DC Motors : Characteristics

Dr. Bharat Singh Rajpurohit

Objectives Torque equationRelations between Torque, Speed, Flux, Armature Current, back emfCharacteristics of Motors (for shunt, series and compound)

Armature current Vs TorqueArmature current Vs SpeedTorque vs Speed

Few Numerical

DC motors are the machine that convert electrical energy intomechanical energy and works on the principal that when currentcarrying conductor is placed in a magnetic field it experience amechanical force.

DC Motor : Torque EquationTorque is defined as tendency to produce rotation.

Φ = Flux per pole in weber,P = Number of poles,N = Speed in rpm,Z = Total number of armature conductors, two cond. per turnA = Number of parallel pathsTa= Armature torqueEb= Voltage generated in armature opposes the armature current

Power developed = Ta * 2ΠN watt,Electrical Power = Eb* Ia = ΦPNZ/A * Ia

→Ta = 0.159 * ΦPZ/A * Ia

1. A motor develops a torque of 150N.m and is subjected to a 10 % percent reduction in field flux, which produces a 50% increase in armature current. Find the new torque produced as a result of this change in field flux.

Solution: - ф Ia TOriginal condition 1.0 1.0 150N.mNew condition 0.9 1.5 ?

T=K’фIaUsing the ratio method, the new torque is the product of two new ratio changes: T= 202 Nm.

Conclusion: Torque depends upon flux and armature current but independent of speed. But speed depends upon torque (not vice-versa).

Back (or Counter ) EMF2. A dc shunt motor having a armature resistance of 0.25 Ω and a brush contact voltage drop of 3V receives an applied voltage across its armature terminals of 120V. Calculate the armature when:• The speed produces a counter EMF of 110V at a given load.• The speed drops (due to application of additional load) and the

counter EMF is 105V. • Compute the percentage of change in counter EMF and in

armature current.

• Solution: Ia= (V- Eb - BD )/ Ra =28AAt increased load,Ia=( V- Eb - BD )/ Ra = 48AδEc=4.54 % and δIa=71.4 %

Conclusions: Small dec. in Eb resulted in much larger inc. in Ia. Consequently, Small changes in N (& Eb) corresponds to large changes in motor current .

Motor Speed: Back EMF and Flux Ia= (V- Eb - BD )/ Ra and Eb = kфN

Hence N = (V- Ia Ra - BD ) / kфFundamental equation for speed relation of DC motor.

Cases: 1. ф tends to zero, N will shoot-up.2. Ia and ф const. and than N depends upon V only.3. ф and V const., inc. in load i.e. Ia, N will change accordingly to Eb.

Motor Speed: Back EMF and Flux 3. A 120V dc shunt motor having an armature circuit resistance of 0.2Ω and a field circuit resistance of 60Ω, draws a line current of 40 A at full load. The brush voltage drop is 3V and rated full-load speed is 1800rpm. Calculate: •The speed at half load (1863 rpm)•The speed at an overload of 125 percent (1769 rpm).

Sol.: At full load, Ia=Il-If=40A- 120/60 =38A; Eb=Va-(IaRa+BD) =120-(38*0.2+3)=109.4VAt the rated speed of 1800 rpm,Eb=109.4V and Ia=38A (full load)At half speed,Ia= 38/2=19A; Eb=Va-(IaRa+BD) =120-(19*0.2+3) =113.2VUsing the ratio method, half-load speed isN=1800*(113.2/ 109.4)=1863 rpm

Motor Speed: Back EMF and Flux Sol.: At 125 % load,Ia=47.4 A; Eb=Va-(IaRa+BD)=120-(47.5*0.2+3)=107.5 VN5/4=1800 * 107.5/ 109.4 =1769 rpm

Motor Speed: Back EMF and Flux 4. The dc motor of ex. 3 is loaded to a line current of 66A(temporarily) , but in order to produce the necessary torque, thefield flux is increased by 12 % by decreasing the field circuitresistance to 50Ω. Calculate the speed of the motor.

Ia=Il-If=66- 120/50=63.6 AEb=Va-(IaRa+BD)=120-(63.6*0.2+3)=104.3VN= k Eb / ф =1800* (104.3/109.4) * (1.0/1.12)=1532 rpm

Back EMF and Mechanical Power DevelopedEb=Va-IaRa

EbIa=VaIa-I2aRa

(Power Mech. = Electrical power – Losses)•Now ratio, EbIa/VaIa which is same as Eb/Va and will decide electrical power available for conversion to mechanical power.•Higher the ratio Eb/Va for higher the efficiency. •Also for max. Eb = kфN, ф & N can be increased.•But also, N = (V- Ia Ra - BD ) / kф, so ф inc. means N dec.•Conclusion: For a given mech. load and resulting IL and Ia, there is particular N and ф where max. power can be produced. (From ex. 4)

Back EMF and Mechanical Power Developed5. Calculate the armature power developed, Pd, for each of the loadsof Ex. 3 and Ex. 4 and tabulate all results for ready reference andcomparison.Sol.:

Conclusion: For inc. in Ia or motor load.•The Eb dec.•The N dec.•The motor armature power developed inc.

Torque and SpeedN = (V- Ia Ra - BD ) / kф (1)T = k ф Ia (2)Ia= (V- Eb - BD )/ Ra (3)

Relation (2) says that inc. in ф will inc. T (also possibly N).But relation (1) says that inc. in ф will dec. the N. Inconsistency?Now, relation (3) can be used to explain this inconsistency.•The ф can be reduced by reduction in If.•The Eb (= фN) will dec. instantaneously (but not N due to inertia).•The dec. in Eb cause an inc. in Ia instantaneously (relation 3).•But small dec. in ф produces and large inc. in Ia.•Relation 2, small dec. in ф will be balanced by large inc. in Ia. Nownew torque has increased value than previous torque value.•This inc. in torque produces an inc. in N.

