Number Of Slots Per Pole Per Phase Formula

Number Of Slots Per Pole Per Phase Formula Rating: 4,2/5 1335 votes

There are six poles and three phases, so 2 slots per pole per phase requires 2×3×6 =36 slots; Each coil requires two slots (one for each side of the coil). In a single layer winding (which is all you need to know about in this course) 36 slots will be filled by 18 coils. There are three phases, so each phase will have 6 coils.

  1. Number Of Slots Per Pole Per Phase Formula Sheet
  2. Number Of Slots Per Pole Per Phase Formula Cost
  3. Number Of Slots Per Pole Per Phase Formula Calculator
  • The number of slots should be selected to give an integral number of slots per pole per phase. The stator slot pitch at the air gap surface should be between 1.5 to 2.5 cm. Stator slot pitch at the air gap surface = τ.
  • The stator of a 3 - Φ IM has a 3 slots per pole per phase. If supply frequency is 50 Hz, then the number of stator poles produced 4 poles.
  • An 8 pole machine produces 4 full cycles per revolution. So, in the original problem: Frequency = (8 poles x 1/2 cycle/pole) x 1500 rev/minute x 1min/60 seconds Poles & minutes cancel, leaving 100 cycles per sec. Each phase is separate, so number of phases doesn't affect the calculation.
  • If the number of slots per pole per phase 𝒒𝒒= 𝑺𝑺 𝒎𝒎𝑷𝑷. Is an integer, then the winding is called an. Integral-slot winding in case the number of slots per pole per phase, q is not an integer, the winding is called. Fractional-slot winding. For example a 3-phase winding with 36 slots and 4 poles is an integral slot.

Assignment—5

Sub : Electrical Engg (EEE-101/201)

B.Tech. First year

(Direct Current Generator)

Formula:

Eg = for LapWinding ; Eg = for Wave WindingEg = Vt + Ia ra + 2Vb ; Ia =Ish + IL ; Ish = ; IL = ; PL= Vt .IL

Where Eg = Generated emf, φ = Flux per pole, Z = total no. of conductors

N = Speed in rpm, P = no. of poles, Vt = terminal voltage, Ia = armature current, IL = Line(load) current , Ish = Shunt field current, Rsh = Shunt Field resistance , Vb =Brush drop, P L= Poweroutput , RL = Loadresistance

1. A 4 - pole, wave generatedarmature in a dc generator has 51 slots with 12 conductors per slot. It isdriven at 900rpm. If the useful flux per pole is 25mWb, calculate the value ofthe generated emf. Ans: 459V

2. An 8 -pole, dc generator running at 1200 rpm and with a flux of 25mWb per polegenerates 440V. Calculate the numbers of conductors, if the armature is i) lap wound ii) wave wound.

Ans: i) 880 ii) 220

3. A 4-poledc shunt generator has a useful flux per pole of 0.07Wb. The armature has 400lap wound conductors, each of resistance 0.002 Ω ohm and is rotated at 900 rpm.If the armature current is 50 A, calculate the terminal voltage.

Ans: 417.5V

Formula

4. An 8-pole dc generator has 500armature conductors and a useful flux per pole of 0.05Wb. i) What will be theemf generated, if it is lap-connected and runs 1200 rpm? ii) What must be the speed at which it is tobe driven to produce the same emf, if it is wave wound?

Ans: i) 500V ii) 300 rpm

5. A separately excited 4 - pole, 900rpm wave wound dc generator has an induced emf of 240V at rated speed and ratedfield current. When connected to a load, the terminal voltage is observed to be200V. If the armature resistance is 0.2 Ω and the flux per pole is 10 m Wb,compute the armature current and the number of conductors.

Ans 200A, 800

6. A 4- pole dc shunt generator with lapconnected armature has field and armature resistance of 80 Ω and 0.1 Ωrespectively. It supplies power to 50 lamps rated for 100Volts, 60 Watts each.Calculate the armature current and the generated emf by allowing contact dropof 1V per brush.

