Simple loop generator:Consider a single turn loop ABCD rotating clock wise in a uniform magnetic field with a constant speed as shown in figure. As the loop rotates, the flux linking the coil sides AB and CD changes continuously. Hence the emf induced in these coil sides also changes but the emf induced in one coil side adds to induced in the other.

(i) When the loop is in position 1, the generated emf is zero because the coil sides (AB and CD) are cutting no flux but are moving parallel to it.(ii) When the loop is in position 2, the coil side is moving at an angle to the flux and there fore a low emf is generated as indicated by point 2.(iii) When the loop is in position 3, the coil sides are at right angle to the flux and are, therefore cutting the at a maximum rate. Hence at this instant, the generated emf is maximum.(iv) At position 4, the generated emf is less because the coil sides are cutting flux at an angle.(v) At position 5, no magnetic lines are cut and hence induced emf is zero.(vi) At position 6, the coil sides move under a pole of opposite polarity and hence the direction of generated emf is reversed. The maximum emf in this direction will be when the loop is at position 7 and zero at position 1.

Action of a commutator: (i) In figure the coil sides AB and CD are under N pole and S pole respectively. Note that segment c1connects the coil side AB to point P of the load resistance R and the segment c2 connects the coil side CD to point Q of the load . Also note the direction of current through load. It is from Q to P.(ii) After a half revolution of the loop (i.e. 180 rotation), the coil side AB under S pole and the coil side CD under N pole. The current in the coil side now flow in the reverse direction but the segment C1 and C2 have also moved through 180 i.e. segment C1 is now in constant with (+ve) brush and segment C2 in contact with (-ve) brush. Note that commutator has reversed the coil connections to the load i.e. coil side AB is now connected to point Q of the load and coil side CD to the point P of the load but direction of current remain same as before.

Construction of DC Generator: Field system: The function of field system is to produce uniform magnetic field within which armature rotates.Armature: The armature core is keyed to machine shaft. It is consisting of slotted soft iron lamination. The purpose of laminating the core is to reduce eddy current loss.Armature Winding: The slots of the armature core hole insulated conductors that are connected in a switch manner is known as armature winding . This is the winding in which working emf induces. The armature conductors are connected in series-parallel.The conductors being connected in series so as to increase voltage and in parallel so as to increase current.Commutator: A commutator is a mechanical rectifier which converts the alternating voltage generated in the armature winding into direct voltage across the brushes. Commutators are made of copper and are insulated from each other by mica sheet. Depending upon the manner in which armature conductor are connected to commutator segment three are two types of armature winding :(a) Lap winding (Z/P)(b) Wave winding

Simple lap winding:

(a) There are as many as parallel path as the no. of pole (p).(b) Each parallel path has Z/p conductors in series where z and p are the total no. of armature conductors and poles respectively.(c) Total armature current ,Ia = p*current / no, of parallel path.(d) Emf generated =emf / parallel path.

Simple wave winding: (a) There are tow parallel paths irrespective of no. of poles of machine(b) Each parallel path has Z / 2 conductors in series .(c) Emf generated = emf / parallel path.(d) Total armature current ,Ia = 2*current / no. of parallel path.

EMF equation a d.c generator: Let, =Flux/pole in webers(wb) Z=Total number of armature conductor. P=Number of poles. A=Number of parallel paths=2 for wave winding =p for lap winding. N=Speed of armature in r.p.m. Eg=emf of the generatorFlux cut by one conductor in one revolution of the armature, D=p webersSince the armature rotates N times in 60secarmature rotates 1 times in 60/N=dt secEmf gerated/conductor=d/dt = p/(60/N)= pN/60 volts.EMf of generator Eg =Emf per parallel path =Emf per conductor*no. of conductors in series per parallel path. = pN/60 *Z/A = pZN/60AFor wave winding= pZN/60*2For lap winding = pZN/60*pTypes of DC generator:i. Separately excited dc generatorsii. Self excited dc generators

Separately excited dc generators: A dc generator whose field magnet winding is supplied from an independent external dc source is called a separately excited generator. Armature current, Ia=ILTerminal voltage,V= Eg-IaRaElectric power developed= EgIaPower delivered tothe load= (Eg-IaRa)Ia=VIa Self excited generated:(i) Series generator(ii) Shunt generator(iii) Compound generator

Series generator: In the series wound generator the field winding is connected in series with armature winding so that whole armature current flows through the field winding as well as the load. Armature current, Ia=Ise=IL=ITerminal voltage,V=Eg I(Ra+Rse)Power developed in armature= EgIa Power delivered to load= [Eg I(Ra+Rse)]Ia=VIaShunt generator:In a shunt generator[Fig.1.35],the field winding is connected in parallel with the armature winding so that terminal voltage of the generator is applied across it. The shunt winding has many turns of fine wire having high resistance. Shunt field current, Ish =V/ Rsh Armature current Ia = IL + Ish Terminal voltage,V= Eg IaRaPower developed in armature= EgIaPower delivered to load=VIL Compound generator: (a) Short shunt Series field current, Ise = IL Shunt field current, Ish=V+ IseRse/ Rsh Terminal voltage,V= Eg IaRa- IseRsePower develop in armature,P= EgIaPower delivered to load=VIL(b)Long shunt:Series field current Ise = Ia= IL+ IshShunt field current Ish=V/ RshTerminal voltage,V= Eg-Ia(Ra+ Rsh)-brush dropPower developed in armature = EgIaPower delivered to load=VILProblem 1: A shunt generator delivers 450 A of 230 V and the resistor of the shunt field and armeture are 50 ohm and 0.03 ohm respectively. Calculate the generated emf.Solve:

Terminal Voltage,V= Eg - Ia.Ra Or, Eg=V + Ia.Ra Shunt field current , Ish=v/Rsh = 230/50 =4.6 A. Armature current , Ia = IL + Ish =454.6 Generated emf, Eg= 230 +(454.6*0.03) = 243.64 VProblem 2: A long shunt compound generator delivers a load current of at 500 V and has armeture series field and shunt field resistance of 0.05 ohm, 0.03 ohm and 250 ohm respectively. Calculate generator voltage and ac current. Allow 1V perbrush for contact drop.Solve:

Shunt field current, Ish=V/Rsh=500/250 = 2 Aseries field current, Ise=Ia=Il+Ish=50+2 =52 AArmature voltage drop, IaRa=52x0.05 =2.6 VSeries winding voltage drop, IseRse=52x0.03 =1.56 V Brush drop=2x1=2 VGenerated emf, Eg=V+IaRa+IseRse+Brush drop =500+2.6+1.56+2 =506.16 VProblem-3: A short shunt compound generator deliver a load current of 30 A at 220 V and has armature,series field and shunt field resistances of 0.05 ohm, 0.30 ohm and 200 ohm respectively. Calculate the induced emf and the armature current. Allow 1.0 V per brush for contact drop.

