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Chapter 3 induction machine

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  • 1. CHAPTER 3 INDUCTION MACHINE3.1 INTRODUCTION• Induction motor is the common type of AC motor.• Induction motor was invented by Nicola Tesla (1856-1943) in 1888.• Also known as asynchronous motor.• It has a stator and a rotor mounted on bearings and separated from the stator by an air gap.• It requires no electrical connection to the rotating member.• Such motor are classified induction machines because the rotor voltage (which produce the rotor current and the rotor magnetic field) is induced in the rotor winding rather than being physically connected by wires.• The transfer of energy from the stationary member to the rotating member is by means of electromagnetic induction.• This motor is widely used by the industries because: - Rugged. - Simple construction. - Robust. - Reliable. - High efficiency. - Good power factor. - Require less maintenance - Easy to start. - Rotates itself without external assistant. - Less expensive than direct current motor of equal power and speed.• The weaknesses of this machine are: - Low starting torque if compared to dc shunt motor. - Speed will be reduced when load increased. - Speed can’t be changed without reducing efficiency.• Small single phase induction motors (in fractional horsepower rating) are used in many household appliances such as: - Blenders - Lawn mowers - Juice mixers - Washing machines - Refrigerators• Two phase induction motors are used primarily as servomotor in control system. 49
  • 2. • Large three phase induction motors (in ten or hundreds of horsepower) are used in: - Pumps - Fans - Compressors - Paper mills - Textile mills, and so forth.3.2 INDUCTION MOTOR CONSTRUCTION• Unlike dc machine, induction machine have a uniform air gap.• Composed by two main parts: - Stator - Rotor• Figure 4.1 and 4.2 show the inside of induction machine. Figure 3.1 50
  • 3. Figure 3.2Stator ConstructionThe stator and the rotor are electrical circuits that perform as electromagnets. The statoris the stationary electrical part of the motor. The stator core of a NEMA motor is made upof several hundred thin laminations. Figure 3.3:Stator coreStator WindingsStator laminations are stacked together forming a hollow cylinder. Coils of insulated wireare inserted into slots of the stator core. 51
  • 4. Figure 3.4:Stator windingEach grouping of coils, together with the steel core it surrounds, form an electromagnet.Electromagnetism is the principle behind motor operation. The stator windings areconnected directly to the power source.Rotor Construction• The rotor also consists of laminated ferromagnetic material, with slot cuts on the outer surface.• The rotor are of two basic types : - Squirrel cage - Wound rotorSquirrel cage rotor• It consist of a series of a conducting bars laid into slots carved in the face of the rotor and shorted at either ends by large shorting ring.• This design is referred to as squirrel cage rotor because the conductors would look like one of the exercise wheels that squirrel or hamsters run on.• Small squirrel cage rotors use a slotted core of laminated steel into which molten aluminums cast to form the conductor, end rings and fan blades.• Larger squirrel cage rotors use brass bars and brass end rings that are brazed together to form the squirrel cage. 52
  • 5. • Skewing the rotor slots help to: - Avoid crawling (locking in at sub-synchronous speeds) - Reduce vibration• Squirrel cage rotor is better than wound rotor because it is: - Simpler - More rugged - More economical - Require less maintenance Figure 3.5:Squirrel cage Rotor Figure 3.6 : Rotor core 53