Ans: 105.12V

7. A 4- pole dc shunt generator witha shunt field resistance of 100 ohm and armature resistance of 1 Ω has 120 waveconnected conductor in its armature. The flux per pole is 60mWb. If a loadresistance of 20Ω is connected across the armature terminals and armature isdriven at 1000 rpm, calculate the voltage across the load terminals.

Ans: 226.41Volts

8. Find the flux per pole of a 50 KW dc shuntgenerator having 4 poles and a lap wound armature with 380 conductors. Themachine is run at a speed of 800 rpm and generates 460V. Resistance of thearmature and the shunt field are 0.5 Ω and 300Ω respectively. Find the currentflowing in the armature at the full load and the terminal voltage.

Ans: Ф = 0.0908Wb, Vt = 396.2V, Ia = 127.43 A

9. A 4 pole lap connected armature ofa dc shunt generator is required to supply the following loads: i) 5KW geyser at 250 V dc ii) 2.5 KW lightingload at 250VDC. Generator has an armature resistance of 0.2 Ω and a fieldresistance of 250 Ω. Armature has 120 conductors in slots and runs at 1000 rpm.Allowing 1V per brush for contact drop and neglecting armature reaction,find i) flux per pole ii) current per parallel path.

Ans: i) 0.129 Wb ii) 7.75 A

10. A 200 KW, 600 V DC, 6 pole dcshunt generator has an armature with 18 slots per pole. It has a simple lapwinding with 4 conductors per slot. The armature and shunt field winding hasresistance of 0.1 Ω and 50 Ω respectively. The flux per pole is 0.07 Wb.Calculate the speed at which it is to be driven for rated load operation.

Ans: Eg = 634.53V , N= 1259 rpm

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DIRECT CURRENT MOTOR

Formula:

Eb = for Lap winding; Eb = for Wave winding

Vt = Eb + Iara + 2Vb; Ia = IL - Ish ; Ish = ; PL= Vt .IL

Where Eb =Back emf, φ = Flux per pole, Z = total no. of conductors

N=Speed in rpm, P = no. of poles, Vt = terminal voltage, Ia = armature current, IL = Linecurrent , Ish = Shunt field current, Rsh= Shunt Field resistance , Vb = Brushdrop, P L= Power output

1. A DC shunt motor operating at 200Vdc has armature resistance of 0.5 Ω. If its armature current is 25 A, calculatethe back emf. Ans: 187.5 V

2. A 4- pole lap wound armature DCshunt motor has flux per pole of 25 mWb. The number of armature conductor is200. The motor draws armature current of 20A when connected across a 200V dcsupply. Calculate the back emf and the speed of the motor if the armatureresistance is 0.4 Ω. Ans: 192V, 2304 rpm.

3. A 4- poles, lap wound armature DCshunt motor draws 40A armature current. The armature conductor is 400. Flux perpole is 20mWb. Calculate the gross torque developed by the armature of themotor. Ans : 50.83 Nm.

4. A 4 - pole lap wound DC motor has480 conductors. The flux per pole is 24mWb and the armature resistance is 1 Ω.If the motor is connected to a 200V dc supply and running at 1000 rpm on noload, calculate

i) backemf ii)armature current iii)power output iv) lost torque

Ans: i) 192V ii) 8A iii) 1536W iv) 14.67 Nm

5. A 60HP, 110VDC series motor has armaturewinding resistance of 0.18 Ω and series winding resistance of 0.13 Ω. The ratedcurrent is 45 A at 450 rpm. Determine the speed of the motor when the motorcurrent is 50 A. Ans: 398.6 rpm

6. A 200V DC shunt motor has armatureresistance 0.2 Ω and shunt field resistance 200Ω. If the no load and the fullload current drawn by the motor are 5A and 40 A respectively then calculate thefull load speed, assuming that the no load speed is 1000 rpm. Ans: 964.85rpm

.7. A 250 V DC shuntmotor having armature resistance of 0.25 Ω carries an armature current of 50 Aand runs at 750 rpm. If the flux is reduced by 10 %, find the speed. Assumethat the torque remains the same. Ans: 828.46 rpm.

