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Coupling losses in inhomogeneous cores of superconducting cables
Coupling losses in inhomogeneous cores of superconducting cables S. T a k f i c s Institute of Electrical Engineering, Slovak Academy of Sciences, 842 39 Bratislava, Czechoslovakia
Received 9 July 1991; revised 23 September 1991 The coupling losses are calculated for round and flat cables whose cores consist of two layers (an inner insulating layer for flat cables and a thin outer shell for round cables, which is a substitute for the solder connecting the superconducting strands to the core). The previously developed method based on the solution of the Laplace equation and its extension to flat cables is used for calculating the different contributions to coupling losses. It is shown that for flat cables it is necessary to include an insulation layer through the whole cable to suppress totally the transverse losses. Otherwise, the circular losses are of the same order as the parallel losses between neighbouring strands. Decreasing coupling losses by using highly resistive solder for round cables is effective, but not very radical in its effects. To decrease the coupling losses by about one order of magnitude, with realistic solder thickness (about one-tenth of the core radius), one has to use a solder with a resistivity about 100 times higher.
Keywords: a.c. losses; superconducting cables; coupling losses
The losses encountered in changing electromagnetic fields represent one of the main obstacles to more extensive practical applications of type II superconductors. These losses consist of three main components (see, for example, References 1 - 4 ) . Due to the recent development of very fine filament superconductors 5-8, the hysteresis losses can be decreased substantially. In addition, reversible flux motion can brinAg about another significant decrease in these losses 6'9A°. The eddy current losses in the matrix can also be lowered markedly by decreasing the twist pitch Ip and introducing a highly resistive matrix between the filaments, such as C u - N i 5-7 or C u - M n 8. The coupling losses, which are actually a type of eddy current losses, are caused by currents between strands and other substructures of the cables or braids. Because the structure of conductors used for producing high current densities is often very complicated, the calculation of coupling losses is, in the most cases, not a simple task. The cables are produced in multiple stage processes, with different twist and cabling lengths. The differences in these lengths can lead to additional a.c. losses ll'12. Analytical calculations are hard to realize and even the formulation of problems is often difficult. In the 'history' of coupling loss calculations, many surprising effects have appeared, partly due to the unexpected order of magnitude of these effects. As an example, we could mention the size effects in finite cables 13 or the behaviour of cables in spatially O011 - 2275/92/030258 ©
1992. Butterworth
258
- 07 - Heinemann
Ltd
Cryogenics 1992 Vol 32, No 3
inhomogeneous applied fields 14'15. In addition, some 'new' (forgotten?) contributions to losses have appeared (currents in finite samples, currents at finite magnetic field range, 'circular' currents ~6'~7, etc.). Maybe even more exist. Hence, it would be very useful to clarify some problems involving simplified or model structures as well as to perform more precise calculations on problems already assumed to be solved. All these questions can be solved only by comparing different methods of loss calculation; the methods used previously for coupling loss calculations have been different in several ways (calculations of currents or voltages between the strands 13-t~, Fourier transforms, numerical solutions '8-2°, etc.). We developed an alternative procedure by solving the Laplace equation, knowing the voltage distribution on the cable surface 16'17. Here, we use this method for solving the Laplace equation and calculating the coupling losses in two cases. The first case involves a cylinder geometry with different resistivity of the cylinder sheath (which cart correspond to the solder connecting the strands to the cable) and its inner part (which can even be insulating). The second case is connected with the (sometimes questioned) necessity for the central insulating layer in flat cables. In all our calculations we assume an infinitely long fiat or round cable with single layered strands in a transverse (spatially homogeneous) applied a.c. magnetic field B. The cabling length L is always supposed to be much
Coupling losses in superconductor cables: S. Takdcs longer than the twist pitch lp. The number of strands should be sufficient to enable us to perform integration instead of summation of the different fields and voltages from the individual strands 13-15.
Solution of the Laplace equation The general solution of the Laplace equation for cylinder geometry is oo
U= ~
(a. r" + b. r-") [ c. cos(n~o) + d. sin(n~o) ]
n=l
+ const Figure 1 Round cable with surface layer of different resistivity
For an applied magnetic field perpendicular to the x axis (~ = 0, 70, we can assume, without loss of generality, that U = 0 on the x axis (due to the symmetry of the problem). The solution is then
(due to the existence of some solder, for example)
co
U= E
(a.r" + b.r-")sin(n~o)
2 Due to the finite nature of
n=l
The boundary conditions are given by the following relations: On the surface U, = U(~) is given by external sources; the 'peripheral' currents between neighbouring strands are closed, leading to parallel losses Wp. For single-layered strands, one has~3-~7
BbL us =
_ _
4r
BbL us =
--
16
U2=(Ay+B)sin~o at the centre, we have B = 0. 3 Due to the continuity of U at the interface, we obtain
al z + _b~ = Az Z
sin ~o = Um sin ~o sin 3 ~o =Um sin 3 ~o
for the round and flat cable, respectively. 2 The current density normal to the interfaces should be continuous. 3 The potential at metallic interfaces should be continuous. 4 The potential should be finite everywhere (this leads to b, = 0 for solid cylinders).