  • 6. Figure 3.7Wound rotor• Has a complete set of three phase insulated windings that are mirror images of the winding on stator.• Its three phase winding are usually wye connected and ends of three rotor wires are tied to a slip rings on the rotor shaft.• The rotor winding are shorted through carbon brushes riding on the slip rings.• The existence of rheostat enable user to modify the torque speed characteristic of the motor. It is used to adjust the starting torque and running speed.• The three phase rheostat is composed of three rheostat connected in wye with a common lever.• Lever is used to simultaneously adjust all the three rheostat arms. Eg: Moving rheostat to the zero resistance position shorts the resistor and simulates a squirrel cage motor.• Are rarely used because: - More expensive than squirrel cage induction motor. - Larger than squirrel cage induction motor with similar power. - Require frequent maintenance due to wear associates to brushes and slip ring. 54
  • 7. Figure 3.8 Wound rotor induction motor showing rheostat connections Figure 3.9:Wound rotor3.3 ROTATING MAGNETIC FIELD• When a three phase stator winding is connected to a three-phase voltage supply, three-phase currents will flow in the winding which induce three-phase flux in the stator.• These flux will rotate at a speed called as Synchronous Speed, ns.• The flux is called as rotating magnetic field. 55
  • 8. 120 f• The equation is:- n s = where f = supply frequency , p = no. of poles p• Rotating magnetic field will cause the rotor to rotate the same direction as the stator flux.• Torque direction is always the same as the flux rotation.• At the time of starting the motor, rotor speed is 0.• The rotating magnetic field will cause the rotor to rotate from 0 speed to a speed that is lower than the synchronous speed.• If the rotor speed is equal to the synchronous speed, there will be no cutting of flux and rotor current equals zero. Therefore, it is not possible for the rotor to rotate at ns.3.4 SLIP AND ROTOR SPEED• Slip is defined as : ns − n s= ns where ns = synchronous speed in rpm n = rotor speed in rpm• Slip can also represented in percent.• The frequency of the rotor, fr is: f r = sf where s = slip f = supply frequencyExample 1Calculate the synchronous speed of a 3-phase induction motor having 20 poles when it isconnected to a 50 Hz source.Solution 120 f 120(50)ns = = = 300rpm p 20Example 2 56
  • 9. A 0.5 hp, 6-pole induction motor is excited by a 3 –phase, 60 Hz source. If the full-loadspeed is 1140 rpm, calculate the slip.Solution 120 f 120(60)ns = = = 1200rpm p 6 n − n 1200 − 1140s= s = = 0.05 = 5% ns 1200Example 3The 6-pole,wound-rotor induction motor is excited by a 3-phase, 60 Hz source. Calculatethe frequency of the rotor current under the following conditions:(i) at standstill(ii) motor turning at 500 rpm in the same direction as the revolving field(iii) motor turning at 500 rpm in the opposite direction to the revolving field(iv) motor turning at 2000 rpm in the same direction as the revolving fieldSolutionns = 120f / p = 120(60/6) = 1200 rpm(i) n=0 ns − n 1200 − 0s= = =1 ns 1200fr = sf = 1 x 60 = 60Hz(ii) n = +500 ns − n 1200 − 500s= = = 0.583 ns 1200fr = sf = 0.583 x 60 = 35 Hz(iii) n = -500 ns − n 1200 − (−500)s= = = 1.417 (s>1 motor is operating as a brake) ns 1200fr = sf = 1.417 x 60 = 85 Hz 57
