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RAM FREQUEHY HEATERS
APPLEED Ti} MAGNETE AW NW MRENEUC
EWNERWAL EDHQEENBRS

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ALDO A. CACAVELOS
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3 can‘t

 

 

 

This is to certify that the
thesis entitled

Radio Frequency Heating
Applied to Magnetic and Non Magnetic

Cylindrical Conductors
presented by

Aldo A. Cacavelos

has been accepted towards fulfillment
of the requirements for ‘

£4080 degree in EOE.

 

Date May 14; 1948

THESKS

RADIO FREQUENCY HEATING APPLIED T0
MAGNETIC Ann non MAGNETIC cmmucm.
commons.

By
ALDO A. CACAVEIDS.

A THESIS
Submitted to the wheel-of Graduate Studies ..
of Michigan State College of Agriculture and"
Applied Science in partial fulfillment of the
requirements for the degree of

MASTER OF SCIENCE.

Department of Electrical Eng.
1948

THESIS

The objective of this thesis is the study of the
application of induction heating principles to the
heating of several metallic pieces to obtain the desired
results. By means of the formulas that will be presented
later, the circuit requirements are calculated and
suitable coils designed to accomplish a determined
purpose, namely, to heat the material up to Tdegrees
Farenheit in t seconds.

One of the sources of trouble in induction heating
is the variation of the resistivity of metals with
temperature. This effect causes the load power to vary
and therefore the reflected impedance due to the load,
and the Q of the work coil. As a consequence, the Q of
the-tank circuit, the tank voltage, and current change
too. This can be best understood by inspection of the

circuit diagram.

 

 

03
We
W

Considering the circuit shown as equivalent to

the actual circuit, we can write

I \l\

 

x. ;;I:
Q —— —- ——~ w-w— —-—---——~~—~ --~ ..;v
work Metal
coil to be heated.
2
R9 = R1 . WLEEI (1)

9
R2” e (WL2)2

X _ x2r<wm2

X 3
Egg 0 (WL2)2

e

 

Therefore, there is always an increase in
resistance due to the secondary circuit and a decrease
in reactance. It must be well understood that the
original reactance, 22, to be considered, has the same
magnetic circuit as the actual one with the only
difference that the secondary circuit is open circuited.

Thus, in the case of heating a cOpper tubing, Xé
is measured with the tubing inside the coil, and X1
without the tubing. The experiment shows that Xe and
Re are 1.93.11-and 0.161~le X1 and R1 being 2.#l and
0.160JL. ‘

The problems dealing with magnetic loads are

somewhat more complicated, although the variation of R

follows the same behavior. In this case El would be the
coil with the magneticlc core inside but not allowing any
secondary current to flow. This is certainly not the
same reactance as that measured with the coreless coil
and can be thought of as the same coil with a laminated
core of the same permeability . This way, starting from
the coreless coil reactance § we can obtain Xe in the
following steps:

1) - Coreless coil X5

2) - Same coil with ideal laminated: core

X1' X0)”

3) - Same coil with actual core:

X2 (WM)2 X (WM)2
x = x - . - x - 2
e "1 R22, X2? 0/1 m2 0 X2

 

The following eXperiments show that Xe :2 x0 and of course
that Re 7- Bl.

In all the problems on induction heating, the
variation of the resistivity of the load plays an
important part. While the computations on'r - f - power
deal with the resistivity that corresponds to the aver-
age temperature of the material to be heated, the design
of the tank circuit must be made considering the minimum
Q, that is the maximum resistance R6 of the circuit.
Formula (1) says that this corresponds to the maximum
resistivity of the core. During the Operation, changes
in R2 due to the temperature, produce changes in Re and

therefore in the Q of the main tank. This, in turn,

causes the tank voltage and current to change and as a
consequence the D.C. plate voltage must be adjusted
continuously to keep a constant tank current and so a
constant magnetizing force._ All this will be illustrated

'later with the aid of vector diagrams.

CALCULATION OF REQUIRED MAGNEI;ZING FORGE

"Radio Frequency Heating" by Brown, Hoyles and
Blerwith, on page 29, gives the following general
formula for heating cylindrical conductors.

«. 2 ?“ H~ ‘ y‘ ;« c.g.s. rationalized

- “ g -—-T practical system of
I} ' “: units.

 

When: P- watts per unit length

HO : magnetizing force of the coil without
the core.

sf = conductivity. .
(o - radius of cylinder

skin depth in the cylinder

F = a function of several Bessel's
functions plotted in the following graph:

!n l

-- » _7»1'

it.

.’ 9
l l

’ 1

 

4‘
'

.. ‘ ‘45,}. .3 ~ , CI /
4 + - - - —w_::. "I / ‘_“

 

C:
In all the experiments performed in connection with

this thesis a/s was much higher than 5 and so we can set

F equal to one.

P = §;E;Eo3 , .2. watts per cm length.
0.

The power density (watts per cm cube), becomes:

- . 2 2
(1) PD = §417EQ_ .—§« 3 §_EQ__ watts/cm3

 

 

V ’1 n" S ’1‘ a S
S - ,,r g-- cm (Brown's page 17)
214/109“. .. t {
Therefore;-
" ~ .‘e' --L -. :._____
. ‘- q . f 1 ’ .\ ll; \ . .'.
(3) P.D = .________/;.--—-— \/ ,,‘ , .1 .‘ w. ‘1.“ . ..._____J__,r___ _.,.. ‘ q \‘j in; M "/

From here on we will analyze separately the

magnetic and non-magnetic materials.

MAGNETIC MATERIALS
By definition

Therefore:

5 .
‘1‘/‘

I

But since induction heating is accomplished under
sa turation conditions -

B = constant
and: A. ,lfiiu Where i1

;,u~'
Hr ~' A“). — H Iv‘t‘

 

Therefore: P.D = -;5‘ “:sfi?4*"*- J

where P.is the resistivity of the
material

and 2 I‘ x ‘ f )1. L}

The variation of P with the temperature causes the energy
input to vary continuously so that the total energy input

per cubic cm is: 7

Assuming now that the temperature increases linearly
with time and neglecting the initial value, we can write:
tO - K1 t sec.

Assuming also that the resistivity increases with temper-
ature following a straight line through the origin, we
can write:

I" 3 K11 t° " K1 ’5 sec

Where K1 8 K11 x K1
This assumption is not very far from the actual condflsions
as can be seen from the corresponding graph of}? =&.(t0)
Fig. (1). Besides, as far as power computations are
concerned, it yields acceptable results since the main
contributions of energy are effected at higher values
of P when it approaches the straight line closely enough.

Under these conditions we can write:

.
"mm... A- sh A . if y». .._,_ _.