Solve:

Shunt field current, Ish=V+IseRse/Rsh =220+30x0.30/200 =1.145 A Armature current, Ia=Il+Ish =30+1.145=31.145 A Armature voltage drop, IaRa=31.145X0.05 =1.56 V Series winding voltage drop, IseRse=30x0.30 =9 V Brush drop=2x1=2 V Generated emf, Eg=V+IaRa+IseRse+Brush drop =220+1.56+9+2 =232.56 VProblem-4: In a long shunt compound generator,the terminal voltage is 230 V when generator delivers 150 A. Determine i. Induced emf ii. Total power generated and iii. Distribution of this power.Given that shunt field,series field,diverter and armature resistance are 92 ohm,0.015 ohm,0.03 ohm and 0.032 ohm respectively.Solve:

Shunt field current, Ish=V/Rsh=230/92 =2.5 AArmature current, Ia=Il+Ish =150+2.5=152.5 ASince series field resistance and diverter are at parallel, so their combined RT=Rse.Rd/Rse+Rd =0.015x0.032/0.015+0.032 Total armature circuit resistance is, =0.032+0.01=0.042 ohmi. Induced emf, Eg=V+Ia(Ra+RT) =230+152.5x0.042= 236.405 Vii. Total power generated, EgIa=236.405x152.5 =36051.76 Wiii. In armature, I2aRa=(152.5)2x0.032=744.2 W In series field and diverter, I2seRT=(152.5)2x0.01 =232.56 W In shunt field, I2shRsh=(2.5)2x92=575 W Power delivered to load, VIL=230x150 =34500 WP-4:- The following information is given for a 300-kw,600-v,long-shunt compound generator: Shunt field resistance=75,armature resistance including brush resistance =0.03,commutating field winding resistance =0.011,series field resistance =0.012,divertor resistance =0.036. When the machine is delivering full load, calculate the voltage and power generated by the armature.Solution:- Power output= 300,000wOutput current=300000/600=500AIsh =600/75=8AIa =500+8=508ASince the series field resistance and divertor resistance are in parallel their combined resistance is-=(0.012*0.036)/0.048=0.009Total armature circuit resistance =0.03 -0.011+0.009=0.05Voltage drop=508*0.05=25.4v Voltage generated by armature=600+25.4=625.4VPower generated=625.4*508=317,700w=317.7kw P-5:- A four-pole generator, having wave-wound armature winding has 51 slots, each slot containing 20 conductors. What will be the voltage generated in the machine when driven at 1500 rpm assuming the flux per pole to be 7.0mWb?Solution:- Now Eg =ZNP/60A volt Here = 7*10^-3Wb,Z = 51*20=1020, A= P= 4, N= 1500 rpm.Eg bbnbbb = ( 7*10^-3 * 1020*1500*4)/(60*2)=178.5V P -6:- A 10Kw, 250V, d.c., 6-pole shunt generator runs at 1000 rpm, when delivering full-load. The armature has 534 lap-connected conductors. Full-load Cu loss is 0.64 Kw. The total brush drop 1 volt. Determine the flux per pole. Neglect shunt current.Solution:- Since shunt current is negligible, there is no shunt Cu loss. The copper loss occurs in armature only.I=Ia=10,000/250=40A; (Ia)^2Ra=Arm.Cu loss or 40^2*Ra=0.64*10^3; Ra=0.4IaRa drop = 0.4*40=16V; Brush drop= 2*1 =2V Generated emf. Eg=250+16+1=267V

Now, Eg=(ZNP)/60A volt 267=(*534*1000*6)/60*6 =30*10^-3Wb =30 mWb.

P-7: A shunt generator delivers195A at terminal p.d. of 250V. The armature resistance and shunt field resistance are 0.02 and 50 respectively. The iron and friction losses equal 950W. Find a. E.M.F. generated b. cu losses c. output of the prime mover d. commercial, mechanical and electrical efficiencies.Soln: a. Ish =250/50=5A; Ia 195+5=200AArmature voltage drop=Ia Ra=200*0.02=4VGenerated e.m.f=250+4=254Vb.Armature cu loss= Ia2 Ra=(200)2*0.02=800Wshunt cu loss=V Ish =250*5=1250WTotal cu loss = 1250+800=2050Wc.Stray loss=950W;Total losses=2050+950=3000Woutput=250*195=48750W;Input=48750+3000=51750WOutput of prime mover=51750Wd.Generated input=51750W;Stray losses=950WElectrical power produced in armature=51750-750=50800m =(50800/51750)*100=98.2%Electrical or cu loss=2050We=[48750*100/948750+2050)]=95.9%and c=(48750/51750)*100=94.2%P-8: A 500V d.c shunt motor draws a line current of 5A on light load . If armature resistance is 0.15 and field resistance is 200 , determine the effiency of the machine running as a ganarator, delivering a load current of 40A.Soln: i. As a motor , on light load out of 5A of line current 2.5A are required for field circuit and 2.5A are required for armature. Neglecting cu loss in armature at no load,the armature power goes towards armature coreloss and mechanical loss at the rated speed , This amounts to (500*2.5)=1250Wii. As a generator,for a line current of 40A , total current for the armature is 42.5A . output of generator =500*40=20kwTotal losses as a generator =1250+field cu loss+arm. Cu loss=(1250+1250+42.52*0.15)w=2.771kwEfficiency=[20*100/(20+2.771)]=87.83%P-9:The armature of a four pole d.c. shunt generator is lap wound and generates 216v when running at600rpm . Armature has 144 slots, with 6 conductors per slot.If this armature is rewound, wave connected, find the e.m.f generated with the same flux per pole but running at 500rpm. Soln: Total no. of armature conductors=Z=144*6=864For a lap winding no. of parallel paths in armature=no. of poles.In the e.m.f. equation, E=(ZN/60)(p/a)P=aE= ZN/60=216/8640=25 milli-webersIf the armature is rewound with wave connection, no. of parallel paths=2Hence at 500r.p.m. with 25mwb as the flux per poleThe armature emf=(25*103*864*500/60)*2=360V Losses of d.c machineThe losses in a dc machine may be defined into three phases.1. Copper losses2. Iron or core losses3. Mechanical losses1. Copper losses_a. Armatur copper loss = Ia2Rab. Shunt field copper loss =Ish2Rshc. Series field copper loss = Ise2Rse2. Iron or core losses_a. Hysteresis losses: Hysteresis loss occurs in the armature of the dc machine since any given part of the armatureis subject to magnatic riversals as it passes under successive poles. It is given by_Ph=B1.6max f v watt Bmax =maximum flux density in the armature f=frequency of the magnatic revarsal=NP/120 N= rpm, p= no. of poles , V = volume of armature in m3 =steinetz hystersis co efficient.b. Eddy current loss_When armature rotates in the magnatic field of the poles, an emf is induced in it which circulates eddy current loss. In order to reduce this lossthe armature core is build up of thin laminations insulated from each other by a thin layer of varnish.Pe= KeB2maxf2t2v wattKe= constant,Bmax =maximum flux density in the core, f= frequency of the magnatic reversals,t= thickness of lamination,v= volume of core in m33. Mechanical losses:a. Frictional losses: for example( bearing friction, brush friction etc)b. Windage loss: (air friction of rotating armature) Iron loss and mechanical loss are togther called stray losses. Constant and variable loss. Constant loss Variable loss

