Charles Proteus Steinmetz. # Engineering mathematics; a series of lectures delivered at Union college online

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harmonics; that is, ig.

Furthermore, in the sum of even harmonics, 1*2 may again

be separated from its second harmonic, i 4 , and its multiples,

and therefrom, ig, and its third harmonic, i' 6 , and its multiples,

thus giving all the harmonics up to the 9th, with the exception

of the 5th and the 7th. These latter two would require plotting

the curve and taking numerical values at different intervals,

so as to have a number of numerical values divisible by 5 or 7.

It is further recommended to resolve this unsymmetrical

exciting current of Table XI into the trigonometric series by

calculating the coefficients a n and &, up to the 7th, in the man-

ner discussed in paragraphs 6 to 8.

TABLE XI

9

io

8

io

e

io

d

io

+ 95.7

90

-26.7

180

-34.3

270

- 3.3

10

+ 78.7

100

-27.3

190

-27.3

280

- 1.8

20

+ 53.7

110

-28.1

200

-16.8

290

+ 1.2

30

+ 23.7

120

-28.8

210

-11.3

300

+ 4.7

40

- 2.3

130

-29.3

220

- 8.3

310

+ 10.7

50

-16.3

140

-29.8

230

- 7.3

320

+ 22.7

60

-22.8

150

-31

240

- 6.3

330

+ 41.7

70

-24.3

160

-32.6

250

- 5.3

340

+ 65.7

80

-25.8

170

-33.8.

260

- 4.3

350

+ 85.7

TRIGONOMETRIC SERIES. 139

D. CALCULATION OF TRIGONOMETRIC SERIES FROM

OTHER TRIGONOMETRIC SERIES.

94. An hydraulic generating station has for a long time been

supplying electric energy over moderate distances, from a num-

ber of 750-kw. 4400- volt 60-cycle three-phase generators. The

station is to be increased in size by the installation of some

larger modern three-phase generators, and from this station

6000 kw. are to be transmitted over a long distance transmis-

sion line at 44,000 volts. The transmission line has a. length

of 60 miles, and consists of three wires No. B.. & S. with 5

ft. between the wires.

The question arises, whether during times of light load the

old 750-kw. generators can be used economically on the trans-

mission line. These old machines give an electromotive force

wave, which, like that of most earlier machines, differs con-

siderably from a sine wave, and it is to be investigated, whether,

due to this wave-shape distortion, the charging current of the

transmission line will be so greatly increased over the value

which it would have with a sine wave of voltage, as to make

the use of these machines on the transmission line uneconom-

ical or even unsafe.

Oscillograms of these machines, resolved into a trigonomet-

ric series, give for the voltage between each terminal and the

neutral, or the Y voltage of the three-phase system, the equa-

tion :

e = e |sm0-0.12sin (30- 2. 3) -0.23 sin (50-1.5)

+0.13 sin (70-6.2)!. . (1)

In first approximation, the line capacity may be considered

as a condenser shunted across the middle of the line; that is,

half the line resistance and half the line reactance is in series

with the line capacity.

As the receiving apparatus do not utilize the higher har-

monics of the generator wave, when using the old generators,

their voltage has to be transformed up so as to give the first

harmonic or fundamental of 44,000 volts.

44,000 volts between the lines (or delta) gives 44,000 ~ Vs =

25,400 volts between line and neutral. This is the effective

140 ENGINEERING MATHEMATICS.

value, and the maximum value of the fundamental voltage

wave thus is: 25,400 X V2 = 36,000 volts, or 36 kv.; that is,

eo = 36, and

e = 36{sin 0-0.12 sin (30-2.3)-0.23 sin (50-1.5)

+ 0.13 sin (70-6.2)}, . (2)

would be the voltage supplied to the transmission line at the

high potential terminals of the step-up transformers.

From the wire tables, the resistance per mile of No. B. & S.

copper line wire is TO = 0.52 ohm.

The inductance per mile of wire is given by the formula :

L = 0.7415 log Y + 0.0805mh, .... (3)

IT

where l s is the distance between the wires, and l r the radius of

the wire.

In the present case, this gives l s = 5 ft. = 60 in. l r = . 1625 in.

LO = ! .9655 mh., and, herefrom it follows that the reactance, at

/= 60 cycles is

XQ = 27r/L = . 75 ohms per mile (4)

The capacity per mile of wire is given by the formula :

0.0408

hence, in the present case, C = 0.0159 mf., and the condensive

reactance is derived herefrom as :

x < - ^n~ = 16,6000 ohms ; (6)

ZTT/C o

60 miles of line then give the condensive reactance,

x c = |^=2770 ohms;

30 miles, or half the line (from the generating station to the

middle of the line, where the line capacity is represented by a

shunted condenser) give: the resistance, r = 30r = 16.6 ohms

TRIGONOMETRIC SERIES. 141

the inductive reactance, x=30x = 22.5 ohms, and the equiva-

lent circuit of the line now consists of the resistance r, inductive

reactance x and condensive reactance x c , in series with each

other in the circuit of the supply voltage e.

95. If i= current in the line (charging current) the voltage

consumed by the line resistance r is rL

The voltage consumed by the inductive reactance x is x -j

do

the voltage consumed by the condensive reactance x c is x c ( idO,

and therefore,

idO (7)

Differentiating this equation, for the purpose of eliminating

the integral, gives

de d 2 i di

or

. . (8)

The voltage e is given by (2), which equation, by resolving

the trigonometric functions, gives

e = 36 sin 6-4.32 sin 30-8.28 sin 50+4.64 sin 76

+0.18 cos 30+0. 22 cos 50-0. 50 cos 70; . (9)

hence, differentiating,

de

^ = 36 cos 0-12.96 cos 30-41.4 cos 50 + 32.5 cos 70

-0.54 sin 30-1.1 sin 50 + 3.5 sin 70. . (10)

Assuming now for the current i a tiigoriometric series with

indeterminate coefficients,

i = ai cos 0+aa cos 30+a 5 cos 50+ ay cos 70

H-&I sin 0+63 sin 30 +6 5 sin 50+6 7 sin 70, . (11)

142

ENGINEERING MATHEMATICS.