;. '7
“ 4", ‘ \H- V AM"

a
.
, - .1
v

But Pmax = Kl Tsec' Therefore:
rsp_ii
E = G. qumax _§_ Tsec

If‘P is a linear function of temperature, as has been

assumed, then:

 

 

 

Pmax ' 2 Paverage

Therefore: W
' w ' , 3/2 ' “

E - 6XO°9H5'g/Pave Tsec. ‘ 0°9HSK H0 1(PavetiTsecs
However:

E 3 (PD)ave x Tsecs.
So: 3/2 fl_

= 0.9%; Kng ’JTP
Hence:
2/3
HO = E (P°D)ave a ; cgs rationalized
r: . s ste of units
(Oo9li-S.K. \V’J Pave) “ y m

 

The value of K = 159 x 10'5‘. depends upon the

so

magnetic induction of the material at saturation and,
of course, it changes if we change the system of units.
For ferrous materials, "Electronic for Industry"

by Bendz gives: /
2 3
_ PD x d
Ho ' E 3

..-. .—

( 001*38 \r ,’ Pave )

oersteds

Where:
' ~ watts/cubic inch

P D

diameter in inches

resistivity in In inches

”'0
II

HO peak magnetizing force (oersteds)

f‘
B

relative permeability

magnetic induction (Gausses)

The condition
§.% 7/ Lv.25 ( to make F .3} 1) expressed

in these units becomes
6. inches 3d70 \/‘
sginches 5170 \J pm """' Z 1525
ff

PM
d-JflL/D 23:. 13h00

The value of//« Pto Mbe used in these formulas was found

from experimentation (Bendz's, page 298) to be:

 

 

/K = 1.8 Bfizaturation ' 1.8 x 18000 = 32u00
Ho «H

NON-MAGNETIC MATERIALS

Letting /pnf equal unity, equation (3) becomes:

= 159 x 10-5 x H02
P.D a “f)xi

Under the same assumptions done before, the following

 

final formula canbe obtained (Bendz's 30%)

P D x d
H —~— t
0 =Jz. 59 x 10-3137" ““3 “5

 

 

Where, as before:

diam. in inches

resistivity, .A. inches

f = frequency, cpo.

The condition g C? H.25 becomes:

Hw._..—.—_—_-——-
'_’

’ I
I -\ ,

L ' ‘-‘ '4. - L... "3
// , l / g I
v -- .“__.____ ,

I r, -‘

SELECTION OF FREQUENCY
The original formula
.7) _‘ --- .. i 3 T 3333

and the plot of F as a function of 3 shown before, show
that increasing 3 in the range O-2.25(by increasing
frequency) causes a high increment of F and thereforeP.

When § reaches 2.25, F reaches 0.8 and approaches
saturation. Therefore, we can define a minimum fre-
quency (for g - 2.25) for which the coupling of energy
is efficient.

In fact, let's analyze the variation of the ratio
"power absorbed in the core to power lost in the exciting

solenoid" as a function of F.

Bower core
E=IZR
solenoid

 

But

Rsolenoid = ‘2‘ 5° (Brown's, page 22)

 

Besides from (2)

(‘_ " Where the A's are constant
' ~ for a given material

Therefore:

 

 

Rsole A2 3} f
Similarly:
a core = A ”“E_
3 core 3 i!
So:
R801. : Ana core
5 core
and .
Power core Power core
L 12A“ a core = A a core H 2
5 core 5“ 3 core 0
substituting E 2
o
and from ( H ) gives:
E = L;, F If g > 2.25 F 35 constant
Therefore i“ 5 constant.

With higher values of frequency (a/s 3’ 2.25) we
couple, of course, more energy into the core (for same
H0) but the ratio E is very nearly constant and so is,
of course, the efficiency.

The minimum frequency for efficient heating is,

therefore:

 

So 3 = a 2ff./'.L~VTV;” - 2.25 (cgs rationalized
. / ' ‘ units)

lg8.5x 106

, . 8 ‘
1 mln q )1\,1
I \

In the Specific case of the piece of iron that
will be heated in the lst experiment we have:

’—

*;- = 26000 A_L_L (to obtain the higher
min

’0

..—-.\- -..'...-.-A. 3; min)

«a 938
d O. 5" = 1.25

Therefore:

5 min 938 x 1.55

The critical frequency to heat a 0.5" diam. c0pper'

cylinder is: . .3; ,u_ w , _ t

. ~ .-' ‘ \
j .; . ' t '1
. 3 ‘ I '- fl .
J I ‘ .\ ’1) Ir] ./ '3 J.,' I I I

The equipment used to conduct the experiments
has a frequency of 350 KC, thus obtaining the highest
possible efficiency for a given material and work
coil. The object of using so high a frequency is to

‘couple more power into the core for a given circuit.

EQUIPMENT AND‘VECTORIAL DIAGRAMS

The equipment used during the experiments was a
1 KW, 20 amp. tank current, westinghouse Generator.
The equipment is so designed that we can obtain two
different frequencies by changing the connections.

a) Hartly_ circuit - For dielectric heating
with a frequency of 10 M c/s.

b) Feedback oscillator - For induction circuit
with a rated frequency of 300 KB.

L11-

The experiments were run with the oscillator set
in (b) and the frequency was checked to be 350 KC.
Essentially the circuit is as shown below:

‘ !.L»
I r- «' 1‘
.——-L——__—_

1 "1 1 ‘ '
/” '-

‘1;
',.. .
y a
‘-

 

 

The ectorial diagram can be drawn for the tank

circuit as follows:

 

 

t"

When the material to be heated is introduced in the work
coil,it has been found that Q decreases due to the
reflected resistance as has already been explained.
This produces a slight shift in the vector diagram but
the main effect is that the current 31_ decreases.

In fact, the impedance of the tank circuit at

resonance is:
so, if R increases RT
decreases and, for '

, 7 , H the same plate current,

‘1 p .; 1E7 decreases too. This
" 3‘ causes I, and I, to
decrease: ”

When the material is being heated the reflected
resistance changes continuously due to the variation
of resistivity of the core with temperature which causes
the above mentioned change to take place. As a result,
if a constant magnetizing force (constant IL) is
desired, the anode supply voltage Ebb must be
continuously adjusted.

The measurements of temperature were made with
two iron constanton thermocouples whose calibration
curves are attached, Fig.(2). The lst couple was used

during the lst experiment and the 2nd one was used in

the others.

-13-

 

 

HEATING §E_VERAL SAMPLES WOF LOW CARBON STEEL

-11+-

The following specific heat and resistivity data
was obtained from the "Handbook for Metallurgical

Engineers".

fiflEDL§H IRON

RESISTIVIE:

9;; 3 HEAT DENSITY

68 .1075 70°F - 7.85

77 .108 600 - 7.78 00 0F P
10h .1097 1200 - 7.63 0 32 13.2
140 .112 100 212 20.1
212 .11k 200 390 27.4
302 .121 300 570 36.
392 .128 #00 750. #6.2
M82 .138 500 930 58.3
572 .1no 600 1110 72.
752 .151

932 .163

1022 .172

1112 .188

1200 .208

So the following average specific heat values were
obtained for different ranges:
700F -
7O - 200 - .111

.1075 BTU/pound 0F a small c014 ,.“

I
1

7O "' 600 " 012)”?
70 - 1200- olh’3

-15;

EXPERIMENT I

Piece of iron (0.2% Carbon) 5.25" long,
0.5" diam. to be heated from 70°F up to
600°F in 1 minute.