1. Iron loss1. copper loss in armature winding

2. Mechanical loss2. copper lossin series field winding

3. Shunt field loss

Total loss= constant loss+variable lossPower Stages:A) Mechanical power input > Iron and friction lossesB) Electrical power developed in armature EgIa > Cu lossesC) Electrical power output VIL

i) Mechanical Efficiency: m = = ii) Electrical Efficiency: e = = iii) Commercial or Overall Efficiency: c == = Condition for Maximum Efficiency:The efficiency of a d.c. generator is not constant but varies with load. Consider a shunt generator delivering a load current IL at a terminal voltage V.Generator Output = VILGenerator Input = Output + Losses = VIL + I2Ra + Wc = VIL + (IL + ISh)2 Ra+ WcThe shunt field current ISh is generally small as compared to IL and therefore can be neglected.Generator Input = VIL + I2Ra + Wc = = = ----------(1)Now, =0 =0 IL2Ra = Wc Variable loss = Constant loss.Open Circuit Characteristic of a D.C. Generator

The O.C.C. for a d.c. generator is determined as follows. The field winding ofthe d.c. generator (series or shunt) is disconnected from the machine and isseparately excited from an external d.c. source as shown in Fig. (3.1) (ii). Thegenerator is run at fixed speed (i.e., normal speed). The field current (If) isincreased from zero in steps and the corresponding values of generated e.m.f.(E0) read off on a voltmeter connected across the armature terminals. On plottingthe relation between E0 and If, we get the open circuit characteristic as shown inFig. (3.1) (i).

Fig. 3.1

The following points may be noted from O.C.C.:(i) When the field current is zero, there is some generated e.m.f. OA. This isdue to the residual magnetism in the field poles.(ii) Over a fairly wide range of field current (upto point B in the curve), thecurve is linear. It is because in this range, reluctance of iron is negligible ascompared with that of air gap. The air gap reluctance is constant and hencelinear relationship.(iii) After point B on the curve, the reluctance of iron also comes into picture. Itis because a t higher flux densities, r for iron decreases and reluctance of iron is no longer negligible. Consequently, the curve deviates from linear relationship.(iv) After point C on the curve, the magnetic saturation of poles begins and E0tends to level off.The reader may note that the O.C.C. of even self-excited generator is obtainedby running it as a separately excited generator.

Critical field resistance of a shunt-generator:

Fig.17.29The voltage build up in a shunt generator depends upon field circuit resistance. If the field circuit resistance is r1 (line OA), then generator will build up a voltage OM as shown fig 17.29. If the field circuit resistance is increased to r2 (line OB), the generator will build up a voltage OL, slightly less than OM, as the field circuit resistance is increased, the slop of resistance line also increase. When the field resistance line becomes tangent (line OC) to O.C.C. the generator would just excite. If the field circuit resistance is increased beyond this point (say line OD), the generator will fail to excite. The field circuit resistance is represented by line OC (tangent to O.C.C.) is called critical field resistance Rc for the shunt generator. If may be defined as under: The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical circuit.Characteristics of Series GeneratorFig. 17.30, shows the connections of a series wound generator. Since there isonly one current (that which flows through the whole machine), the load currentis the same as the exciting current. Fig 17.30(i) O.C.C. Curve 1 shows the open circuit characteristics (O.C.C) of a series generator. It can be obtained experimentally by disconnecting the field winding from the machine and exciting it from a separate d.c source as discussed in section 22.(ii) Internal characteristics: curve 2 shows the total or internal characteristics of a series generator. It gives the relation between the generated e.m.f. E on load and armature current. Due to armature reaction, the flux in the machine will be less than the e.m.f.E0 generated under no load conditions. Consequently, internal characteristics curve lies below the O.C.C. curve; the difference between them representing the effect of a armature reaction (see fig.17.30).

(iii) External characteristics: curve 3 shows the external characteristics of a series generator. It gives the relation between terminal voltage v and load current IL.. V = E Ia (Ra + Rse)

Fig.17.31Therefore; external characteristics curve will lie below internal characteristics curve by an amount equal to ohmic drop [i.e.; Ia (Ra+ Rse ) ] in the machine as shown in fig.17.30(ii). The internal and external characteristics of a dc series generator can be plotted from one another as shown in fig. 17.31. Suppose we are given the internal characteristics of the generator. Let the line OC represent the resistance of the whole machine i.e; Ra +Rse. if the load current is OB, drop in the machine is AB i.e. AB = ohmic drop in the machine =OB ( Ra + Rse )Now raise a perpendicular from point B and make a point b on this line such that ab = AB. Then point b will lie on external characteristics of the generator. Following similar procedure, other points of external characteristics can be located. It is easy to see that we can also plot internal characteristics from the external characteristics. CHARACTERISTICS OF A SHUNT GENERATOR:Fig. 17.32 (i) shows the connections of a shunt wound generator. The armature current Ia splits up into two parts small fraction Ish following through shunt field winding while the major part IL goes to the external load.