substitution of (10) and (11) into equation (8) must give an

identity, from which equations for the determination of a n and

b n are derived; that is, since the product of substitution must

be an identity, all the factors of cos 6, sin 0, cos 30, sin 30,

etc., must vanish, and this gives the eight equations :

36 =2770ai+ 15.66i- 22.5ai;

= 27706i- 15.6oi- 22.5&i;

-12.96 = 2770a 3 .+ 46.86 3 - 202. 5a 3 ;

- 0.54 = 27706 3 - 46.8o 3 - 202. 56 3 :

-41.4 = 2770a 5 + 786 5 - 562. 5a 5 ;

1.1 =27706 5 - 78a 5 - 56.25& 5 ;

32.5 = 2770a 7 + 109.2& 7 -1102.5a 7 ;

3.5 = 27706 7 - 109. 2a 7 - 1102.56 7 . J

Resolved, these equations give

ai= 13.12;

61= 0.07;

a 3 = - 5.03;

6 3 = - 0.30;

a 5 = -18.72;

65= 1.15;

a 7 = 19.30;

6 7 = 3.37;

hence,

i = 13. 12 cos 0-5. 03 cos 30- 18. 72 cos 50 + 19. 30 cos 70

+0.07 sin 0-0.30 sin 30-1.15 sin 50+3.37 sin 70

= 13.12 cos (0-0.3)-5.04 cos (30-3.3)

-18. 76 cos (50- 3. 6) +19. 59 cos (70-9.9).

(12)

(13)

.(14)

TRIGONOMETRIC SERIES. 143

96. The effective value of this current is given as the square

root of the sum of squares of the effective values of the indi-

vidual harmonics, thus :

As the voltage between line and neutral is 25,400 effective,

this gives Q = 25,400X21. 6 = 540,000 volt-amperes, or 540 kv.-

amp. per line, thus a total of 3Q = 1620 kv.-amp. charging cur-

rent of the transmission line, when using the e.m.f. wave of

these old generators.

It thus would require a minimum of 3 of the 750-kw.

generators to keep the voltage on the line, even if no power

whatever is delivered from the line.

If the supply voltage of the transmission line were a perfect

sine wave, it would, at 44,000 volts between the lines, be given

by

ei = 36sin 0, (15)

which is the fundamental, or first harmonic, of equation (9).

Then the current i would also be a sine wave, and- would be

given by

i\ = a\ cos +bi sin 6,

-13.12 cos 6 +0.07 sin 0,

= 13.12 cos (0-0.3),

and its effective value would be

13.12

/i = =- = 9.3 amp (17)

This would correspond to a kv.-amp. input to the line

3d = 3 X 25.4 X 9.3 = 710 kv.-amp.

The distortion of the voltage wave, as given by equation (1),

thus increases the charging volt-amperes of the line from 710

(16)

144

ENGINEERING MATHEMATICS.

kv.-amp. to 1620 kv.-amp. or 2.28 times, and while with a sine

wave of voltage, one of the 750- kw. generators would easily be

able to supply the charging current of the line, due to the

FIG. 47.

wave shape distortion, more than two generators are required.

It would, therefore, not be economical to use these generators

on the transmission line, if they can be used for any other

purposes, as short-distance distribution.

FIG. 48.

In Figs. 47 and 48 are plotted the voltage wave and the

current wave, from equations (9) and (14) repsectively, and

TRIGONOMETRIC SERIES. 145

the numerical values, from 10 deg. to 10 deg., recorded in

Table XII.

In Figs. 47 and 48 the fundamental sine wave of voltage

and current are also shown. As seen, the distortion of current

is enormous, and the higher harmonics predominate over the

fundamental. Such waves are occasionally observed as charg-

ing currents of transmission lines or cable systems.

97. Assuming now that a reactive coil is inserted in series

with the transmission line, between the step-up transformers

and the line, what will be the voltage at the terminals of this

reactive coil, with the distorted wave of charging current

traversing the reactive coil, and how does it compare with the

voltage existing with a sine wave of charging current?

Let L= inductance, thus x = 2nfL= reactance of the coil,

and neglecting its resistance, the voltage at the terminals of

the reactive coil is given by

Substituting herein the equation of current, (11), gives

sin 6+30,3 sin 30+5a 5 sin 50+7a 7 sin 76 }

(19)

hence, substituting the numerical values (13),

135.1 sin Id }

.6 cos 76\

(20)

x{ 13.12 sin (9-15.09 sin 30-93,6 sin 56 + 135.1 sin 76 }

-0.07 cos 0+0.90 cos 36 +5.75 cos 50-23.6 cos 76}

= x\ 13.12 sin (0-0.3) -15.12 sin (30-3.3)

-93.8 sin (50 -3.6) +139.1 sin (70-9.9)}.

This voltage gives the effective value

while the effective value with a sine wave would be from (17),

hence, the voltage across the reactance x has been increased

12.8 times by the wave distortion.

146

ENGINEERING MATHEMATICS.

The instantaneous values of the voltage e f are given in the

last column of Table. XII, and plotted in Fig. 49, for z = l.

As seen from Fig. 49, the fundamental wave has practically

FIG. 49.

vanished, and the voltage wave is the seventh harmonic, modi-

fied by the fifth harmonic.

TABLE XII

e

c

t

e'

e

i

<-'

10

20

-0.10

+ 2.23

3.74

+ 8.67

+ 5.30

- 0.86

- 17

+ 46

+ 3

90

100

110

27.41

31.77

40.57

- 4.15

+ 26.19

+ 24.99

-200

-106

+ 119

30

40

50

7.47

17.35

31.70

+ 7.39

+ 30.39

+ 38.58

+ 131

-116

+ 36

120

130

140

42.70

33.14

18.03

- 8.10

-38.79

-36.65

+ 182

+ 93

- 96

60

70

80

42.06

40.33

32.87

+ 15.66

-19.01

-29.13

+ 167

+ 159

- 54

150

160

170

6.99

2.88

1.97

-13.41

+ 2.43

- 1.00

-138

- 31

+ 54

90

27.41

- 4.15

-200

180

+ 0.10

- 8.67

+ 17

CHAPTER IV.

MAXIMA AND MINIMA.