1) Calculation of Thermal power required:

TP I 17.6 MCT\T where:

M I rate of material to be heated per minute
(in pounds/minute).

C

average specific heat in BTU/pound 0F

T temperature variation in 0F

In this case it is:

\‘ t- (. = 1003 C11. inCh
o . g; . ound
Density at 70 F 7.85 cm3 .28# cu . nch
Hence:
M = 1.03 x 0.28%.: 0.292 ggggge
Besides

C70 - 600 0F = 0.12%
Therefore:

Thermal power = 17.6 x 0.292 x 0.12% x 530
- 338 watts.

2) Radiation, convection and conduction losses.

piece

Conduction losses are very small because the

is held by two 0.1" diam. pivots fixed one at

each end.

The radiation and convection lossess are less

than 10 watts (from Bendz's, page 301) and therefore

are negligible.

-16...

3) Power density.

P.D 3 lgtal input (watts) a 338 : 328 Watts

Volume of metal in 1.03 Cub.inch
coil

h) Peak magnetizing force:
The corresponding formula for a solid magnetic

cylinder is:

 

H -’ PD x d I 2 valid where d .
peak - ( 0.1.‘38 I" t ) /3 r * /
Where H - c.f«fi.
d a diameter (inches)
« - relative permeability
- resistivity at average temp (3350F) =
;}= frequency in C/S = 350,000 0/3

As already explained, the permeability to be used in

these formulas is:

/*= - 32—899

The magnetizing force therefore becomes

I

4‘
h

Wa - " 2/3
H a K 0. 3 ‘/3 x 10J x 10 x 10“°‘= 200 =
V .

peak

 

— 38.5

v

The condition

938

 

is satisfied.

5) C011 design:

The coil was designed to be 6.7" long so as to
cover the material thoroughly. With a 19 min. interior
diameter this allowed a clearance of 32 mm around the
work piece. The coil was made of 19 turns of copper
tubing 6 +v>n diameter. The outside diameter became
31

The sketch below shows the coil with the iron

piece inside:k;___ ‘( n

.4; [21l_l-5 VI
. ____“l« 7 . . a»
g a
: l

'1

 

 

 

 

“AL—.—
6) Coil current.

The values of H in ;\1».‘,, and I in amperes

are related by the following formula:

, _

( '\\\\‘ > " Ltl lull. )

0-» aw «mm.

 

If H is to represent the peak magnetizing force and I

is expressed in terms of (anq ) of a '.,n: wave, then:

Therefore:

 

To take account of the coil-core spacing the factor

K, is added:

I ' ‘ . v ~
I J 4 I‘ 4... . . . >‘ I
v . - n ~

.1: ' \’ QM“ ‘~-, (Bendz 312)

 

coil length
Spacing

 

Kl being a function of

In this example the current will be:

 

Where for a %%§§I%g ratio equal to 45, K1 is very
nearly 1.02. Hence:

I 8 amp. 18.0
As the rating of the equipment to be used is 1 KW,
20 amp., this experiment can be run without further

complications.

The table shows the experimental results where the
values of temp, I and V were measured, thus allowing

the calculation of the impedance to be made:

-19-

gABLE I
Temp.(over
room temp.

 

Th.coup1e
; Secs. I 31:1 OF v z
350 KC 10 as 2.5 95 61 3.3
2 20. 5 :30 23 3.9
O 1 Temp. in 1
" 35 gig 33(5) 3? 3. 37 minute -
"- 29 ll #20 66 515 o 70 :
" 0 13—3 51—5 6673 . 9 585°F
" .90 18.2 .700 69. 3.73

 

To separate both, the resistive and reactive
components of Z , measmurements of the Q of the
circuit were done by means of a Q meter. The material
was heated up to a certain temperature and quickly
connected to the Q meter and disconnected from the
heating equipment. The same operation was repeated
at different maximum temperatures.

The following data was obtained:

 

 

 

 

 

 

TABLE II
0F over
room. -_ -,
r _ temp. 0 ‘*—--- -11., i.—
350 KC 70 9 ”Qt" s” .2? J 3 fl" J o) 5'
I! 2 O 5 L" ~..- 4, --./
" ‘h 0 L; 1 u__1
" ‘ 700 I... S. .-..__. .
" 800 3.3
" ‘200 ,3.5

 

Plotting Z and Q as functions of temperatures
in Fig.(h) provides a means of determining the values
of R and X independently. The separation of R and X

was made by means of:

-20-

E22 = 32 o X2
(Q - X

( ._.-

Solving the above equations gives:

2
22 - X2 c %7 = X2 (l 0 $2)

Hence

(

( X = -;TWMI = f;:;_:::

E 11 . 52 ~.Q? . 1
- .3.

ER" Q

The following table gives the results obtained for

the first sample tested:

 

TABLE III

0F over

‘gggm temp. Z Q11» X R
100 3.3 8.7 3.28 0.377
200 3.43 7.6 3.903 0.887
300 3. 52 6. 55 3.98 0. 532
#00 ‘3.56 5.5 3.50 0.636
500 3.60 9.73 3.53 0.786
600 3.69 8.23 3.59 0.837
700 3.73 3.9 3.61 0.927

The same measurements were repeated for the coil

alone over the range of temperature from room tempera-

ture up to the temperature that the coil attains when

the material has been heated.

It was found that Z and

Q are constants, the values being:

-21-

TABLE IV

Coil
Alone Q. Z X R

The same
range of
variations

of temper-
ature 15 2.H2 2.91 0.16

Where X and B have been calculated as before.

These curves are drawn on the same graph as the
parometers of the cored coil (Fig. H). The curves of
Fig. k show to what extent the resistance and

reactance of the coil increases when the core is set

inside.

CALCULATION OFfiCOIL VOLTAGE

The following vector diagram shows the core and
coil currents and voltages and the distribution of

fluxes through the core and air space:

n:

 

\
1,-7.4

 

'Where:

§w, Ew, flux through the work piece and corresponding
emf in the coil -

TA, EA, flux through the air space and corresponding
emf in the coil -

coil current

H
O
p

core (secondary) current -

[11 H
B
I I

2 secondary emf (core)

:11
I

magnetizing force dueta the coil current.

The coil current produces the magnetizing force
H and the flux TA in phase with it. The flux in the
core 8w is in phase with the magnetizing current Im.

The emf induced in the core E2 is 909 lagging
9w and the core current I2 is lagging due to the
reactive characteristic of the circuit.

The primary emel is made up of two parts: EA
due to the flux in the air space, and Ew due to the
flux in the work piece. Both these vectors are leading
the corresponding fluxes.

The variation of H (peak magnetizing force) vs.
the diameter of the coil and work piece, is shown to

be (Brown, p. 28) roughly:

 

 

D

 

 

 

, 3.- 1
__ I'L {4.4. 1., L “,3- -__

y

;;__,_.. 1.1.V _ 1 <1
.1. .
\V/

v
I
I

~

‘ ‘9'
CC 7 i 1"” 1".