Fig. 17.32(i) O.C.C. The O.C.C. of a shunt generator is similar in shape to that of a series generator shown in fig 17.32 (ii). The line OA represents the shunt field circuit resistance. When the generator is run at normal speed, it will up a voltage OM. At no load, the terminal voltage of the generator will be constant (=OM ) represented by the horizontal dotted line MC(ii) Internal characteristic: When the generator is loaded, flux per pole is reducing due to armature reaction. Therefore e.m.f. E generated on load is less than the e.m.f. generated at no load. As a result, the terminal characteristics (E/Ia) drop down slightly as shown in fig. 17.32 (ii).(iii) External characteristic: Curve 2 shows the external characteristics of a shunt generator. It gives the relation between the terminal voltage V and load current IL.V = E - Ia Ra =E ( IL + LSh ) RaTherefore, external characteristics curve will lie below the internal characteristics curve by an amount equal to drop in the armature circuit [i.e.; (IL + Ish ) Ra] as shown in fig.17.32 (ii).Note: It may be seen from the external characteristics that change in terminal voltage from no-load to full load is small. The terminal voltage can always be maintained constant by adjusting the field rheostat R. CHARACTERISTICS OF COMPOUND GENERATORS:In a compound generator, both series and shunt excitation are combined as shown in fig. 17.33. The shunt winding can be connected either across the armature only or across armature plus series field (long-shunt connection G). T he compound generator can be cumulatively compounded or differentially compounded generator. The latter is rarely used in partice. Therefore we shall discuss the characteristics of cumulatively compounded generator. It may be noted that external characteristics of long and short shunt compounded generators are almost identical.

External characteristic: Fig. 17.34 shows the external characteristics of a cumulatively compound generator. The series excitation aids the shunt excitation. The degree of compounding depends upon the increase in series excitation with the increase in load current.(i) The series winding turns are so adjusted that with the increase in load current the terminal voltage increases, it is called over-compounded generator. In such a case as the load current increases, the series m.m.f. increases and tends to increase the flux and hence the generated voltage. The the increase in generated voltage isgreater than the Ia Ra drop so that instead of decreasing, the terminal voltage increases as shown by curve A in fig. 17.34.(ii) If series winding turns are so adjusted that with the increase in load current, the terminal voltage substantially remains constant, it is called flat-compounded generator. The series winding of such a machine has lesser number of turns than the one in over-compounded machine and, therefore, does not increase the flux as much for a given load current. Consequently, the full-load voltage is nearly equal to the no-load voltage indicated by curve B is in fig. 17.34.(iii) If series field winding has lesser number of turns than for a flat compounded machine, the terminal voltage falls with increase in load current as indicate by curve C in fig. 17.34. Such a machine is called under-compounded generator.

Theory of an Ideal transformerAn ideal transformer is one that has, (i) no winding resistance (ii) no leakage flux i.e the same flux links both the windings (iii) no iron losses(i.e eddy current and hysteresis losses) in the core.Consider an ideal transformer on no load i.e. secondary is open-circuited as shown in Fig.19.2(i). Under such conditions, the primary is simply a coil of pure inductance. When an alternating voltage V1 is applied to the primary, it draws a small magnetising current Im which lags behind the applied voltage by 900. This alternating current Im produces an alternating flux which is proportional to and in phase with it. The alternating flux links both the windings and induces e.m.f. E1 in the primary and e.m.f. E2 in the secondary. The primary e.m.f E1 is, at every instant, equal to and in opposition to V1. Both e.m.f E1 and E2 lag behind flux by 900 which is shown in the phasor diagram in Fig.19.2(ii). However, their magnitudes depend upon the number of primary and secondary turns. TransformerA transformer is a static piece of equipment used either for raising or lowering the voltage of an a.c. supply with a corresponding decrease or increase in current. It essentially consists of two windings, the primary and secondary, wound on a common laminated magnetic core as shown in Fig.19.1. The winding connected to the a.c. source is called primary winding and the one connected to load is called secondary winding. Working: When an alternating voltage V1 is applied to the primary, an alternating flux is set up in the core. This alternating flux links both the windings and induces e.m.f.s E1 and E2 in them according to Faradays laws of electromagnetic induction. The e.m.f. E1 is termed as primary e.m.f. and e.m.f E2 is termed as secondary e.m.f. Clearly, E1 = - N1 And E2 = - N2 E2/E1 = N2/N1 E.M.F EQUATION OF A TRANSFORMER Consider that an alternating v1 of frequency f is applied to the primary as shown in fig.(1). The sinusoidal flux produced by the primary can be represented as: = msint The instantaneous e.m.f e1 induced in the primarty, e1 = -N1 = -N1 (msint) = N1 m cost = -2fN1 m cost = 2fN1 m sin(t-900)..(i)It is clear from the above equation that maximum value of induced e.m.f in the primary is: Em1 = 2fN1 mThe r.m.s value E1 of the primary is: E1 = = = = 4.44f N1 mSimilarly E2 = 4.44f N2 mIn an ideal transformer, E1 = V1 and E2 = V2 Voltage transformation ratio (k)From the above equations of induced e.m.f , we have E2/E1 = N2/N1 = KThe constant k is called voltang transformation ratio. Thus if k=5(i.e. N2/N1=5),then,E2 =5E1 For an ideal transformer:(i) E1= V1 and E2 = V2 as there is no voltage drop in the windings. E2/E1 = V2/V1 = N2/N1 = K(ii) There are no losses. Therefore , vot-amperes input To the primary are edual to the output volt-amperes i,e V1 I1 = V2 I2 ; I1/I2 = V1/V2 = 1/kHence,currents are in the inverse ratio of voltage transformarmation ratio. This simply means that if we raise the voltage,there is a corresponding decrease of current.

PRACTICAL TRANSFORMERA practical transformer differs from the ideal transformer in many respects. The practical transformer has (i) iron losses (ii)winding resistances (iii) magnetic leakage,giving rise to leakage reactances.(i) Iron losses: Since the iron core is subjected to alternating flux, there occurs eddy current and hysteresis losses in it. These two losses together are known as iron losses or core losses.the iron losses depend upon the supply frequency, it may be noted that magnitude of iron losses is quite small in a pratical transformer.(ii) winding resistances: since the windings consist of copper conductor ,it immediately follows that both primary and secondary will have winding resistance.the primary resistance R1 and secondary resistance R2 act in series with the respective windings as shown in fig.19.3. Leakage reactances: Both primary and secondary currens produce flux. The flux which links both the windings is the useful flux and is called mutual flux .however,primary current would produce some flux 1 which would produce some flux 1 that would not link the primary winding. The flux such as 1 or 2 which links only one winding is called leakage flux . the leakage flux paths are mainly through the air . The effect of these leakage fluxes would be the same as though inductive reactance were connected in series with each winding of transformer that had no ieakage flux as shown in fig 19.3 . in other words ,the effect of primary leakage flux 1 is to introduce an inductive reactance X1 in series with the primary winding as shown in fig 19.3. Similarly,the secondary leakage flux 2 introduces an inductive reactance X2 in series with the secondary winding. PRACTICAL TRANSFORMER ON ON LOAD Consider a partical transformer on on load,I,e. secondary on open-circuit as shown in fig.19.5(i).the primary will draw a small current I0 to supply(i)the iron losses and (ii) a very small amount of copper loss in the primary.hence the primary no load current I0 is not 900 behind the applied voltage V1 but lags it by an angle 0900 as shown in phasor diagram in fig.19.5(ii). No load input power, W0 =V1I0cos0 (i) The component Iw in phase with the applied voltage V1 . This is known as active component and supplies the iron and a very small primary copper loss. Iw=I0 cos0(ii) The component Im lagging behind V1 by 900 and is known as magnetising component. It is this component which produces the mutual flux in the core. Im = I0 sin0

Therefore, No load p.f., cos0 = IW/I0

Ideal Transformer on Load:- Let connect a load ZL across the secondary of an ideal transformer as shown in figure. The secondary e.m.f E2 will causes a current I2 to follow through the load.

