calculation of hysteresis losses in hard super-...

Calculation of hysteresis losses in hard superconductors carrying ac: isolated conductors and edges of thin sheets

W. T. NORRIS

Central Electricity Research Laboratories of the Central Electricity Generating Board, Leatherhead, Surrey

MS. receiwd 19th May 1969, in revised form 28th Nocember 1969

Abstract. Two methods of calculating hysteresis losses in hard superconductors are described. The London model is assumed in which the critical current density is taken independent of magnetic field. Losses in isolated wires of different cross section are considered but it is found that losses for solid wires vary by at most a factor of 3 for different shaped wires of the same current-carrying capacity. The loss at saturation current is usually 0.4-0.6 I c z p ~ / n .

Losses at theedges of thin sheets are also calculated and a fourth-power dependence on current (for low current) is found. Three systems are examined: a slit parallel to the current in a wide sheet (Le - poj2g2r3F4/24), one pair of the edges of two wide strips set back-to-back and carrying antiparallel currents (Le 2 ponj2s2F4/6) and a long thin wall parallel to the current flow on a wide sheet ( L c ~ p ~ . r r 3 j 2 a 2 F ' i / 3 ) . Le is the loss per cycle per unit length, F is the current peak as a fraction of saturation current, g the width of the slit, s the spacing of strips, a the height of the asperity and j the critical current density per unit width. All in MKS units.

1. Introduction Any theoretical estimation of ac losses in a superconductor is unlikely to accord with

measurements very accurately. The materials are variable and we are still uncertain what are the important physical qualities which influence losses. This is particularly so for type I or surface superconductors, run below their first critical field.

For worked type 11 materials, the so-called hard superconductors (HSC), run at high currents, an approximate method of calculation due to London (1963) is available. An idealized behaviour is assumed. The superconductors can carry current up to some maximum critical current density and then become resistive. The resistance is assumed to rise very steeply as the current tries to increase above the critical value and the resistance is such that ohmic voltage drop exactly balances the driving emf with the current density remaining constant.

I t is assumed for the simplest form of the model, and in this paper, that the critical current density is independent of ambient magnetic field although it is well known that critical currents depend not only on the magnitude but also the direction of the field. Such field dependence can be incorporated in calculations of losses and this has been done by Hancox (1966) and Dunn and Hlawiczka (1968) (who also consider surface currents) for electric circuits of simple geometric configuration, e.g. round wires, infinite sheets.

In real materials the resistance rises only to a finite value and not then as a sharp step function. The error introduced by neglecting the second of these deficiencies in the model is insignificant in comparison with the assumption of constant critical current and the error introduced by the first deficiency is not significant at frequencies of up to 1 kHz.

An important characteristic of current distributions in hard superconductors is that as conditions change, e.g. as the current in a wire increases, the current distribution begins altering near the outside first and only later in deeper regions. This principle needs careful

489

490 W. T. Norris

application. One simple example is a system consisting of a long bar of superconductor concentric with a long evenly and closely wound solenoid in which the current is slowly increasing (figure 1). When the field due to this exciting current is small currents flow,

__c Depth into surface

Sample o f hard ( j J y superconductor

i --Current o f exciting 1 col1

Figure 1. Current and magnetic field variations in a round bar excited with a concentric solenoid. The two rows of graphs show how the distribution varies as the applied field rises from zero and falls back through zero. In the case of a wire carrying current the distributions are similar but the current is axial and the field circumferential. (The drawings are made assuming the penetration of

flux is slight : deeper penetration would distort the distributions slightly.)

according to Lenz’s law, in the surface round the bar of superconductor to produce a field counter to the exciting field and to prevent the penetration of flux into the bar. As the exciting field increases further the outer regions reach the critical current density and deeper regions attempt to exclude flux from the centre of the specimen. There is a gradual penetration of flux into the bar. Flux passing over the currents in the outer layers attempts to increase these currents but is prevented from doing so by the resistance which appears and which also leads to the ohmic dissipation of heat. This is one way of looking at the loss.

When the exciting field is reversed Lenz’s law again shows that the current reversals occur in outer regions first and then deeper in the specimen. The distribution of flux and current is shown in figure 1. It will be observed that when the external magnetic field has returned to zero there is still flux in the specimen. This trapped flux is typical of a hysteresis effect and the loss is indeed a hysteresis loss in a diamagnetic specimen. According to the theory the power dissipation is directly proportional to the frequency of an alternating effect, the energy loss per cycle being constant.