Therefore, the flux through the air space is

,' - (f B and H)
I\ ' éoth (Acoil ' Aw) ( cgs units )

Where
A coil 8 cross sectional area coil in inchesL
Aw . cross sectional area work piece

The flux through the work piece is calculated

assuming that it penetrates the material at a constant

value B (phase and magnitud constants) to a depth

/

equal to at below the surface. It is further assumed
that the material is saturated and B - 18000. (See
transactions A1EE Vol. 63, June 19HH, p. 273).

On this basis the total flux through the material

is approximately:

 

 

 

 

6w ' 18000 If .d . 5 X 6.45 ( «I and J. in inches)

Where
5 8 17.6\/ 11?... i l7.6\/3+°5 x 1%6 3‘ 10-6 inches

 

 

 

f 35 x 10‘F
Where
P s 37 filo-6.110% = 11+.6 x 10'6 _{\. inch

(Corresponds to 6000F)

So:

 

= lZL6 x 3‘8 = 67 x 10‘s inch.

5} 105

Under the assumptions made Qw is in phase with
B in the surface which is, in turn, in phase with I

coil. Therefore, the diagram becomes:

Ew

In this ideal diagram we are neglecting the in
phase component of Ew that is small compared with the
reactive component.

Now we can compute Ew and EA‘

Ew - 328.6 x r x T x 18000 If 0. 8f x 10-8 volts.
= 3 0.0162 x f x T x d x 8f = J 36 volts

Where T = coil turns = 19. .d - core diam = 0.5"

f = frequency.

The value of EA must be calculated by means of
the equivalent ideal coil. The same coil shall be
discussed in connection with the heating of a non-
magnetic material.

From there we get:

A coil - Aw = 0.67 - 0.196 - 0.478 sq. inch.
Therefore:
EA = J 28.6 x r x r x H x 0.974 x 10"8 = 331 V°1ts

Neglecting the internal resistance and reactance
voltage dr0ps we have:

ET = 131 0 336 = J 67 volts.
This value corresponds to 600OF and coincides

with the one measured in the experiment.

From there 2 =7%— = gg = 3.73nu (6000F)
When the core is taken out, all the voltage drop
is due to the flux in the air and, assuming roughly

the same current distribution in the coil, we get:

-26..

(A coil 0.67) inch

(85w s 0
(EA ' 6.1+5H X Acoil B H X 0.67 X 6.’+5

Therefore; if H is kept constant (I = 18.0'emp) we get:
Ew= 0
EA = j 28.6 x f x T x H x 0.67 x 10"8 = jhh volts

and the reactance of the coil would be:

E ##
xcoil = ‘_ ' ' 2°h4-Jl-
alone I i3

This coincides with the value obtained from the
experiment.

Therefore, the introduction of the iron core
increases the reactance of the coil from 2.#4_n_to
3.73.9Lat 6000F and also reflects some resistive-
component that can be calculated by means of the power

factor:

Input to the core (watts)-
Internal Power factor 1 --r .1 -
S. VI (work coil)

Where V does not take account of the internal

1R1 and IXi drops.

0 ,I_ 338 watts -
°s%‘67x18 “0°28

- 0

cosm(;0.2s

Where W6 is the angle shown in the diagram

 

 

 

Therefore:
RrZriéZéeé 0.28 = 1.095 .11, (at 6000F)

The total resistance of the coil with the core
inside is the sum of this term plus the coil
resistance. This was measured and was found to be:

R1 = 0.16 4\.
Therefore:
Ema, = 0.16 . 1.045 =- 1.205 .n. (at. 6000F)

This figure differs from 0.837 .rL found in
the experiment by 30%. This is understandable if we
consider how indirectly this experimental value was
obtained.

The efficiency of the coil is:

2
effy - I 1 0” 100 = 87%.
12 1.205

-28-

and the total power factor

005 10,. = fig? = 0.323

The above results have been plotted in graphs

2, 3 and 9.

EXPERIMENT #2

Piece of iron (0.2% carbon) 6" long, 0.37"

diameter to be heated from 70°F up to 1000°F in one

minute.

1)

2)

3)

Calculation of required thermal power:

.Again:

TP 8 17.6 x M x C xAT (units as explained before)
In.this case:

Volume = 0.695 cub. inches

M = 0.6,L 95 x .289 = 0.l,_83 pounds/minute.
c = average 70 - 10000F = 0.137 BTU/pound xOF
Therefore:

TP = 17.6 x 0.183 x 0.137 x 930 = 910 watts.
Radiation, convection and conduction losses at
maximum temperature anzl7 watts (9%) and will be
neglected during the calculations.

Power density
watts

- 910 -
PD ' 0761:?“ ‘ 637 mu . nch

-29-

9)

5)

3|

Peak magnetizing force:

The same formula used before applies in this

problem:

 

2 __-
H peak = (PD x d ___, ) /3 valid when d\&@j;213,900
(0.938 \[fp ) [3

The resistivity is:
p = 953501., - 39/..{H/m = 13.9 x 10"6 JL inch

Therefore:

H . (637.): 0.37 - “- ”3
peak 00.938 \/35 x 107x13.9x10‘6;

- (299)2/3 = 39.6

 

The condition

 

u
I818 x 35 X 10 = 0.37 x 105 x 96
V 13.9 x 10"6

= 1,700,000 2 13,900

Where: /k = 3%%Q% = 818 is satisfied.

 

d 1%}: 0.37

Coil design.
The same work coil employed in the preceding
problem was also used for this experiment. A

sketch of the coil and core is shown below.

 

  
 

M £1 gures
in mm

 

Coil length = 6,7", made of 19 turns of copper
tubing 6 mm in diameter.
0) C011 current:

As in the preceding example:

I = 1.93 x Hpeak x (,(inch) K1
N

 

Where R1 for a ratio:

Coil length - 169 . 35
Spacing 9.8

is:
Therefore:

I = 19591 x139°6 K 6:1 1.03 = 20 .6 amp.

The experiment was run at 20 amp, this being the

 

maximum rating of the equipment used.
The following table gives the eXperimental

results obtained:

TABIEJ

f secs I temp(over V 2

room temp)

/““’ °
350 10 20 3.5 150 63 3.15
350 20 20 6 250 66 3.3
350 30 2O 9 375 68 3.9
350 90 20 12.2 520 70 3.5
350 50 20 16 660 75 3.75
350 60 2O 19 790 78 3.9

a

9' V-‘l Temperature in
1 minute = 790

0 70 = 860°F

-31-

Voltage data has also been obtained to calculate
the impedance of the cored-coil.

Following the same procedure as in the first
experiment, the 3:,of the cored coil was measured at

different temperatures, the following results being

obtained:
Team-LI
0F over
5 room temp. Q
350 700 9.5
350 300 7.
350 ‘HDO 5.
350 500 5.
350 700 3.5

Both Q and Z were plotted on the same graph as
a function of temperature (Fig. 5). Therefrom pairs
of values at the same temperatures were obtained. As

a next step X_ and 3_ were calculated as before:

 

WEI

0F over

room temp.- Z 30_ X R
100 3.05 9.3 3.035 0.327
200 3.25 8.1 3.22 0.397
300 3.35 6.8 3.32 0.988
900 3.9 5.6 3.35 0.6
500 3.5 9.7 3.93 0.73
600 3.63 9.2 3.53 0.89
700 3.8 3.7 3.68 0.995
800 3.95 3.9 3.79 1.11

-32-

As already stated the coil used in this experiment
is the same as the one used before; its reactance and

resistance being constant and of value:

Z - 2.92 JL
X = 2.91 _;L
R= 0.16 ,0,-

7These curves have been plotted on the same graph
as the curves for the cored coil. The curves indicate
to what extent the resistance and reactance of the coil
increased when the core is set inside.