Losses in round cables Usually the strands (for simplicity we assume singlelayered strands) are connected to the central part of the cable by some solder whose resistivity can be very different from that of the central core. We replace this by assuming a surface layer with thickness d and resistivity 01, with the resistivity of the central core being P2 and the radius of the cylinder R (Figure 1). Then, both solutions of the Laplace equation should obey the following boundary conditions:
with z = R~/R. The normal component of the current density at the interface should be continuous too and, therefore
(a-,
Pl
P2
These conditions lead to al =Um - - +, 1 g N
bl = U m
z- 2 (- 1 -- g ) ,
A -
N
2Urn N
with g=P_L,N=l+g+z2(l_g)=l
+z2+g(1-z
2)
P2
The total transverse loss of the cable (per cabling length L) is then
1 Taking Us = Um sin ~o (other values of Us(~o) could be included as well), we have
U~ = (aly + ~ ) s i n ¢ withy=r/R
and
a+b=Um.
j0 p2 L\ay# N 2 Pl
\a
{[(l + g ) 2 + z 2 ( l - g ) e ] (
yj] l _ z 2 ) + 4 z 2}
Cryogenics 1992 Vol 32, No 3
259
Coupling losses in superconductor cables: S. Takdcs dx
Here we give the results for two special cases in more detail. For the homogeneous cylinder (g = 1) one has
wtl L
"$r--
p
Comparing this with the result for a thin hollow cylinder (or with the insulated inner part, i.e. P2 ~ 01) with thickness d one obtains Wt2 _ r U2m d -
L
o R
7r B2Rd
160
The ratio Wt2/Wtl is proportional to d / R (as we shall see later, this ratio is nearly quadratic in d/R for the flat cable). In addition, Wt2 for the hollow cylinder is comparable with the parallel losses Wp between neighbouring strands, as stated earlier 16'~7. For practical purposes, it could be important to know whether using a highly resistive solder can decrease the losses considerably 2~. We take as an example the case where the thickness of the solder is about one-tenth that of the cylinder, but its resistivity is increased 10 times (d/R = 1 - z = 0.1, g = 10). The transverse losses are then decreased by about a factor of 1/2. In addition, the parallel losses are decreased by nearly the same factor (see Appendix).
Transformation: f l a t - round - flat cable We shall consider an infinitely long flat cable with single-layered strands (on both sides of the flat core with thickness c) in a perpendicular harmonic applied magnetic field. The resistivity of the core tape P is assumed to be homogeneous. The solution of the Laplace equation is comparatively simple for the cylinder geometry; therefore, by transforming the flat cable to the round cable problem one obtains many advantages. The principles of the transformation can be seen in Figure 2. The insulating core is transformed to the hollow part of the cylinder and the thickness of the round cylinder is half the core layer thickness. The 'geometrical' factor b - 7rR (or R - b/Tr) has to be taken into account in both transformations, although this is not a completely precise procedure. The induced transverse voltage on the flat cable is given by 11,17 BLb G = sin2 ~o 8 with sin2 ~o = 1 - ( ~ - ) 2
where b is the cable width and L the cabling length. Because dx = dl sin ~o (see Figure 2), the corresponding voltage on the 'transformed' round cable is
260
Cryogenics 1992 Vol 32, No 3
Figure 2 Transformation of flat cable with central insulating layer to hollow cylinder. Note the change of surface element during the transformation
Ut = U, sin ,: = -
B/b 8
sin 3 ~p
Then the solution of the Laplace equation in the cylinder geometry can also be used for the fiat cable, with this boundary condition at the cable surface. Using the relation 3 1 sin 3 ~o = - sin ~o sin (3~p) 4 4 one has
U--- U m
1 -.{-z 2
'{-
sin
( y 3 + z ~ ) sin (3~o)] 1 1 41+z 6
Flat cable with insulation layer It is certain 13-17 that with a metallic core the coupling losses in flat cables are mainly caused by the induced transverse currents, because the thickness of the layer c is much smaller than b, the cable width. The losses between neighbouring strands ('parallel' losses) are due to an induced voltage U. between strands on the same side. These are given ~y13-17 Wp _ B2bcL 2
L
48p
Coupling losses in superconductor cables: S. Takdcs
c[I _b_ 2
0
b
x
Figure 3 Flat cable, including central insulation layer with thickness c~ and resistivity Pi
flat cable to an inhomogeneous cylinder, would lead to the case already treated in the previous section, but with position-dependent resistivity of the inner part. This is not a practicable method. Instead of this, one can return to the flat cable and take into account the continuity of the normal component of the current density at the interface. One then has 2U(z) _ 1 aU(z) p
2dip i
OR
It is assumed that the induced currents between neighbouring strands are closed ('short circuited') by these strands. This assumption is well fulffiled for currents flowing in strands always below the critical current density; a realistic situation at frequencies used for most practical applications (accelerators, fusion, etc.). For the flat cable these losses are much lower than the transverse coupling losses (between strands on different sides of the cable)
