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### Transcript of Machines Ty s

• Electrical EngineeringElectrical Machines

WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK

2016

Detailed Explanations ofTry Yourself Questions

T1 : Solution

(c)(c)(c)(c)(c)As shown in figure below,

A A

0.3

0.2

0.2

0.3

Ia1 Ia2

If1 If2

300 V300 V

500 AA

B

300 0.2 I = 300 0.3

I

I (0.2) =

I (0.3)

II

=

=I I

+I I = 500

+I I = 500

I =

= 200

I = 1.5

I = 1.5 200 = 300 A

andI (0.3) =

I (0.2)

II

=

I = 1.5

I

I +

I = 500 A

I + 1.5

I = 500

I =

= 200

DC Machines1

3Try yourself

T2 : Solution

(34)(34)(34)(34)(34)

= 500 0.5 50 = 475 ...(1)

Let the current when joined in series = I2 AThe resistance = 0.5 2 = 1

= 500 I2 1 = 500 I2 ...(2)

=

= = ...(3)

where, k =

From (1), (2) and (3)

I=

500 I2 =

...(4)

=

= =

I II

( T2 = 2T1)

100 = kI2

k =

I...(5)

From (5) and (4),

500 I2 =

=

I I I2 = 34 A

T3 : Solution

(0.85)(0.85)(0.85)(0.85)(0.85)

Input power =

=

Input line current =

=

Shunt field current =

=

Back emf,

= 500 0.4 85.7 = 466.5 V

Since the load and the constant losses are constant

I =

I = 466.5 83.7 = 39046

Total resistance after the series field is added= 0.4 + 0.1 = 0.5

• 4 Electrical Engineering Electrical Machines

(500 0.5 I2)I2 = 39046

I 1000 I2 + 78092 = 0

This gives, I2 = 85 AThe input current = 85 + 2 = 87 A

The efficiency =

=

= 0.85

T4 : Solution

(403)(403)(403)(403)(403)

T =

I

209 =

I = 10.34 Ia

Ia = the line current =

Eb =

=

=

440 =

or, N =

=

T5 : Solution

(1989)(1989)(1989)(1989)(1989)

Eg =

200 =

or, = ! !

The flux density, B = "#

=

=

\$

The ampere-turns required for 0.8 cm gap

=

l =

= 1989

T1 : Solution

(d)(d)(d)(d)(d)Maximum allowable flux, max = Bmax core area

= 1.2 0.025= 0.03 Wb

Voltage on primary (high voltage) side,V1 = 6000 V

Number of turns of (low voltage) side of main transformer

N2 =

= = 60

Number of turns on low voltage side of teaser transformer= 0.866 N2 = 0.866 60 = 52

Point to be tapped =

=

T2 : Solution

(c)(c)(c)(c)(c)

pu resistance =

=

pu reactance =

=

Voltage regulation will be maximum, when phase angle,

=

\$

\$

II

= "#"#

"%%"#

=

=

Power factor, cos = cos 63.435 = 0.447 lagging (for maximum regulation)Maximum regulation = pu resistance cos + pu reactance sin

= 0.025 0.447 + 0.05 0.8944

= 0.0559 or 5.59%

Transformers2

• 6 Electrical Engineering Electrical Machines

T3 : Solution

(b)(b)(b)(b)(b)At 3/4 Full load and unity power factor output

=

=

Maximum efficiency, max = 97%

Input =

= =

Total loss = Input output = 386.6 375 = 11.6 kW

This loss is equally divided between iron and copper loss, so

= Pi = =or, Pc =

=

Percentage resistance =

I

=

II

= &

=

Percentage impedance = 10%

Percentage reactance = = 9.786%

Percentage regulation =

#% %

+ I I

= 2.06 0.8 + 9.786 0.6 = 7.52%

T4 : Solution

(c)(c)(c)(c)(c)Rated output, S = 25 kVASecondary voltage, V2 = 220 VFull-load (or rated) secondary current,

I2 =

= = 113.64 A

Total resistance referred to secondary,

= K2R1 + R2 =

+

= 0.02

Full-load copper loss, Pc = I = (113.64)2 0.02

7Try yourself

= 258 W or 0.258 kWIron loss, Pi = 80% of full load copper loss

=

= 206.4 W or 0.2064 kW

Full-load output at 0.8 p.f. = 25 cos = 25 0.8= 20 kW

Full-load efficiency at 0.8 p.f., = '

+ i

=

+ +

= 97.93%

T1 : Solution

(a)(a)(a)(a)(a)Since the stator resistance is neglected

Stator out-put = Rotor out-put

=( !

