EE 6604 DESIGN OF ELECTRICAL MACHINES

UNIT-1 INTRODUCTIONPART-A

1. Give the expression for temperature rise.

θ = θf – ( θf – θ1 ) e(-t/τ)

where, θ = Temperature rise during the time ‘t’ (heating), θf = Final steady temperature rise,

θ1 = initial temperature rise, τ = Heating time constant.

2. Give expression for cooling.

θ‘ = θf‘ + (θ2 – θf’ ) e(-t/ τ

c)

where, θ‘ = Temperature rise during the time ‘t’ (cooling)

θf‘ = Final temp. rise reached during cooling, θ2 = Initial temp. rise at which cooling is started

τc = Cooling time constant.

3. What are the major design considerations? (Dec 2011) (May 2013)

a. Cost b. Durability c. Performance.

4. What are the limitations in design?

a. Saturation b. Temperature rise c. Insulation d. Efficiency.

5. What are the causes of failure of insulation?

a. Weak points in the insulator b. Effect of aging and mechanical fatigue

c. Mica migration d. Tracking.

6. Give the expression for heat flow in 2-D.xp

°C

where, Q = Total heat produced in coil, l = Length of coil in m, w = Width of coil, t = Thickness of

coil, ρx, ρy = Thermal resistivity in x and y directions.

7. Give the expression for temperature difference between center and overhang of a conductor.

°C

Where, θ = Temperature difference in °C δ = Current density in A/m2

ρ = electrical resistivity in Ωm ρc = Thermal resistivity in Ωm

L = Length of embedded conductors.

8. Express heating time constant in terms of rate of temperature difference.prprprprprpr

; T – heating time constant; – Final temperature rise during heating;

θ – Temperature rise during time ‘t’.

9. What are the factors that affect the size of rotating machine? (Dec 2013) (May 2015)

The factors affecting the size of the rotating machines are speed, electric and magnetic loadings.

10. What is specific Electric loading? (May 2011) (May 2012) (May 2013) (May 2014)(May 2015)

The total amount of ampere conductors available at the armature periphery per unit length is

called specific electric loading. It is given by ac = Izz/πD, A/m

11. What are the different types of magnetic materials according to their degree of magnetism?

(May 2011 & Dec 2011) (May 2014)

a. Ferro magnetic materials - µr >>1 b. Para magnetic materials - µr > 1

c. Dia magnetic materials - µr < 1.

12. What is real and apparent flux density?

The real flux density is due to the actual flux through a tooth. The apparent flux density is due to

total flux that has to be passing through the tooth. Since some of the fluxes pass through slot so that

the real flux density is always less than the apparent flux density.

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PART B

1. Various factors to be considered for choice of specific electric and magnetic loadings SPECIFIC ELECTRIC & MAGNETIC LOADINGS:

1. Total magnetic loadings = PΦ

2. Total electric loadings= IZZ.

3. Specific Magnetic loadings, Bav= PΦ/ πDL.

4. Specific Electric loadings, ac = IZZ / πD.

Where P= Number of Poles

Φ= Flux per pole in wb,

IZ= Current through each conductors, A

Z= Total number of armature conductors.

D&L = Main dimensions of rotating machines.

Choice of specific magnetic loadings:

1. Maximum flux density in iron parts of machine,

2. Magnetizing current &

3. Iron losses.

Choice of specific electric loadings:

1. Permissible temperature rise.

(i) Temperature rise

(ii) Cooling coefficient.

2. Voltage

3. Size of the Machine

4. Current density

2. Heat flow in 1-D & 2-D and Temperature rise-time curves. (Heating & Cooling Curves)

CALCULATION OF TEMPERATURE GRADIENT:

Let l= length of plate in m.

w= width of the plate in m.

ρ= thermal resistivity of the material in Ωm.

q= Heat produced per unit volume in W/m3.

Q= Heat produced in W.

θ = Temperature rise in C.

Temperature difference between the centers of the plate to the outer surface

θo= q ρ t2 C

8

Temperature of the hottest spot will be θm= θo + θs

Where θs= temperature of the outer surface.