In a wire carrying transport current, the flux also penetrates from the outside, although the current flow is now parallel to the axis of the wire and the magnetic field circumferential.

2. Calculation procedures for losses Since infinitely long straight conductor systems are to be discussed exclusively, it will be

useful to have the word ‘fibre’ to mean an infinitely long straight element of the conductor parallel to the conductor itself. The position of the fibre will be specified by a two- dimensional vector giving the point of intersection of the fibre with the cross-sectional plane in which the vector lies. The complete specification of a fibre will include its (infinitesimal) cross-sectional area. I t will also be useful to consider lines outside but parallel to the conductor : these lines are similarly specified.

Calculation of hysteresis losses in hard superconductors 49 1

Magnetic fluxes and losses are always understood to be expressed per unit length of conductor. MKS units are used throughout the paper unless it is stated otherwise.

2.1. A general method One method of calculating losses is to introduce the wire into a situation where from

first principles we can calculate the loss in terms of the magnetic field produced and currents flowing.

I t is useful to notice that in the case of an isolated wire there is usually some fibre, which we call the kernel, which carries no current at any time (except perhaps during an infinitesimal period at peak current flow), across which no flux ever passes, and along which the electric (electrostatic plus magnetically induced) field is zero since the resistance and the current are both zero.

Imagine a long wire carrying current, the return being through a perfectly conducting cylinder concentric with the wire, but of sufficiently large radius that the current is uniformly distributed around its periphery. To achieve this condition the radius R of the return cylinder is allowed to approach infinity and the loss in the wire is calculated at this limit. A generator giving an emf E (per unit length) is used to drive the current. Let r , referred to the kernel, define a line parallel to the conductor. The flux per unit length between the kernel fibre and the outer cylinder is

R += lim !” B.dr 0

R-+ m

where B is the magnetic field (induction) at r and R is a point on the outer cylinder. Since, as we have observed, there is no current

and no resistive voltage drop along the kernel, the emf induced by this changing flux is balanced by the emf of the generator driving current round the circuit of the wire and the cylindrical return path. The emf (per unit length) required from the generator is

As the current changes, this flux changes.

R

The instantaneous power input from the supply during current changes is the product of this emf and the instantaneous current :

R

where l i s the current in the wire, and so through the generator. Let Zm be the peak current magnitude under alternating conditions. The net energy input from the generator as the current changes from - Zm to + Zm is all dissipated since the magnetic fields at the end of the process are the reverse of those at the beginning and the energy stored in the magnetic field is the same. The energy lost in half a cycle (per unit length) is then

4c 2 = T E Z d t = r z d+. ( 1 4 - Im - Im

It is often useful to introduce a dummy variable especially if I and B can be expressed in terms of it. Suppose S is such a variable and S- and S+ are the limiting values of this variable during a half cycle from peak current to negative peak current. The loss in half a cycle

R

492 W. T. Norris

Provided a kernel can be found in the wire, this method is quite general and independent of how the material resistance varies with I or E.

If two conductors carrying antiparallel currents are under consideration and a kernel can be found in each, then a similar argument shows that, if 0 is the flux between the kernels when current I is flowing,

Lc= 2 1 I d @ . (IC)

& cycle

2.2. Alternative method The existence of a kernel can also be used for a more direct evaluation of the loss by

summing dissipation in individual fibres. In this case more must be assumed about the properties of the conductor material. Consider a fibre of area 6az at r2 relative to the kernel. Since there is no electric field along the kernel the electric field along this fibre is &$,/at where 4 2 is the flux between the kernel and YZ. Thus the energy dissipation rate in the fibre J'Z is

u (a$) 6a2.

where U is the current density. This expression can be integrated both over time and the cross section to give the total

loss in the wire. In the London model adopted here, 5 is taken as constant. The current at any point is

+ 5, - U, or zero, depending on the stage in the cycle or whether during a cycle flux penetrates to the fibre at all. In following the further development of the argument it may help if the reader has in mind a wire of circular cross section and then subsequently reviews wires of other cross section.