98. In engineering investigations the problem of determin-

ing the maxima and the minima, that is, the extrema of a

function, frequently occurs. For instance, the output of an

electric machine is to be found, at which its efficiency is a max-

imum, or, it is desired to determine that load on an induction

motor which gives the highest power-factor; or, that voltage

V

FIG. 50. Graphic Solution of Maxima and Minima.

which makes the cost of a transmission line a minimum; or,

that speed of a steam turbine which gives the lowest specific

steam consumption, etc.

The maxima and minima of a function, y=f(x), can be found

by plotting the function as a curve and taking from the curve

the values x, y, which give a maximum or a minimum. For

instance, in the curve Fig. 50, maxima are at PI and P 2 , minima

at P 3 and P 4 . This method of determining the extrema of

functions is necessary, if the mathematical expression between

147

148

ENGINEERING MATHEMATICS.

x and y, that is, the function y=f(x), is unknown, or if the

function y=f(x) is so complicated, as to make the mathematical

calculation of the extrema impracticable. As examples of

this method the following may be chosen:

10

2-

10 12 14 16 18 20 22 24

28 30

FIG. 51. Magnetization Curve.

Example i. Determine that magnetic density (B, at which

the permeability n of a sample of iron is a maximum. The

relation between magnetic field intensity 3C, magnetic density

05 and permeability /* cannot be expressed in a mathematical

equation, and is therefore usually given in the form of an

1400

^

*^~

^ "

KSr^

X

100TV

/

^

N

\

.X

/

x

\

-600-

z

/

\

\

f

\

7^00

$>

i

J !

*

i

> (

{ t

!

) 1

1

Kilo-lines

1 1^ 13 1

4 1

r >

FIG. 52. Permeability fcurve.

empirical curve, relating (B and 3C, as shown in Fig. 51. From

this curve, corresponding values of (B and 3C are taken, and their

/D

ratio, that is, the permeability / = , plotted against as abscissa.

uC

This is done in Fig. 52. Fig. 52 then shows that a maximum

MAXIMA AND MINIMA.

149

occurs at point fj^ &x , for (B = 10.2 kilolines, /* = 1340, and minima

at the starting-point P 2 , for (B = 0, ju = 370, and also for (B=oo ,

where by extrapolation /* = !.

Example 2. Find that output of an induction motor

which gives the highest power-factor. While theoretically

an equation can be found relating output and power-factor

of an induction motor, the equation is too complicated for use.

The most convenient way of calculating induction motors is

to calculate in tabular form for different values of slip s, the

torque, output, current, power and volt -ampere input, efficiency,

power-factor, etc., as is explained in " Theoretical Elements

of Electrical Engineering," third edition, p. 3G3. From this

__ -

D

-^

R

T

Cos0/

0.90

0.88

0.86

0.84.

0.82

/

X

X

/

\

\

1

r

\

\

20

00

3C

X)

40

P

00

50

00

60

30 V

/atts

FIG. 53. Power-factor Maximum of Induction Motor.

table corresponding values of power output P and power-

factor cos 6 are taken and plotted in a curve, Fig. 53, and the

maximum derived from this curve is P = 4120, cos = 0.904.

For the purpose of determining the maximum, obviously

not the entire curve needs to be calculated, but only a short

range near the maximum. This is located by trial. Thus

in the present instance, P and cos 6 are calculated for s = 0.1

and s = 0.2. As the latter gives lower power-factor, the maximum

power-factor is below s = 0.2. Then s = 0.05 is calculated and gives

a higher value of cos 6 than s = 0.1; that is, the maximum is

below s = 0.1. Then s = 0.02 is calculated, and gives a lower

value of cos 6 than s = 0.05. The maximum value of cos 6

thus lies between s = 0.02 and s = 0.1, and only the part of the

curve between s = 0.02 and s = 0.1 needs to be calculated for

the determination of the maximum of cos 6, as is done in Fig. 53.

99. When determining an extremum of a function y=f(x).

by plotting it as a curve, the value of x, at which the extreme

150 ENGINEERING MATHEMATICS.

occurs, is more or less inaccurate, since at the extreme the

curve is horizontal. For instance, in Fig. 53, the maximum

of the curve is so flat that the value of power P, for which

cos became a maximum, may be anywhere between P = 4000

and P = 4300, within the accuracy of the curve.

In such a case, a higher accuracy can frequently be reached

by not attempting to locate the exact extreme, but two points

of the same ordinatc, on each side of the extreme. Thus in

Fig. 58 the power PO, at which the maximum power factor

cos = 0.904 is reached, is somewhat uncertain. The value of

power-factor, somewhat below the maximum, cos # = 0.90,

is reached before the maximum, at P\ = 3400, and after the

maximum, at P 2 = 4840. The maximum then may be calculated

as half-way between PI and P 2 , that is, at P = i{Pi+P2! =

4120 watts.

This method gives usually more accurate results, but is

based on the assumption that the curve is symmetrical on

both sides of the extreme, that is, falls off from the extreme

value at the same rate for lower as for higher values of the

abscissas. Where this is not the case, this method of inter-

polation does not give the exact maximum.

Example 3. The efficiency of a steam turbine nozzle,

that is, the ratio of the kinetic energy of the steam jet to the

energy of the steam available between the two pressures between

which the nozzle operates, is given in Fig. 54, as determined by

experiment. As abscissas are used the nozzle mouth opening,

that is, the widest part of the nozzle at the exhaust end, as

fraction of that corresponding to the exhaust pressure, while

the nozzle throat, that is, the narrowest part of the nozzle, is

assumed as constant. As ordinates are plotted the efficiencies.

This curve is not symmetrical, 'but falls off from the maximum,

on the sides of larger nozzle mouth, far more rapidly than on

the side of smaller nozzle mouth. The reason is that with

too large a nozzle mouth the expansion in the nozzle is carried

below the exhaust pressure p2, and steam eddies are produced

by this overexpansion.

The maximum efficiency of 94.6 per cent is found at the point

PO, at which the nozzle mouth corresponds to the exhaust

pressure. If, however, the maximum is determined as mid-

way between two points PI and P 2 , on each side of the maximum,

MAXIMA AND MINIMA.