Note :

1. Consider all the heat flows along the direction OX.

The temperature difference between O & A

θoA= q ρx x2 C

2

2

2. Simillarly, the temperature difference between O & B

θoB= q ρy y2 C

2

3. Heat produced per unit Volume, q= Q/ volume in W/m3

Alsoq= Core loss/Kg * Density * Stacking factor

4. Volume of the coil = Area * length

5. ρe = ρi(1- sf1/2

)

HEAT FLOW IN TWO DIMENSIONS:

Let l = length of coil, m

w= width of the coil, m

t= thickness of the coil, m

ρx= thermal resistivity along bb’, in Ωm.

ρy =thermal resistivity along aa’in Ωm.

q= Heat produced per unit volume in W/m3.

Q= Heat produced in W.

θ = Temperature rise in C.

θ =

][8yx w

t

t

wi

Q

rr+

in C.

TEMPERATURE RISE-TIME CURVE:

Let Q= Power loss or heat developed in W or J/s.

G= weight of active parts of he machine in Kg,

h= Specific Heat, J/ Kg -C.

S= Cooling surface, m2.

λ= Specific heat dissipation, W/ m2-C.

c= 1/λ= cooling coefficient, C-m2/ W.

θ = Temperature rise at any time t in C.

θm= Final steady temperature rise while heating in C.

θn = Final steady temperature rise while cooling in C.

θi = initial temperature rise over ambient temperature in C.

Th= Heating time constant, sec.

Tc= Cooling time constant, sec.

t= time in sec.

Heating curve:

1. Heat energy developed in the machine = Heat energy stored + Heat energy dissipation.

Q dt= G h dθ + S λ θdt

2.Final steady temperature rise, θm= Q

Sλ

3. Heating time constant, Th= Gh

Sλ

4. θ = θm[ 1- e –t/ Th

] + θie –t/ Th

.

If the machine starts from cold conditions, therefore te initial temperature of the machine is

considered as zeroie.,θi = 0.

θ = θm[ 1- e –t/ Th

]

3

Cooling curve:

1. Cooling time constant, Tc= Gh

Sλ

2. θ = θn[ 1- e –t/ Tc

] + θie –t/ Tc

.

If the machine is shutdown, no heat is produced and so its final steady temperature rise when

cooling is zero ie.,θn=0.

θ = θie –t/ Tc

.

UNIT 2 DESIGN OF DC MACHINES PART-A

1. Define copper space factor for coil.(May 2015)

It is the ratio of area of copper in the field coil to the area of cross section of field coil.

2. Give the expression for gap contraction factor for Slots (Dec 2014) (May 2015)

Kgs= Ys/(Ys – Kcs Ws)

Where; Ys – Slot pitch; Ws – width of slot Kcs – carter’s gap coefficient.

3. What is slot loading? (May 2011)

The number of ampere conductors per slot is known as slot loading. It should not exceed 1500A.

IzZs≠1500A.

4. Name any two methods to reduce armature reaction. (May 2011)

(i) Increasing the length of air gap at pole tips (ii) Increasing reluctance of pole tips

(ii) Compensating winding (iv) Interpoles

5. Show how the specific magnetic and electric loadings are interdependent. (Dec 2011)

If one value is chosen higher, the other value of other has to be assumed lower.

6. What is meant by peripheral speed? Write the expression for peripheral speed of a rotating

machine.(May 2012) (Dec 2013)

The peripheral speed is a translational speed that may exist at the surface of the rotor, while it is

rotating. It is translational speed equivalent to the angular speed at the surface of the rotor.

peripheral speed Va=πDrn in m/sec.

7. What is meant by magnetic loading? (May 2012)

Total amount of flux available in air gap of armature pheriphery is called magnetic loading.

8. Define field form factor. (May 2012)

Field form factor (KF) is defined as the reatio of average gap density over the pole pitch to

maximum flux density in the air gap. KF =(Bav/Bg) also KF=(Pole arc/ Pole pitch).

9. What is meant unbalanced magnetic pull? (May 2012)

The unbalanced magnetic pull is the radial force acting on the rotor due to non uniform air-gap

around armature periphery.

10. Write down the Output equation of a DC Machines. (May 2013)

LnDCP oa

2= Where 32 10* -= acBC avo p

11. Why square pole is preferred?(Dec 2013) (May 2015)

Square pole is preferred to reduce copper requirements.