The angle at which I2 lead or lags V2 depends upon the resistance of the load. In the present case we have considered inductive load so that current I2 lags behind V2 by 2The secondary current I2 sets up an m.m.f N2I2 which produce a flux in the opposite direction to the flux originally set up in the primary by magnetizing current. This will change the flux in the core from the original value. However the flux in the core should not the change in the original value. In ordered to fulfill this condition the primary must developed an m.m.f which exactly counterbalances the secondary m.m.f N2I2 . Hence a primary current I1 must follow such that. N1I1 = N2I2 I1 = (N2 / N1) I2 = K I2 Phasor diagram:- Shows the Phasor diagram of an ideal transformer on load. Note that in drawing the Phasor diagram, the valu of K has been assumed unity so that primary Phasor are equal to secondary Phasors. The secondary current I2 lags behind V2 by 2. It causes a primary current I1 = K I2 = I I2 = I2 which is in antiphase with it. 1 = 2 Cos 1 = cos 2Thus, power factor on the primary side is equal to the p.f on the secondary side.

Transformer with resistance and leakage resistance:- Fig. shows a practical transformer having winding resistance and leakage resistance. This is the actual condition that exit in a transformer. This is voltage drop in R1 and X1 so that prima n y e.m.f E1 is less than the applied voltage V1. Similarly there is voltage drop in R1 and X1 so that secondary terminal voltage V2 is less than the secondary e.m.f E2. Let us take the usual case of inductive load which causes the secondary current I2 to lag behind the secondary voltage V2 by 2. The total primary current I1 must two requirements viz.1. It must supply the no load current I0 to meet the iron losses in the transformer and to provide flux in the core. 1. It must supply a current I2 to counteract the demagnetizing effect of secondary current I2. The magnitude of I2 will be such that:

Or

The total primary current I1 will be the Phasor sum of and I1 = I2+I0

Where = - E1 + I1 Z1

Phasor diagram:- Fig. shows the Phasor diagram of a practical transformer for the usual case of inductive load. Both E1 and E2 lag the mutual flux by 900. Note that current e.m.f that opposes that applied voltage V1 is -E1. Therefore of we add I1R1 and I1X1 to -E1. We get the applied primary voltage V1.

Load power factor=

Primary power factor=

Input power to transformer

Output power of transformer Impedance Ratio:- Consider a transformer having impedance Z2 in the secondary as shown in figure.

Or Note the importance of above relation. We can transfer the parameters from one winding to the others, thus;1. A resistance R1 in the primary becomes K2R1 when transferred to the secondary. 1. A resistance R2 in the secondary becomes R2/ K2 when transferred to the primary. 1. A resistance X1 in the primary becomes K2 X1 when transferred to the secondary.1. A resistance X2 in the secondary becomes X2/K2 when transferred to the primary.Shifting Impedance In a Transformer: Fig. shows a transformer where resistance and reactance are shown external to the winding. The resistance and reactance of one winding can be transferred to the other by appropriately using the factor K2. This makes the analysis of the transformer a simple affair because then we have to work in one winding only. 1. Referred to primary:- when secondary resistance or reactance is transferred to primary it is divided by K2 . it is then called equivalent secondary resistance or reactance referred to primary and is denoted by .

Equivalent resistance of transformer referred to primary,

Equivalent reactance of transformer referred to primary,

Equivalent impedance of transformer referred to primary, 1. Referred to secondary:- when primary resistance or reactance is transferred to the secondary it is multiplied by K2 . it is then called equivalent primary resistance or reactance referred to the secondary and its denoted by .

Equivalent resistance of transformer referred to secondary

Equivalent reactance of transformer referred to secondary

Equivalent impedance of transformer referred to secondary, Approximate Equivalent Circuit of A Transformer:The no load current I0 is only 1 3% of the rated primary current and may be neglected without any serious error. The transformer can then show as in figure.This is an approximate representation because no load current had been neglected. Note that all the circuit elements have been shown external so that the transformer is the ideal one. 1. Equivalent Circuit Of Transformer To Primary:- All the secondary quantities are referred to primary ,we get the equivalent circuit of the transformer referred to primary as shown in figure. The equivalent circuit in fig. is an electrical circuit and can be solved for various current voltages. Thus if we find V2 and I2 , then actual secondary values can be determind as under:

Actual secondary voltage,

Actual secondary current, 1. Equivalent Circuit Of Transformer To Secondary:- All the primary quantities are referred to secondary ,we get the equivalent circuit of the transformer referred to secondary as shown in figure.

The equivacircuit circuit shown in fig . is an electrical circuit and can be solved for various voltages and currents . thus is we find V1 and I1 , then actual primary values can be determind as under:

Actual secondary voltage,

Actual secondary current,

Voltage regulation : The voltage regulation of a transformer is a arithmatic difference between the no load secondary voltage (0V2) and the secondary voltage V2 on load expresed as percetage of no load voltage i.e% age voltage regulation = Where, 0V2 = No-load secondery voltage =KV1: V2 = Secondary voltage on load We know , 0V2 V2 = I2 R02 cos 2 I2X02 sin 2The +ve sign is for lagging p.f and ve sign is for leading p.f .

