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Chapter 1 dc machines new


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Chapter 1 dc machines new

  1. 1. CHAPTER 1 DC MACHINES1.1 INTRODUCTIONEnergy is needed in different forms:• Light bulbs and heaters - electrical energy• Fans and rolling miles - mechanical energy• Need for energy converters Figure 1.1• DC generator – supply current to the load• DC motor – need current from the supply• AC electric supply - AC machines (synchronous and asynchronous)• DC electric supply - DC machines• Advantages of DC Machine (i) Adjustable motor speed over wide ranges (ii) Constant mechanical output (torque) (iii)Rapid acceleration or deceleration (iv) Responsive to feedback signals 1
  2. 2. Figure 1.2 : DC machine Construction1.2 GENERATION OF AC SIGNAL• Generator needs something to rotate or move itself.• It is called the prime mover.• The shaft of the prime mover will be coupled to the shaft of the generator so that the generator can rotate once the prime mover rotates.• In a big system, they normally used 3-phase induction motor.• But, in small scale machine, they can use a mechanism to rotate the generator which then generated electricity. coil Slip ring:used to generate ac signal Carbon brush: used to collect output voltage (ac signal) carbon brush Figure 1.3: Physical Arrangement of AC Generating Method 2
  3. 3. Figure 1.4 : Emf generated at various instant (angle position)1.3 GENERATION OF DC SIGNAL• The generating of DC signal can be done by replacing the slip ring with commutator so that rectification process can happen.• Commutator can generate dc signal.• This is done through rectification or commutation process; which converts ac signal into dc mechanically.• Therefore a commutator is called a mechanical rectifier. Carbon commutator:used to brush generate dc signal Carbon brush: used to collect output voltage (dc signal) commutator Figure 1.5 : Physical Arrangement of DC Generating Method 3
  4. 4. Figure 1.6: AC is converted into DC Signal through Commutation Process1.4 EMF GENERATED ON DC MACHINEThe EMF generated:- zNP EMF = 2 φ c60where z = no of conductors in the armature circuit c = no. of parallel path N = speed in rpm Φ = flux/pole (Wb) P = no. of pole pair Figure 1.7 : Permanent Magnet Dc Machine 4
  5. 5. Example 1If the no-load voltage of a separately excited generator is 135 V at 850 rpm, what will bethe voltage if the speed is increased to 1000 rpm? Assume constant field excitation.Solution 2 zPNφEMF = c60 2 zPK= c60Therefor ,E1 = KN 1φ1E 2 = KN 2φ 2Constant field excitation:- i1 = i2 , ⇒ φ1 = φ 2E1 KN 1φ1 N = = 1E 2 KN 2φ 2 N 2 E1 N 2 1000rpm(135V )E2 = = = 158.8V N1 850rpmExample 2A separately excited generator has no-load voltage of 140 V when the field current isadjusted to 2 A. The speed is 900 rpm. Assume a linear relationship between the fieldflux and current. Calculate:(i) the generated voltage when the field current is increased to 2.5 A. Assume N1=N2.(ii) the terminal voltage when the speed is increased to 1000 rpm with the field current set at 2.2 A.SolutionV1 = 140V , I F 1 = 2 A, N 1 = 900rpmφαI F(i) I F 2 = 2.5 A E1 KN 1φ1 I = = F1 E 2 KN 2φ 2 I F 2 E1 I F 2 140(2.5) E2 = = = 175V I F1 2(ii) N 2 = 1000rpm, I F 2 = 2.2 A, V2 = ? 5
  6. 6. E1 KN 1φ1 NI = = 1 F1 E 2 KN 2φ 2 N 2 I F 2 E1 N 1 I F 2 140(1000)(2.2) E2 = = = 171.11V N 2 I F1 (900)21.5 DC MACHINE CONSTRUCTION rotor/armature pole Figure 1.8 : DC Machine ConstructionDC motor principles• DC motors consist of rotor-mounted windings (armature) at the rotor side and stationary windings (field poles) at the stator side. 6
  7. 7. • In all DC motors, except permanent magnet motors, current must be conducted to the armature windings by passing current through carbon brushes that slide over a set of copper surfaces called a commutator, which is mounted on the rotor.• The commutator bars are soldered to armature coils.• The brush/commutator combination makes a sliding switch that energizes particular portions of the armature, based on the position of the rotor.• This process creates north and south magnetic poles on the rotor that are attracted to or repelled by north and south poles on the stator, which are formed by passing direct current through the field windings.• Its this magnetic attraction and repulsion that causes the rotor to rotate. Figure 1.9 : DC Machine Construction1.6 MACHINE WINDINGS• Machine winding can be divide into 2:- (i) armature winding (rotor side) (ii) field winding (stator side) 7