151

at which the efficiency is the same, 93 per cent, a point PO' is

obtained, which lies on one side of the maximum.

With unsymmetrical curves, the method of interpolation

thus does not give the exact extreme. For most engineering

purposes this is rather an advantage. The purpose of deter-

mining the extreme usually is to select the most favorable

operating conditions. Since, however, in practice the operating

conditions never remain perfectly constant, but vary to some

extent, the most favorable operating condition in Fig. 54 is

not that where the average value gives the maximum efficiency

88-I3

B*H

82^

80-

00

V

P'

Nozzle Opening

08 09 10

11

12

1.3

FIG. 54. Steam Turbine Nozzle Efficiency; Determination of Maximum.

(point P ), but the most favorable operating condition is that,

where the average efficiency during the range of pressure, occurr-

ing in operation, is a maximum.

If the steam pressure, and thereby the required expansion

ratio, that is, the theoretically correct size of nozzle mouth,

should vary during operation by 25 per cent from the average,

when choosing the maximum efficiency point PO as average,

the efficiency during operation varies on the part of the curve

between PI (91.4 per cent) and P 2 (85.2 per cent), thus averaging

lower than by choosing the point P</(6.25 per cent below PO)

as average. In the latter case, the efficiency varies on the

part of the curve from the Pi'(90.1 per cent) to P 2 '(90.1 per

cent). (Fig. 55.)

152

ENGINEERING MATHEMATICS.

Thus in apparatus design, when determining extrema of

a function y=f(x), to select them as operating condition,

consideration must be given to the shape of the curve, and

where the curve is unsymmetrical, the most efficient operating

point lies not at the extreme, but on that side of it at which

the curve falls off slower, the more so the greater the range of

variation is, which may occur during operation. This is not

always realized.

100. If the function y=f(x) is plotted as a curve, Fig.

50, at the extremes of the function, the points PI, P 2 , PS, P

of curve Fig. 50, the tangent on the curve is horizontal, since

90-?

88 -

Q_

80

OG

p;

&J

0.7

08

Nozzle Oper

09

mg

10

1.1

X*

1.2

FIG. 55. Steam Turbine Nozzle Efficiency; Determination of Maximum.

at the extreme the function changes from rising to decreasing

(maximum, PI and P^), or from decreasing to increasing (min-

imum, PS and P), and therefore for a moment passes through

the horizontal direction.

In general, the tangent of a curve, as that in Fig. 50, is the

line which connects two points P' and P" of the curve, which

are infinitely close together, and, as seen in Fig. 50, the angle

6, which this tangent P'P" makes with the horizontal or X-axis,

thus is given by:

tan 6 =

P"QJy

P'Q dx

MAXIMA AND MINIMA. 153

At the extreme, the tangent on the curve is horizontal,

that is, 4-0 = 0, and, therefore, it follows that at an extreme

of the function,

= (2)

dx

The reverse, however, is not necessarily the case; that is,

if at a point x, y : "p^O, this point may not be an extreme;

that is, a maximum or minimum, but may be a horizontal

inflection point, as points P 5 and P 6 are in Fig. 50.

With increasing x, when passing a maximum (Pi and P 2 ,

Fig. 50), y rises, then stops rising, and then decreases again.

When passing a minimum (P 3 and P 4 ) y decreases, then stops

decreasing, and then increases again. When passing a horizontal

inflection point, y rises, then stops rising, and then starts rising

again, at P 5 , or y decreases, then stops decreasing, but then

starts decreasing again (at PQ).

The points of the function y=f(x), determined by the con-

du

dition, j- = 0, thus require further investigation, whether they

represent a maximum, or a minimum, or merely a horizontal

inflection point.

This can be done mathematically: for increasing x, when

passing a maximum, tan 6 changes from positive to negative;

that is, decreases, or in other words, -7- (tan 0)<0. Since

tan 6 =-r, it thus follows that at a maximum -7-^ < 0. Inversely,

ctx dx

at a minimum tan 6 changes from negative to positive, hence

increases, that is, -7- (tan #)>0; or, -r-|>0. When passing

a horizontal inflection point tan 6 first decreases to zero at

the inflection point, and then increases again; or, inversely,

tan 6 first increases, and then decreases again, that is, tan =

~ has a maximum or a minimum at the inflection point, and

therefore, -7- (tan } = Tp = Q at the inflection point.

154 ENGINEERING MATHEMATICS.

In engineering problems the investigation, whether the

solution of the condition of extremes, ~f~. = 0, represents a

minimum, or a maximum, or an inflection point, is rarely

required, but it is almost always obvious from the nature of

the problem whether a maximum of a minimum occurs, or

neither.

For instance, if the problem is to determine the speed at

which the efficiency of a motor is a maximum, the solution

speed = 0, obviously is not a maximum but a mimimum, as at

zero speed the efficiency is zero. If the problem is, to find

the current at which the output of an alternator is a maximum,

the solution ^ = obviously is a minimum, and of the other

two solutions, i\ and 12, the larger value, 12, again gives a

minimum, zero output at short-circuit current, while the inter-

mediate value i\ gives the maximum.

101. The extremes of a function, therefore, are determined

by equating its differential quotient to zero, as is illustrated

by the following examples :

Example 4. In an impulse turbine, the speed of the jet

(steam jet or water jet) is Si. At what peripheral speed S 2 is

the output a maximum.

The impulse force is proportional to the relative speed of

the jet and the rotating impulse wheel; that is, to (81-82).

The power is impulse force times speed $2; hence,

(3)

? r>

and is an extreme for the value of $2, given by -TO- =0; hence,

Si-2S 2 = and & = ; ..... (4)

that is, when the peripheral speed of the impulse wheel equals

half the jet velocity.

Example 5. In a transformer of constant impressed

e.m.f. e = 2300 volts; the constant loss, that is, loss which

is independent of the output (iron loss), is P t . = 500 watts. The

internal resistance (primary and secondary combined) is r = 20

MAXIMA AND MINIMA. 155

ohms. At what current i is the efficiency of the transformer

a maximum; that is, the percentage loss, ^, a minimum?