We suppose during a half cycle from peak current to negative peak current that flux, and so current reversal, penetrates evenly from the outside. During the half cycle, once flux has started to cross a fibre and the current has reversed, flux continues to cross in the same sense and so also the current in that fibre remains constant for the remainder of the half cycle. Integrating expression (2) over a half cycle, the loss in a half cycle in a fibre becomes

4 2 Sa2 ( 2 4 where A42 is the change in 4 2 . The flux changes monotonically and U is assumed constant. Since at the end of the half cycle the magnetic fields are simply reversed, A+z is twice the flux 02 at peak current between the kernel and the fibre ~ 2 .

If the peak current does not saturate the conductor it is useful to observe that there is a permanent field-free core : 0 2 is the flux between the fibre YZ and this inactive core and the kernel may be taken as any fibre within the core.

The loss per cycle can be found by integrating expression (2a) to give

L 4 - 4uO2daz. (3)

To calculate the loss, all it is required to know is the flux distribution within the conductor at peak current. Choosing between the method of this section and that of $2.1 is a matter of deciding whether 0 2 for all points in the section at peak current, or B along a radius at all times in a half cycle, is easier to determine.

The flux 0 2 for a wire carrying saturation current is given by simple double integration of the Biot-Savart law (figure 2) :

Calculation of hysteresis losses in hard superconductors 493

Figure 2. Illustrating calculation of loss when peak current is the saturation current.

where dal is the area of a fibre at YI. Thus the loss per cycle (per unit length), for a conductor whose peak current saturates the conductor, is

--- - 2p0ze2 (in Rs - In Re) 7T

where Rs is the geometric mean distance (gmd) of the current-carrying part of the wire cross section from itself and Rc is the gmd of the wire cross section from the kernel. IC is the critical saturation current. (poIC/2r) In Rc is the value of the vector potential of the magnetic field at the magnetic centre when IC is flowing ; In Rs is related to the ‘internal’ inductance of the wire (Gray 1967).

For currents Zm below IC the values of Rc and Rs should be those for the section of wire actually carrying current : the kernel for calculation of R, can be at any convenient point in the inner current and field-free core ; Im replaces IC in equation (3b).

3. Losses in wires carrying saturation peak current 3 , l . Round- or elliptic-sectioned wires

The method of 42.2 can be applied straightforwardly to the case where wires carry the full saturation current. The required gmd’s can be found in texts such as that of Grover (1962). Thus the loss per cycle per unit length in a wire of circular or elliptical cross section is

3.2. Round wires side by side

of n wires are given in table 1. A strip could be formed of round wires layed side by side. The losses in strips composed

3 .3. Rectangular conductors carrying saturation current For a rectangle of sides a and b the exact expressions for the gmd‘s Rc and Rs are rather

long strings of logarithmic and inverse tangent terms (Gray 1967) but the formulae are well approximated by

In Rs=ln(a+b)-3+ln E1 In Re = In &(a + b) - 1 +In E2

where E1 and E2 are very close to unity and depend only on the ratio a/b (Grover gives

494 W. T. Norris

Table 1. Losses in strip composed of n round wires layered side by side. (The loss per cycle is A(p0lc2/n) where A is tabulated and IC

is the critical current of the arrangement)

Number of wires Loss factor A

1 0 .5 2 0.445 3 0.474 4 0.434 5 0.452

Infinite 0.386 3 wires in cluster 0.472

values of In El). Thus the loss per cycle becomes :

= (0.40 & 0.035). ir

Values of In E3 are given in table 2.

Table 2. Correction factor for losses in rectangular strip

alb 0.0 0.1 0 . 2 0 .4 0 .6 0.8 1.0 h E 3 0.0 -0.023 +0.034 +0*052 +0.059 $0,063 $0,063

The case a/b =O*O corresponds to the thin strip. Rectangular conductors show slightly lower losses than round wires.

3 . 4 . Cruciform-sectioned conductor

good method of reducing losses. like a Greek cross, each arm being a thin sheet. Let a be the length of each arm:

The strip has lower losses than the round wire : angularity may at first sight seem to be a Consider therefore a conductor with a cross section

In Rs=ln a-;++ In 2+$ir In Rc=ln a- 1

and the loss per cycle is Le=- poL2 (+ In 2 + t n - 1)

7T

which is rather higher than that for a round wire.

