4.4. DESIGN OF SOLENOID MAGNETS 281
coolant in removing heat at conventional temperatures. The oils, for example,
have heat-transfer characteristics about one-fifth that of water [D11]. Coils
with low resistance, operating at low voltage, are almost always cooled with
deionized or demineralized water, mains water causing electrolysis even at low
voltage. The cooling water is circulated through the coil, a heat-exchanger, a
filter and a deionizer. In the heat-exchanger the heat is carried o by tap water.
The flowrate of water T (in cm
3
/s or m
3
/h) required to restrict the temper-
aturerisetoW (in
0
C) in a magnet with power input of Z (in watts) is given
[M22] by:
T =
Z
4=186W
(fjv)=
Z
1163=7W
(VL) (4.54)
For instance, for a 250 kW magnet, with temperature rise of W =30
0
C, the
required flowrate is approximately 2000 cm
3
/s, or 7.2 m
3
/h.
The operating temperature of a magnet aects the resistance of the winding
and thus the power needed to achieve the required magnetic field. The hotter
the coil, the higher the resistance will be and the more power will be needed.
The temperature dependence of resistivity of conductors can be written
(
W
= (
20
[1 + (W 20
0
F)] (4.55)
where (
W
and (
20
are resistivities at temperature W and 20
0
C, respectively
and is the temperature coe!cient. For copper (
20
=1.72×10
8
mand
= 0.0041/
0
C, while for aluminium (
20
= 2.7×10
8
mand = 0.0045/
0
C.
Therefore, a copper conductor operating at 60
0
C has resistivity of 2×10
8
m,
while an aluminium conductor has resistivity of 3.2×10
8
m.
Whether or not the amount and flow velocity of the coolant, as determined
by eq. (4.54), can be forced through the winding of the magnet depends on the
pumping power available and on the detailed structure of magnet. The pressure
drop across a tube is the sum of the entrance and exit losses and frictional drop
in the tube. The total pressure drop S (in Pa) becomes
S =(0=74 + 0=015
O
g
)10
3
y
2
(4.56)
where O and g are the length and diameter of the cooling path (in m), and
y is the coolant velocity (in m/s). A typical pressure drop for copper hollow
conductors is 0.5 to 0.8 MPa.
4.4.4 Thickness of the iron cladding
The basic requirement for the iron cladding is that it must not be saturated.
The magnetic flux in the cladding is M
v
D
f
where D
f
is the cross-sectional area
of the cladding. In order to avoid oversaturation of the cladding, the following
inequality must be satisfied:
E
j
D
j
5 M
v
D
f
(4.57)
where E
j
is the magnetic induction in the air gap of cross-sectional area of D
j
.
The thickness of the cladding can thus be found from this equation.