Charles Proteus Steinmetz.

# Engineering mathematics; a series of lectures delivered at Union college online

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Online LibraryCharles Proteus SteinmetzEngineering mathematics; a series of lectures delivered at Union college → online text (page 9 of 17)
Font size 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)

Online LibraryCharles Proteus SteinmetzEngineering mathematics; a series of lectures delivered at Union college → online text (page 9 of 17)