All the above results have been plotted

graphically in Figs. N0. 5 and 6.

EXPERIMENT3#3

Piece of iron 1.98" long, 0.75" diameter to be
heated from 70°F up to 600°F in one minute.
The procedure is the same followed before.
1) Calculation of required thermal power:
T.PIl7.6xMxCxAT
In this case:
Volume = 0.655 cub. inch.
M - 0.655 x 0.289 I 0.186 pounds/minute.
C I average 70-6000F I 0.129 BTU/pound 0F.
Therefore:

T.P I 17.6 x 0.186 x 0.129 x 530 = 215 watts.

-33-

2)

3)

9)

5)

Radiation, convection and conduction losses are
negligible.

Power density.

. 215 = Watt
P D 0.655 329 cub. inch

Peak magnetizing force:
The formula to be applied is the same as in the

preceding problems:

 

I (PD x d g2/3 valid when

H
peéE___£9:”38 (3???
d 772%.— 7, 13900

)3: 633501,. = 25/0fiC/‘M: 10 x 10'6 J1, inch.
Therefore:

H = E x 0075 2/3 :
peak q -6
0.938 \/35 x 10 x 10 x 10

3002/3 = 95.

 

Hence:

The condition:

720 x 35 x 109 - _
d f - 0. - 000 0. -
\17‘: ‘ 75/ 10 x 10-6 75 x 5 3

3,770,000 7 13900
is satisfied.

 

 

Coil design:

The coil was made of 6 turns of 6 mm diameter

00pper tubing and was designed 5,7 0%» (2.29 inches)

in length and 39 mm outside diameter, thus allowing a

clearance of 1.5 mm around the material.

 

 

 

.Figures in mm.

 

 

6) Coil current:
I = 1.93 x Hpeak x 1 (inch) K1 .
N
1.93 x%+5 x 2.29 K1

 

Where Kl correspondends to a ratio:

coil length = g . =
Spacing 52.5 38 So Kl 1.03.

 

Therefore:
I = 29.7
To get this current through the work coil with
the 20 amps rated equipment, the work coil was shunted
with a high voltage '075/u3f mica condenser. The

following data was obtained:

-35-

TABLE VIII

 

OF over

room temp.

 

 

 

 

 

 

 

4. secs Icoil ramp. 0F V Z
3 0 10 25’ 2.5 105 22.3 0.893
350 20 25 9 170 23 0.92
350 3O 25 5. 5 230 23 0.92
350 90 25 7.5 310 23.8 0.953
350 50 25 9 375 23.8 0.953
350 60 25 10.2 930 29.1 0.969
350 80 25 19 580 25. l.
The scheme shows
the parometers
measured.
+————9- l
‘ vxé: \kflf
.075}.

 

f

951:9:

I’

 

Final temperature in 1 minute = 930 0 70 = 5000F

—36-

The Q readings gave the following results:

°F over

TABLE IX

room temp.

700
1000

Finally, computations were made for the reactance

.F‘mflflro

n

and resistance of the coil as before.

 

IA§--I

°F over

room temp, Z 0 X R
100 0.89 7.35 0.885 0.12
200 0.92 6.7 0.911 0.136
300 0.995 6.2 0.939 0.151
900 0.96 5.6 0.996 0.169
500 0.975 5 0.956 0.191
600 1. 9.9 0.975 0.221

-37-

II

The correSponding readings for the coil alone are:

 

TABLE XI
t
over roomfi, Q. -__ Z X R
70° 10 0.8 0.796 0.0796
600 10 0.8 0.796 0.0796

Measurements of the temperature were made at
both the surface and the center of the piece under
identical conditions and running the experiment with

20 amp. The following results were obtained:

 

 

 

TABLs_XII
°F over room
temperature
ICOil Center w______Surface Secs
20 amp. 130 130 20
" 170 170 30
" 210 228 90
" 263 278 I 50
" 317 333 60
" 970 990 90

It can be seen that the temperature is fairly
uniform due to the small size of the piece and the
fact that the heating was rather slow.

All of the above results have been plotted
graphically in Figs. 7, 8 and 9.

-38—

QQIL VQLTAGE COMPUTATIONS

Following the same procedure as in the first

experiment and neglecting the in phase component:

Ew

 

01»

C

EW= 1 0.016.: d x; x T x 5; volts
( 3‘ and 6L in inches)

Where: -

: ’HXPT : 95xl5x10’6'
3% 17.6 & max. 17.6 V 35 x 109 inches
Where
P - P = 15 10'6 JL 1 h

max 600°F x no

Hence -

5

g (6000)

 

 

= 77.9 x 10-5 inches.

Therefore: .
EN (6000) I f 0.0162 x 0.75 x 35 x 107 x 6 x 77.9 x
10‘5 = a 19.75 volts

Besides:
EA = 5" 28.6 x I x T x H (10 - Aw) x 10'8 volts.
Ac is the cross sectional area of the equivalent ideal

coil. As will be explained later in connection with

v
the heating of a 00pper tubing, the spacing (L of

this fictitious coil would be: (Brown's Chapter VIII)

I," If \I I.
/

actual current density

 
  

g
i _.

;:>\\\\\\‘\\._. equivalent

 

ideal coil
1 - ( a )2 -
h-h 1-(-—--h) Wherea-3mm
h=9.5mm

 

hi 9.5 Ml- 0.995 = 3.33 mm.
Therefore the diameter of the ideal coil is:

de = 1.01" and the area: A0 = 0.8 sq. inch.

Hence :
13A = 4: 28°6 x 35 x 10% x 6 X 1+5 x 0.358 x 10"8 =

‘1 9.66 volts

Therefore

V 2 EA + Ew = 3' 9.66 + J 19.75 = 3’ 29.91 volts(6000F)

And:)( '-‘ 29.91 = 1.19%
24.7

 

Actually Ew has an "in-phqse component" that provides
the power input to the load.
We can thus calculate the internal power factor

(not including the Tax, I‘Yg voltage drops in the coil):

 

internal power gr c.59L _ TNput to the core (watts)
factor b ‘ V I (work coil)

Where V 3 EA + E

00$ I};

Therefore:

II

R)
H
U1

= 0.296

 

29.7 x 29.91

1.19 x 0.296 3 0.35352

Rreflejoda

 

The total resistance of the coil with the core
inside is the sum of this plus the coil resistance.
{This was measured and was found to be:

R1 = 0.0796

Rtotal : 0.0796 + 0.353 = 0.9326 41(6000F)
Tfiiis value is high compared with that found in the

experiment (0.221). This disagreement is due partially

to the fact that the power was less than 215 watts as
can be observed by looking at the heating curve. How-
ever the power could not be measured directly to check
this.