ThelossisP = P t +n 2 = 500+20t 2 (5)

The power input is PI =ei = 2300i; .... (6)

Furthermore, in the sum of even harmonics, 1*2 may again

be separated from its second harmonic, i 4 , and its multiples,

and therefrom, ig, and its third harmonic, i' 6 , and its multiples,

thus giving all the harmonics up to the 9th, with the exception

of the 5th and the 7th. These latter two would require plotting

the curve and taking numerical values at different intervals,

so as to have a number of numerical values divisible by 5 or 7.

It is further recommended to resolve this unsymmetrical

exciting current of Table XI into the trigonometric series by

calculating the coefficients a n and &, up to the 7th, in the man-

ner discussed in paragraphs 6 to 8.

TABLE XI

9

io

8

io

e

io

d

io

+ 95.7

90

-26.7

180

-34.3

270

- 3.3

10

+ 78.7

100

-27.3

190

-27.3

280

- 1.8

20

+ 53.7

110

-28.1

200

-16.8

290

+ 1.2

30

+ 23.7

120

-28.8

210

-11.3

300

+ 4.7

40

- 2.3

130

-29.3

220

- 8.3

310

+ 10.7

50

-16.3

140

-29.8

230

- 7.3

320

+ 22.7

60

-22.8

150

-31

240

- 6.3

330

+ 41.7

70

-24.3

160

-32.6

250

- 5.3

340

+ 65.7

80

-25.8

170

-33.8.

260

- 4.3

350

+ 85.7

TRIGONOMETRIC SERIES. 139

D. CALCULATION OF TRIGONOMETRIC SERIES FROM

OTHER TRIGONOMETRIC SERIES.

94. An hydraulic generating station has for a long time been

supplying electric energy over moderate distances, from a num-

ber of 750-kw. 4400- volt 60-cycle three-phase generators. The

station is to be increased in size by the installation of some

larger modern three-phase generators, and from this station

6000 kw. are to be transmitted over a long distance transmis-

sion line at 44,000 volts. The transmission line has a. length

of 60 miles, and consists of three wires No. B.. & S. with 5

ft. between the wires.

The question arises, whether during times of light load the

old 750-kw. generators can be used economically on the trans-

mission line. These old machines give an electromotive force

wave, which, like that of most earlier machines, differs con-

siderably from a sine wave, and it is to be investigated, whether,

due to this wave-shape distortion, the charging current of the

transmission line will be so greatly increased over the value

which it would have with a sine wave of voltage, as to make

the use of these machines on the transmission line uneconom-

ical or even unsafe.

Oscillograms of these machines, resolved into a trigonomet-

ric series, give for the voltage between each terminal and the

neutral, or the Y voltage of the three-phase system, the equa-

tion :

e = e |sm0-0.12sin (30- 2. 3) -0.23 sin (50-1.5)

+0.13 sin (70-6.2)!. . (1)

In first approximation, the line capacity may be considered

as a condenser shunted across the middle of the line; that is,

half the line resistance and half the line reactance is in series

with the line capacity.

As the receiving apparatus do not utilize the higher har-

monics of the generator wave, when using the old generators,

their voltage has to be transformed up so as to give the first

harmonic or fundamental of 44,000 volts.

44,000 volts between the lines (or delta) gives 44,000 ~ Vs =

25,400 volts between line and neutral. This is the effective

140 ENGINEERING MATHEMATICS.

value, and the maximum value of the fundamental voltage

wave thus is: 25,400 X V2 = 36,000 volts, or 36 kv.; that is,

eo = 36, and

e = 36{sin 0-0.12 sin (30-2.3)-0.23 sin (50-1.5)

+ 0.13 sin (70-6.2)}, . (2)

would be the voltage supplied to the transmission line at the

high potential terminals of the step-up transformers.

From the wire tables, the resistance per mile of No. B. & S.

copper line wire is TO = 0.52 ohm.

The inductance per mile of wire is given by the formula :

L = 0.7415 log Y + 0.0805mh, .... (3)

IT

where l s is the distance between the wires, and l r the radius of

the wire.

In the present case, this gives l s = 5 ft. = 60 in. l r = . 1625 in.

LO = ! .9655 mh., and, herefrom it follows that the reactance, at

/= 60 cycles is

XQ = 27r/L = . 75 ohms per mile (4)

The capacity per mile of wire is given by the formula :

0.0408

hence, in the present case, C = 0.0159 mf., and the condensive

reactance is derived herefrom as :

x < - ^n~ = 16,6000 ohms ; (6)

ZTT/C o

60 miles of line then give the condensive reactance,

x c = |^=2770 ohms;

30 miles, or half the line (from the generating station to the

middle of the line, where the line capacity is represented by a

shunted condenser) give: the resistance, r = 30r = 16.6 ohms

TRIGONOMETRIC SERIES. 141

the inductive reactance, x=30x = 22.5 ohms, and the equiva-

lent circuit of the line now consists of the resistance r, inductive

reactance x and condensive reactance x c , in series with each

other in the circuit of the supply voltage e.

95. If i= current in the line (charging current) the voltage

consumed by the line resistance r is rL

The voltage consumed by the inductive reactance x is x -j

do

the voltage consumed by the condensive reactance x c is x c ( idO,

and therefore,

idO (7)

Differentiating this equation, for the purpose of eliminating

the integral, gives

de d 2 i di

or

. . (8)

The voltage e is given by (2), which equation, by resolving

the trigonometric functions, gives

e = 36 sin 6-4.32 sin 30-8.28 sin 50+4.64 sin 76

+0.18 cos 30+0. 22 cos 50-0. 50 cos 70; . (9)

hence, differentiating,

de

^ = 36 cos 0-12.96 cos 30-41.4 cos 50 + 32.5 cos 70

-0.54 sin 30-1.1 sin 50 + 3.5 sin 70. . (10)

Assuming now for the current i a tiigoriometric series with

indeterminate coefficients,

i = ai cos 0+aa cos 30+a 5 cos 50+ ay cos 70

H-&I sin 0+63 sin 30 +6 5 sin 50+6 7 sin 70, . (11)

142

ENGINEERING MATHEMATICS.