  8. 8. Machine Winding Field winding Armature Winding Separately excited **Self excited -no direct connection -direct connection between between armature circuit armature circuit and the and the field circuit field circuit Series Compound Shunt excitation excitation excitation Figure 1.10: Winding Connection in DC MachineSELF-EXCITED FIELD WINDING• In self-excited dc machine, there are three types of excitation method namely: (i) Series Excitation: the field winding is connected in series with the armature circuit. (ii) Shunt Excitation: the field winding is connected in parallel with the armature circuit. (iii) Compound Excitation: the field winding are connected in series and parallel with the armature circuit.• The schematic diagrams for the three types of these machines are illustrated in Figure 1.11.• The difference between dc motor and dc generator is in terms of the current direction.• In dc generator: armature current, Ia is supplied by the armature. In dc motor: armature current, Ia is received by the armature. 8
  9. 9. (a) DC Series machine Rf Rf IL IL Ra Ra dc supply DC loa d Ia VT Ia VT 1.0 k 1.0 k 1.0m 1.0m 5.0 + + Eg EC - -Generator: Eg = VT + Ia(Ra + Rf) Motor: Ec = VT - Ia(Ra + Rf)(b)DC Shunt machine IL If IL If Ra Ra DC loa d 1.0 k dc supply 1.0 k 1.0 m 1.0 m Ia Rf VT 5.0 Ia Rf + VT + Eg Ec - -Generator: Eg = VT + IaRa Motor: Ec = VT - IaRa(c) DC Compound machine Rf2 Rf2 IL IF IL Ra IF Ra Ia D Cload Ia Rf1 VT Rf1 dc supply 1,0m 1,0 1,0m DCM1 1,0m + VT + Eg Ec D CM1 - 1,0m 1,0m -Generator: Eg = VT + Ia(Ra + Rf 2) Motor: Ec = VT - Ia(Ra + Rf 2) Figure 1.11 ** Eg = generated emf Ec = counter emf 9
  10. 10. 1.7 POWER FLOW DIAGRAMPower flow diagram is normally represented as a fish bone. Input Power = Output + Losses Losses can be divided into 2:- (i) Copper losses (Armature copper loss , Pca) and (Field copper loss,Pcf) (ii) Iron losses, Pµ (friction, stray, windage,mechanical losses) For DC Generator: Pin = Pout + Total losses where Pout = VTIL For DC Motor: Pin = Pout + Total losses where Pin = VTIL Total losses = Pca + Pcf + Pµ 10
  12. 12. POWER FLOW DIAGRAM FOR DC MOTOR DC SERIES Pca Pµ Pm Pout Pin=VTIL Pcf DC SHUNT Pca Pµ Pm Pin=VTIL Pout Pcf DC COMPOUND Pca Pcf2 Pin = VTIL Pm Pout Pµ Pcf11.8 MOTOR TORQUEFor load torque@shaft torque@net torque@output torque: 60 PoutTo = 2πNFor mechanical torque: 60 PmTm = 2πNFor loss torque: 60 PµTL = 2πN 12
  13. 13. 1.9 EFFICIENCYGenerally efficiency is: Poutη= PinFor dc generator: Pout VTILη= = Pin VTIL + LossesFor dc motor: Pout Pm − Pµη= = Pin VTILExample 3A short-shunt compound generator delivers 50 A at 500 V to a resistive load. Thearmature, series field and shunt field resistances are 0.16, 0.08 and 200 Ω, respectively.Calculate the generated EMF and armature current, if the rotational losses are 520 W,determine the efficiency of the generator.Solution 0.08 ohm IF IL 50 A 0.16 ohm load D CM1 1,0m 200 ohm 1,0m 500V 1,0m - Eg +Pµ = 520WPout = VI = 500(50) = 25000W 13
  14. 14. 500VIF = = 2.5 A 200ΩI a = I F + I L = 2.5 + 50 = 52.5 AE g = VT + I a ( Ra + R f )E g = 500 + (52.5)(0.08 + 0.16) = 512.6VPin = Pout + PlossPloss = Pca + Pcf + Pµ = (52.5 2 )(0.16) + (52.5 2 )(0.08) + (2.5 2 )(200) + 520 = 2431.5W Pout 25000η= = x100% = 91.13% Pin 25000 + 2431.5Example 4A 150 V shunt motor has the following parameters:Ra = 0.5 Ω, Rf = 150 Ω and rotational loss 250 W. On full load the linecurrent is 19.5 A and the motor runs at 1400 rpm. Determine:(i) the developed power/developed mechanical power(ii) the output power(iii) the output torque(iv) the efficiency at full loadSolutionIL = 19.5AN = 1400rpm(i) Pm = E c I a Ia = IL − IF I L = 19.5 A VT 150 IF = = =1 R F 150 I a = 19.5 − 1 = 18.5 A E c = VT − I a Ra = 150 − 18.5(0.5) = 140.75V Pm = E c I a = (140.75)(18.5) = 2603.88W 14
  15. 15. OR :- Pm = Pin − Pca − Pcf Pm = VT I L − I a Ra − I F RF 2 2 Pm = 150(19.5) − (18.5 2 )(0.5) − (12 )(150) = 2603.88W(ii) Pout = Pm − Pµ = 2603.88 − 250 = 2353.88W 60 Pout 60(2353.88)(iii) To = = = 16.06 Nm 2πN 2π (1400) Pout P 2353.88(iv) η= = out = = 80.47% Pin VT I L 150(19.5) 15
  16. 16. Tutorial 11. A 300 V compound motor has armature resistance 0.18 Ω, series field resistance 0.3 Ωand shunt field resistance 100 Ω. The rotational losses are 200 W. On full load the linecurrent is 25 A and the motor runs at 1800 rpm. Determine:(i) the developed mechanical power(ii) the output power(iii) the output torque(iv) the efficiency at full load2. A 120 V series motor has 0.2 Ω field resistance. On full load, the line current is 16.5A. The output power is 1500 W and rotational loss is 150 W. Find the value of armatureresistance.3. Briefly explain the difference between motor and generator of DC machine.4. A compound DC motor rated at 415 V, 6 HP, 2000 rpm has armature resistance 0.18Ω, series field resistance 0.3 Ω and shunt field resistance 100 Ω. The rotational lossesare 200 W. The full load line current is 40 A.(i) Find the developed mechanical power.(ii) Find the output power.(iii) Find the load torque.(iv) Find the efficiency of the motor.(v) Draw the power flow diagram for this type of motor.5. A DC series generator delivers 100 kW at 10 kV to a load. The armature resistance is20 Ω and the field resistance is 50 Ω. Calculate:(i) the generated emf, Eg(ii) the input power if the stray and friction losses are 400 W. 16