Transformer test:The circuit constants, eficiency and voltage regulation of a transformer can be determined by two simple test 1. Open circuit test and 2. Shot circuit test. These tests are very convenient as they provide the required information without actually loading the transformer .1. Open circuit or no load test:In this test, the rated voltage is applied to the primary (usually low voltage winding ) while the secondary is left open circuited. The applied primary voltage V1 is measured by the voltmeter, the no load current I0 by ammeter and no-load input power W0 by wattmeter as shown in fig. As the normal rated voltage is applied to the primary, therefor normal iron losses will occur in the transfomer core. Hence wattmeter will record the iron losses and small copper loss in the primary. Since no load current I0 is very small , Cu losses in the primary under no load condition are negligible as compared with iron losses. Hence wattmeter reading practically gives the iron losses in the transformer. It is reminded that iron losses are the same at all load . Iron losses, P1 = Wattmeter reading = W0 No load current = ammeter reading = I0 Applied voltage = voltmeter reading = V1 Input power , W0 = V1I0 cos 0 No load p.f. cos 0 = W0/V1I0 Iw = I0 cos 0 ; Im = I0 sin 0Thus open circuit test gives Pi , I0, cos 0 , Iw , and Im.2. Short-Circuit or impedance test.In this test, the secondary in short-circuited by a tick conductor and variable low voltage is applied to the primary as shown in fig. The low input voltage is gradually raised till at voltage Vsc, full-load current I1 flows in the primary. Then I2 in the secondary also has full-load value since I1/I2 = N2/N1 . Under such conditions, the copper loss in the windings is the same as that on full load. There is no output from the transformer under short circuit conditions. Therefore , input power is all lost and this loss is almost since the voltage Vsc is very small. Hence the wattmeter will practically register the full-load copper losses in the transformer windings. Fig 2shows the equivalent circuit of a transformer on short circuit as referred to primary; the no load current being neglected due to its smallness.Full load Cu loss, Pc = Watt meter reading = Ws Applied voltage, = Voltmeter reading = Vsc F.L primary current = Armature reading = I1

Where is the total resistance of transformer referred to primary. Total impedance referred to primary.

Total leakage reactance referred to primary,

Short circuit power factor

The short circuit test gives full load Cu loss,. LOSSES IN A TRANSFORMER The power losses in a transformer are of two types, namely ;1. Core or Iron losses.2. Copper losses .These losses appear in the form of heat and produce 1. An increase in temperature and 2. A drop in efficiency 1. Core or Iron losses (Pi) : These consist of hysteresis and eddy current losses and occur in the transformer core due to the alternating flux. These can be determined by open-circuit test .Hysteresis loss = Khf Bm1.6 watts/m3Eddy current loss = Ke f2Bm2 t2 watts/m3Both hysteresis and eddy current losses depend upon (i) maximum flux density Bm in the core and (ii) supply frequency f. Since transformer are connected to constant-frecuency, constant voltage supply, both f and Bm are constant . Hence, core or iron losses are practically the same at all loads.Iron or core losses, Pi = Hysteresis loss + Eddy current loss = Constant losses .The hysteresis loss can be minimised by using steel of high silicon content whereas eddy current loss can be reduced by using core of thin laminations .

2. Copper losses : These losses occur in both the primary and secondary windings due to their ohmicresistence. These can be determined by short-circuit test.Total Cu losses PC = I12 R1+ I22 R2 = I12 R01 Or I22 R02 It is clear that copper losses vary as the square of load current . Thus if copper losses are 400W at a load current of 10A, Then they will be () 400 =100 w at a load current of 5A .Total losses in a transformer = Pi +PC = Constant losses + Variable losses It may be noted that in a transformer, copper losses account for about 90% of the total losses .EFFICIENCY FROM TRANSFORMER TEST.

F.L Iron loss = Pi F.L Cu loss = PC Total F.L losses = Pi + PC We can now find the full-load efficiency of the transformer at any p.f without actually loading the transformer.F.L efficiency F.L = Also for any load equal to x full-loadCorresponding total losses = Pi + x2 PCCorresponding x = Note that iron losses remains the same at all loads. Condition for maximum efficiency :

Output Power = V2 I2 cos 2 If R02 is the total residence of the transformer referred to secondary, then ,Total Cu loss PC = I22 R02Total losses = Pi + PC

So , Transformer = = ..(i)For a normal transformer, V2 is approximately constant. Hence for a load of given p.f efficiency depends upon load current I2. It is clear from exp(i) above that numerator is constant and for the efficiency to be maximum, the denominator should be minimum i.e = 0 Or, ( V2 cos2 + Pi/I2 + I2 R02 ) = 0 Or, 0- + R02 = 0 Or, Pi = I22 R02 (ii)i.e Iron losses = Copper losses Hence efficiency of a transformer will be maximum when copper losses are equal to constant or iron losses .From equ,,..(ii) above ,the load current I2 corresponding to maximum is given by ;

I2 = Output kVA Corresponding To Maximum Efficiency .

Let, PC = copper losses at full-load kVA Pi = Iron losses x = Fraction of full-load kVA at which efficiency is maximumTotal Cu losses = x2 PCso, x2 PC = Pi .for maximum efficiency

Or, x = = So, output kVA corresponding to maximum efficiency = x full load kVA = full load kVAIt may be noted that the value of kVA at which the efficiency is maximum is independent of p.f of the load.Example 32.1. The maximum flux density in the core of a 250/3000 volts, 50-Hz single phase transformer is 1.2 Wb/m2. If the e.m.f. per turn is 8 volt, determine (1) primary and secondary turns (2) area of the core.

Solution. (1)E1 = N1e.m.f. induced/ turnN1=250/8=32; N2=3000/8=375(2)E2=4.44 f N2 Bm A3000=4.44503751.2A; A=0.03 m2

Example 32.2.The core of a 100-Kva, 11000/550 v, 50-Hz, 1-ph, core type transformer has a cross section of 20 cm 20 cm. Find (1) the number of H.V. and L.V. turns per phase and (2) the e.m.f. per turn if the maximum core density is not to exceed 1.3 tesla. Assume a stacking factor of o.9. What will happen if its primary voltage is increased by 10% on no- load?Solution. (1) Bm=1.3 T, A=(0.20.2)0.9=0.036 m211000=4.4450N11.30.036, N1=1060550=4.4450N21.30.036, N2=53Or,N2=KN1=53(2) e.m.f./ turn=11000/1060=10.4 VExample 32.3. A single phase transformer has 400 primary and 1000 secondary turns The net cross sectional area of the core is 60 cm2.If the primary winding be connected to a50 hz supply at 520 v, calculate(1)the peak value of flux density in the core (2)the voltage induced in the secondary winding.Solution.(1)K=N2/N1=1000/400==2.5E2/E1=K ; E2=KE1=2.5520=1300 v(2)E1=4.44 f N1 Bm A500=4.4450400Bm(6010 -4) ;Bm =0.976 Wb/m2

