(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt

d Calculate the reactive power consumed or supplied by this load.. c The power factor of this load is PF=cos − ° =45 0.707 leading d This load supplies reactive power to the source.. a

Time

dt

d

0 < t < 2 s

( )0.010 Wb

500 t

2 s

2.50 V

2 < t < 5 s

( ) 0.020 Wb

500 t

3 s

5 < t < 7 s

( )0.010 Wb

500 t

2 s

2.50 V

7 < t < 8 s

( )0.010 Wb

500 t

1 s

5.00 V

The resulting voltage is plotted below:

1-17 Figure P1-13 shows the core of a simple dc motor The magnetization curve for the metal in this core is

given by Figure 1-10c and d Assume that the cross-sectional area of each air gap is 18 cm2 and that the width of each air gap is 0.05 cm The effective diameter of the rotor core is 4 cm

SOLUTION The magnetization curve for this core is shown below:

The relative permeability of this core is shown below:

Note: This is a design problem, and the answer presented here is not unique Other

values could be selected for the flux density in part (a), and other numbers of turns

could be selected in part (c) These other answers are also correct if the proper steps

were followed, and if the choices were reasonable

(a) From Figure 1-10c, a reasonable maximum flux density would be about 1.2 T Notice that the

saturation effects become significant for higher flux densities

(b) At a flux density of 1.2 T, the total flux in the core would be

(1.2 T)(0.04 m)(0.04 m) 0.00192 Wb

BA

(c) The total reluctance of the core is:

TOT = stator+ air gap 1+ rotor+ air gap 2

At a flux density of 1.2 T, the relative permeability µr of the stator is about 3800, so the stator reluctance

is

stator

stator stator

0.48 m

62.8 kA t/Wb

3800 4 10 H/m 0.04 m 0.04 m

l A

× R

At a flux density of 1.2 T, the relative permeability µr of the rotor is 3800, so the rotor reluctance is

rotor

stator rotor

0.04 m

5.2 kA t/Wb

3800 4 10 H/m 0.04 m 0.04 m

l A

× R

The reluctance of both air gap 1 and air gap 2 is

air gap

air gap air gap

0.0005 m

221 kA t/Wb

4 10 H/m 0.0018 m

l A

×

Therefore, the total reluctance of the core is

TOT = stator+ air gap 1+ rotor+ air gap 2

TOT =62.8 221 5.2+ + +221=510 kA t/Wb⋅ R

The required MMF is

TOT =φ TOT = 0.00192 Wb 510 kA t/Wb⋅ =979 A t⋅

Since F=Ni , and the current is limited to 1 A, one possible choice for the number of turns is N = 1000

1-18 Assume that the voltage applied to a load is V=208∠ − °30 V and the current flowing through the load is

5 15 A

= ∠ °

(a) Calculate the complex power S consumed by this load

(b) Is this load inductive or capacitive?

(c) Calculate the power factor of this load?

(d) Calculate the reactive power consumed or supplied by this load Does the load consume reactive power

from the source or supply it to the source?

SOLUTION

(a) The complex power S consumed by this load is

208 30 V 5 15 A 208 30 V 5 15 A

S VI

1040 45 VA

S

(b) This is a capacitive load

(c) The power factor of this load is

PF=cos − ° =45 0.707 leading

(d) This load supplies reactive power to the source The reactive power of the load is

sin 208 V 5 A sin 45 735 var

1-19 Figure P1-14 shows a simple single-phase ac power system with three loads The voltage source is

120 0 V

V , and the three loads are

1= ∠ ° Ω5 30

Z Z2= ∠ ° Ω5 45 Z3= ∠ − ° Ω5 90

Answer the following questions about this power system

(a) Assume that the switch shown in the figure is open, and calculate the current I, the power factor, and

the real, reactive, and apparent power being supplied by the source

(b) Assume that the switch shown in the figure is closed, and calculate the current I, the power factor, and

the real, reactive, and apparent power being supplied by the source

(c) What happened to the current flowing from the source when the switch closed? Why?