Besides, the value of R from the experiment was
found through a very indirect way basing our compu-
tations on the measurements of Q. done with an inad—
equate instrument and for this reason it cannot be
considered accurate.

The total power factor:
COS $0. = 0.9326 = 0.363
T19—
And the efficiency of the coil:

effeciency = I2 x 0.353 x 100 : 87.8%
12 x 0.9326 ‘

Condénsercomputations.

In the last experiment of this paper there is a
complete discussion about the application of condensers
in parallel with the work coil. .

‘ Starting from V = 29,91 volts and selecting I con-
denser so that:

Icond : Icoil - Irated = 29.7 — 20 = 9.7 amp
equip _

liven though this formula is not estrictly’correétit can
'beeapplied successfully due to the fact that 10 is al-

nuast an entirely resistive current.

-42-

 

 

 

We can write

Icond : V0 Co We

Icond = 9.7 x 106 I

l.‘

C
V W 29.91 x 2tfx 35 x 101+ I J

O
I!

o .073 {~-

I

EXPERIMENT # 9

from tubing 3.22" long, 0.11" thick and 0.89"
external diameter to be heated from 7 OF up to 600°F
in one minute.

{The calculations follow the same procedure as before

1) Thermal power:

T? = 17.6 x M x C x ACT
Where
Volume : 319x'07892 - 07622 X 3.22 :
9 ‘9
0.805 Cub. inch

M = 0.805 x 0.289 : 0.229 pounds/minute

-43-

C = average 70 - 6000F : .129
a 530

Thence:

2)

3)

4)

TP ; 17.6 x 0.229 x 0.129 x 530 z 265 watts.
Radiation, correction, and conduction losses
are negligible:

Power density.

 

PD : 265 : 265 : 149 watts
total 1.78 cub. inch
volume

Peak magnetizing force. The same equation
for solid magnetic cylinders applies for the
hollow ones if the wall thickness is greater

than the depth of current penetration or:

F'”7§
\:

i /
Therefore: 1 g L
H :’ PD x d \2/3 =/ H+5x 0 37
0181.99???) \ ’
f'
1 (3350F) :10 x 103$L.1n0h
2/

H =( 285 9 = 28.6
.0{

So:

Wall thickness 2; 3170

 

    
 

 

DJ

71 a: 32900
/ O 0 Z 1130

The maximum skin depth takes place at maximum
tempgerature. It is:

9W) - 30/nh 1h“: 15 X 10- 6..-AL~ 111011

 

Skin depth

3170] 15 x 10‘6 44.3170x1. 95
1130 X 39 X 10q 107

6.18 x 10—4 inches

-44-

5)

The condidion
Wall thickness 3 0.11") 6.18 x 10'” inch
is satisfied.
Coil design.
The coil was made of 10 turns of 6 turn
diameter copper tubing and was designed
3.5" (8.9cw9 long and 38 inmhas outside diam-
eter thus allowing a clearance of 2 innhns
around the core.

Coil current

For coil length :.§2 z 45 K1 2 1.02
spacing 2 ( Bendz page 312)

 

Therefore:

I 1 = 1.43 x 28.6 X 3.5 1.02 : 14.5 amp.
0011 10

 

The following data was obtained:

TABLE XIII

 

t

secs. 1 coil OF over room

 

 

 

Temp. V .Z
10 15.5 1.3 60 36 2.32
20 " 3.2 140 38 2.45
30 " 5.4 230 42 2.7
40 ” 7. 290 42 2.7
50 " 8.7 360 '2 2.7
60 " 10.5 35 42 2.7
90 " 15 620 h2 2.7

 

Efiinal temp. in 1 minute = 435 + 70 = 505

Following the same procedure a

the 0's

C)
U'
(I)
*5
O
W
(0

Vfiire measured at different temperature.

TABLE XIV

 

 

 

Temp 0
70 7.5
350 5
60 ’4
800 4

 

From the graph N0. 11 the following table can be

constructed:

 

 

 

TABLE XV
OF over room
temp . Z Q X R.
100 2.4 7.6 2.j8 0.313
200 2.6 6.8 2.57 0.378
300 2.7 6 2.665 0.444
400 2.7 5.2 2.65 0.51
500 2.7 4.5 2.f4 0.587
600 2.7 4.2 2.63 0.603

 

The following readings were obtained taking the

core out of the coil:

TABLE XVI

 

 

Coil Temp.
alone over room 7L ‘1 )< FL
70 1.75 15 1.715 0.114
”50 H n I. u
700 M It It It

 

All these results are plotted in figs. 10 and 11.
Cczhdlusions are similar to those drawn from experiment

NC). 1.

EVr

BRIDE

e.—

if

The same iron tubing

inside, ant

parallel.

41-“
a nigher

TABLE XVII

£2
.J

was heated with a coil

current using a condenser in

 

 

 

 

 

Coil current secs. Temp over room

I )1. amp UT?
34 10

" 20 1 40
” 30 1 . 5 c5
” 40 1.7 72
" 50 2. 85
” 60 2.7 115
” 90 4.5 190

 

The coil had the following character

external diameter
internal diameter

 

length = 8.5

1’10.

.0
01

turns

24?

13 +nahns
: Inohas

3.35 inches

15

1:” “2:3Lfdfi.~.

J

4:4
L goikx€r C01 1
. -1

, I _ -_-_ ._O..-__.__ . _.:T——

istics:

(The final AT in one minute was only 115019. To
get a temperature of 6000F in one minute (AYTz 530)
the current should have been 12, and the peak magne-
tizing force He.
But:

H = K1(P.D)2/3 : K2 x (Thermal power)2/3 2K3 xeng/3.

The ratio between the actual H1 and H2 is:

43/2 7 0 4 *1 13/2
‘ ~ = 33 : .O = 2
H1 liS 1372
1

'1 3 2
/2 c
12 : (3400)3/2 X 4061 : 138 X 4061 = 915
Hence:

I — 94.5 amp and the value of H2 peak magneti—
zing force for such a coil to heat the material- up
to 6000F in one minute, becomes:

ngeak : 0.70 X 15iX 94.5 : 296.

 

3-33
~ - ° °‘ ‘ 3 P .
Therefore in this case, tne ratio 01 Hinner
Houter

‘to accomplish the same purpose is:

Hinner = 295 : 10-35

__1___. k
Houter ‘0

vThe results are shown in graph No. 10.