8. A 500 V DC shunt motor runs at a speed of 250 rpm whennormally excited and taking a armature current of 200 A. The resistance ofarmature is 0.12 Ω. Calculate the speed of the motor when the flux is reducedto 80% of the normal value and the motor is loaded for an armature current of100A. Ans. 320 rpm.

9. A 10 HP, 230V DC shunt motor hasan armature resistance of 0.5 ohm and field resistance of 115 ohm. At no loadthe rated speed is 1200 rpm and the armature current is 2 ampere. If the loadis applied the speed drops to 1100 rpm. Determine the armature current and theline current.

Ans: Ia = 40.16 A, IL = 42.16 A

10. A DC shunt machine connected to230 V DC supply has resistance of armature as 0.115 Ω and of field winding as115 Ω. Find the ratio of the speed as generator to the speed as a motor withthe line current in each case being 100 A. Ans : 1.105: 1

11. A DC series motor with seriesfield and armature resistance of 0.06 Ω and 0.04 Ω respectively is connectedacross 220 V dc mains. The armature takes 40 A and its speed is 900 rpm.Determine its speed when the armature takes 75A and excitation is increased by15% due to saturation. Ans: 770 rpm.

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THREEPHASE INDUCTION MOTOR

Formula:

·Synchronousspeed Ns = rpm

·Rotor speed Nr = Ns ( 1 – s) rpm

·Slip Speed = Ns– Nr rpm

·Absoluteslip s =

·Percent Slip( %s) =

·Rotor Frequency =f ‘ = s .f Hz

·Air Gap Power Pg= Pi – Stator Loss

·Rotor copper loss= s. Pg

·Mechanical powerdeveloped in rotor Pm = Pg (1 – s)

·Output power Po = Pm – Mechanical loss

·BHP = HP

·Total Stator Input Pi =

Where VL =Line Voltage, I L = Linecurrent , cos Ф = power factor

·Efficiency

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1. A threephase, 50 Hz induction motor has a full load speed of 960 rpm. Calculate i) slip ii) number of poles iii) frequencyof rotor induced emf

iv) speed of rotor field with respect to rotorstructure v) speed of rotor field with respect to stator field vi) speed ofrevolving magnetic field in the air gap (synchronous speed) Ans: 4%, 6, 2Hz, 40rpm, 1000rpm, 1000rpm

2. A 20 HP, 440V, 50Hz 3 - phaseinduction motor works on 50 Hz supply. Determine i) speed of revolving magneticfield ii) motor speed at a slip of 5% iii) frequency of current in rotorcircuit when slip is 7.5% iv) speed of revolving magnetic field with respect torotor when slip is 4. 5%.

Ans: i) 1500 rpm ii) 1425 rpm iii) 3.75Hz iv) 1440rpm

3. The rotor of a 3- phase 6 - poles,400V, 50 Hz induction motor alternates at 3 Hz. Compute the speed and % slip ofthe motor. Find the rotor Cu loss per phase if the full load input to the rotoris 111.9 KW.

Ans: Speed Nr = 940 rpm, % slip = 6%, Rotor Cu loss per phase =2.238KW

4. The power input to a 500V, 50Hz, 6-poles,3- phase squirrel cage induction motor running at 975 rpm at 40 KW. The statorlosses are 2KW. Calculate

i) slip ii) rotor cu loss iii) BHP iv) efficiency.