+

-I

V

+

-+

-+

-1

3

Z

120 0 V

V

SOLUTION

(a) With the switch open, only loads 1 and 2 are connected to the source The current I in Load 1 is 1

1

120 0 V

24 30 A

5 30 A

∠ °

∠ °

I

The current I in Load 2 is 2

2

120 0 V

24 45 A

5 45 A

∠ °

∠ °

I

Therefore the total current from the source is

1 2 24 30 A 24 45 A 47.59 37.5 A

The power factor supplied by the source is

PF=cosθ=cos −37.5° =0.793 lagging

The real, reactive, and apparent power supplied by the source are

cos 120 V 47.59 A cos 37.5 4531 W

cos 120 V 47.59 A sin 37.5 3477 var

(120 V 47.59 A)( ) 5711 VA

(b) With the switch open, all three loads are connected to the source The current in Loads 1 and 2 is the

same as before The current I in Load 3 is 3

3

120 0 V

24 90 A

5 90 A

∠ °

∠ − °

I

Therefore the total current from the source is

1 2 3 24 30 A 24 45 A 24 90 A 38.08 7.5 A

The power factor supplied by the source is

PF=cosθ=cos − ° =7.5 0.991 lagging

The real, reactive, and apparent power supplied by the source are

cos 120 V 38.08 A cos 7.5 4531 W

( )( ) ( )

cos 120 V 38.08 A sin 7.5 596 var

(120 V 38.08 A)( ) 4570 VA

(c) The current flowing decreased when the switch closed, because most of the reactive power being

consumed by Loads 1 and 2 is being supplied by Load 3 Since less reactive power has to be supplied by the source, the total current flow decreases

1-20 Demonstrate that Equation (1-59) can be derived from Equation (1-58) using simple trigonometric

identities:

( ) ( ) ( ) 2 cos cos

( ) cos 1 cos 2 sin sin 2

SOLUTION

The first step is to apply the following identity:

1

2

The result is

( ) ( ) ( ) 2 cos cos

p t =v t i t = VI ωt ω θt− )

1

2

p t = VI ω ω θt t ω ω θt t

( ) cos cos 2

p t =VI θ ω θt Now we must apply the angle addition identity to the second term:

cos α β− =cos cosα β+sin sinα β The result is

( ) cos cos 2 cos sin 2 sin

p t =VI θ+ ωt θ+ ωt θ Collecting terms yields the final result:

( ) cos 1 cos 2 sin sin 2

p t =VI θ + ωt +VI θ ωt

1-21 A linear machine has a magnetic flux density of 0.5 T directed into the page, a resistance of 0.25 Ω, a bar

length l = 1.0 m, and a battery voltage of 100 V

(a) What is the initial force on the bar at starting? What is the initial current flow?

(b) What is the no-load steady-state speed of the bar?

(c) If the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new

steady-state speed? What is the efficiency of the machine under these circumstances?

SOLUTION

(a) The current in the bar at starting is

100 V

400 A 0.25

B V i

R

Ω Therefore, the force on the bar at starting is

( ) (400 A 1 m 0.5 T)( )( ) 200 N, to the right

i

(b) The no-load steady-state speed of this bar can be found from the equation

vBl e

V B = ind =

(0.5 T 1 m100 V)( ) 200 m/s

B V v

Bl

(c) With a load of 25 N opposite to the direction of motion, the steady-state current flow in the bar will

be given by

ilB F

Fapp = ind =

50 A 0.5 T 1 m

F i

Bl

The induced voltage in the bar will be

ind B 100 V - 50 A 0.25 87.5 V

and the velocity of the bar will be

(0.5 T 1 m87.5 V)( ) 175 m/s

B V v

Bl

The input power to the linear machine under these conditions is

in B 100 V 50 A 5000 W

The output power from the linear machine under these conditions is

out B 87.5 V 50 A 4375 W

Therefore, the efficiency of the machine under these conditions is

out in

4375 W

5000 W

P P

1-22 A linear machine has the following characteristics:

0.33 T into page

0.5 m

(a) If this bar has a load of 10 N attached to it opposite to the direction of motion, what is the steady-state

speed of the bar?

(b) If the bar runs off into a region where the flux density falls to 0.30 T, what happens to the bar? What

is its final steady-state speed?

(c) Suppose V is now decreased to 80 V with everything else remaining as in part (b) What is the new B

steady-state speed of the bar?

(d) From the results for parts (b) and (c), what are two methods of controlling the speed of a linear

machine (or a real dc motor)?