3 .5 . Box-sectioned conductor

methods of $2.2, is Consider a hollow conductor of square section and very thin walls. The loss, using the


This low loss may be understood by regarding the section as an only slightly distorted annulus.

4. Variation of loss with changes in peak current If the peak current is greater than the saturation current the material goes ‘normal’ and

losses increase greatly. If the peak current is less than the saturation value the first question is to determine the parts of the wire which remain free of current. These are delineated by contours reached by current reversal at intermediate stages during the full cycle and are such that current flowing in the conductor cross section outside them produces no magnetic field within the contour. They are the same contours reached by current and field for lower peak currents in the wire. Thus an intermediate stage is described by superposition of the current distribution at peak current and minus twice the distribution of some lower peak current to achieve reversed current in the outer regions. For a strip conductor some contours are shown schematically in figure 3.

Y

t X

Figure 3. Contours of current penetration in a rectangular conductor shown schematically.

The determination of the contours for simple cross sections is comparatively trivial. For infinitely wide flat-sheet conductors the contours are straight lines parallel to the conductor surface and the current ; for circular- or annular-section conductors the contours are concentric circles.

4.1. Losses in wires in elliptical section An attractive concise, but generally insoluble, method for finding contours in wires of

any section arises directly from a paper by Beth (1967). Consider the section of the wire represented by a contour C1 in the complex z(=x+iy) plane. Then the magnetic field H,+iH,, at Z=X+iY, is given by

for Z inside CI C,

(7)

where the asterisk denotes the complex conjugate, and cr is the current density in the wire. Thus if CZ is a contour internal to C1 such that, with current flowing in the region between C1 and CZ there is no magnetic field inside CZ, then CZ must be such that, with Z inside CZ,

496 W. T. Norris

magnetic field (expressed now as B,+iB,) in and around an elliptical wire is No general solution for CZ is known. However for an ellipse Beth observes that the

B= (bX- ia Y) inside a+b (9)

It is to be noted that, as for a round wire, the field inside the wire depends on the position the current density and the eccentricity of the ellipse (i.e. on the ratio m=b/a) and not on the size of the wire. Thus it is clear that an inner contour C2 as defined above is simply a similar concentric ellipse. The simplicity of the formulae suggests the method of 52.1 will conveniently allow prediction of losses.

Consider a wire whose section is an ellipse of semi-major axis a and which carries alternating current, and consider the condition when the current reversal penetrates at peak current to an ellipse of semi-major axis ao. Let a1 be the semi-major axis of the current reversal contour of some intermediate state so that the current is then

I= 7ToLu(2a12-uo~-a2). (10) The magnetic field on the x axis is obtained from (9) setting Y=O. Taking into account

the different regions (see figure 4)

Figure 4. Contours of current penetration in an elliptical wire.

X<Ul

Thus for x > a l x -a ( lim B dx) = ap0o(-2a1 In a1 + a1 (constant)}.

aai X+m

Substituting equations (10) and (1Oc) in equation (lb), the loss per cycle 0

L c = 2 J 1 4 ( aa1 x - t m lim f B d x ) dal U 0

=npo,2,2 (I - - - - -2 a 2 ao21n (:)I.

Calculation of hysteresis losses in hard superconductors

If the current is FZc, where Zc is the critical current,

and Z, = o m a 2

Lc=-- ((1 - F ) In (1 -F)+(2-F) Fj2) P p o 7 7

497

(1 1)

which is independent of the eccentricity and, thus, is the same formula as for a round wire ; L1 is given as a function of F i n table 3.

Table 3. Loss factors for various systems carrying peak alternating current

For small values of F : L12: F 3 / 6 r , LP 2: F4/6r , L3 N F4r3/24, L4 N F47r/6, LS N F4r3 /3 .

less than the saturation value

F= I/Ic. L, is loss per cycle per unit length ; j is current density per unit width on sheets ; SI units.