where 2di = c i is the thickness of the insulation and pi is its resistivity (the potential difference between both sides of the cable is 2U due to the symmetry, as U = 0 for , p = 0 , ~-). As already mentioned, on the surface we have
W~
B2L2b3
and the solution of the Laplace equation is, therefore
L
120pc
U =Um sin 3 ~o = ~Ur, - [3 sin ~p- sin(3p) ]
U = (alY + ~)sin ~o+ (a2y3 + ~)sin(3~ °) Then Wt_ 48b 3
Wp
_ 2 {b'~ 2
120cbc! 5
\c)
which is about 40 for b/c -~ 10. Therefore the central problem for the flat cable was to decrease substantially the transverse losses. The procedure involving an insulating central core seemed to be a solution, although its usefulness has sometimes been questioned (partly due to the inconvenience of the technology involved in the procedure). To be an adequate solution, the transverse currents should be suppressed almost totally. Otherwise, the currents are forced to flow near the edges. The corresponding losses can also be calculated from the solution of the Laplace equation. If the insulating layer suppresses the transverse currents totally, the transverse losses diminish and the total coupling losses decrease considerably. However, in many cases this would be very difficult, so, for example, the strands should be connected to the central part by some solder near the edges too. This produces the necessary path for the induced transverse currents. Here, for simplicity, we take the case where the resistivity of the core near the edges is the same as that elsewhere (Figure 3). The general case of 'restricted' resistivity due to narrowing (or other effects) would be too complicated. For this case, the transformation to the round cable geometry is straightforward, as given in the preceding section. In addition to the previous calculations 13-17, here we also include the currents through the insulation layer (which were completely neglected previously). The most difficult problem is 'adequate' transformation of the insulating layer. One has to assure its constant resistivity, independent of the angle ~o. This can be done in two ways. The first, the total transformation of the
with 3 a, + b l = = 4 Um,
1 a2 +b2 = - ~ Um
I Z Jr-
= /9
ciP i
2Z 3 +
a I -Z2 }
=_
a2z 2 --
ciPi
P
Z ~-
This leads to 3 l+k al = ~ Um k(1 + z 2) + (1 - Z2)
3 bl = -. UmZ2 k(1 4
k-1 +z 2)+(1-z
4 Um 3 1 + z:
2)
3 r
Um
4
F
l+z2
11 +3k
1 a2 =
_ _
3k(1 + z 6) + (1 - z 6)
4 1
Um
41+Z
1
b2 = - 4
,
-4
fl+l~ 6
Umz 6
3k} 3k-
1
3k(1 + z 6 ) + ( 1 - z
z6(
Um 1 +Z 6
)
6)
1-
Cryogenics 1992 Vol 32, No 3
261
Coupling losses in superconductor cables: S. Takdcs
if pi/p ~ (b/c)(b/q). The ratio between the circular and parallel losses is then
where
k = dipi Ro
Wc 277r2 ~ 1.04 Wp 256
The circular losses in the core layer are then given by W~ =
0
d~p
y dy (E 2 + E2)/p
It is worthwhile mentioning that the circular losses do not change substantially with other potential distributions on the cable surface. Here we give the results for Us ~- sin so and
with Us~ Er =
±sin 2so=-
(al-yb~E~)sinso +3(a2y 2 -yb~24)sin(3so)
8 7r
After integrating we obtain 9 /.fl~r ~ l - z 2 ( 1 ) p ~l-+~Z: 1 + ~ -
1 --Z6 ( 3(1 + z')
~k~-) 1+
+ 9
C 16 c -3 b
Wc=
U~m -
for Us -- sin so and ± sin2so, respectively.
Discussion
I --~/
(a2z 2 _ b2"~q ~,] J [1
- ( ~ - ) 2 ] dx
5~mpb 3 p piCi
The total losses induced by the transverse voltage are then Wet_ 9 r2 U2mC [ l + p (2CiC+ 8 0 bc~c)] L 16 p b Pi 27r2 9 r2U2mc ( 1 + 80 0 b~ic) 16 p b 27r2 Pi because b ~> d. Therefore, we can neglect the additional losses caused by the finite value of the insulating layer, 262
1
The circular losses in these cases (for Pi ~ OO) are given by
For pid~ ~.pd, one obtains the result obtained previously by neglecting the currents through the insulator 17. The second term is due to the currents in the core, which are partially flowing through the insulator. The losses in the insulating layer are
~
sin(3so)
\1~z6/3
9 7r2 U2mc ( 1 + 2 P ~i) 16 p b Pi
,: '[l(a
3.5
]
2 (1 --Z6~2~ +
2]
- sin (5so) - sin(7so) 5.21 7.45
2 "1 z z\2 +-~ (1-~Z2)
= 9 U 2 7rd ( l + 2 P ~ i ) 8 m pR Pi
Wi L --PiCid_b/2 ;
sin so 1
L-=16
1 I)[4-(2n+1)
x sin(2n + 1)so
= (al +yb~) cOs so+3(a2y2+yb~24)cos(3so)
W~
8 ~* ~, a" ~ = ( 2 n +
Cryogenics 1992 Vol 32, No 3
There are many results concerning coupling losses which have undoubtedly been proved experimentally ~8-25, among them some unexpected effects as mentioned in the introduction. Concerning the necessity of having an insulating layer in fiat cables, both experimental results 26 and our calculations show that the transverse losses Wt~ can be orders of magnitude larger than other coupling loss terms, e.g. with respect to the parallel losses Wp
wp
-
5
~>1
If the insulating layer passes through the whole cable, the transverse losses can be decreased considerably. They are essentially given by
Wit-
U2m5b p~ 3d~
Coupling losses in superconductor cables: S. Tak~cs
Appendix
or, with respect to Wp
Wit -
5
Wp
16 p~ ci c
Let us suppose that the currents between two neighbouring strands flowing through two paths in series have a resistance which is effectively given by
p b2
The transverse losses can be neglected for
(Oi/P)
D+27 RI = Pt - dl
b2/ci c.