SOLUTION

(a) With a load of 20 N opposite to the direction of motion, the steady-state current flow in the bar will

be given by

ilB F

Fapp = ind =

60.5 A 0.33 T 0.5 m

F i

Bl

The induced voltage in the bar will be

ind B 120 V - 60.5 A 0.50 89.75 V

and the velocity of the bar will be

ind 89.75 V

544 m/s 0.33 T 0.5 m

e v

Bl

(b) If the flux density drops to 0.30 T while the load on the bar remains the same, there will be a speed

transient until Fapp=Find =10 N again The new steady state current will be

F =F =ilB

66.7 A 0.30 T 0.5 m

F i

Bl

The induced voltage in the bar will be

ind B 120 V - 66.7 A 0.50 86.65 V

and the velocity of the bar will be

ind 86.65 V

577 m/s 0.30 T 0.5 m

e v

Bl

(c) If the battery voltage is decreased to 80 V while the load on the bar remains the same, there will be a

speed transient until Fapp=Find =10 N again The new steady state current will be

F =F =ilB

66.7 A 0.30 T 0.5 m

F i

Bl

The induced voltage in the bar will be

( )( ) ind B 80 V - 66.7 A 0.50 46.65 V

and the velocity of the bar will be

ind 46.65 V

311 m/s 0.30 T 0.5 m

e v

Bl

(d) From the results of the two previous parts, we can see that there are two ways to control the speed of

a linear dc machine Reducing the flux density B of the machine increases the steady-state speed, and reducing the battery voltage V B decreases the stead-state speed of the machine Both of these speed control

methods work for real dc machines as well as for linear machines

Chapter 2: Transformers

2-1 The secondary winding of a transformer has a terminal voltage of ( )v t s =282.8 sin 377 Vt The turns

ratio of the transformer is 100:200 (a = 0.50) If the secondary current of the transformer is

( ) 7.07 sin 377 36.87 A

s

i t = t− ° , what is the primary current of this transformer? What are its voltage regulation and efficiency? The impedances of this transformer referred to the primary side are

eq 0.20

R = Ω R C=300 Ω

eq 0.750

X = Ω X M =80 Ω SOLUTION The equivalent circuit of this transformer is shown below (Since no particular equivalent circuit was specified, we are using the approximate equivalent circuit referred to the primary side.)

The secondary voltage and current are

282.8

0 V 200 0 V 2

V

7.07

36.87 A 5 -36.87 A 2

S = ∠ − ° = ∠ °

I

The secondary voltage referred to the primary side is

100 0 V

S′ =a S = ∠ °

The secondary current referred to the primary side is

10 36.87 A

S S

a

′ = = ∠ −I °

I

The primary circuit voltage is given by

( eq eq)

P = S′+ S′ R + jX

100 0 V 10 36.87 A 0.20 0.750 106.2 2.6 V

V

The excitation current of this transformer is

EX

106.2 2.6 V 106.2 2.6 V

0.354 2.6 1.328 87.4

300 80

C M

j

EX =1.37∠ −72.5 A°

I

Therefore, the total primary current of this transformer is

EX 10 36.87 1.37 72.5 11.1 41.0 A

The voltage regulation of the transformer at this load is

106.2 100

100

P S S

V aV aV

The input power to this transformer is

IN P P cos 106.2 V 11.1 A cos 2.6 41.0

P =V I θ=

IN 106.2 V 11.1 A cos 43.6 854 W

The output power from this transformer is

OUT S S cos 200 V 5 A cos 36.87 800 W

Therefore, the transformer’s efficiency is

OUT IN

800 W

854 W

P P

2-2 A 20-kVA 8000/480-V distribution transformer has the following resistances and reactances:

Ω

= 32

P

Ω

= 45

P

Ω

=250k

C

The excitation branch impedances are given referred to the high-voltage side of the transformer

(a) Find the equivalent circuit of this transformer referred to the high-voltage side

(b) Find the per-unit equivalent circuit of this transformer

(c) Assume that this transformer is supplying rated load at 480 V and 0.8 PF lagging What is this transformer’s input voltage? What is its voltage regulation?

(d) What is the transformer’s efficiency under the conditions of part (c)?