~48-

EXPERIMENT # 6

copper tubing to be hosted from 70 to 320°F in
two minutes.
The dimensions of the copper tubing are:
outer diameter a 0.5 inches
length : 6.2 inches
thickness = 0.018 inches
weight - 0.056 pounds
external volume : 1.215 cub. inches

The specific heat and resistivity are for copper:

 

 

 

TABLE XVIIIA
OF S. heat F/‘tJlC/u.
- 1.72
70 0.09 -
400 0.096 3.1
600 0.099 -
800 0.101 4.8
1200 0.106 6.6

 

‘ggggg, average s. heat (70-320) = 0.093
1) Thermal power
TP = 17.6 x 0.056 x 0.093 x 239 = 11.5 watts
2) Radiation, convection and conduction losses
are negligible.
3) Power density

PD 3 11.5 e 11.; ; 9.46 watts
.2 5

ernaICVOlume cub. inch
4) Peak magnetizing force:
For this hollow cylinder to be considered

-49-

equivalent to a thick-walled one, the following
condition must be satisfied:
wall thickness é; skin depth
That is: _._i. .
Wall thickness Z:3170VL@21= (Bendz's p.305)
When ' g

(3 EN : 2.8 x'I0gum\: 1.1 x'106_fL inch

WA}: 310
So:

 

Wall thickness 2:31 0 .liIU6= 1 0 ~ ,
7 \é’FiTOT' 106-1-77=5.6xm31n

The required condition is satisfied.
Under these conditions we can apply the following

formula developed in this paper previously:

 

 

n :1 PD x d valid 1: 0‘151313400

. I
V2.59 x 1'03 5pm 4“
Where (3 : ;3 :2.35 x 101131:0.93 x 106.JL inch
Mt .' If! T

The condition

cuff" z 0.5 [35:90" =0.5x105x6.14:3.07x10+5>13400
,9
'“VL 0.93iIU

is also satisfied.

 

 

 

 

Therefore: 1m1111111
H : 39.46 x 0.5I =J6-2 X 103 : 55-5
- —— *6
2.59dE9J55x10 xo.93iIU

5) 0011 design.
The same coil used in experiment No. 1, was
used for this experiment, its characteristics being:
19 turns of copper tubing 6 inches in diameter

-50-

6.7" long
31 We outside diameter
19 W inside diameter
The clearance was 3.2 inches and the factor’Kl
nearly unity.

6) 0011 current

I : 1.43 x 56.5 x 6.1 a 28.5 amp.

 

Coil voltage
The diagram shows the relative sizes of coil and

core 3

 

3|

 

 

 

 

It is easily seen that we have to consider two
different fluxes, the one through the work piece Twand
the one through the air space surrounding it, (.5 .

Let a vector diagram be constructed. The coil

current Ic produces the flux @Ain phase with it. The

-51-

#14.

"d.

fluxffiNis lagging due to the secondary current
(12 : Inork)r£}3 yet to be found.

As a second step let's find the emf's. The emf
induced in the core E2 is 961agging ii and the core
current is lagging'iodue to the resistive and inductive
characteristic of it, considered as a secondary circuit
short circuited. Thus, 32 is made up of IgRg and 12X2-
Nov adding up IO and I2 referred to the primary, we.
obtain the total core magnetizing current In. The
primary emf E1 is made up of 2 parts: Ea due to the
flux in the air and EW due to the flux in the work
piece. Each of these vectors is leading 90° the

corresponding flux.

F-

    

 

 

/ x
f j?! 7’;1w
, z;
17“
L: Ex—
Ikxz

'; is the internal power factor.

-52-

 

1"

 

Shifting the diagram so as to make Ic be in the

x axes we can write:

QA
@A = H 3: Area air space #

_w
é = H (A0011 " A!) V92 5:; > 6‘
@w '-' Aw ' 3(P'3 ) 3%,

Where Acoil and Aw represent the area of the coil
and work piece and H the peak magnetizing force of the
coil above; that is, the air space. P and Q therefore,
take care of any change in the magnitude and phase of
H while coming into the core.

The emf's core then, can be written:

Ea = 328.6 xix T x H (A0011 - A“) 1'68 volts.
E" a 328.6 1/1 T x A. H (P-J (9)1138 volts.

T represents the number of coil turns.
In the present experiment at a temperature of

320°? we have:

inch .111
P = 1 o 1 I V
8.: 320 “6

d a diameter : 0.5 inch.

Hence:

-__—._..——_.—.__—..

d‘rfi- 0. 5\ x 10 - _0.5 x 105 x 5.63 = 282,000
\ \1.1x10

The parameters 3 and g vary as shown in the
following graph. Taken from AlEE transactions V01. 63 ,
$1123; 19%. p. 273. ‘

-53..

 

 

 

The valueqM-z—{i )- 9 on the X axes corresponds to

up; = 13400

Hence in the present case we are beyond that

point and:
-2
PaQ: 6310 :2.2hx10
Therefore:

E. . J 28.6x35x104x19x0.19x56.5 (2.24-12.24)10'1°volts
\, o
E. = .1 (471-,1n71)10"3. 0.n7l+30.n7l volts -.- 0.665 L115.
We can see that, in all the theoretical cases very
small and the flux them: be neglected, the internal
power factor angle is 45°.
The value of Ea is analyzed by considering an
equivalent ideal coil. ‘
Due to the currents flowing in the cylinder the
current density in the coil increases at the inntr
boundary and decreases at the outer one as the diagram

shows :

  

ouiv.udwe]

 

In Brown S. Hoiler's Chapter VIII there is a
complete analysis of the matter where for a single
turn coil they conclude that for the equivalent ideal
0011 the following equation applies:

h':h\/l-(_fi_\\2

In this case: ‘

 

h : 6 innhvs
a .73 imvhes

Therefore:

 

h' = 6 1:0.25 : 6.: 0.865 - 5.2 inches
and the diameter of this ficticious coil

80:

-55...

128.6 x 35 x 10" x 19 x 565 x (0.67-0.196) x 10"8
J 51 volts = 51 ng° volts.
"He must add to this internal voltage (emf) the

In
m
u

voltage drops in the coil itself due to its internal
impedance.
From (Brown's page 22)
“1'“ Fallen?
Theoretical calculations are not accurate, whereas
a simple reading in the experiment gives (neglecting X1)
R1 : 0.16-JL
Hence
1R1 = 28.5 x 0.16 a #.5 volts
We can see that in this case the total voltage isvery

nearly the same as Ea and our \0; becomes nearly 90°. The

total coil voltage is shown in the following diagram:

 

 

 

 

 

 

IX{(M9.8‘€QT€ cl) IQC
(1‘ EA/d”
Lro+§i
QTbT—é“
EW
Ox)T of R08“

 

-56-

Hence:

Total coil voltage: 52 volts 85°
Where 8&0 is the totaleince it comprises also the
131, 1X1, drops on the coil. The efficiency of the
coil is:

Effy -_-. Efi (in Jhase componenfl I 3 El! 100

12111 + E,',I MW 1

: 0.421 100 = 0.871 100 m 9.6%.
. + . logiI

Total power factor = Coa‘€,= 0.1

In the experiment the voltage through the coil
was 7% higher than this value, due to the fact that a
slight difference in the measure of the coil diameter
increases greatly the area (A0011 - Aw) and as the coil
was made by hand its diameter was not ‘werye : uniform.
- The variation of the internal power factor with

the ratio Acoil has been shown to be (AlEE transac-

 

 

.Awork
tions'Vol. 53, June, 1944, page 273)
fitosWC
;] 1. , ____1_1, 2_11__._U
rift/I”,,,ii
"‘L /’

W J:-
\
\
\
/

 

 

.1 o-
’W
\ /,»
O _J__._ 1.. i i. l - L _..e \.__.1 1 - 11.4. .4 L._>
2 H x c a

-57...