However, if the insulation between strands on opposite sides is not good (e.g. by including some solder at the cable edges), one has additional losses due to circular currents. For both round and fiat cables, the circular losses are about the same as the parallel losses (and one could assume that the parallel losses would be doubled). There is some experimental evidence for this 21. As we have shown by our examples, the method of coupling loss calculation using solutions of the Laplace equation can be very useful for solving some problems already examined by other methods more precisely, as well as for considering some effects which could not be treated using other methods. In addition, the transformation procedure, r o u n d - f l a t - r o u n d cable gives us the advantages of having relatively simple analytical solutions for the cylindrical geometry. The results for losses in the outer normal layers 27 can also be used.
References 1 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Bean, C.P. Rev Mod Phys (1964) 36 31 Carr, W. J. J Appl Phys (1974) 45 929 Kwasnitza, K. Cryogenics (1977) 17 616 Campbell, A.M. Cryogenics (1982) 22 3 Cave, J.R., F6vrier, H.G., Ky, Y. and Laumond, Y. IEEE Trans Magn (1987) MAG-23 1732 and (1989) MAG-25 1945 Sumiyoshi, F., Matsuyama, M., Noda, M., Matsushita, T., Funaki, K., Twakuma, M. and Yamafuji, K. Jpn J Appl Phys (1986) 25 L148 Collings, E.W. Adv Cryog Eng (1988) 34 867 Matsumoto, K., Akita, S., Tanaka, Y. and Tsukamoto, O. Appl Phys Lett (1990) 57 816 Takdes, S. and Campbell, A.M. Supercond Sci Technol (1989) 1 53 Takdes, S. Czech J Phys (1988) !138 899 Campbell, A.M. Cryogenics (1980) 20 651 Sumiyoshi, F., Nagaishi, H., Oehimaru, O. and Yamafuji, K. Cryogenics (1986) 26 2639 Ries, G. and Takdcs, S. IEEE Trans Magn (1981) MAG-17 2281 Takdcs, S. Cryogenics (1982) 22 661 Talulcs, S. Cryogenics (1984) 24 237 Takdcs, S. IEEE Trans Magn (1991) MAG-27 2206 Takdcs, S. Proc Int Syrup on AC Superconductors and Devices Smolenice, Czechoslovakia, VEDA Publishing Co., Bratislava (1991)5 Hartmann, R.A., Rein, P.C. and van de Klundert, L.J.M. IEEE Trans Magn (1987) MAG-23 1584 Roovers, A.J.M. and van de Klundert, L.J.M. IEEE Trans Magn (1989) MAG-25 2127 Bottura, L. and Minervini, J. IEEE Trans Magn (1989) MAG-25 1730 Kwasnitza, K. and Bruzzone, P. Proc ICEC 11 Butterworths, Guildford, UK (1986) 741 Sehmidt, C. IEEE Tram Magn (1983) MAG-19 707 Kwasnitza, K. and Bruzzone, B. Cryogenics (1984) 24 271 Marken, K.R., Markworth, A.J., Sumption, M.D., Collings, E.W. and Seanlan, R.M. IEEE Trans Magn (1991) MAG-27 1791 Niessen, E.M.J. and van de Klundert, L.J.M. IEEE Trans Magn (1991) MAG-27 1896 Pohik, M., Krempask#, L., Hhisnik, I. and Perot, J. IEEE Trans Magn (1981) MAG-17 2035 Turck, B. J Appl Phys (1979) 50 5397
2RI' + R~ = 2pl d~ D + 2r 2r 2 + P2 d ~ where d2 is the effective thickness for the induced currents (dE = D + 2r). The total resistance is then (see Figure 4) 1
l
-
R,
R;
1
+
2R;'+R;
and the correction factor for the losses (with respect to the case without solder, R~) is given by R~ Rt
-
P2 dl
1
+
Pl d2
Pl 1+--× 02
did2 ?(D + 2r)
For D = 2 7 (as for d2 : D + 27, one obtains R2, _ P2 dl +
Rt
Pl d2
the
LCT
cable 26) and
1
Pl dl l+---P2 P
For the parameters assumed for the inhomogeneous round cable in this paper (pl/P2 = 10) and for a realistic
/
b
~ d*27 (:dz)
R;
Figure 4
(a) Effective resistivity between neighbouring strands by including solder (S) on core (C) and (b) the corresponding compensation scheme
Cryogenics 1992 Vol 32, No 3
263
Coupling losses in superconductor cables: S. Tak~cs solder thickness (dl/r = 0.2) we then obtain
R~
1
R,
21
__R~ I Rt
3
However, for pl/P2 = 100, one obtains
264
Cryogenics 1992 Vol 32, No 3
which means that the parallel losses are decreased by a factor of 21. For the same parameters the circular losses are decreased about 10 times.