Ans: % slip = 2.5%, Rotor Cu loss = 0.95KW, BHP = 47.88 HP

Efficiency = 92%,

5. A 440V, 6- poles, 50Hz, 3-phaseinduction motor develops 20KW including mechanical loss when running at 980rpm. The power factor being 0.85 lag. Calculate i) slip ii) rotor frequency iii)total stator input if stator losses are 1500 Watts iv) line current

Ans: % slip = 2%, Rotor Frequency =1Hz, Total stator input = 20.9KW

Line current = 37.2A

6. A 6 poles, 50 Hz, 500V, 50Hz, 3 -threephase induction motor running on full load with 4% slip develops 14. 92 KW withpower factor being 0.86 lag. The friction and windage losses are 200 W andstator copper and stator iron losses amounts to 1620 W. Calculate i) rotorcopper loss ii) efficiency

iii) line current iv) rotor frequency.

Ans: Rotor Cu loss = 620 W, Efficiency = 85.7%, Line current= 23.04A

Rotor Frequency = 2Hz

7. A 3-phase, 6-poles induction motordevelops 30HP including 2HP mechanical loss at a speed of 950 rpm on 550V, 50Hzmains. The power factor being 0.88 lag. Calculate i) slip ii) rotor copper loss iii)total stator input if stator losses are2KW iv)line current v) motor output

vi) efficiency.

Ans: % slip= 5%, Rotor Cu loss = 1.177K W, Total stator input = 25.55KW,

Line current = 30.41A, Motor output = 20.88 KW, Efficiency = 81.7%,

8. A 3- phase delta connected 440V,50 Hz, 4 pole induction motor has a rotor standstill emf per phase of 130 V. Ifthe motor is running at 1440 rpm. Calculate i) the slip ii) the rotorfrequency iii) the value of the rotorinduced emf per phase iv) stator to rotor turns ratio.

Ans: i) % slip = 4% ii) 2Hz iii) 5.2V/phase iv) 3.38:1

Number Of Slots Per Pole Per Phase Formula

9. The power input to rotor of 440 V,50 Hz, 3- phase, 6- pole induction motor is 50 KW. The rotor emf makes 120cycles per minute. Stator iron and stator copper losses are 1KW and frictionand windage losses are 2KW respectively. Calculate i) slip ii) rotor speed iii) rotor cu loss

iv) Mechanicalpower developed v) output torque

Ans: % slip= 4%, Rotor speed = 960 rpm, Rotor Cu Loss =2KW, Mechanical Power Developed =48KW, Output torque = 457.80Nm

10. A 3 - phase, 50 Hz, 50 KWinduction motor has an efficiency of 90% at rated output. At this load thestator cu loss, the rotor cu loss & stator core loss are equal. Thefriction and windage loss (mechanical loss) is equal to one third of the statorcore loss. Calculate i) rotorcu loss ii) airgap power iii)slip

Ans : i) 1665W ii) 52220W iii) 0.031 (3.1%)

11. The power input to the rotor of a 400V, 50Hz, 6-pole, 3 - phase induction motor is 75KW. The rotor electromotive force isobserved to make 100 complete alternations per minute. Calculate i) slip ii) rotor speed

iii) rotor cu losses per phase iv) mechanical power developed v) therotor resistance per phase if the rotor current is 60 A

Ans: i) 3.3% ii)967rpm iii)825W iv) 72.525KW v) 0.23 Ω

12. The shaft torque or useful torqueof a three phase, 50 Hz, 8 pole induction motor is 190Nm. If the rotorfrequency is 1.5Hzand mechanical losses are 700 Watts, Calculate i) slip ii) power output iii) mechanical power developed iv) rotor copper loss

Ans: 3%, power output = 14474.9 W

mechanical power developed =15174.9W, rotor copper loss = 469.326W

13. An induction motor is running atfull load and develops a torque of 180 Nm when the rotor makes 120 revolutionsper minute. If number of poles be 4 and supply frequency is 50 Hz, calculatethe shaft power.

Ans: shaftpower = 27.13KW.

14. Thepower input of a three phase induction motor is 90 KW. Find the mechanicalpower developed if the stator losses are 1.5 KW. Also find the rotor copperloss. It is given that slip of the induction motor is 3%.

