufcturer s lrge numbers of similr mchines re mnufctured. Version EE IIT, Khrgpur
9 rocedure Connect the two similr (sme rting) coupled mchines s shown in figure With switch S opened, the first mchine is run s shunt motor t rted speed. It my be noted tht the second mchine is operting s seprtely excited genertor becuse its field winding is excited nd it is driven by the first mchine. Now the question is wht will be the reding of the voltmeter connected cross the opened switch S? The reding my be (i) either close to twice supply voltge or (ii) smll voltge. In fct the voltmeter prcticlly reds the difference of the induced voltges in the rmture of the mchines. The upper rmture terminl of the genertor my hve either + ve or negtive polrity. If it hppens to be +ve, then voltmeter reding will be smll otherwise it will be lmost double the supply voltge. Since the gol is to connect the two mchines in prllel, we must first ensure voltmeter reding is smll. In cse we find voltmeter reding is high, we should switch off the supply, reverse the rmture connection of the genertor nd strt fresh. Now voltmeter is found to red smll lthough time is still not ripe enough to close S for prlleling the mchines. Any ttempt to close the switch my result into lrge circulting current s the rmture resistnces re smll. Now by djusting the field current I fg of the genertor the voltmeter reding my be djusted to zero (E g E b ) nd S is now closed. Both the mchines re now connected in prllel s shown in figure Version EE IIT, Khrgpur
10 40.5. oding the mchines After the mchines re successfully connected in prllel, we go for loding the mchines i.e., incresing the rmture currents. Just fter prlleling the mmeter reding A will be close to zero s E g E b. Now if I fg is incresed (by decresing R fg ), then E g becomes greter thn E b nd both I g nd I m increse, Thus by incresing field current of genertor (lterntively decresing field current of motor) one cn mke E g > E b so s to mke the second mchine ct s genertor nd first mchine s motor. In prctice, it is lso required to control the field current of the motor I fm to mintin speed constnt t rted vlue. The interesting point to be noted here is tht I g nd I m do not reflect in the supply side line. Thus current drwn from supply remins smll (corresponding to losses of both the mchines). The loding is sustined by the output power of the genertor running the motor nd vice vers. The mchines cn be loded to full lod current without the need of ny loding rrngement Clcultion of efficiency et field currents of the mchines be re so djusted tht the second mchine is cting s genertor with rmture current I g nd the first mchine is cting s motor with rmture current I m s shown in figure Also let us ssume the current drwn from the supply be I 1. Totl power drwn from supply is VI 1 which goes to supply ll the losses (nmely Cu losses in rmture & field nd rottionl losses) of both the mchines, Now: ower drwn from supply VI 1 Field Cu loss for motor VI fm Field Cu loss for genertor VI fg Armture Cu loss for motor Imr m Armture Cu loss for genertor Igr g VI 1 VI fm + VI fg + Imrm+ Igrg (40.1) Rottionl losses of both the mchines ( ) Since speed of both the mchines re sme, it is resonble to ssume the rottionl losses of both the mchines re equl; which is strictly not correct s the field current of the genertor will be bit more thn the field current of the motor, Thus, Rottionl loss of ech mchine, rot ( fm fg m m g g) VI1 VI + VI + I r + I r Once rot is estimted for ech mchine we cn proceed to clculte the efficiency of the mchines s follows, Efficiency of the motor As pointed out erlier, for efficiency clcultion of motor, first clculte the input power nd then subtrct the losses to get the output mechnicl power s shown below, Version EE IIT, Khrgpur
11 Totl power input to the motor power input to its field + power input to its rmture inm VI fm + VI m osses of the motor VI fm + Imrm + rot Net mechnicl output power outm ( inm VI fm + Imrm + rot ) Efficiency of the genertor η m outm inm For genertor strt with output power of the genertor nd then dd the losses to get the input mechnicl power nd hence efficiency s shown below, Output power of the genertor, outg VI g osses of the genertor VI fg + Igrg + rot Input power to the genertor, ing ( outg + VI fg + Igrg + rot ) η g outg 40.6 Condition for mximum efficiency ing We hve seen tht in trnsformer, mximum efficiency occurs when copper loss core loss, where, copper loss is the vrible loss nd is function of loding while the core loss is prcticlly constnt independent of degree of loding. This condition cn be stted in different wy: mximum efficiency occurs when the vrible loss is equl to the constnt loss of the trnsformer. Here we shll see tht similr condition lso exists for obtining mximum efficiency in d.c shunt mchine s well Mximum efficiency for motor mode et us consider loded shunt motor s shown in figure The vrious currents long with their directions re lso shown in the figure. I I + I I + I f I f M V (supply) M R n rps Figure 40.8: Mchine opertes s motor   n rps Figure 40.9: Mchine opertes s genertor Version EE IIT, Khrgpur