  • 10. (iv)n = +2000 ns − n 1200 − 2000s= = = −0.667 ns 1200fr = sf = -0.667 x 60 = -40 Hz (-ve means that the phase sequence of the voltages inducedin the rotor winding is reversed)Example 4A 3-phase, 4 pair of poles, 400kW,400V,60Hz induction motor is 780 rpm full-loadspeed. Determine the frequency of the rotor current under full load condition.Solutionf rotor = sf 120 f 120(60)n s= = = 900rpm p 8 ns − n 900 − 780s= = = 0.133 ns 900 f r = sf = 0.133(60) = 8Hz3.5 PER-PHASE EQUIVALENT CIRCUIT OF THREE-PHASE INDUCTIONMOTORThe per-phase equivalent circuit of a three-phase induction motor is just like a single –phase transformer equivalent circuit. The difference is only that the secondary winding is short-circuited unlike in thetransformer it is open-circuited as a load is to be connected later.Complete Equivalent Circuit For Induction Machine Referred To The Stator Circuit 58
  • 11. I1 R1 X1 I2 X2 1,0k Rm Xm R2 1,0m V input 1,0m 1,0k 1,0m s Figure 3.10The subscript ‘1’ is refering to the stator side while ‘2’ is referring to the rotor sideR1, X1, R2, Rm , Xm are value perphaseInput Power, Pin = 3V1I1cosθStator copper loss, Pscl = 3I12R1Core Loss, Pcl = 3V12/Rm (always neglected because too small)Power across the air-gap, Pag = 3I22R2 /s = Pin - Pscl - PclRotor copper loss, Prcl = 3I22R2Mechanical power/gross output power/converted power, P mech = Pag – Prcl = 3I22R2 /s - 3I22R2 = Pag (1-s)Net power output, Poutput = P mech – P friction & windage lossFor Torque: 60 Pm Pm Pag Tmechanical / induced = Tm = = = 2πn wr ws 60 P0 Pout Toutput / load = To = = 2πn wr 59
  • 12. 3(Vsup ply − p ) 2 R2Tstarting = ws [ (R 1 + R2 ) 2 + (X1 + X 2 )2 ] 3(Vsup ply − p ) 2 1Tmax = 2 ws [ 2 R1 + R1 + ( X 1 + X 2 ) 2 ] R2Maximum Slip: S max = 2 R1 + ( X 1 + X 2 ) 23.6 POWER FLOW OF AN INDUCTION MOTOR P CONV = P MECH PAG=(3I22R2)/s =PAG-PRCL = Pin-PSCL-PCL PRCL=3I22R2 PSCL=3I12R1 PCL= = sPag 3V12/RM Figure 3.11 60
  • 13. Example 5A 10 poles, 50 Hz, Y connection 3-phase induction motor having a rating of 60kW and415V. The slip of the motor is 5% at 0.6 power factor lagging. If the full load efficiencyis 90%, calculate:(i) Input power(ii) Line current and phase current(iii) Speed of the rotor (rpm)(iv) Frequency of the rotor(v) Torque developed by the motor (if friction and windage losses is 0)Solution Pout(i) η = Pin Pout 60kWPin = = = 66666.7W η 0.9 VL(ii) Y connection, IΦ = IL, VΦ= 3P in =3VΦIΦcos= 3VLIL cos θ Pin 66.67 kWIL = = = 154.59 A 3VL cos θ 3 (415)(0.6)IΦ = IL=154.59 ∠ cos −1 0.6 = 154.59∠ − 53.13° A 120 f 120(50)(iii) n s = = = 600rpm p 10 ns − n s= ns n = n s (1-s) = 600 (1-0.05) = 570 rpm(iv) fr = sf = 0.05(50) = 2.5Hz Pm Pout 60kW(v) T = = = = 1005.2 Nm wm 2πnm / 60 2πnm / 60 Or 2πf 2πf ws= = = 62.83rad / s p/2 5 wm = ws(1-s) = 62.83(1-0.05) = 59.69 rad/s 60kWT= = 1005.2 Nm 59.69rad / s 61
  • 14. Example 6A 3-phase, delta connection, 4 pole, 440V, 60 Hz induction motor having a rotor speed1200rpm and 50kW input power at 0.8 power factor lagging. The copper losses and ironlosses in the stator amount to 2kW and the windage and friction losses are 3kW.Determine:(i) Net output power(ii) Efficiency(iii) Input currentSolution(i) ns = 120f/p = 120(60)/4 = 1800 rpm ns − n 1800 − 1200 s= = = 0.33 ns 1800 P input(to stator) P input rotor /Pag Pmech=Pout rotor P out net =50kW =50kW-2kW =(48-16)kW =(32-3)kW =48kW =32kW =29kW P stator losses P rotor losses P wind and fric =2kW =s(P input rotor) losses = 0.33(48kW) =3kW =16kW P net output = 29kW Pout 29kW(ii) η = = = 58% Pin 50kW(iii)Δ connection Pin = 3V p I p cos θ Pin 50kW Ip = = = 47.35∠ cos −1 0.8 = 47.35∠ − 36.87° A 3V p cos θ 3( 440)(0.8) I L = 47.35 3 = 82 A 62