12. What is mean by magnetic circuit calculations? (Dec 2013)

Magnetic circuit calculations involve computation of magnetic flux, mmf and reluctance of the

magnetic circuit.

13. What are the factors to be considered in the design of commutator of a DC machine?

(May 2014)

(i) Number of segments (ii) Commutator Diameter (iii) Length of Commutator

14. Mention any two guiding factors for the choice of number of poles. (Dec 2011) (May 2013, 2014)

Frequency of flux reversal (ii) Weight of iron (iii) Weight of copper (iv) Length of commutator.

15. Mention the factors governing the length of armature core in a dc machine? (Dec2014)

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Cost and Ventilation

16. Why dc motors are preferred in general? (Dec2014)

i. Speed control over a wide range both above and below the rated speed

Read ii. High starting torque iii. Accurate steep less speed with constant torque iv. Quick starting,

stopping, reversing and acceleration and v. Free from harmonics, reactive power consumption

PART B

1. Output equation of DC Machine

Output Equation: Let,

D-- Armature diameter, m

L—Length of armature, m

N—Speed in rpm

n— Speed in rps

P--- Rating of DC Machine, kW

Pa--Power developed by the armature, kW

p—No of poles,

a—No of parallel paths,

Z—Total number of armature conductors,

Iz – Current in each conductors, A

Ia – Armature current, A

Iph – Current per phase, A

E – Back emf, v

Eph – Induced emf /Phase,v

Tph – Turns/ Phase

Bav = DL

p

pf

Specific magnetic loadings, wb/m2.

ac = D

ZI z

pSpecific electric loadings A/m

2

t --

p

Dp pole pitch

Power developed by the armature Pa =E Ia X10

-3 kW

Also, LnDCP oa

2= kW

Where oC is called output coefficient of DC Machine.

oC = 32 10-acxBavp

D & L are called as Main Dimensions of DC Machine

Note: For Motor, Pa = P

For Generator, Pa= hP

For Square pole face,

1=polearc

L

5

Separation of D & L:

· In general, a ratio of L///// or L/D is assumed.

· Using the output equation and the knowledge of kW rating, specific loadings and speed, the value

of LD2is estimated.

· Then by solving the two equations (ie the equation of L/L/L/ or L/D and LD2), the values of D &

L are estimated.

2. Design procedure for Armature, Commutator & Brushes

Design of armature: Guiding factor for selecting no of armature slots.

i) Slot pitch: It lies between 25mm to 35mm. For small machine it should be less than or equal to 20mm.

ii) Slot loading: Slot loading should not exceed 1500 ampere conductors.

Slot loading = No of conductors in slot X Current/ conductor.

iii) Flux pulsation: If the no of slots per pole is an odd integer then flux pulsation losses can be minimized.

Normally, Slots/pole = odd integer ±1/2

iv) Commutation: To avoid sparking, the no of slots/pole should have a minimum value of 9. The

no of slots/pole normally lies between 9 to 16.

In case of small machine, it can be 8.

v) Suitability of winding: The no of slots selected should be suitable for type of winding.

In case of lap winding, no of slots should be a multiple of pole pair.

In case of wave winding, no of slots should not be a multiple of pole pair.

Design of commutator & Brushes:

1. Number of commutator segments. 1

2C uS= ,

where u® no of coil sides per slot

S® no of slots

2. Minimum No of segment is min

15

EpC =

3. Commutator dia lies between 0.6 to 0.8 times the armature diameter.

If voltage of machine is 350 to 700v ® Commutator diameter is 62% of armature

diameter.

For 200 to 250 v ® Commutator diameter is 68% of armature diameter.

For 100 to 125V ® Commutator diameter is 75% of armature diameter.

4. Peripheral speed = . It should be kept below 15m/s. For high speed machines, it

should be around 30m/s.

5. Commutator segment pitch. bccD

C

p= . Where

cD ® diameter of commutator

The thickness of commutator segment should not be less than 3mm.

Commutator segment pitch should not be less than 4mm.

6. Length of commutataor ( ) 1 2c b b bL n w c c c= + + + ,

6

where bn ® Number of brushes,

bw ® width of brushes,

1c 2c bc ® Clearances.