5
O
\,
A
\

r
\
~

,

The experiment was run by connecting a condenser
in parallel with the work coil as shown. The table

'gives the results obtained.

 

 

 

 

TABLE XIX
Wcoil secsu over room temp V Z
29 52 *10 '148 75 59 1.235
I! I! 30 30 130 n n
s ” 40 305 150 " n
w " 50 3'8 160 " n
n I 60 4.2 175 n u
w " 90 502 220 " n
n n 120 5.7 240 n "

 

 

 

Final temperature in 2 minutes I 240 + 70 : 310°F
The value of Q was found to be constant through

this temperature range and equal to Q = 11.

-58-

Therefore, R and X were calculated with the result

listed below:

 

TABLE xx
0011 with temp.
hollow cop- ' over room Q 2. X A?
per core. - . . .

 

For the coil alone the corresponding parameters

are also constants over the same range of temperature.

 

 

TABLE XXI
Coil alone Q (z, X R
same ra e

 

It is seen that the copper core acts as an induc-
tive secondary circuit increasing the total resistance
and decreasing the reactance.

The fact that the resistance of the secondary is
very small compared with the reactance causes the total
reactance and resistance to be fairly constant.

In fact, the total impedance is:

R]. + quleRZ . J (1 _. W 2L
R22V+ (me)2 + w 1 R22 + 041.2)5

That in this case reduces to:

 

R1 + M2 + Jw(L1 - M2) where M2 2 is small.
IT'
1722 2 2

 

Condenser‘fiamputations
In applying a condenser in parallel with the work

0011 care must be taken to insure that this combination

—RCI_
all

will not form a tank circuit that can take control of
the frequency of oscillation. ‘ I

To help avoid this undesired possibility the
following condition will be imposed.

Qloadié 80% Q main tank circuit with work coil
This can be written as: (Bendz's p.319)

 

 

 

 

 

 

 

Coil volta e x coil current Hufi

 

 

oad‘watts
enerator rated volt-am
4§HKOZ:EL‘ ’ load watts ’62
or vc x 10 5 8075 VtIt

In addition, the Westinghouse Mfg. Co. recommends
(lkw generator catalogue) that: Icoil élIT to keep
the load tank circuit away from resonance.

In this range the condenser current is to be chosen

-60..

so that

I00nd ' Icoil ‘ Itrated

In this case:

Icond : 29 - 20 = 9 amp
This formula is only correct in the total absence of
power in the circuit but in most practical problems
it yields approximate results because ‘Qr is very

nearly 900 as analyzed before.

1\ Emma

”'5
’f‘v
WT , T
41‘0ch f

ITD'T'CH

 

 

  
 

Inn“)

Finally:

9x105 -4173”!—

C=Im
1V3; 56 x 21V; 35 x 10h /

The results of this experiment are shown in graphs 12 and 13.

("I

 

Conclusions

Developing the expression for 'E' found at the
beginning of this paper we get:

 

 

 

For F £3 1
E = P(power core) g 8113 2 Acore 1 2 Ssol
IZdeeu, 'core Score 12 Rdcusol. Asol
For a given solenoid this becomes:
”________.
E = K Acore Ssol a K Acore V/‘core wcore
u"core core u"core T301

E a K Acore W'core Pcore independent of freq.

This formula shows that for a given , solenoid, E

 

is much smaller when heating a piece of non magnetic
material than heating a magnetic one, no matter what
the frequency is, as long, of course, that we are over

2025’

Assuming for instance:

.3...
a-

Amagcore = Anon mag core
[Armag m 800
amag core ' 10 Pnon mag core

there results:

 

Eng; .-_-\/800 x 10 : 90
Enon mag

If we also assume an efficiency of 85% for the

heating of the magnetic core we can write:

effxmag : E? = 85% where P335: = 85
12Rsol 3 15

-62-

 

 

 

c 1
s
1 n
1 ‘ '
c
. 1
r ‘ . I
o l
‘ m
.
. a
u
‘ u
1 ' 1 ,
i m
u ' l
, , Q 1 .
.
p
c
h
b
. _ . 1
' . n . 1
o
1 _2

For the same 12R the power that gets to the

sol’
load for non magnetic material would be:

Powernon mag : g8 : 0.95

and the effy:

We can conclude, therefore, that to keep the
efficiency of this method between reasonable values
‘while heating non magnetic materials much larger coils
must be used.

High frequency heating, then, is especially
convenient to heat magnetic materials but in spite
of its low efficiency is used many times to heat non
magnetic ones due to its Outstanding characteristics
such as:

1) Heating can be accomplished faster than with

other methods for most metal pieces.

2) Heating can be confined to the desired area

which may be a small fraction of the total.

3) Heating starts instantly when radio frequency

power is applied and stops instantly when
radio frequency power is removed.

These features frequently outweigh the relatively

high cost of the necessary equipment.

List of the equipment used in these experiments

-Westinghouse industrial radio-frequennggenerator3
E v 020, Style No. 867692, m, 300 n c/s low
frequency, and 10 M c/s high frequency.

-Radio frequengyg'Q' meter: Type 160 A, Serial
No. 2632, Boonton Radio Corp.

evacuum tube voltmeter; Type 726A, Serial No. 3091
General Radio Co.

-Iron constantan thermocouple with galvanometer
Weston model 301, EE 2504, 30 micro amp

-Ammeters. GE Type CG 22037, 200 amps (rf)

EEAA124
GE Type 00 22041, 50 amps (rf)
EEAA125
-High voltage condensers (rf)
Cornell - Dubilier GD 48449-5
EE 2819-0.03 mf
Cornell - Dubilier GD 48996-5
EE 2818 - 0.045 mi

-54-

 

 

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REARVIEW

66

 

COILS AND CYLINDRICAL CONDUCTORS TO BE HEATED.

67

References

-Brown, Hoyler, Blerwith, "Radio Frequency Heating"
van Nostrand Company, 1947.

éBendz, w. I. "Electronics for Industry" (Wiley &
Sons), 1947.

-Ramo & Whinnery "Fields and waves in.Modern Radio"
Wiley & Sons, 1944.

-M-I-T Staff "Applied Electronics". Wiley & Sons,
1943.

-Harkus & Feluff "Electronics for Engineers" MCGraw
3111, 1945.

-H. B. Dwight "Electric Coils and Conductors", McGraw
3111, 1945.

-Terman, F. E. ”Radio Engineers' Handbook" McGraw Hill,
1943. ‘

-Baker, R. M. ”Heating of non magnetic electric con-
ductors by magnetic induction - longitudinal flux”
AlEE Transactions,‘Vo1. 63, June, 1944, P. 273.

-Baker, R. M. ”Induction heating of moving magnetic
strip" AlEE Transactions, vol. 64, April, 1945,
P. 184.

-Stance1, N. R. "Induction Heating - Election of
Frequency" AlEE Transactions, Vol. 63, October,
1944, P. 755.

-53-

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