  • 15. Example 7A 3-phase induction motor, delta connection,5 pair of poles, 60 Hz is connected to a440V source.The slip is 3% and the windage and friction losses are 3kW. Theequivalent circuit perphase referred to the stator circuit is:- R1 = Stator resistance = 0.4Ω X1 = Stator leakage inductance = 1.4Ω R2’ = Rotor resistance = 0.6Ω X2’ = Rotor leakage inductance = 2Ω Rm = no-load loses resistance = 150Ω Xm = magnetizing reactance = 20ΩCalculate:(i) Input power(ii) Speed of the rotor(iii) Mechanical power(iv) Developed torque(v) EfficiencySolution 0.4Ω j1.4Ω j2Ω I1 I260Hz, 1.0m 1.0k440v, V2 1.0k 1.0m 1.0m 10 S j20Ω R2/s=0.6/0.03=20Ωpoles, Δ(i) P in =3VIcosθ V = 440V 20 j ( 20 + 2 j ) Z total = 0.4 + 1.4 j + = 9.45 + 11.45 jΩ 20 j + 20 + 2 j V 440 I1 = = = 18.87 − 22.86 j = 29.64∠ − 50.46° A Ztotal 9.45 + 11.45 j Pin = 3(440)(29.64)cos(-50.460) = 24907.5W 120 f 120(60)(ii) ns = = = 720rpm p 10 ns − n s= , ns n = ns(1-s) , n = 720(1-0.03) = 698.4 rpm 63
  • 16. (iii) Pm = 3(I22R2/s – I22R2) V2 = 440 – I1Z1 = 440 – (18.87-22.86j)(0.4+j1.4) = 400.45-17.27j V V 2 400.45 − j17.27 I2 = = = 19.74 − j 2.84 = 19.94∠ − 8.19° A Z2 20 + j 2  0.6  Pm = 3 [19.942   - 19.942(0.6)] = 23140.5W  0.03  Pag Pm(iv) T dev = = ws wm Pag = 3I22R2/s = 3(19.94)2(0.6)/(0.03) = 23856.2W 2πf 2π (60) ws = = = 75.4rad / s p 2 5 23.856.2 T = = 316.4 Nm 75.4 Poutput 23140.5W − 3kW(v) η = = = 81% Pin 24901.5WExample 8A 3-phase induction motor, wye connection, 60 Hz is connected to a 220V source.Theslip is 5% and rotor speed is 855 rpm. The equivalent circuit perphase is:- R1 = Stator resistance = 0.4Ω X1 = Stator leakage inductance = 1Ω R2’ = Rotor resistance = 0.8Ω X2’ = Rotor leakage inductance = 3.5Ω Rm = no-load loses resistance = 150Ω Xm = magnetizing reactance = 10ΩCalculate:(i) Number of poles(ii) Input power(iii) Mechanical power(iv) Developed torque(v) Efficiency 64
  • 17. Solution 0.4Ω j1Ω j3.5Ω60Hz, 1.0m 1.0k220V, 1.0k 1.0m 1.0m S j10Ω R2/s=0.8/0.05=16Ω Y(i) n = 855rpm s = 0.05 ns = 120f/p ns − n s= , sns = ns – n , n = ns – sns ns = ns(1-s) n ns = 1− s 855 = = 900rpm 1 − 0.05 ns = 120f/p 120 f 120(60) p= = = 8 pole ns 900rpm(ii) Pin = 3V L I L cos θ (16 + j 3.5)( j10) Z total = 0.4 + j1 + = 4.05 + j 7.9 Ω 16 + j 3.5 + j10 220 3 I1 = = 6.53 − j12.73 = 14.3∠ − 62.84° A 4.05 + j 7.9 Pin = 3 (220)(14.3)(cos− 62.84°) = 2487.36W(iii) Pm = 3(I22R2/s – I22R2) 220 V2 = – I1Z1 3 220 = – (6.53-j12.73)(0.4+j1) 3 = 111.68-1.438j V V 2 111.68 − j1.438 I2 = = = 6.64 − j1.54 = 6.8∠ − 13° A Z2 16 + j 3.5 65
  • 18.  0.8  Pm = 3 [6.82   - 6.82(0.8)] = 2108.54W  0.05  Pag Pm(iv) T dev = = ws wm Pag = 3I22R2/s 2πf 2π (60) ws = = = 94.25rad / s p 2 4 T = (3I22R2/s) / ws =[3(6.8)2(0.8)/(0.05)] / 94.25 = 23.55Nm Poutput 2108.54(v) η= = = 0.85 = 85% Pin 2487.363.7 TORQUE SPEED CHARACTERISTICS Figure 3.12There are 3 regions involve in a 3-phase induction motor:-(i) Braking/Plugging Braking process occurs at s>0(positive slip). In this case the motor acts as a brake where it rotates in opposite direction respect to the rotor.(2<slip<1). 66
  • 19. (ii) Motoring Motoring is the region where induction motor acts as a motor. Slip is reducing from 1 into 0. Slip equals to 0 at synchronous speed,ns.(1<slip<0).(iii) Generating Generating region is a region where motor acts as a generator. During this time the slip is negative. At this time, the motor acts as a generator.(slip<0)3.8 DETERMINATION OF CIRCUIT MODEL PARAMETERThe parameter of the equivalent circuit can be determined from the results of a:- No load test- Blocked rotor test-.DC testThe blocked rotor test:- To determine X1 and X2- When combines with DC test, it also determines R2- Test is performed by blocking the rotor so that it cannot turn and measuring the line voltage, line current and three phase power input to the stator- Connection for the test is shown in Figure 4.13 Figure 3.13 Basic circuit for blocked rotor test and no load testThe no load test:- To determine magnetizing reactance, Xm, and the combined core, friction and windage losses (these losses are essentially constant for all load condition)- The connection for the no load test are identical to those shown in Figure 12- However, the rotor is unblocked and allowed to run unloaded at rated voltage and rated frequency 67