1c ®10mm for small machine & 30mm for large machine.

2c ® lies between 10 to 25 mm & bc ® 5mm.

7. Area of brush under one spindle is 2/ a

b

a a

Icurrent brusharmA

pd d= = .

Also b b b bA n w t=

8. Area of each brush b b ba w t= .

Note: No of brushes should be choosen such that current/brush arm should not exceeding

70A.

9. Brush losses = Brush contact loss + Brush friction loss.

(1) Brush contact loss= Brush contact drop X Armature current.

(2) Brush friction loss bf b b cP P pAVm= .

Where m®Co-efficient of friction,

P® Brush pressure KN/m2.

p® No of poles,

bA ® Area of brush under one spindle,

cV ® Peripheral speed

UNIT – III – DESIGN OF TRANSFORMERS PART A

1. Give expression for EMF / turn.

Et = 4.44 f Bm Ai; where f - frequency; Bm - Max flux density; Ai - net core area

2. Give O/P equation of 1- φ Transformer.

Q = 4.44 f Bm Ai δ Kw Aw X 10-3

KVA

Where f - Frequency; Bm - Max flux density; Kw - window space factor; Aw- window area; δ- current

density.

3. Give O/P equation for 3- φ Transformer.

Q = 3.33 f Bm δ Kw Aw Ai * 10-3

KVA

4. Give expression for heat dissipated with cooling tubes.

Heat dissipated =12.5 St θ + (6.5 At θ) 1.35 watts; where St=Cooling surface of tank;

θ=Maximum allowable temperature; At=Area of cooling tubes.

5. Give temperature rise Q in terms of tank area.

Q =(Total Losses to be dissipated)/(12.5 X tank area)

6. Give expression of Et in terms of KVA.(May 2011)

KVA/PhKtE = ; Where3

4.44fr10K = ; r = Φm /AT, Φm – Maximum flux; AT – Ampere

conductor /m

7. What are the factors affecting the choice of flux density of core in a transformer?

(May 2011)

(i) Core and yoke area (ii) Overall size and weight of the transformer (iii) Magnetizing current

(iv) Iron loss (v) saturation and temperature rise.

8. What is meant by stacking factor? (May 2012) (Dec 2014)

It is defined as the ratio of iron area to total area is called stacking factor. The usual value of

stacking factor is 0.9.

7

9. What are the cooling methods used for dry type transformers? (May 2013)

(i) Air Natural (AN) Cooling & (ii) Air Blast (AB) Cooling

10. Define Window Space Factor. (May 2013)

The window space factor is defined as the ratio of copper area in the window to the total window

area.

11. Distinguish between core and shell type transformers. (Dec 2013)

Core Type Shell Type

Easy in design and construction Comparatively complex

Has low mechanical strength due to non-bracing of windings High mechanical strength

21. Why are the cores of large transformers built-up of circular cross-section? (Dec 2013, 14)

It has the smallest perimeter for a given area and, therefore, requires less copper than rectangular

cross section

22. Define the term voltage regulation. (Dec 2011) (May 2014)

Voltage Regulation= x100%voltageLoad

voltageLoadvoltageloadNo

þýü

îíì -

23. What are the methods by which heat dissipation occurs in a transformer? (Dec 2011, May

2014)

(a) Radiation (b) Convection.

24. What are the advantages of stepped cores?(May 2015) (Dec 2014)

For same area of cross section the stepped cores will have lesser diameter of circumscribing circle

than square cores. This results in reduction in length of mean turn of the winding with consequent

reduction in both copper cost and copper loss.

PART B

1. Output Equation: The equation which relates the rated KVA output of a transformer to

the area of its core and window is called the output equation of the transformer.