substitution of (10) and (11) into equation (8) must give an

identity, from which equations for the determination of a n and

b n are derived; that is, since the product of substitution must

be an identity, all the factors of cos 6, sin 0, cos 30, sin 30,

etc., must vanish, and this gives the eight equations :

36 =2770ai+ 15.66i- 22.5ai;

= 27706i- 15.6oi- 22.5&i;

-12.96 = 2770a 3 .+ 46.86 3 - 202. 5a 3 ;

- 0.54 = 27706 3 - 46.8o 3 - 202. 56 3 :

-41.4 = 2770a 5 + 786 5 - 562. 5a 5 ;

1.1 =27706 5 - 78a 5 - 56.25& 5 ;

32.5 = 2770a 7 + 109.2& 7 -1102.5a 7 ;

3.5 = 27706 7 - 109. 2a 7 - 1102.56 7 . J

Resolved, these equations give

ai= 13.12;

61= 0.07;

a 3 = - 5.03;

6 3 = - 0.30;

a 5 = -18.72;

65= 1.15;

a 7 = 19.30;

6 7 = 3.37;

hence,

i = 13. 12 cos 0-5. 03 cos 30- 18. 72 cos 50 + 19. 30 cos 70

+0.07 sin 0-0.30 sin 30-1.15 sin 50+3.37 sin 70

= 13.12 cos (0-0.3)-5.04 cos (30-3.3)

-18. 76 cos (50- 3. 6) +19. 59 cos (70-9.9).

(12)

(13)

.(14)

TRIGONOMETRIC SERIES. 143

96. The effective value of this current is given as the square

root of the sum of squares of the effective values of the indi-

vidual harmonics, thus :

As the voltage between line and neutral is 25,400 effective,

this gives Q = 25,400X21. 6 = 540,000 volt-amperes, or 540 kv.-

amp. per line, thus a total of 3Q = 1620 kv.-amp. charging cur-

rent of the transmission line, when using the e.m.f. wave of

these old generators.

It thus would require a minimum of 3 of the 750-kw.

generators to keep the voltage on the line, even if no power

whatever is delivered from the line.

If the supply voltage of the transmission line were a perfect

sine wave, it would, at 44,000 volts between the lines, be given

by

ei = 36sin 0, (15)

which is the fundamental, or first harmonic, of equation (9).

Then the current i would also be a sine wave, and- would be

given by

i\ = a\ cos +bi sin 6,

-13.12 cos 6 +0.07 sin 0,

= 13.12 cos (0-0.3),

and its effective value would be

13.12

/i = =- = 9.3 amp (17)

This would correspond to a kv.-amp. input to the line

3d = 3 X 25.4 X 9.3 = 710 kv.-amp.

The distortion of the voltage wave, as given by equation (1),

thus increases the charging volt-amperes of the line from 710

(16)

144

ENGINEERING MATHEMATICS.

kv.-amp. to 1620 kv.-amp. or 2.28 times, and while with a sine

wave of voltage, one of the 750- kw. generators would easily be

able to supply the charging current of the line, due to the

FIG. 47.

wave shape distortion, more than two generators are required.

It would, therefore, not be economical to use these generators

on the transmission line, if they can be used for any other

purposes, as short-distance distribution.

FIG. 48.

In Figs. 47 and 48 are plotted the voltage wave and the

current wave, from equations (9) and (14) repsectively, and

TRIGONOMETRIC SERIES. 145

the numerical values, from 10 deg. to 10 deg., recorded in

Table XII.

In Figs. 47 and 48 the fundamental sine wave of voltage

and current are also shown. As seen, the distortion of current

is enormous, and the higher harmonics predominate over the

fundamental. Such waves are occasionally observed as charg-

ing currents of transmission lines or cable systems.

97. Assuming now that a reactive coil is inserted in series

with the transmission line, between the step-up transformers

and the line, what will be the voltage at the terminals of this

reactive coil, with the distorted wave of charging current

traversing the reactive coil, and how does it compare with the

voltage existing with a sine wave of charging current?

Let L= inductance, thus x = 2nfL= reactance of the coil,

and neglecting its resistance, the voltage at the terminals of

the reactive coil is given by

Substituting herein the equation of current, (11), gives

sin 6+30,3 sin 30+5a 5 sin 50+7a 7 sin 76 }

(19)

hence, substituting the numerical values (13),

135.1 sin Id }

.6 cos 76\

(20)

x{ 13.12 sin (9-15.09 sin 30-93,6 sin 56 + 135.1 sin 76 }

-0.07 cos 0+0.90 cos 36 +5.75 cos 50-23.6 cos 76}

= x\ 13.12 sin (0-0.3) -15.12 sin (30-3.3)

-93.8 sin (50 -3.6) +139.1 sin (70-9.9)}.

This voltage gives the effective value

while the effective value with a sine wave would be from (17),

hence, the voltage across the reactance x has been increased

12.8 times by the wave distortion.

146

ENGINEERING MATHEMATICS.

The instantaneous values of the voltage e f are given in the

last column of Table. XII, and plotted in Fig. 49, for z = l.

As seen from Fig. 49, the fundamental wave has practically

FIG. 49.

vanished, and the voltage wave is the seventh harmonic, modi-

fied by the fifth harmonic.

TABLE XII

e

c

t

e'

e

i

<-'

10

20

-0.10

+ 2.23

3.74

+ 8.67

+ 5.30

- 0.86

- 17

+ 46

+ 3

90

100

110

27.41

31.77

40.57

- 4.15

+ 26.19

+ 24.99

-200

-106

+ 119

30

40

50

7.47

17.35

31.70

+ 7.39

+ 30.39

+ 38.58

+ 131

-116

+ 36

120

130

140

42.70

33.14

18.03

- 8.10

-38.79

-36.65

+ 182

+ 93

- 96

60

70

80

42.06

40.33

32.87

+ 15.66

-19.01

-29.13

+ 167

+ 159

- 54

150

160

170

6.99

2.88

1.97

-13.41

+ 2.43

- 1.00

-138

- 31

+ 54

90

27.41

- 4.15

-200

180

+ 0.10

- 8.67

+ 17

CHAPTER IV.

MAXIMA AND MINIMA.