  • 20. DC test:- To determine R1- Accomplished by connecting any two stator leads to a variable voltage DC source as shown in Figure 4.14.- The DC source is adjusted to provide approximately rated stator current, and the resistance between two stator leads is determined from voltmeter and ammeter reading Figure 3.14 Basic circuit for DC testExample 9The following test data were taken on a 7.5hp, four pole, 208 V, 60 Hz Y connecteddesign A induction motor having a rated current of 28A.DC test : Vdc = 13.6V Idc = 28 ANo-load test: Vt = 208 V f = 60 Hz Ia = 8.12 A P in = 420 W Ib = 8.2 A Ic = 8.18 ALocked rotor test: Vt = 25 V f = 15 Hz Ia = 28.1 A P in = 920 W 68
  • 21. Ib = 28 A Ic = 27.6 ASketch the per-phase equivalent circuit for this motor.SolutionFrom the DC test, VDC 13.6VR1 = = = 0.243Ω 2 IDC 2( 28.0 A)From the no-load test, 8.12 A + 8.2 A + 8.18 AIL = = 8.17 A 3 208Vφ = = 120V 3 120VZ= = 14.7Ω = X 1 + X 2 8.17 APscl = 3I12R1 = 3(8.17A)2(0.243Ω) = 48.7WPag = Pin –Pscl = 420W- 48.7W = 371.3WFrom the locked-rotor test, 28.1A + 28 A + 27.6 AIL = = 27.9 A 3 Vφ 25 3Z= = = 0.517Ω I 27.9 APin = 3VLIL cos θ Pin 920Wθ = cos-1 = = 40.4° 3VLIL 3 (25V )(27.9 A)R1 + R2 = 0.517(cos 40.4°) = 0.394Ω 69
  • 22. DC test, R1 = 0.243Ω, R2 = 0.394Ω – 0.243Ω = 0.151ΩAt 15 Hz, X = 0.517(sin 40.4°)=0.335Ω 60 HzAt 60 Hz, X = (0.335Ω) = 1.34Ω = X1 + X2 15HzFor class A induction motor, this reactance is assumed to be divided equally between therotor and stator, X1 = X2 = 0.67 Ω Xm = 14.7 -0.67 = 14.03Ω3.9 STARTING OF INDUCTION MOTOR(i) Direct On Line Starter(DOL)- A widely-used starting method of electric motors.- The simplest motor starter.- A DOL starter connects the motor terminals directly to the power supply.- Hence, the motor is subjected to the full voltage of the power supply.- Consequently, high starting current flows through the motor.-This type of starting is suitable for small motors below 5 hp (3.75 kW).- Most motors are reversible or, in other words, they can be run clockwise and anti- clockwise.- A reversing starter is an electrical or electronic circuit that reverses the direction of a motor automatically.- Logically, the circuit is composed of two DOL circuits; one for clockwise operation and the other for anti-clockwise operation.- It takes a starting current 6(six) times the full load current.- For large motor the high starting current causes voltage drop in the power system which may trip other motors in the systems.(ii)Star-Delta Starter- For star-delta connection the motor windings are connected in star during starting.- The connection is changed to delta when the motor starts running.- The starting current and starting torque of DOL started and start-delta connected motorsare as follows:Example:DOL -6I and 2TStar-delta - 2I and 2T/3 70
  • 23. - Thus it can be seen that the starting current and starting torque are both reduced.- The motor should be capable to start at such reduced torque with load.- The Star Delta starter can only be used with a motor which is rated for connection in delta operation at the required line voltage(iii)Autotransformer starter- An Auto transformer starter uses an auto transformer to reduce the voltage applied to a motor during start.- The auto transformer may have a number of output taps and be set-up to provide asingle stage starter, or a multistage starter.- Typically, the auto transformer would have taps at 50%, 65% and 80% voltage, enabling the motor to be started at one or more of these settings.- As the motor approaches full speed, the auto transformer is switched out of the circuit 71