F L1 Lz L3 Elliptical Strips Gap, g ,

wires Lc = Ic2poL2 in sheet Le = Ze‘poL1 Lc= poj2g2L3

Equation (1lb) Equation (16a) Equation (22a) 0.1 5.6 10-5 5 .4 10-6 1.33 10-4 0.2 4.7 10-4 8 .6 10-5 2.05 10-3

0-2 0-2 0-1 0-1

0 .3 1 . 7 0 .4 4.3 10-3 0 .5 9 .1 10-3 0 . 6 1 .7 0 .7 3 .0 0 .8 5.0 0 .9 8 .4 10-2 0.95 1.11 10-1 0.98 1.34 10-1 1 1 . 6 10-1

4.44 1.45 3.7 8 .4 1.62 3.06 5.7 8 .0 1 .0 1.23

10-4 10-3 10-3 10-3 10-2 10-2 10-2 10-2 10-2 10-1

1.22 4.7 1.45 4 .2 1.23 4 .3

26.0 128 910 cc

L4 Pair of edges of antiparallel strips, s apart Le= poj2s2L4 Equation (30a)

5.25 10-5 8 . 5 10-4 4 . 4 10-3 1.45 10-2 3.7 10-2 8.25 1.68 10-1 3 .4 10-1 6.9 10-1 1.09 1.79

LS Asperity height a

Lc= poj2a2Ls

Equation (32a) 9 .8 10-4 1-36 5.6 1.36 10-1 2.54 10-1 4 .0 10-1 5.7 10-1 7.5 10-1 9 .3 10-1 1.03 1.09 1.13

5. Losses in a thin strip of finite width For a rectangular sectioned conductor a number of current reversal contours are shown

schematically in figure 3. A current- and field-free region in the centre is shown also. It is especially to be recollected that no flux crosses this region. If the rectangle is very elongated so that the conductor approximates a thin strip at peak current the magnetic field is then very closely perpendicular to the strip in those parts where current occupies the full thickness and very small and parallel to the strip in the other central part. For the thin strip we may talk then about the current per unit width which has a saturation valuej, where current occupies the full thickness, and a lower value elsewhere. At these other points the magnetic field perpendicular to the strip is zero.

Figure 5(a) shows the current and perpendicular field distributions across the strip at peak current, with the strip saturated at the edges and carrying lower currents in the middle. The preliminary problem is to find this distribution of current across the middle parts. The central currents are, of course, confined to the faces of the strip, the very centre being current-free.

Figure 5(b) shows current and field distributions at an intermediate stage during current reversal, found as before by superposition of the peak current conditions and minus twice

498 W. T. Norris

4 Current density

Current density t Transverse

t

Figure 5 . (a) Current and field distribution across a superconductor strip at peak current ; (b) at lower current after the first current maximum.

the conditions corresponding to some lower peak current (see 94). Losses will be calculated by the method of 92.2 for which it is necessary to know only the fields at peak current.

This can be seen two ways. Discontinuities on the strip correspond to a contour within the thickness having an angular kink which is impossible with finite current densities without flux cross the boundary contour. More convincingly perhaps, discontinuity in current across the strip would lead to infinite fields at the discontinuity with positive fields one side, negative on the other; and so again there would be flux penetration into a region which must be field-free.

It is to be noticed that current density is continuous across the strip.

5.1. Current andjield distributions ut peak current The initial problem is to find the current distribution in the central region at peak current :

thence the field distribution at peak current is found and so the losses. Consider a strip of width 2a : let j be the saturation current density per unit width and 2b

the width of the field-free region. Formally then, if J is the current per unit width at a distance t from the centre :

J= j a b It1 > b (12)

= J ( t ) I t( <b. ( 1 W Consider first a strip of width 2b with its section in the complex w (= u+iv) plane (figure 6).

Let there be a current filament I passing through W = U O outside the strip and determine

“ ‘ } / 2 4 b 1 +s Current f i lament ,. L U

z plane w plane

Figure 6. Conformal transformation used to estimate current distribution in the central region of a strip.


current distribution on the strip to exclude magnetic field from it. Transform the strip conformally to a circle of radius $b in the z(=x+iy) plane by w=z+b2/4z. The current filament is then at x = (a0 + (u02- b2)1/2)/2, The current distribution round the cylinder to exclude flux can be found by the image method. The current density at a point on the circle z= (b/2) eiS (corresponding to U = b cos 8) is

Z (U$ - b2)'j2 7T b(u0 - U cos 8). -

Transforming back to the w plane the current density on one face of the strip at U is

I(uo2 - b 2 ) 1 / 2

2r(b2 - u ~ ) ~ / ~ (UO - U)'

This quantity must be doubled to account for both faces. When a second filament at U'= -UO is added, the current density on both faces of the strip at U is

- 2UOZ(U02 - b2)1/2 (b2 - u2)1/2 (U02 - U 2)'

This expression can be integrated to give the effect of a distributed current external to I U I < b of j per unit width in the range b < I uo I < a yielding a current on the strip at U of

A uniform current density on the periphery of the circle in the z plane also produces no internal magnetic field and is equivalent to a distribution proportional to 1@2- u2)l/2 on the strip. From the earlier discussion a condition in which current is continuous over the range I U I < a is required and can be reached by adding in a distribution proportional to 1/(b2 -U')'/'.