12 We ssume tht field current I f remins constnt during chnge of loding. et, rot constnt rottionl loss V I f constnt field copper loss Constnt loss const rot + V I f Now, input power drwn from supply V I ower loss in the rmture, I r Net mechnicl output power so, efficiency t this lod current η m VI I r ( VI + ) f rot VI I r const VI I r VI const Now the rmture copper loss I rcn be pproximted to I rs I I. This is becuse the order of field current my be 3 to 5% of the rted current. Except for very lightly loded motor, this ssumption is resonbly fir. Therefore replcing I by I f in the bove expression for efficiency η m, we get, η m VI I r VI const Ir 1 V VI Thus, we get simplified expression for motor efficiency η m in terms of the vrible current (which depends on degree of loding) I, current drwn from the supply. So to find out the condition for mximum efficiency, we hve to differentite η m with respect to I nd set it to zero s shown below. d const I r dηm di 0 const or, di V VI 0 r or, + V VI const Condition for mximum efficiency is I r I r const So, the rmture current t which efficiency becomes mximum is I const r Mximum efficiency for Genertion mode Similr derivtion is given below for finding the condition for mximum efficiency in genertor mode by referring to figure We ssume tht field current I f remins constnt during chnge of loding. et, Version EE IIT, Khrgpur
13 rot constnt rottionl loss V I f constnt field copper loss Constnt loss const rot + V I f Net output power to lod V I ower loss in the rmture, I r Mechnicl input power VI + I r + ( VI + ) f rot VI + I r + const VI so, efficiency t this lod current η g VI + I r + const I As we did in cse of motor, the rmture copper loss I rcn be pproximted to I rs I. So expression for ηg becomes, VI η g VI + I r + const Thus, we get simplified expression for motor efficiency ηg in terms of the vrible current (which depends on degree of loding) I, current delivered to the lod. So to find out the condition for mximum efficiency, we hve to differentite ηg with respect to I nd set it to zero s shown below. dη di g 0 or, d VI di 0 VI + Ir + const Simplifying we get the condition s I r I r const So, the rmture current t which efficiency becomes mximum is I const r Thus mximum efficiency both for motoring nd generting re sme in cse of shunt mchines. To stte we cn sy t tht rmture current mximum efficiency will occur which will mke vrible loss constnt loss. Eventully this leds to the expression for rmture current for mximum efficiency s I const r Tick the correct nswer 1. A 5 kw, 30 V, d.c shunt motor is operting t rted condition with 8% efficiency. ower drwn form the supply nd the totl loss re respectively (A) 5 kw nd 900 W (C) 6.1 kw nd 900 W (B) 6.1 kw nd 1100 W (D) 5 kw nd 1100 W. 10 kw, 30 V, d.c shunt genertor is operting t rted condition with 80% efficiency. Input power nd the totl loss re respectively Version EE IIT, Khrgpur
14 (A) 5 kw nd.5 kw (C) 1.5 kw nd.5 kw (B) 1.5 kw nd 1 kw (D) 5 kw nd 1 kw 3. A 10 kw, 0 V, d.c shunt motor is drwing A from supply mins t rted condition. The mchine hs r 0.3 Ω nd R f 0 Ω. The core nd friction loss together is (A) 0 W (B) W (C) W (D) W 4. A 10 kw, 0 V, d.c shunt genertor is supplying lod t rted condition. Input mechnicl power is found to be 1.5 kw. The mchine hs r 0.3 Ω nd R f 0 Ω. The core nd friction loss together is (A) W (B) W (C) W (D) 500 W 5. During no lod test on 0 V, shunt motor hving r 0.7 Ω nd R f 0 Ω, the rmture current is recorded to be 4 A. The rottionl loss of the motor is (A) 880 W (B) 11. W (C) W (D) W 6. A d.c shunt motor with r 0.8 Ω, hs 00 W field circuit loss nd 50 W rottionl loss. Motor will operte mximum efficiency when loded to crry n rmture current of bout (A) A (B) 3.7 A (C) A (D) A 40.8 Solve the following 1. A 0 V d.c shunt motor hs rmture nd field resistnce s 0.8 Ω nd 00 Ω. During Swinburne s test, current drwn from the supply is found to be.5 A. Estimte the efficiency of the mchine, (i) When it is running s motor drwing line current of 40 A from the 0 V supply. (ii) When it is running s genertor delivering lod current of 40 A t 0 V.. Two similr coupled mchines of sme rting, ech hving n rmture resistnce of 0.5 Ω re connected for Hopkinson s test. Test dt recorded s follows: Supply voltge 30 Volts Totl line current drwn from the supply 8 A Field current of the mchine running s genertor 3 A Field current of the mchine running s motor A Genertor rmture current 17 A (i) (ii) (iii) Estimte the rottionl loss of ech mchine. Estimte the efficiency of the genertor. Estimte the efficiency of the motor. Version EE IIT, Khrgpur
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