  • 24. Tutorial 31. A 3 phase induction machine 373kW, 6 poles is connected to a 440V, 50 Hz, has a fullload speed of 950 rpm. If the machine is comprised of 6 poles, calculate the frequency ofthe rotor current during full load.2. Determine the synchronous speed of a six pole 460V 60 Hz induction motor if thefrequency is reduced to 85 % of its rated value.3. A 4 pole induction machine is supplied by 60 Hz source and having 4% of full loadslip. Calculate the rotor frequency during:(i) Starting(ii) Full load4. A 3-phase induction motor, delta/star connection, 2 poles, 50 Hz is connected to a 410Vsource .The rotor speed is 2880 rpm and the windage and friction losses are 600 W. Theequivalent circuit perphase referred to the stator circuit is:- R1 = 0.4Ω X2 =2Ω X1 =2Ω Rm = 150Ω R2 =2Ω Xm = 20ΩCalculate:(i) Input power(ii) Air-gap power(iii) Mechanical power(iv) Developed torque/torque induced(v) Efficiency5. A 440V, 50Hz, 10 pole, delta/Y connected induction motor is rated at 100kW. Theequivalent parameter for the motor are: Rs = 0.08Ω / phase RR = 0.1Ω / phase X s = 0.3Ω / phase X R = 0.2Ω / phase X m = 6Ω / phase Rc = ∞At full load condition , the friction and windage losses are 400W, the miscellaneouslosses are 100W and the core losses are 1000W. The slip of the motor is 0.04.(i) Calculate the input power(ii) Calculate the stator copper loss(iii) Calculate the air gap power(iv) Calculate the converted power(v) Calculate the torque induced by the motor(vi) Calculate the load torque(vii) Calculate the starting torque(viii) Calculate the maximum torque and slip(ix) Calculate the efficiency of the motor 72
  • 25. 6. Squirrel cage and wound rotors are the two common types of rotor used ininduction machines. Give four(4) advantages of squirrel cage rotor.7. A 4 pole induction machine is supplied by 50 Hz source and having 4% of full loadslip. Find the rotor frequency during:(i) Starting(ii) Full load8. A 3-phase, Y-connected, 50 Hz, 4 pair of poles, induction motor having 720 rpm fullload speed. The motor is connected to a 415 V supply. The machine has the followingimpedances in ohms per phase referred to the stator circuit: R1 = 0.2 Ω X1 = 2.0 Ω R2 = 0.9 Ω X2 = 4.0 Ω Xm = 60 ΩIf the total friction and windage losses are 200 W,(i) Find the slip, s.(ii) Find the input power, Pin.(iii) Find the air gap power, Pag.(iv) Find the mechanical power, Pm.(v) Find the torque induced by the motor, τ ind.(vi) Find the efficiency of the motor.9. Induction machine is a common type of AC machine. State three weaknesses of theinduction machine.10. A 3-phase, delta-connected, 50 Hz, 2 pair of poles, induction motor having 1455rpm full load speed. The motor is connected to a 415 V supply. The machine has thefollowing impedances in ohms per phase referred to the stator circuit: R1 = 0.2 Ω X1 = 0.6 Ω R2 = 0.9 X2 = 0.4 Ω Xm = 20 Ω If the total friction and windage losses are 1000 W, calculate:(i) Slip(ii) Input power, Pin(iii)Air gap power, Pag(iv) Mechanical power, Pconv(v) Torque induced by the motor, τ ind(vi)Efficiency of the motor11. A 3-phase, Y-connected, 6 poles, 415 V, 50 Hz induction motor having a rotor speed950 rpm. The input power is 100 kW at 0.85 power factor lagging. The copper and ironlosses in the stator are 4 kW and the windage and friction losses are 4 kW. Determine theoutput power of the motor. 73