Received 9 July 1991; revised 23 September 1991 The coupling losses are calculated for round and flat cables whose cores consist of two layers (an inner insulating layer for flat cables and a thin outer shell for round cables, which is a substitute for the solder connecting the superconducting strands to the core). The previously developed method based on the solution of the Laplace equation and its extension to flat cables is used for calculating the different contributions to coupling losses. It is shown that for flat cables it is necessary to include an insulation layer through the whole cable to suppress totally the transverse losses. Otherwise, the circular losses are of the same order as the parallel losses between neighbouring strands. Decreasing coupling losses by using highly resistive solder for round cables is effective, but not very radical in its effects. To decrease the coupling losses by about one order of magnitude, with realistic solder thickness (about one-tenth of the core radius), one has to use a solder with a resistivity about 100 times higher.
Keywords: a.c. losses; superconducting cables; coupling losses
The losses encountered in changing electromagnetic fields represent one of the main obstacles to more extensive practical applications of type II superconductors. These losses consist of three main components (see, for example, References 1 - 4 ) . Due to the recent development of very fine filament superconductors 5-8, the hysteresis losses can be decreased substantially. In addition, reversible flux motion can brinAg about another significant decrease in these losses 6'9A°. The eddy current losses in the matrix can also be lowered markedly by decreasing the twist pitch Ip and introducing a highly resistive matrix between the filaments, such as C u - N i 5-7 or C u - M n 8. The coupling losses, which are actually a type of eddy current losses, are caused by currents between strands and other substructures of the cables or braids. Because the structure of conductors used for producing high current densities is often very complicated, the calculation of coupling losses is, in the most cases, not a simple task. The cables are produced in multiple stage processes, with different twist and cabling lengths. The differences in these lengths can lead to additional a.c. losses ll'12. Analytical calculations are hard to realize and even the formulation of problems is often difficult. In the 'history' of coupling loss calculations, many surprising effects have appeared, partly due to the unexpected order of magnitude of these effects. As an example, we could mention the size effects in finite cables 13 or the behaviour of cables in spatially O011 - 2275/92/030258 ©
1992. Butterworth
258
- 07 - Heinemann
Ltd
Cryogenics 1992 Vol 32, No 3
inhomogeneous applied fields 14'15. In addition, some 'new' (forgotten?) contributions to losses have appeared (currents in finite samples, currents at finite magnetic field range, 'circular' currents ~6'~7, etc.). Maybe even more exist. Hence, it would be very useful to clarify some problems involving simplified or model structures as well as to perform more precise calculations on problems already assumed to be solved. All these questions can be solved only by comparing different methods of loss calculation; the methods used previously for coupling loss calculations have been different in several ways (calculations of currents or voltages between the strands 13-t~, Fourier transforms, numerical solutions '8-2°, etc.). We developed an alternative procedure by solving the Laplace equation, knowing the voltage distribution on the cable surface 16'17. Here, we use this method for solving the Laplace equation and calculating the coupling losses in two cases. The first case involves a cylinder geometry with different resistivity of the cylinder sheath (which cart correspond to the solder connecting the strands to the cable) and its inner part (which can even be insulating). The second case is connected with the (sometimes questioned) necessity for the central insulating layer in flat cables. In all our calculations we assume an infinitely long fiat or round cable with single layered strands in a transverse (spatially homogeneous) applied a.c. magnetic field B. The cabling length L is always supposed to be much
Coupling losses in superconductor cables: S. Takdcs longer than the twist pitch lp. The number of strands should be sufficient to enable us to perform integration instead of summation of the different fields and voltages from the individual strands 13-15.
Solution of the Laplace equation The general solution of the Laplace equation for cylinder geometry is oo
U= ~
(a. r" + b. r-") [ c. cos(n~o) + d. sin(n~o) ]
n=l
+ const Figure 1 Round cable with surface layer of different resistivity
For an applied magnetic field perpendicular to the x axis (~ = 0, 70, we can assume, without loss of generality, that U = 0 on the x axis (due to the symmetry of the problem). The solution is then
(due to the existence of some solder, for example)
co
U= E
(a.r" + b.r-")sin(n~o)
2 Due to the finite nature of
n=l
The boundary conditions are given by the following relations: On the surface U, = U(~) is given by external sources; the 'peripheral' currents between neighbouring strands are closed, leading to parallel losses Wp. For single-layered strands, one has~3-~7
BbL us =
_ _
4r
BbL us =
--
16
U2=(Ay+B)sin~o at the centre, we have B = 0. 3 Due to the continuity of U at the interface, we obtain
al z + _b~ = Az Z
sin ~o = Um sin ~o sin 3 ~o =Um sin 3 ~o
for the round and flat cable, respectively. 2 The current density normal to the interfaces should be continuous. 3 The potential at metallic interfaces should be continuous. 4 The potential should be finite everywhere (this leads to b, = 0 for solid cylinders).