Example 32.8.A single phase, 50 hz, core type transformer has square cores of 20 cn side. Permissible maximum flux density is 1Wb/m2.calculate the number of turns per limb on the High and Low voltage sides for a 3000/220 V ratio.Solution. Starting with calculation for L.V.turns, T2,4.4450[(202010-4)1]T2=220T2=220/8.88=24.77T2=26T1/T2=V1/V2; T1=263000/220=354Number of turns on eacg Limb =177Number of turns on each Limb=13

Example 32.7. A 25 KVA single phase transformer has 250 turns on the primary and 40 turns on the secondary winding .The primary is connected to 1500 volt,50 Hz mains.(1)primary and secondary currents on full load,(2)secondary e.m.f.,(3)maximum flux in the core.Solution.(1) If V2= secondary voltage rating=secondary e.m.f. V2/1500=40/250; V2 =240 volts(2) primary currents=25000/1500=16.67 ampSecondary currents =25000/240=104.2 amp(3) If m is the maximum core-flux in Wb,1500=4.4450m250, m=0.o27 Wb or 27 mWbExample 32.6.A single phase transformer has 500 turns in the primary and 1200 turns in the secondary. The cross- sectional area of the core is 80 sq. cm. If the primary winding is connected to a 50Hz supply at 500 v, calculate (1)Peal flux density, and (2) Voltage induced in the secondary.Soolution. From the e.m.f. equation for transformer ,500=4.4450m500 ; m=1/222 Wb(1) peak flux density, Bm=m/(8010-4)=0.563 wb/m2(2) V2/V1=N2/N1; V2=5001200/500=1200 voltsExample 32.4. A 25 KVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to 3000 V,50 Hz supply. Find the full load primary and secondary currents, the secondary e.m.f. and the maximum fiux in the core. Neglect leakage drops and no load primary current.Solution. K=N2/N1=50/500=1/10 Now, full load=I1=25000/3000=8.33 A,I2=I1/K=108.33=83.3Ae.m.f. per turn on primary side=3000/500=6V Secondary e.m.f.=650=300 V E1=4.44 f N1 m3000=4.4450500m; m= 27mWb

Dc motor principle: A machine that converts dc power into mechanical power is known as a dc motor. The principle of dc motor can be stated as When a current carrying conductor is placed in a magnetic field, the conductor experiences a mechanical force. The direction of this force is given by fleming left hand rule and its magnitudes is given by F = BIL

Voltage equation of a dc motor: Let in a d.c. motor(Fig.) V = applied voltage Eb = back e.m.f Ra = armature resistance Ia = armature currentSince back e.m.f Eb acts in opposition to the applied voltage V, the net voltage across the armature circuit is V-Eb. The armature current Ia is given by : I a = V = E b+ I a Ra.(i)Power equation of d.c. motor:Multiplying both sides of equ. (i), V I a = EbIa+I2aRa Eb Ia = V I a - I2aRaCondition for maximum power:The mechanical power developed by the motor is, Pm= Eb IaFor maximum power, = 0 V -2 IaRa = 0 V = 2IaRa IaRa = V/2From equation (i) V = E b+ I a Ra = Eb+ V/2 Eb = V/2

Armature torque of a dc motor:Let in a dc motor , r = average radius of armature in (m) l = effective length of each con doctor in (m) Z = total no. of armature conductors A = number of parallel path i = current in each conductor = Ia/A B = average flux density in Wb/m3 = flux per pole in Wb P = no. of polesTorque in one conductor, T = Fr =B i lrTotal torque in Z no. of conductors Ta = Z F r = Z B i lr = Z (/a)(Ia/A)lr = (N-m) = 1.59 ZIa (P/A)Since Z, P and A are fixed Ta Ia(i) For shunt motor, Ta Ia [ is constant ](ii) For series motor, Ta Ia2

Prob-1: A 220 V d.c. machine has an armature resistance of 0.5 ohm. If the full-load armature current is 20 A, find the induced e.m.f. when the machine acts as (i) generator (ii) motor.Solve: Generator MotorFrom the Fig.1, shunt current is considered negligible because its value is not given.(i) As generator[Fig.i] Eg = V+IaRa =220+(0.5x20)=230 V(ii) As motor [Fig.ii] Eb = V-IaRa = 220-(0.5x20)=210 V

Prob-2: A separately excited D.C. generator has armature circuit 1000 r.p.m, it delivers a current of 100 A at 250 V to a load of constant resistance. If the generator speed drop to 700 r.p.m, with field current unaltered, find the current delivered to load.

Solve: RL = 250/100= 2.5 ohms.Eg1 = 250+(100x0.1)+2 = 262 VAt 700 r.p.m., Eg2 = 262x(700/1000) = 183.4 V If Ia is the new current, Eg2-2-(Iax0.10)= 2.5 Ia This gives Ia = 96.77 amp.

Prob-3: A 220 V, shunt motor has armature resistance of 0.8 ohm and field resistance of 200 ohm. Determine the back e.m.f. when giving an output of 7.46 kw at 85 percent efficiency.

Solve: Motor input power = (7.46X103)/0.85 WMotor input current = (7460/0.85)/440 = 19.95 A Ish = 440/200 = 2.2 A Ia = IL-Ish = 19.95-2.25 =17.75 A Now Eb = V-IaRa = 440-(17.75x0.8) =425.8 V

Prob-4: A 25 kw, 250 V, d.c. shunt motor has armature resistance of 0.06 ohm and field resistance of 100 ohm. Determine the total armature power developed when working-(i) as a generator delivering 25 kw output and(ii) as a motor taking 25 kw input.

Solve: Generator Motor

As Generator[Fig.i]Output current = 25000/250 =100A Ish = 250/100 = 2.5 A; Ia = 102.5 AGenerated e.m.f. = V+IaRa = 250+(102.5x0.06) = 256.15 V Power developed in armature = EbIa = (256.15x102.5)/1000 = 26.25 kwAs motor [Fig.ii], Motor input current = 100 A ; Ish = 2.5 A ; Ia = 97.5 A Eb = 250-(97.5x0.06) = 244.15 VPower developed in armature = EbIa = (244.15x97.5)/1000 = 23.8 kw

Prob-5: A d.c. motor takes an armature current of 110 A at 480 V. The armature circuit resistance is 0.2 ohm. The machine has 6-poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate- (i) the speed and (ii) the gross torque developed by the armature.

Solve:Here Eb = 480-(110x0.2)= 458 V, =0.05 W, Z = 864Now Eb = (ZNP)/60A Or 458=(0.05x864xNx6)/(60x6) N = 636 r.p.m. Ta = 0.159xZIa x(P/A) = 0.159x0.05x864x110x(6x6) = 756.3 N-m

Prob-6: A 250 V, 4-pole, wave-wound d.c. series motor has 782 conductors on its armature. It has armature and series field resistance of 0.75 ohm. The motor takes a current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb.