98. In engineering investigations the problem of determin-

ing the maxima and the minima, that is, the extrema of a

function, frequently occurs. For instance, the output of an

electric machine is to be found, at which its efficiency is a max-

imum, or, it is desired to determine that load on an induction

motor which gives the highest power-factor; or, that voltage

V

FIG. 50. Graphic Solution of Maxima and Minima.

which makes the cost of a transmission line a minimum; or,

that speed of a steam turbine which gives the lowest specific

steam consumption, etc.

The maxima and minima of a function, y=f(x), can be found

by plotting the function as a curve and taking from the curve

the values x, y, which give a maximum or a minimum. For

instance, in the curve Fig. 50, maxima are at PI and P 2 , minima

at P 3 and P 4 . This method of determining the extrema of

functions is necessary, if the mathematical expression between

147

148

ENGINEERING MATHEMATICS.

x and y, that is, the function y=f(x), is unknown, or if the

function y=f(x) is so complicated, as to make the mathematical

calculation of the extrema impracticable. As examples of

this method the following may be chosen:

10

2-

10 12 14 16 18 20 22 24

28 30

FIG. 51. Magnetization Curve.

Example i. Determine that magnetic density (B, at which

the permeability n of a sample of iron is a maximum. The

relation between magnetic field intensity 3C, magnetic density

05 and permeability /* cannot be expressed in a mathematical

equation, and is therefore usually given in the form of an

1400

^

*^~

^ "

KSr^

X

100TV

/

^

N

\

.X

/

x

\

-600-

z

/

\

\

f

\

7^00

$>

i

J !

*

i

> (

{ t

!

) 1

1

Kilo-lines

1 1^ 13 1

4 1

r >

FIG. 52. Permeability fcurve.

empirical curve, relating (B and 3C, as shown in Fig. 51. From

this curve, corresponding values of (B and 3C are taken, and their

/D

ratio, that is, the permeability / = , plotted against as abscissa.

uC

This is done in Fig. 52. Fig. 52 then shows that a maximum

MAXIMA AND MINIMA.

149

occurs at point fj^ &x , for (B = 10.2 kilolines, /* = 1340, and minima

at the starting-point P 2 , for (B = 0, ju = 370, and also for (B=oo ,

where by extrapolation /* = !.

Example 2. Find that output of an induction motor

which gives the highest power-factor. While theoretically

an equation can be found relating output and power-factor

of an induction motor, the equation is too complicated for use.

The most convenient way of calculating induction motors is

to calculate in tabular form for different values of slip s, the

torque, output, current, power and volt -ampere input, efficiency,

power-factor, etc., as is explained in " Theoretical Elements

of Electrical Engineering," third edition, p. 3G3. From this

__ -

D

-^

R

T

Cos0/

0.90

0.88

0.86

0.84.

0.82

/

X

X

/

\

\

1

r

\

\

20

00

3C

X)

40

P

00

50

00

60

30 V

/atts

FIG. 53. Power-factor Maximum of Induction Motor.

table corresponding values of power output P and power-

factor cos 6 are taken and plotted in a curve, Fig. 53, and the

maximum derived from this curve is P = 4120, cos = 0.904.

For the purpose of determining the maximum, obviously

not the entire curve needs to be calculated, but only a short

range near the maximum. This is located by trial. Thus

in the present instance, P and cos 6 are calculated for s = 0.1

and s = 0.2. As the latter gives lower power-factor, the maximum

power-factor is below s = 0.2. Then s = 0.05 is calculated and gives

a higher value of cos 6 than s = 0.1; that is, the maximum is

below s = 0.1. Then s = 0.02 is calculated, and gives a lower

value of cos 6 than s = 0.05. The maximum value of cos 6

thus lies between s = 0.02 and s = 0.1, and only the part of the

curve between s = 0.02 and s = 0.1 needs to be calculated for

the determination of the maximum of cos 6, as is done in Fig. 53.

99. When determining an extremum of a function y=f(x).

by plotting it as a curve, the value of x, at which the extreme

150 ENGINEERING MATHEMATICS.

occurs, is more or less inaccurate, since at the extreme the

curve is horizontal. For instance, in Fig. 53, the maximum

of the curve is so flat that the value of power P, for which

cos became a maximum, may be anywhere between P = 4000

and P = 4300, within the accuracy of the curve.

In such a case, a higher accuracy can frequently be reached

by not attempting to locate the exact extreme, but two points

of the same ordinatc, on each side of the extreme. Thus in

Fig. 58 the power PO, at which the maximum power factor

cos = 0.904 is reached, is somewhat uncertain. The value of

power-factor, somewhat below the maximum, cos # = 0.90,

is reached before the maximum, at P\ = 3400, and after the

maximum, at P 2 = 4840. The maximum then may be calculated

as half-way between PI and P 2 , that is, at P = i{Pi+P2! =

4120 watts.

This method gives usually more accurate results, but is

based on the assumption that the curve is symmetrical on

both sides of the extreme, that is, falls off from the extreme

value at the same rate for lower as for higher values of the

abscissas. Where this is not the case, this method of inter-

polation does not give the exact maximum.

Example 3. The efficiency of a steam turbine nozzle,

that is, the ratio of the kinetic energy of the steam jet to the

energy of the steam available between the two pressures between

which the nozzle operates, is given in Fig. 54, as determined by

experiment. As abscissas are used the nozzle mouth opening,

that is, the widest part of the nozzle at the exhaust end, as

fraction of that corresponding to the exhaust pressure, while

the nozzle throat, that is, the narrowest part of the nozzle, is

assumed as constant. As ordinates are plotted the efficiencies.

This curve is not symmetrical, 'but falls off from the maximum,

on the sides of larger nozzle mouth, far more rapidly than on

the side of smaller nozzle mouth. The reason is that with

too large a nozzle mouth the expansion in the nozzle is carried

below the exhaust pressure p2, and steam eddies are produced

by this overexpansion.

The maximum efficiency of 94.6 per cent is found at the point

PO, at which the nozzle mouth corresponds to the exhaust

pressure. If, however, the maximum is determined as mid-

way between two points PI and P 2 , on each side of the maximum,

MAXIMA AND MINIMA.