FOR SINGLE PHASE TRANSFORMER :

· Induced emf in transformer = 4.44f ΦmT volts

· Emf/turn = Et/T = (4.44f ΦmT)/T = 4.44f Φm volts

· Q = 2.22 f Bm Ai Aw Kw δ x 10-3

KVA

………………………………………(1)

FOR THREE PHASE TRANSFORMER :

Q = 3.33 f Bm Ai Aw Kw δ x 10-3

KVA ………………………………………(2)

OVERALL DIMENSIONS OF TRANSFORMER : · H -> overall height of the transformer frame(m)

· D -> distance between adjacent core centres(m)

· d -> diameter of circumscribing circle(m)

· Wt -> width of tank(m)

· Ww-> width of window(m)

· Ht -> height of tank(m)

· Hw -> height of window(m)

· Hy -> height of yoke(m)

8

· Ap -> area of cross section of primary winding(m2)

· As -> area of cross section of secondary winding(m2)

· dy -> depth of yoke(m)

· a -> width of largest stamping

For single phase core type transformer: · D= Ww+d

· Dy=a

· H= Hw+2 Hy

· W= D+a

For single phase core type transformer: · H= Hw+2 Hy

· Dy=a

· W=2D+a

· D= Ww+d

2. DESIGN OF COOLING TUBES :

Total dissipating surface area of tank = St = 2 x Ht x (Lt + Wt) m2

Total dissipating surface area of tube = X x St m2

where,

X = (1/8.8) x [ (( Pi + Pc)/θ St) – 12.5] for 35% improvement in convection

X = (1/9.1) x [ (( Pi + Pc)/θ St) – 12.5] for 40% improvement in convection

Where Pi + Pc -> total losses occurring in tank

Θ -> temperature rise in tank

Area of each cooling tube = ∏ x Lt x Dt m2

Number of cooling tubes required

= Total dissipating surface area of tube/ Area of each cooling tube

= (X x St ) / ( ∏ x Lt x Dt )

Note : · A standard diameter of cooling tubes is 50mm.

· The length of the tube depends on the height of the tank, if not given it is 1m.

· Also, the spacing between each cooling tube is normally taken as 75mm.

· If Ht ,Lt and Wt values are not given, calculate using the following formulae

Ht -> height of tank = H + C3 +C4 Wt -> width of tank= 2D + De+ 2C1 ( for 3 Phase)

= D + De+ 2C1 ( for 1 Phase)

Lt-> length of tank= De + 2C2

Where C1 C2 C3 and C4 are clearances.

9

UNIT- IV DESIGN OF INDUCTION MOTOR PART A

1. What is slot space factor?

The slot space factor is the ratio of conductor area per slot and slot area. It gives as indication of the

space occupied by the conductors and the space available for insulation. The slot space factor for

induction motor varies from 0.25 to 0.4.

2. Write the expression for length of mean turn of stator winding.

Length of mean turn of stator Lmts = 2L + 2.3ґ + 0.24

3. Which part of induction motor has maximum flux density?

The teeth of the stator and rotor core will have maximum flux density.

4. What are the methods adopted to reduce harmonic torques?

The methods used are chording, integral slot winding, skewing and increasing the length of air- gap.

5. Write the expression for O/P equation and output co-efficient of induction motor.(May 2013)

The input KVA Q = Co D2L ns in KVA

Output coefficient Co = 11 Kws Bav ac x 10 -3

in KVA/m3- rps.

6. Show a relation between D and L for best power factor.

pLP

D18.0

=

7. State some methods to reduce the harmonics torque in induction motors.

1.Chording 2. Integral slot windings 3. Skewing 4. Increase in air-gap length

8. Define dispersion coefficient of an Induction motor?(May 2011)

It is defined as the ratio of magnetizing current to the ideal short circuit current of IM.

9. How crawling can be prevented by design in an Induction motor? (May 2011)

Crawling can be prevented by selecting number of rotor slots which is 15 to 30 percent larger or

smaller than the number of stator slots.

10. Define Stator slot pitch. (Dec 2011) (May 2014)

Stator Slot pitch =

þýü

îíì

SlotsStatorofNumberTotal

surfaceGap =

sS

ΠD

11. Write down the equation for output co-efficient in an induction motor.

(Dec 2011) (May 2013) (May 2014)

so LnDCQ 2= where Co be the output co-efficient =310*11 -acBK avw .

12. What are the advantages and disadvantages of larger air gap length in induction motor?

(May 2012)

Advantages: A large gap length results in higher overload capacity better cooling, reduction in

noise and reduction in unbalanced magnetic pull.

Disadvantage: High valve of magnetizing current.