Doing this and reverting to equation (12) it follows that with t = u

The magnetic field at to is given by

- a

If J were still unknown in the range - b < t o < b this could be used as an integral equation for 4 since B is zero in this range. Swann (1968) has used this method in another very similar situation but one which cannot be correctly applied to losses in a superconducting strip.

The integral, after inserting the values of J and some manipulation, gives

(14) B(to)=-{1n(a2-to2)-2 w j In (A-(to2-b2)1/2) ltol > b 2rr

where A2=a2-b2 and a>to>b.

t = to, is The flux crossing the strip between t = b and t = t o , or for that matter between t = 0 and

t o

O(to)= J B(t0) dt. b

40

500 W. T. Norris

5.2. Estimation of losses

Combining equations ( 1 5 ) and (3), the loss in a cycle becomes Equation (1 5) gives the A ux between any point and the central field-free region or kernel.

a

Le= 1 4@(to)j dto. b

Evaluating the integrals,

(16)

= Ic2poLz(F) ( 1 6 4

Lc=-{(l Ic2po - F ) In ( 1 - ~ ) + ( 1 + F ) In (1 + F ) - F ~ } n-

where Ic=2ja is the critical current of the strip and Im=FZc is the peak current under consideration. LZ as a function of F is given in table 3. When F= 1 equation (16) gives the same value of loss as the calculations of $3. When F is small the formula gives

6. Losses at a gap in a thin sheet Consider a sheet which carries a critical current density j (per unit width), and has a slit

of width g parallel to the direction of current flow. Let Fj be the peak value of an ac current flowing in the sheet remote from the slit (F< 1). It will be supposed that the main magnetic field is confined to one side of the sheet; if the slit is sufficiently narrow it may be thought of as a slit in one tube of a concentric single-phase cable.

Again the method of 52.2 and equation (3) will be used so that only the conditions at peak current need be found. Current and field distributions in the sheet at peak current are shown in figure 7(a). Figure 8 shows the current reversal contours for the strip when

w plane

A t A'

Image currents (C) _-

(0)

Figure 7. Calculation of losses at a gap in a wide superconducting strip. (U) magnetic field pattern and distributions of current at peak value and transverse magnetic field; (b) z plane

containing the slit ; (c) w plane transformed from z plane.

considered to be of finite thickness. I t will be observed, as before, that there is no current discontinuity at B or D ; and so, as mentioned in $5 the magnetic field at B and D is finite.

Let x= +2b correspond to the points up to which the current density has reached the full critical value. For I X I < 2b the current density is lower and no magnetic field penetrates the strip. Consider the current pattern induced on the field-free part of the strip due to

It is convenient to consider the cross section as in the complex z(=x+iy) plane.


Fiy ld l ine

50 1

-e--

#

U Current reversal

/ Current-free reg ion

contours

Figure 8. Flux and current at penetration contours in a thick sheet.

saturation current at the strip edges (BE1 and EzD in figure 7(b)) and with a current density F j for large values of I x I .

It is useful first to consider currents induced on ABC and CDE due to the currents in BE1 and EzD alone. The conformal transform (Kober 1957, p. 59) z= w+b2/w transforms from the z plane to the upper half of the w(=u+iu) plane. The line BD in the z plane becomes the semicircle w=b eie (see figure 7(b)). Let jw be the current density on the semicircle BD corresponding to j on the line BD in the z plane.

= 2 j sin 8. (184 The field at D due to the current on the semicircle BD and its image has the value in

the 1v plane of

where w = b e'en corresponds to E2. Current continuity at D in the z plane requires a finite magnetic field at this point in the

z plane and so zero magnetic field at D in the w plane (since dz/dw=O). Superpose a uniform current in the w plane of 2j80/7r to achieve this condition. By transforming back to the z plane the full field and current patterns can now be calculated: however, the mathematics is a little cumbersome and we are only interested at the moment in computing losses so the calculation will be pursued with only this goal in mind.