151

at which the efficiency is the same, 93 per cent, a point PO' is

obtained, which lies on one side of the maximum.

With unsymmetrical curves, the method of interpolation

thus does not give the exact extreme. For most engineering

purposes this is rather an advantage. The purpose of deter-

mining the extreme usually is to select the most favorable

operating conditions. Since, however, in practice the operating

conditions never remain perfectly constant, but vary to some

extent, the most favorable operating condition in Fig. 54 is

not that where the average value gives the maximum efficiency

88-I3

B*H

82^

80-

00

V

P'

Nozzle Opening

08 09 10

11

12

1.3

FIG. 54. Steam Turbine Nozzle Efficiency; Determination of Maximum.

(point P ), but the most favorable operating condition is that,

where the average efficiency during the range of pressure, occurr-

ing in operation, is a maximum.

If the steam pressure, and thereby the required expansion

ratio, that is, the theoretically correct size of nozzle mouth,

should vary during operation by 25 per cent from the average,

when choosing the maximum efficiency point PO as average,

the efficiency during operation varies on the part of the curve

between PI (91.4 per cent) and P 2 (85.2 per cent), thus averaging

lower than by choosing the point P</(6.25 per cent below PO)

as average. In the latter case, the efficiency varies on the

part of the curve from the Pi'(90.1 per cent) to P 2 '(90.1 per

cent). (Fig. 55.)

152

ENGINEERING MATHEMATICS.

Thus in apparatus design, when determining extrema of

a function y=f(x), to select them as operating condition,

consideration must be given to the shape of the curve, and

where the curve is unsymmetrical, the most efficient operating

point lies not at the extreme, but on that side of it at which

the curve falls off slower, the more so the greater the range of

variation is, which may occur during operation. This is not

always realized.

100. If the function y=f(x) is plotted as a curve, Fig.

50, at the extremes of the function, the points PI, P 2 , PS, P

of curve Fig. 50, the tangent on the curve is horizontal, since

90-?

88 -

Q_

80

OG

p;

&J

0.7

08

Nozzle Oper

09

mg

10

1.1

X*

1.2

FIG. 55. Steam Turbine Nozzle Efficiency; Determination of Maximum.

at the extreme the function changes from rising to decreasing

(maximum, PI and P^), or from decreasing to increasing (min-

imum, PS and P), and therefore for a moment passes through

the horizontal direction.

In general, the tangent of a curve, as that in Fig. 50, is the

line which connects two points P' and P" of the curve, which

are infinitely close together, and, as seen in Fig. 50, the angle

6, which this tangent P'P" makes with the horizontal or X-axis,

thus is given by:

tan 6 =

P"QJy

P'Q dx

MAXIMA AND MINIMA. 153

At the extreme, the tangent on the curve is horizontal,

that is, 4-0 = 0, and, therefore, it follows that at an extreme

of the function,

= (2)

dx

The reverse, however, is not necessarily the case; that is,

if at a point x, y : "p^O, this point may not be an extreme;

that is, a maximum or minimum, but may be a horizontal

inflection point, as points P 5 and P 6 are in Fig. 50.

With increasing x, when passing a maximum (Pi and P 2 ,

Fig. 50), y rises, then stops rising, and then decreases again.

When passing a minimum (P 3 and P 4 ) y decreases, then stops

decreasing, and then increases again. When passing a horizontal

inflection point, y rises, then stops rising, and then starts rising

again, at P 5 , or y decreases, then stops decreasing, but then

starts decreasing again (at PQ).

The points of the function y=f(x), determined by the con-

du

dition, j- = 0, thus require further investigation, whether they

represent a maximum, or a minimum, or merely a horizontal

inflection point.

This can be done mathematically: for increasing x, when

passing a maximum, tan 6 changes from positive to negative;

that is, decreases, or in other words, -7- (tan 0)<0. Since

tan 6 =-r, it thus follows that at a maximum -7-^ < 0. Inversely,

ctx dx

at a minimum tan 6 changes from negative to positive, hence

increases, that is, -7- (tan #)>0; or, -r-|>0. When passing

a horizontal inflection point tan 6 first decreases to zero at

the inflection point, and then increases again; or, inversely,

tan 6 first increases, and then decreases again, that is, tan =

~ has a maximum or a minimum at the inflection point, and

therefore, -7- (tan } = Tp = Q at the inflection point.

154 ENGINEERING MATHEMATICS.

In engineering problems the investigation, whether the

solution of the condition of extremes, ~f~. = 0, represents a

minimum, or a maximum, or an inflection point, is rarely

required, but it is almost always obvious from the nature of

the problem whether a maximum of a minimum occurs, or

neither.

For instance, if the problem is to determine the speed at

which the efficiency of a motor is a maximum, the solution

speed = 0, obviously is not a maximum but a mimimum, as at

zero speed the efficiency is zero. If the problem is, to find

the current at which the output of an alternator is a maximum,

the solution ^ = obviously is a minimum, and of the other

two solutions, i\ and 12, the larger value, 12, again gives a

minimum, zero output at short-circuit current, while the inter-

mediate value i\ gives the maximum.

101. The extremes of a function, therefore, are determined

by equating its differential quotient to zero, as is illustrated

by the following examples :

Example 4. In an impulse turbine, the speed of the jet

(steam jet or water jet) is Si. At what peripheral speed S 2 is

the output a maximum.

The impulse force is proportional to the relative speed of

the jet and the rotating impulse wheel; that is, to (81-82).

The power is impulse force times speed $2; hence,

(3)

? r>

and is an extreme for the value of $2, given by -TO- =0; hence,

Si-2S 2 = and & = ; ..... (4)

that is, when the peripheral speed of the impulse wheel equals

half the jet velocity.

Example 5. In a transformer of constant impressed

e.m.f. e = 2300 volts; the constant loss, that is, loss which

is independent of the output (iron loss), is P t . = 500 watts. The

internal resistance (primary and secondary combined) is r = 20

MAXIMA AND MINIMA. 155

ohms. At what current i is the efficiency of the transformer

a maximum; that is, the percentage loss, ^, a minimum?

ThelossisP = P t +n 2 = 500+20t 2 (5)

The power input is PI =ei = 2300i; .... (6)

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