13. What are the factors to be considered for selecting the number of slots in induction machine

stator? (May 2012)

The factors to be considered for selecting the number of slots is tooth pulsation loss, leakage

reactance, magnetizing current, iron loss and cost. Also the number of slots should be multiple of

slots per pole per phase for integral slot winding.

14. List the advantages of using open slots. (Dec 2014)

1. The coils can be formed, wound and insulated prior to being placed in the slots.

2. It provides facility in removal and replacement maintenance of defective coil.

15. Why fractional slot winding is not used for induction motor? (Dec 2014)

Fractional slot winding creates non uniform flux density distribution in air gap and it leads to torque

ripples.

10

PART B

OUTPUT EQUATION OF AC MACHINE:

11

12

13

2. MAIN DIMENSIONS OF AC MACHINE

14

15

16

17

18

19

20

21

22

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UNIT – V DESIGN OF SYNCHRONOUS MACHINES

PART A 1. Give the output equation.

Q = (11 Bav ac Kw x 10-3

) D2 ns

Bav = Specific Magnetic loading; ac= Specific electric loading

Kw= Winding factor; D= Diameter of stator L= length

2. Define SCR. (Dec 2011) (May 2013) (May 2014)

Ratio of excitation current needed to under rated voltage in the stator winding under no load

conditions to the excitation current needed to circulate rated current under rated short circuit current.

But the speed of the machine remains rated for both the tests.

3. How is cylindrical pole different from salient pole in a synchronous machine? (May 2015) i. Cylindrical pole are non-projecting pole whereas the salient pole machines are projecting pole.

ii. Cylindrical rotor construction is used for turbo alternators which are driven by high speed steam or

gas turbines whereas the salient pole construction is used for generators driven by hydraulic

turbine since these turbines operate at relatively low speeds.

4. What are the disadvantages of low value of SCR?

i. Poor stability limit, ii. Poor voltage regulation, ii. Unsatisfactory parallel operation

5. Define specific magnetic loading of a Synchronous machine. (May 2014)

The average flux density over the air gap of a machine is known as specific magnetic loading.

Bav = Total flux around the air gap/Area of flux path at the air gap = PΦ/πDL

6. What is critical speed of alternator?

The rotor is a structure with mass and velocity and o has a natural frequency of vibration. The

speed of the machine at which this occurs is known as critical speed.

7. Why salient pole construction is rejected for high speed alternators?

1. The rotor part are subjected to very high mechanical stress.

2. Excessive windage loss, 3. The machine would be very noisy

8. How the value of SCR affects the design of alternator? (May 2012)

For high stability and low regulation the value of SCR should be high, which requires larger

airgap. When the length of airgap is large, the mmf requirement will be high and so the field

system will be large. Hence the machine will be costlier.

9. What are the factors to be considered for the choice of specific magnetic loading in

synchronous machine? (Dec 2011) (May 2013)

Iron loss, stability, voltage rating, parallel operation, transient short circuit current.

10. What are the factors to be considered for the choice of specific electric loading in

synchronous machine?

Copper loss, synchronous reactance, temperature rise, stray losses, voltage rating.

11. What is the limiting factor for the diameter of synchronous machines? (Dec 2013)

Peripheral speed. The limiting value of peripheral speed is 175 m / sec for cylindrical rotor

machines and 80 m/sec for salient pole machines.

12. What is run away speed of Synchronous Machine? (May 2011) (May 2012) (Dec 2013)

Speed which the prime mover would have if suddenly unloaded when working at its rated load is

known as runaway speed.

13. Give the need for damper winding in synchronous machine? (May 2011)

i. To damp out hunting, ii. For self-start.

14. How the dimensions of induction generator differ from that of an induction motor?(May 2015)

Dimensions of induction generator and induction motor are same. Energy conversion process is

reversible. Therefore, induction motor can operate as induction generator.

15. How is the efficiency of an alternator affected by load power factor?(Dec 2014)

When the load power factor increases, output power increases in turn efficiency of an alternator

also increases.

16. Mention the factors to be considered for the selection of number of armature slots.(Dec2014)

i. Balanced windings ii. Cost iii. Hot spot temperatures iv. Leakage reactance

v. Torque ripples vi. Flux density in iron

24

PART B

1. Construction details:

25

26

Output equation

27

28

29

30

31

3.

32

33

34

35

36

37

38