Because of the conformal nature of the transform, currents in corresponding regions are the same in both planes, and magnetic flux crossing corresponding lines is the same. The loss calculation may therefore be performed in the w plane to calculate the flux linking any point on the circle BD, say at be'$, and the point D, which is the limit of the field- free region, Owing to the currents on BD and their images this flux is

The flux due to the superposed uniform field is jbp0(-280 sin $)/T and so the total flux becomes

502 W. T. Norris

Thus the loss at the edges in half a cycle for one side of the slit (using equation (3) and observing thatjw is not constant on the semicircle but given by equation (18a)) is

&= 2 f 2 j b sin $( -2%) d$ 0

If Fj is the current density over most of the strip, F=2Bo/i-r and the loss for both sides of the gap in a full cycle is

L c = w (2 In cos +TF+&TF (2-cos &rF) tan 3nF)lcos 3n-F i-r

= pOj2g 2L3(F) since g=4b cos Bo. For small values of F this becomes

yo j 2 g 2i-r3F4 24 *

L, N

Some values of L3 are given in table 3.

7. Losses at the edges of two parallel strips Consider two parallel strips of superconductor carrying antiparallel currents (figure 9(a)),

a configuration often used in measuring ac losses in superconductors. This system can be analysed as it stands, but it is much simpler and usually acceptable to consider the conductors

Current in < I t

Current out 1" ( U ) 4v w plane

I f f ' z plane x-

E y x i x x ~ x ~ 7 E I = - c / ~ T + ic

y!^ z. a tj r=c l f f B ' D

C D A A - I c + I

X r - U = - v cot v A

U

( b ) ( C )

Figure 9. Calculation of field at edge of two wide antiparallel strips (a) the strips; (b) one edge and centre line in the complex 2 plane; (c ) transformed w plane.

as very wide so that only one pair of opposing edges need be considered. The system may be further simplified without further loss of detail by considering one edge of a semi- infinite sheet above an infinite perfectly conducting sheet. The image of the semi-infinite in the infinite sheet is the same as the mating semi-infinite sheet first proposed. In this analysis a method similar to that of 52.1 will be used since that of 52.2 leads to complicated integrals. However, first consider the peak-current condition. The abstracted system now under review is shown in figure 9 with the section of the sheet as a part of the complex z(=x+iy) plane. The part BE carries saturation current; c is the distance apart of the two sheets. The conformal transformation

C ?r

z=- (w+ln w )


C

d M,

takes the system into the w( = U + iv) plane, the line BE becoming the curve

The point B is at w = - 1. U= -U cot 0. (24)

If a is the length of BE in the z plane and E is at a point where z: = 00 in the w plane, then

The current density on BE in the w plane may be expressed per unit change in v along BE :

iu dv = j da so

The field at B in the w plane due to currents on BE and its image in the real axis is given by the integral along BE :

(26) j v du(da/du) cvo j J” n((u + 1)2 +c2) = 2’

0

As in earlier discussion to get the peak field condition there is neither current discontinuity nor infinite field at B in the z plane and so the field is zero at B in the IY plane. Therefore suppose a flux cvo j /n streaming out of C in the w plane which is equivalent to a field v o j j n deep between the sheets in the z plane.

I t will also be necessary to know the total current induced on CA in the w plane by the currents on BE. This is

v j(da/dv) dv I ~ = L 1 dul J’ - 57 U12 - 2u1v cot ti + u 2 C O t 2 v + 212

0 0

(27) - c j -~ vo(1 -vo cot vo). 7T2

This is also equal in magnitude to the sum of the currents on BE and those induced by BE on the upper sheet in the z plane. The total current on the upper sheet within distance r from the end of the unsaturated region due to the flux streaming from C is proportional to this flux, so we will write it

cvo j K - Ti

so where K depends upon r . The condition r + 03 is to be taken. Thus the total current up to a distance r + a0 from E on the upper sheet is

cvoj czioj 77 n2

Io=K- +-- (1 -vo cot vo)

and the flux deep between the sheets is

coo j (Do=- 57

504 W. T. Norris

To get an intermediate condition as before superpose a condition -21s where Is is a lower current than IO. The corresponding flux

where U gives the point E if Is were peak current when also the length of the saturated region is a. Is itself is made up of the currents flowing in the saturated end and the currents they induce, plus the currents due to the streaming flux equal to Os, These latter currents may be considered in two parts ; one part would be currents distant r from the end of unsaturated part proportional to the flux, i.e. K(cvj/n). Since we must count currents up to the same physical point for both IO and 1, we must add in the currents on the region ao - a in figure 10, i.e. (uj/n) (ao-a) on the inside face plus a small current proportional to l /r (and therefore negligible) on the outside face.

