Datasheet

Kanthal Appliance Alloys Handbook 81
10
We assume that a steel tube of initially
9.5 mm (0.37 in) diameter is being used and
can then expect a resistance reduction of
about 30 % upon rolling. The resistance of
the coil should therefore be about 65.3 Ω.
The wire surface prior to compression is
7 % bigger, or 22.5 cm
2
(3.49 in
2
), and the
ratio between wire surface and resistance
0.34 cm
2
/Ω (0.053 in
2
/Ω).
The corresponding wire size is 0.26 mm
(0.01 in). Tests with this wire size have to be
made in order to check the resistance reduc-
tion as a result of compression.
Coil suspended on a Mica-cross, element
for a hair dryer
Rating, P 350W
Voltage, U 55 V
Length of coil, l 250 mm
Coil outer diameter, D 6 mm
For this application a surface load, p, of
7 W/cm
2
is reasonable, using equation [6]
gives a wire surface of:
p =
P
A
c
=
7A
c
350
=
P
p
= 50 cm
2
➔
Assuming a wire temperature of 600 °C and
choosing Kanthal D with an Ct value of
1.04. Next step is to calculate hot- and cold
resistance, according to combining equation
[9], [10] and [2]:
R
20
=
R
T
= 8.31 Ω
C
t
By calculating the surface area to cold
resistance ratio, a suitable wire dimension
is found:
A
C
=
8.31 Ω
50 cm
2
= 6.01
R
20
Ω
cm
2
According to the table in chapter 9, Kanthal
D Ø 0.70 mm has an area to resistance ratio
of 6.27 cm
2
/Ω.
Verifying the geometry of the coil, suitable
values for the D/d ratio are between 6-12.
D/d ratio has to be considered since too
low as well as too high values will create
problems in the coiling process. In this case:
D
=
0.7 mm
6 mm
= 8.6 which is within limits
d
To get the length of wire we have to calcu-
late the ratio between resistance needed
and resistance per meter according to table
chapter 9, KANTHAL D, d = 0.7 mm
R
20/m
= 3.51 Ω/m.
The length of the wire becomes:
Based on [17] the coil pitch, s, is calculated
to:
s
=
0.7
2.09
r =
d
= 2.98
L
2370
and subsequently a relative pitch:
Finally the actual surface load is based on [6]
calculated to:
Combining [1’] and [7’], [8’]
Wire
π · (D – d) · L
e
=
s =
= 2.09 mm
π · (7 – 0.7) · 250
P
=
p =
= 6.7 W/cm
2
350
A
c/m
· L
22 · 2.37
R
T
=
U
2
=
350
P
55
2
= 8.64 Ω
R
20
=
3.51 Ω
8.31 Ω · m
L =
R
20 / m
= 2.367 m