Ans:mechanical power developed = 85.845 KW,

Rotor copperloss = 2.655KW

THE END

As we know that principle of induction motor is very similar to the transformer with some differences. An induction motor at standstill condition is similar to a transformer at no load condition. Therefore, the method of drawing the induction motor phasor diagram is also same as that of a transformer phasor diagram. In this post we will discuss the phasor diagram of induction motor at standstill condition and at full load slip.

Induction Motor Phasor Diagram at Standstill Condition:

Before going into the phasor diagram, there are some important points to be taken care:

  • Per phase value of induced emf E1 in the stator winding is given as below

E1 = √2πf1kw1N1Ø

where f1 = supply frequency

N1 = Number of series turns per phase

Ø = resultant air gap flux per pole

kw1 = Stator winding factor

  • Per phae value of induced emf E2 in rotor winding is given as

E2 = √2πf2kw2N2Ø

where f2 = frequency of induced emf in rotor = sf1

Number Of Slots Per Pole Per Phase Formula Sheet

N2 = Number of series turns per phase

Ø = resultant air gap flux per pole

kw2 = Rotor winding factor

  • Total air gap mmf Fr of induction motor is the sum of stator mmf (F1) and rotor mmf (F2).
  • Magnetizing current Im taken by stator winding from the supply always remains in phase with the resultant flux Ø.
  • The induced emf always lags behind the resultant flux Ø by 90°.

Now we are at a stage to draw the induction motor phasor diagram. Let us take the resultant air gap flux Ø as the reference. This flux Ø will be in phase with the resultant mmf Fr. Also, the induced emf E1 and E2 in stator and rotor winding will lag behind the Ø by 90°. This is shown in the below phasor diagram of induction motor.

Since the rotor mmf counteracts the stator produced mmf as per lenz’s law, therefore stator takes extra current from supply to counterbalance the effect of rotor current. Therefore under normal condition,

Stator mmf = Rotor mmf

N1’I2’ = N2’I2

Number Of Slots Per Pole Per Phase Formula Cost

where N1’ and N2’ are the effective stator and rotor turns per phase.

This component of stator current is called the load component. In addition to load component, stator also takes magnetizing current Im to build magnetic flux in the air gap. Thus the total stator current I1 = I2’ + Im. This is shown in the above phasor diagram. I2’ is shown opposite to the rotor current I2 for the reason discussed above.

At standstill condition, E2 = I2 (r2 + jx2). The core loss component of stator current Ic is in phase phase with V1’ or –E1. At standstill condition, the friction and windage loss is zero, therefore the stator no load current is given as

I0 = Im + Ic

Since stator applied voltage V1 must balance the stator back emf V1’ or -E1, stator impedance drop I1(r1 + jx1), therefore we can write

V1 = V1’ + I1(r1 + jx1) ……(1)

Similar equation exists for rotor circuit and can be written as

E2 = I2 (r2 + jx2) ……..(2)

The above equations are applied for drawing induction motor phasor diagram as shown in above figure. It can be easily seen from the above phasor diagram that, the power factor (cosɵ) of induction motor at starting is very poor as ɵ is large.

Induction Motor Phasor Diagram at Full Load Slip:

At full load, the slip s of induction is low. The stator voltage equation (1) do not changes when the motor is loaded. But the rotor voltage equation changes with slip. The rotor induced voltage at any slip s becomes sE2 and the rotor circuit reactance becomes sx2. Therefore,

sE2 = I2 (r2 + jsx2)

The above rotor equation when implemented, the induction motor phasor diagram will becomes different from the phasor at standstill condition. The induction motor phasor at full load slip s is shown below.

Number Of Slots Per Pole Per Phase Formula Calculator

Since at full load condition, some friction and windage loss will exists. This means that stator will have to draw some extra current from the supply main to provide this additional loss. Therefore the total no load current I0 taken by stator is the sum current Ifc and Im. It can be seen from the above phasor diagram that power factor of induction motor improves. generally at full load the power factor ranges in between 0.8 to 0.9.