Solve: Here Eb = (ZNP)/60A .................(i) Now, Eb = V-IaRa = 250-(40x0.75) = 220 VFrom equation (i), 220 = (25x10-3x782xNx4)/60x2 N = 337 r.p.m Ta = 0.159xZIa x(P/A) = 0.159x25x10-3x782x40x2 = 249 N-m

Prob-7: A d.c. shunt machine develops an a.c. e.m.f. of 250 V at 1500 r.p.m. Find its torque and mechanical power developed for an armature current of 50 A.

Solve:Mechanical power developed in the armature = EbIa = 250x50=12,500 W Now, Ta = 9.55(EbIa/N) = 9.55(12500/1500) = 79.6 N-m

Prob-8: Determine developed torque and shaft torque of 220 V, 4-pole series motor with 800 conductors wave-connected supplying a load of 8.2 kw by taking 45 A from the mains. The flux per pole is 25 mWb and its armature circuit resistance is 0.6 ohm.

Solve: Developed torque, Ta = 0.159xZIa x(P/A) = 0.159x25x10-3x800x45(4/2) = 286.2 N-m

Eb = V-IaRa = 220-(45x0.6) = 193 VNow, Eb = (ZNP)/60A Or 193= 25x10-3x800xN(4/2) N = 289 r.p.mAlso, 2NTsh = OutputOr Tsh(2 x 289) = 8200 Tsh = 4.52 N -m

Prob-9: A 220-V, d.c. shunt motor runs at 500 r.p.m. when the armature circuit is 50 A. Calculate the speed if the torque is doubled. Given that Ra = 0.2 ohm.

Solve:We know, Ta IaSince is constant, Ta IaNow, Ta1 Ia1 and Ta2 Ia2 Ta2/Ta1 = Ia2/Ia1 Ia2 = 100 ANow, N2/N1 = Eb2/Eb1.................(i) Eb1 = 220- (50x0.2)= 210 V Eb2 = 220-(100x0.2) = 200 VFrom equation (i), N2/500 = 200/210 N2 = 476 r.p.m.

Prob-10: A 500 V, 37.3 KW, 1000 r.p.m. d.c. shunt motor has on full-load an efficiency of 90 percent. The armature circuit resistance is 0.24 ohm and there is total voltage drop of 2V at the brushes. The field current is 1.8 A. Determine - (i) full-load line current (ii) full load shaft torque in N-m and (iii) total resistance in motor starter to limit the starting current to 1.5 times the full load current.

Solve: (i) Motor input = 37300/0.9 = 41444 W F.L. line current = 41444/500 = 82.9 A (ii) Tsh = 9.55 ( Output/N) = 9.55(37300/1000) = 356 N-m (iii) Starting line current = 1.5 x 82.9 = 124.3 A Arm. current at starting = 124.3-1.8 = 122.5 A If R is the starter resistance( which is in series with armature), then- 122.5(R+0.24) +2 = 500 R = 3.825 ohm.

Prob- 11: A 4-pole, 220 V shunt motor has 540 lap-wound conductor. It takes 32 A from the supply mains and develops output power of 5.595 KW. The field winding takes 1 A. The armature resistance is 0.09 ohm and the flux per pole is 30 mWb.

Calculate- (i) the speed and (ii) the torque developed in N-m.

Solve: Ia = IL - Ish = 32-1 =31 A ; Eb = V-IaRa = 220-(31x0.09) = 217.2 VNow, Eb = (ZNP)/60A 217.2 = ( 30x10-3x540xNx4)x(60x4) 217.2 = 0.27 N (i) N = 804. 444 r.p.m.(ii) Tsh = 9.55 ( Output/N) = 9.55(5595/804.444) = 66.42 N-m

Prob- 12: Determine the torque established by the armature of a four-pole D.C. motor having 774 conductors, two paths in parallel, 24 mWb of pole- flux and the armature current is 50 Amps.

Solve: We know, Torque, T = 0.159xZIa x(P/A) Nw-mTwo paths in parallel for a 4-pole case means a wave winding. T = 0.159x(24x10-3)x774x50x(4/2) = 295.36 Nw-m

Prob- 13: A 500 V D.C. shunt motor draws a line-current of 5 A on light-load. If armature resistance is 0.15 ohm and field resistance is 200 ohm, determine the efficiency of the machine running as a generator delivering a load current of 40 Amps.

Solve: (i) No load, running as a motor: Input power = 500x5 = 2500 W Field copper-loss = 500x2.5 = 1250 W (ii) As a generator, delivering 40 A to load: Output delivered = 500x40x10-3 = 20 KW Losses: (a) Field copper-loss = 1250 W (b) Arm. copper-loss = 42.52x0.15 = 271 W (c) No load losses = 1250 W Total losses = 2.771 KW Generator efficiency = (Output power/Input power)x100 % = (20/22.771)x100% = 87.83 %

Prob-14: A d.c. series motor takes 40 A at 220 V and runs at 800 r.p.m. If the armature and field resistance are 0.2 ohm and 0.1 ohm respectively and the iron and friction losses are 0.5 KW, find the torque developed in the armature. What will be the output of the motor ?

Solve: Armature torque is given by, Ta = 9.55 ( EbIa/N) N-mNow, Eb = V-Ia(Ra+Rse) = 220-40(0.2+0.1) = 208 V Ta = 9.55x208x(40/800) = 99.3 N-mCu loss in armature and series field resistance = 402x0.3 = 480 WIron and friction losses = 500 W Total losses = 480+500 = 980 WMotor power input = 220x40 = 8800 WMotor output = 8800-980 = 7820 W = 7.82 KW

Prob-15: A cutting tool exerts a tangential force of 400 N on a steel bar of diameter 10 cm which is being turned in a simple lathe. The lathe is driven by a chain at 840 r.p.m. from a 220 V d.c. Motor size is the motor pulley if the lathe pulley has a diameter of 24 cm ?

Solve:Torque Tsh = Tangential force x radius = 400x0.05 = 20 N-mOutput power = Tshx2N Watt = 20x2(840/60) Watt = 1760 WMotor input = Output power/efficiency = 1760/0.8 = 2200 WCurrent drawn by motor = 2200/220 = 10 A Let N1 and D1 be the speed and diameter of the driver pulley respectively and N2 and D2 the respective speed and diameter of the lathe pulley. Then N1xD1 = N2xD2 Or 1800xD1 =840x0.24 D1 = 201.6/1800 m = 0.112 m

48