Losses in round cables Usually the strands (for simplicity we assume singlelayered strands) are connected to the central part of the cable by some solder whose resistivity can be very different from that of the central core. We replace this by assuming a surface layer with thickness d and resistivity 01, with the resistivity of the central core being P2 and the radius of the cylinder R (Figure 1). Then, both solutions of the Laplace equation should obey the following boundary conditions:
with z = R~/R. The normal component of the current density at the interface should be continuous too and, therefore
(a-,
Pl
P2
These conditions lead to al =Um - - +, 1 g N
bl = U m
z- 2 (- 1 -- g ) ,
A -
N
2Urn N
with g=P_L,N=l+g+z2(l_g)=l
+z2+g(1-z
2)
P2
The total transverse loss of the cable (per cabling length L) is then
1 Taking Us = Um sin ~o (other values of Us(~o) could be included as well), we have
U~ = (aly + ~ ) s i n ¢ withy=r/R
and
a+b=Um.
j0 p2 L\ay# N 2 Pl
\a
{[(l + g ) 2 + z 2 ( l - g ) e ] (
yj] l _ z 2 ) + 4 z 2}
Cryogenics 1992 Vol 32, No 3
259
Coupling losses in superconductor cables: S. Takdcs dx
Here we give the results for two special cases in more detail. For the homogeneous cylinder (g = 1) one has
wtl L
"$r--
p
Comparing this with the result for a thin hollow cylinder (or with the insulated inner part, i.e. P2 ~ 01) with thickness d one obtains Wt2 _ r U2m d -
L
o R
7r B2Rd
160
The ratio Wt2/Wtl is proportional to d / R (as we shall see later, this ratio is nearly quadratic in d/R for the flat cable). In addition, Wt2 for the hollow cylinder is comparable with the parallel losses Wp between neighbouring strands, as stated earlier 16'~7. For practical purposes, it could be important to know whether using a highly resistive solder can decrease the losses considerably 2~. We take as an example the case where the thickness of the solder is about one-tenth that of the cylinder, but its resistivity is increased 10 times (d/R = 1 - z = 0.1, g = 10). The transverse losses are then decreased by about a factor of 1/2. In addition, the parallel losses are decreased by nearly the same factor (see Appendix).
Transformation: f l a t - round - flat cable We shall consider an infinitely long flat cable with single-layered strands (on both sides of the flat core with thickness c) in a perpendicular harmonic applied magnetic field. The resistivity of the core tape P is assumed to be homogeneous. The solution of the Laplace equation is comparatively simple for the cylinder geometry; therefore, by transforming the flat cable to the round cable problem one obtains many advantages. The principles of the transformation can be seen in Figure 2. The insulating core is transformed to the hollow part of the cylinder and the thickness of the round cylinder is half the core layer thickness. The 'geometrical' factor b - 7rR (or R - b/Tr) has to be taken into account in both transformations, although this is not a completely precise procedure. The induced transverse voltage on the flat cable is given by 11,17 BLb G = sin2 ~o 8 with sin2 ~o = 1 - ( ~ - ) 2
where b is the cable width and L the cabling length. Because dx = dl sin ~o (see Figure 2), the corresponding voltage on the 'transformed' round cable is
260
Cryogenics 1992 Vol 32, No 3
Figure 2 Transformation of flat cable with central insulating layer to hollow cylinder. Note the change of surface element during the transformation
Ut = U, sin ,: = -
B/b 8
sin 3 ~p
Then the solution of the Laplace equation in the cylinder geometry can also be used for the fiat cable, with this boundary condition at the cable surface. Using the relation 3 1 sin 3 ~o = - sin ~o sin (3~p) 4 4 one has
U--- U m
1 -.{-z 2
'{-
sin
( y 3 + z ~ ) sin (3~o)] 1 1 41+z 6
Flat cable with insulation layer It is certain 13-17 that with a metallic core the coupling losses in flat cables are mainly caused by the induced transverse currents, because the thickness of the layer c is much smaller than b, the cable width. The losses between neighbouring strands ('parallel' losses) are due to an induced voltage U. between strands on the same side. These are given ~y13-17 Wp _ B2bcL 2
L
48p
Coupling losses in superconductor cables: S. Takdcs
c[I _b_ 2
0
b
x
Figure 3 Flat cable, including central insulation layer with thickness c~ and resistivity Pi
flat cable to an inhomogeneous cylinder, would lead to the case already treated in the previous section, but with position-dependent resistivity of the inner part. This is not a practicable method. Instead of this, one can return to the flat cable and take into account the continuity of the normal component of the current density at the interface. One then has 2U(z) _ 1 aU(z) p
2dip i
OR
It is assumed that the induced currents between neighbouring strands are closed ('short circuited') by these strands. This assumption is well fulffiled for currents flowing in strands always below the critical current density; a realistic situation at frequencies used for most practical applications (accelerators, fusion, etc.). For the flat cable these losses are much lower than the transverse coupling losses (between strands on different sides of the cable)