SOLUTION

(a) The turns ratio of this transformer is a = 8000/480 = 16.67 Therefore, the secondary impedances

referred to the primary side are

2

R′ =a R = Ω = Ω

2

The resulting equivalent circuit is

32 Ω

250 kΩ

j45 Ω

j30 kΩ

j16.7 Ω 13.9 Ω

(b) The rated kVA of the transformer is 20 kVA, and the rated voltage on the primary side is 8000 V, so

the rated current in the primary side is 20 kVA/8000 V = 2.5 A Therefore, the base impedance on the primary side is

Ω

=

=

A 2.5

V 8000

base

base base

I

V Z

Since Zpu = Zactual/ Zbase, the resulting per-unit equivalent circuit is as shown below:

0.01

78.125

j0.0141

j9.375

0.0043 j0.0052

(c) To simplify the calculations, use the simplified equivalent circuit referred to the primary side of the

transformer:

32 Ω

250 kΩ

j45 Ω

j30 kΩ

j16.7 Ω 13.9 Ω

The secondary current in this transformer is

20 kVA

36.87 A 41.67 36.87 A

480 V

I

The secondary current referred to the primary side is

41.67 36.87 A

2.50 36.87 A 16.67

S S

a

I

Therefore, the primary voltage on the transformer is

+

′

8000 0 V 45.9 61.7 2.50 36.87 A 8185 0.38 V

V

The voltage regulation of the transformer under these conditions is

8185-8000

8000

(d) Under the conditions of part (c), the transformer’s output power copper losses and core losses are:

OUT cos 20 kVA 0.8 16 kW

CU S EQ 2.5 45.9 287 W

core

8185

268 W 250,000

S

C

V P

R

′

The efficiency of this transformer is

OUT

16, 000

16,000 287 268

P

2-3 A 1000-VA 230/115-V transformer has been tested to determine its equivalent circuit The results of the

tests are shown below

Open-circuit test Short-circuit test

V OC = 230 V V SC = 19.1 V

I OC = 0.45 A I SC = 8.7 A

P OC = 30 W P SC = 42.3 W All data given were taken from the primary side of the transformer

(a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer (b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8

PF leading

(c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging

SOLUTION

(a) OPEN CIRCUIT TEST:

001957

0 V 230

A 45 0

Y

=

A 45 0 V 230

W 30 cos

OC OC

OC 1

I V

P

θ

mho 0.001873

-0.000567 mho

15 73 001957

0

Ω

=

= 1 1763

C C

G R

Ω

=

= 1 534

M M

B X

SHORT CIRCUIT TEST:

19.1 V

2.2 8.7 A

SC SC

42.3 W

19.1 V 8.7 A

P

V I

Ω +

= Ω

°

∠

= +

= EQ EQ 2.20 75.3 0.558 2.128

Z

Ω

=0.558

EQ

R

Ω

= 2 128

X

To convert the equivalent circuit to the secondary side, divide each impedance by the square of the turns

ratio (a = 230/115 = 2) The resulting equivalent circuit is shown below:

Ω

=0.140

s EQ,

Ω

= 441

,s C

(b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred

to the secondary side The rated secondary current is

A 70 8 V 115

VA 1000

=

=

S

I

We will now calculate the primary voltage referred to the secondary side and use the voltage regulation equation for each power factor

(1) 0.8 PF Lagging:

EQ 115 0 V 0.140 0.532 8.7 36 87 A

VP′ =118.8 1.4 ∠ °V

118.8-115

115

(2) 1.0 PF:

EQ 115 0 V 0.140 0.532 8.7 0 A

116.3 2.28 V

V

115

(3) 0.8 PF Leading:

EQ 115 0 V 0.140 0.532 8.7 36 87 A

VP′ =113.3 2.24 ∠ °V

113.3-115

115

(c) At rated conditions and 0.8 PF lagging, the output power of this transformer is

OUT S S cos 115 V 8.7 A 0.8 800 W

The copper and core losses of this transformer are

2

CU S 8.7 EQ,S A 0.140 10.6 W

2

2 core

118.8 V

441

P

C

V P

R

′

Ω Therefore the efficiency of this transformer at these conditions is

OUT

800 W

800 W 10.6 W 32.0 W

P

2-4 A single-phase power system is shown in Figure P2-1 The power source feeds a 100-kVA 14/2.4-kV

transformer through a feeder impedance of 40.0 + j150 Ω The transformer’s equivalent series impedance referred to its low-voltage side is 0.12 + j0.5 Ω The load on the transformer is 90 kW at 0.80 PF lagging and 2300 V

(a) What is the voltage at the power source of the system?

(b) What is the voltage regulation of the transformer?

(c) How efficient is the overall power system?

SOLUTION

To solve this problem, we will refer the circuit to the secondary (low-voltage) side The feeder’s impedance referred to the secondary side is

2 line

2.4 kV

40 150 1.18 4.41

14 kV