=

=

s

if E2 = rotor stand-still voltage per phase

rotor current at slips =

if s1 and s2 are the slips with two given stator out-puts we have

= 9 ...(i)

and

= 4 ...(ii)

From (i) and (ii)

= 2.25

since, s1 =

)

=

so,

= 2.25 =

2.25 (ns n2) = ns n1 2.25 ns ns = 2.25 n2 n1 1.25 750 = 2.25 n2 n1 ...(iii)

as

= 1.043 (given)

from equation (iii)1.25 750 = 2.25 n2 1.043 n2

n2 = 776.719 777 rpmn1 = 1.043 776.719 810 rpm

Induction Machine3

9Try yourself

T2 : Solution

(c)(c)(c)(c)(c)

Torque at any slip, T =

+...(1)

This will be maximum,when R = sXSince, maximum torque occurs at s = 0.2

R = 0.2 X X = 5 R ...(2)At starting, s = 1

Tst =

+

with R increased to R + 0.5,

=

+

+ +( X = 5R) ...(3)

putting, S =

in (1)

Tmax =

\$

= = ( X = 5R) ...(4)

Tst is 75% of Tmax, from (3) and (4)

+

+ +=

=

38 R2 17R + 0.75 = 0 R = 0.395 or 0.05 and X = (5 0.395) or (5 0.05)or X = 1.97 or 0.25

T3 : Solution

(b)(b)(b)(b)(b)

Synchronous speed, Ns =

=

Slip, s =

=

Since the torque in maximum (stalling)

= s = 0.16

X 2 =

= = 0.44

• 10 Electrical Engineering Electrical Machines

If the total rotor resistance = (in the second case)

= s at starting = 1

= X 2 = 0.44

The resistance to be added = 0.44 0.07 = 0.37

T4 : Solution

(b)(b)(b)(b)(b)

Ns =

=

N1 = 1425 rpm S1 =

=

N2 = 1275 rpm S2 =

=

Slip = # %%

=

(

I

slip R2

=

+ +=

R =

= 0.5 /ph

T1 : Solution

(a)(a)(a)(a)(a)Given data,Given data,Given data,Given data,Given data, z s = (0.5 + j4) = 4.03182.8Vt = 400 V, Ef = 500 V, 2 = 90 2 = (90 82.8) = 7.12

shaft power output = maximum output power friction windage and iron losses

maximum output power =

=

= 41.922 kWshaft power output = (3 42.922 kW) 0.9 kW = 124.867 kW

T2 : Solution

(0.98)(0.98)(0.98)(0.98)(0.98)Taking 11 kV, 1600 KVA as base

The pu synchronous reactance =

= 0.4

if V1 = 10 = terminal voltage and I = 1Then the generated voltage

= + I

or, = + = + + = 1 + 0.4(sin + jcos) = 1 0.04 sin + j0.4cos

For, zero regulation, magnitude of this is 1 (1 0.4 sin )2 + (0.4cos)2 = 1 1 + 0.42 0.8 sin = 1

sin =

= 11.5 power factor, cos = cos11.5 = 0.98

Synchronous Machine4

• 12 Electrical Engineering Electrical Machines

T3 : Solution

(b)(b)(b)(b)(b)

Vt =

= 265.58 V

Ef =

= 317.54 V

zs =

+ = + = 1.67585.44

Gross Pm(output)maximum =

+

=

+ = 45.53 kW

net Pm (output) maximum = 45.53 3 1.4 = 135.19 kWT4 : Solution

(c)(c)(c)(c)(c)

P =

% %

+

=

+ =

+

= 0 for P = Pmax

0 =

+

1.1 cos + cos 2 = 0 2cos2 1 + 1.1 cos = 0

cos =

+=

= 61.12

13Try yourself

T5 : Solution

(b)(b)(b)(b)(b)

Phase voltage =

*

=

I =

=

I

as reference,

drop = I

= 131(0.4 + j6) = 52 + j786The terminal voltage = 3810(0.8 + j 0.6)

= 3048 + j 2286 V The generated voltage = 3100 + j3072 V

= 4.36444.7 kV ...(i )with I as reference, if x kV is the load phase voltage

Terminal voltage = 0.8 x j 0.6 x kV

I

= 0.052 + j 0.786 kV Generated voltage = (0.8 x + 0.052) + j (0.786 0.6 x) ...(ii )From equation (i) and (ii), (0.8 x + 0.052)2 + (0.786 0.6x)2 = 4.3642

x 2 0.86 x 18.424 = 0x = 4.744

line voltage = 4.744 1000 = 8220 V

Single-Phase Motor& Special Machine,

Energy Conversion Principles5T1 : Solution

(5.22)(5.22)(5.22)(5.22)(5.22)

S =

= 0.06

The circuit model is given below,

Im Im 1.8 7.8

1.8 7.8

Im

j48

j48

220 V

3.4 s = 5.67

3.4(2 s) = 1.75

= j48 (1.8 + 56.7 + j7.8) = j48 (58.5 + j7.8)= 3554 = 20.6 + j28.3

= j48 (1.8 + 1.75 + j7.8) = j48 (3.55 + j7.8)= 7.3669.1 = 2.63 + j6.88

= (20.6 + j 28.3) + (2.63 + j6.88) = 42.1656.6

I

=

=

IL = IM = 5.22 A