where 2di = c i is the thickness of the insulation and pi is its resistivity (the potential difference between both sides of the cable is 2U due to the symmetry, as U = 0 for , p = 0 , ~-). As already mentioned, on the surface we have
W~
B2L2b3
and the solution of the Laplace equation is, therefore
L
120pc
U =Um sin 3 ~o = ~Ur, - [3 sin ~p- sin(3p) ]
U = (alY + ~)sin ~o+ (a2y3 + ~)sin(3~ °) Then Wt_ 48b 3
Wp
_ 2 {b'~ 2
120cbc! 5
\c)
which is about 40 for b/c -~ 10. Therefore the central problem for the flat cable was to decrease substantially the transverse losses. The procedure involving an insulating central core seemed to be a solution, although its usefulness has sometimes been questioned (partly due to the inconvenience of the technology involved in the procedure). To be an adequate solution, the transverse currents should be suppressed almost totally. Otherwise, the currents are forced to flow near the edges. The corresponding losses can also be calculated from the solution of the Laplace equation. If the insulating layer suppresses the transverse currents totally, the transverse losses diminish and the total coupling losses decrease considerably. However, in many cases this would be very difficult, so, for example, the strands should be connected to the central part by some solder near the edges too. This produces the necessary path for the induced transverse currents. Here, for simplicity, we take the case where the resistivity of the core near the edges is the same as that elsewhere (Figure 3). The general case of 'restricted' resistivity due to narrowing (or other effects) would be too complicated. For this case, the transformation to the round cable geometry is straightforward, as given in the preceding section. In addition to the previous calculations 13-17, here we also include the currents through the insulation layer (which were completely neglected previously). The most difficult problem is 'adequate' transformation of the insulating layer. One has to assure its constant resistivity, independent of the angle ~o. This can be done in two ways. The first, the total transformation of the
with 3 a, + b l = = 4 Um,
1 a2 +b2 = - ~ Um
I Z Jr-
= /9
ciP i
2Z 3 +
a I -Z2 }
=_
a2z 2 --
ciPi
P
Z ~-
This leads to 3 l+k al = ~ Um k(1 + z 2) + (1 - Z2)
3 bl = -. UmZ2 k(1 4
k-1 +z 2)+(1-z
4 Um 3 1 + z:
2)
3 r
Um
4
F
l+z2
11 +3k
1 a2 =
_ _
3k(1 + z 6) + (1 - z 6)
4 1
Um
41+Z
1
b2 = - 4
,
-4
fl+l~ 6
Umz 6
3k} 3k-
1
3k(1 + z 6 ) + ( 1 - z
z6(
Um 1 +Z 6
)
6)
1-
Cryogenics 1992 Vol 32, No 3
261
Coupling losses in superconductor cables: S. Takdcs
if pi/p ~ (b/c)(b/q). The ratio between the circular and parallel losses is then
where
k = dipi Ro
Wc 277r2 ~ 1.04 Wp 256
The circular losses in the core layer are then given by W~ =
0
d~p
y dy (E 2 + E2)/p
It is worthwhile mentioning that the circular losses do not change substantially with other potential distributions on the cable surface. Here we give the results for Us ~- sin so and
with Us~ Er =
±sin 2so=-
(al-yb~E~)sinso +3(a2y 2 -yb~24)sin(3so)
8 7r
After integrating we obtain 9 /.fl~r ~ l - z 2 ( 1 ) p ~l-+~Z: 1 + ~ -
1 --Z6 ( 3(1 + z')
~k~-) 1+
+ 9
C 16 c -3 b
Wc=
U~m -
for Us -- sin so and ± sin2so, respectively.
Discussion
I --~/
(a2z 2 _ b2"~q ~,] J [1
- ( ~ - ) 2 ] dx
5~mpb 3 p piCi
The total losses induced by the transverse voltage are then Wet_ 9 r2 U2mC [ l + p (2CiC+ 8 0 bc~c)] L 16 p b Pi 27r2 9 r2U2mc ( 1 + 80 0 b~ic) 16 p b 27r2 Pi because b ~> d. Therefore, we can neglect the additional losses caused by the finite value of the insulating layer, 262
1
The circular losses in these cases (for Pi ~ OO) are given by
For pid~ ~.pd, one obtains the result obtained previously by neglecting the currents through the insulator 17. The second term is due to the currents in the core, which are partially flowing through the insulator. The losses in the insulating layer are
~
sin(3so)
\1~z6/3
9 7r2 U2mc ( 1 + 2 P ~i) 16 p b Pi
,: '[l(a
3.5
]
2 (1 --Z6~2~ +
2]
- sin (5so) - sin(7so) 5.21 7.45
2 "1 z z\2 +-~ (1-~Z2)
= 9 U 2 7rd ( l + 2 P ~ i ) 8 m pR Pi
Wi L --PiCid_b/2 ;
sin so 1
L-=16
1 I)[4-(2n+1)
x sin(2n + 1)so
= (al +yb~) cOs so+3(a2y2+yb~24)cos(3so)
W~
8 ~* ~, a" ~ = ( 2 n +
Cryogenics 1992 Vol 32, No 3
There are many results concerning coupling losses which have undoubtedly been proved experimentally ~8-25, among them some unexpected effects as mentioned in the introduction. Concerning the necessity of having an insulating layer in fiat cables, both experimental results 26 and our calculations show that the transverse losses Wt~ can be orders of magnitude larger than other coupling loss terms, e.g. with respect to the parallel losses Wp
wp
-
5
~>1
If the insulating layer passes through the whole cable, the transverse losses can be decreased considerably. They are essentially given by
Wit-
U2m5b p~ 3d~
Coupling losses in superconductor cables: S. Tak~cs
Appendix
or, with respect to Wp
Wit -
5
Wp
16 p~ ci c
Let us suppose that the currents between two neighbouring strands flowing through two paths in series have a resistance which is effectively given by
p b2
The transverse losses can be neglected for
(Oi/P)
D+27 RI = Pt - dl
b2/ci c.