Conclusion: Dec. in ф results in inc. in N.

Torque Characteristics of DC Motors(Torque Vs Armature Current)

Shunt Motor: T = k ф Ia and Eb=Va-IaRaф = const.

Hence, T = k’ Ia (a linear relation)

Series Motor: T = k ф Ia and Eb=Va- Ia (Ra+ Rse)ф proportional to Ia

Hence, T = k”I2a (a quadratic relation)

Compound Motor: T = k ф Iaф = (фsh ± фse)

Hence, T = k (фsh ± фse) Ia or T = (k1 ± k2Ia) Ia = (k1 Ia ± k2I2

a)

Torque Characteristics of DC Motors

6. A cumulative compound motor is operated as a shunt motor(series field disconnected) and develops a torque of 160 lb.ft whenthe armature current is 140A and the field flux is 1.6*106 lines.When reconnected as cumulative compound motor at the samecurrent, it develops a torque of 190 lb.ft. Find•The flux increases due to the series field in percent.•The torque when the compound motor load increases by 10 percent. (assume operation on the linear portion of the saturation curve)Sol.: T Ia фr

In lb.ft in A in linesOriginal 160 140 1.6*106

Added flux 190 140 фfFinal torque Tf 154 1.1*1.9*106

Now, Фf=1.6*106 * 190/160=1.9*106 lines.Percentage of flux increase: 18.8 %The final field flux is 1.1*1.9*106 lines (due to the 10 percent increase in load). The final torque is : 190 * (154/140) * 1.1 = 230 lb.ft

7. A series motor draws a current of 25A and develops a torque of90lb.ft. Calculate•The torque when the current rises to 30A if the field is unsaturated •The torque when the current rises to 30A and the increases in

current produces a 10 percent increase in flux.

Sol.: T=KI2a=90 lb.ft (30/25)2=129.6 lb.ftConclusion: For unsaturation, a 20 % inc. in Ia, gives 44 % inc. in T.

T=KфIa=90lb.ft * (30/25) * (1.1/1.0) =118.8 lb.ftConclusion: For saturation, a 20 % inc. in Ia, gives 32 % inc. in T.

Speed Characteristics of DC Motors(Speed Vs Armature Current)

Shunt Motor: N = (V- Ia Ra - BD ) / kф ф = const.

Hence, N = (V- Ia Ra - BD ) / kф (const.)

Series Motor: N = (V- Ia Ra - Ia Rse- BD ) / kф ф proportional to Ia

Hence, N = (V- Ia Ra - Ia Rse- BD ) / k Ia Note: Ia very small, N very high

Ia very high, N very small Caution: Centrifugal switch and/or coupled load. Fuse in arm. path.

Compound Motor: N = (Eb) / kф (Eb = const.)ф = (фsh ± фse)

Hence, N = (Eb) / k (фsh ± фse) or N = (Eb) / (k1 ± k2Ia)

Speed Characteristics of DC Motors

Torque-Speed Characteristics of DC Motors

Applications of DC MachinesDC Motor Series motor-high starting torque-Traction, cranes, hoists, hybrid vehicle, conveyer belts.Shunt motor - Medium torque, constant speed-Lathe machine, constant speed line-shafting, centrifugal & reciprocating pumps, etc.

We have seen basic characteristics of DC motors.Only DC motor is sufficient to fulfill the loadrequirement related to speed and torque requirements?

A control is required. Speed and Torque control.

8. A 230V,10 hp,1250 rpm cumulative compound has an armatureresistance motor has an Ra =0.25Ω, a combined compensatingwinding and interpole resistance of 0.25Ω, and a brush voltage dropof 5V. The resistance of the series field is 0.15Ω, and the shunt fieldresistance is 230Ω. When the motor is shunt connected, the linecurrent rated load is 55A and the no load line current is 4A. The no-load speed is 1810 rpm. Neglecting armature reaction at ratedvoltage, calculate• Speed at rated load•Internal power in watts and internal horse-power developed.

Sol.: Ia=Il-If=4A-1A=3ANo-load Et=Va-(IaRa+BD)=230-(3*0.5+5)=223.5 at a speed of 1810 rpmFull-load Ec=Va-(IaRa+BD)=230-(54*0.5+5)=198V

N = 1810 (198/223.5) = 1603 rpm.

8. A 230V,10 hp,1250 rpm cumulative compound has an armatureresistance motor has an Ra =0.25Ω, a combined compensatingwinding and interpole resistance of 0.25Ω, and a brush voltage dropof 5V. The resistance of the series field is 0.15Ω, and the shunt fieldresistance is 230Ω. When the motor is shunt connected, the linecurrent rated load is 55A and the no load line current is 4A. The no-load speed is 1810 rpm. Neglecting armature reaction at ratedvoltage, calculate• Internal power in watts and internal horse-power developed.

Sol.: Pd=EbIa=198V*54A=10700Whp = 10700/746 = 14.3 hp

Conclusions: •Full load speed of shunt motor is higher than compound motor.•The internal hp is greater than external hp

ThanksReferences: • I. L. Kosow, Electrical Machinery & Transformers• Clayton & Hancock, The Performance and Design of DC

Machines