However, if the insulation between strands on opposite sides is not good (e.g. by including some solder at the cable edges), one has additional losses due to circular currents. For both round and fiat cables, the circular losses are about the same as the parallel losses (and one could assume that the parallel losses would be doubled). There is some experimental evidence for this 21. As we have shown by our examples, the method of coupling loss calculation using solutions of the Laplace equation can be very useful for solving some problems already examined by other methods more precisely, as well as for considering some effects which could not be treated using other methods. In addition, the transformation procedure, r o u n d - f l a t - r o u n d cable gives us the advantages of having relatively simple analytical solutions for the cylindrical geometry. The results for losses in the outer normal layers 27 can also be used.
References 1 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Bean, C.P. Rev Mod Phys (1964) 36 31 Carr, W. J. J Appl Phys (1974) 45 929 Kwasnitza, K. Cryogenics (1977) 17 616 Campbell, A.M. Cryogenics (1982) 22 3 Cave, J.R., F6vrier, H.G., Ky, Y. and Laumond, Y. IEEE Trans Magn (1987) MAG-23 1732 and (1989) MAG-25 1945 Sumiyoshi, F., Matsuyama, M., Noda, M., Matsushita, T., Funaki, K., Twakuma, M. and Yamafuji, K. Jpn J Appl Phys (1986) 25 L148 Collings, E.W. Adv Cryog Eng (1988) 34 867 Matsumoto, K., Akita, S., Tanaka, Y. and Tsukamoto, O. Appl Phys Lett (1990) 57 816 Takdes, S. and Campbell, A.M. Supercond Sci Technol (1989) 1 53 Takdes, S. Czech J Phys (1988) !138 899 Campbell, A.M. Cryogenics (1980) 20 651 Sumiyoshi, F., Nagaishi, H., Oehimaru, O. and Yamafuji, K. Cryogenics (1986) 26 2639 Ries, G. and Takdcs, S. IEEE Trans Magn (1981) MAG-17 2281 Takdcs, S. Cryogenics (1982) 22 661 Talulcs, S. Cryogenics (1984) 24 237 Takdcs, S. IEEE Trans Magn (1991) MAG-27 2206 Takdcs, S. Proc Int Syrup on AC Superconductors and Devices Smolenice, Czechoslovakia, VEDA Publishing Co., Bratislava (1991)5 Hartmann, R.A., Rein, P.C. and van de Klundert, L.J.M. IEEE Trans Magn (1987) MAG-23 1584 Roovers, A.J.M. and van de Klundert, L.J.M. IEEE Trans Magn (1989) MAG-25 2127 Bottura, L. and Minervini, J. IEEE Trans Magn (1989) MAG-25 1730 Kwasnitza, K. and Bruzzone, P. Proc ICEC 11 Butterworths, Guildford, UK (1986) 741 Sehmidt, C. IEEE Tram Magn (1983) MAG-19 707 Kwasnitza, K. and Bruzzone, B. Cryogenics (1984) 24 271 Marken, K.R., Markworth, A.J., Sumption, M.D., Collings, E.W. and Seanlan, R.M. IEEE Trans Magn (1991) MAG-27 1791 Niessen, E.M.J. and van de Klundert, L.J.M. IEEE Trans Magn (1991) MAG-27 1896 Pohik, M., Krempask#, L., Hhisnik, I. and Perot, J. IEEE Trans Magn (1981) MAG-17 2035 Turck, B. J Appl Phys (1979) 50 5397
2RI' + R~ = 2pl d~ D + 2r 2r 2 + P2 d ~ where d2 is the effective thickness for the induced currents (dE = D + 2r). The total resistance is then (see Figure 4) 1
l
-
R,
R;
1
+
2R;'+R;
and the correction factor for the losses (with respect to the case without solder, R~) is given by R~ Rt
-
P2 dl
1
+
Pl d2
Pl 1+--× 02
did2 ?(D + 2r)
For D = 2 7 (as for d2 : D + 27, one obtains R2, _ P2 dl +
Rt
Pl d2
the
LCT
cable 26) and
1
Pl dl l+---P2 P
For the parameters assumed for the inhomogeneous round cable in this paper (pl/P2 = 10) and for a realistic
/
b
~ d*27 (:dz)
R;
Figure 4
(a) Effective resistivity between neighbouring strands by including solder (S) on core (C) and (b) the corresponding compensation scheme
Cryogenics 1992 Vol 32, No 3
263
Coupling losses in superconductor cables: S. Tak~cs solder thickness (dl/r = 0.2) we then obtain
R~
1
R,
21
__R~ I Rt
3
However, for pl/P2 = 100, one obtains
264
Cryogenics 1992 Vol 32, No 3
which means that the parallel losses are decreased by a factor of 21. For the same parameters the circular losses are decreased about 10 times.