c I I I

r '-* 00 'cI

* c v o / l r c \Saturated region Y

(0)

C \Saturated region __fc

c v / h --- , ( 6 )

Figure 10. Conditions to be superposed for currents at edge of antiparallel strips with less than peak current, (U) Conditions at peak current ; (b) conditions to be superposed to get intermediate

conditions.

Thus, as r -+a, ccj cj cvj

Iq=K- +- (ao-a)+- 77-2 (1 - U cot U ) . n n

Thus the currents at an intermediate state are

I= Io - 21s

Q =Q0--2Qs. and the fluxes between the sheets

Considerations of $2.1 (equation (IC)) show that the energy dissipated in a half cycle at one edge is

The loss can now be calculated. If s (=2c) is the spacing of the double semi-infinite sheet system the sum of the losses on both edges in a full cycle becomes

L " 7 2pos2j2 l"~(1- U cot U) du 0


where 00 = TFand Fjis the current density deep between the sheets. L4 is given also in table 3. For small values of F the losses become

8. Losses at asperities Consider a thin asperity consisting of wall parallel to the current direction on an

sheet as shown in figure 11. Then either adapt the result of Halse (1970) for

(3 1)

infinite a field

Saturated region a t peak current

T A L

Figure 11. A wall-like asperity on an infinitely wide superconductox.

perpendicular to a sheet of finite width which carries no transport current; or apply the procedure of 92.2 using the transform w = z / ( ~ z + b ~ ) ~ / ~ . The loss per cycle becomes

Lc = .~___ (2 In cosh TF- n-F tanh T F ) (32)

= p$jzazL5 ( 3 2 4

2po j zaz 77

where F = J / j ; J is the current density in the sheet remote from the asperity. For small values of F the loss becomes approximately

9. Discussion 9.1. Some examples

It would be exceedingly difficult to find any instance of quantitative agreement between measurement and these calculations. A primary reason for this is the crudeness of the representation of how a superconductor behaves. Nevertheless it is hoped that some useful guide lines are now available. By way of illustration of the use of the formulae some particular cases are calculated below using the equations mentioned and the computed loss factors shown in table 3.

For an elliptical wire with a peak current capacity of 4 0 A carrying 3 2 A peak at 1 Hz the loss, using formula ( l l b ) , is 100 pW m-1. This would be much too high to be acceptable for a superconducting magnet winding.

For a thin strip of the same capacity and carrying a similar current the loss would be about 60 pW m-l, from formula (16a). The strip has lower loss due to transport current but would probably be very much more lossy in an alternating transverse magnetic field.

The case of a gap in a superconducting sheet may be of relevance in constructing a high-power cable. Suppose tubular conductors are formed by seam welding sheet ; further suppose that the weld is not only resistive but very resistive. The seam might be regarded as a gap. Take the saturation current of the sheet as 100 A 105 A m-1, and a gap width of 1 mm. Consider a peak current of 90 A mm-1 at 50 Hz, and the losses become, using equation (22a), 16 W m-1 which would be unacceptable in power cable by a factor of a few hundred. The saturation surface current density has been taken near the critical field of niobium. At a weld the material might have a critical current density of 1011 A m-2 and if m thick would have an effective surface current density of 107 A m-1 leading to

506 W. T. Norris

a loss when carrying the same current as before of 55 pW m-l. This is much more satisfactory.

Consider such a pair 3 mm wide and spaced by 25 pm. If lo5 A m-1 is the saturation surface current density and they carry a peak current of 90 A mm-1 at 50 Hz, equation (30a) gives a loss of 2.7 x 10-5 W m-1 for one pair of edges, equivalent to 0.09 W 111-2 averaged over the two facing current-bearing surfaces. If the edges were cold-worked and had a saturation critical current of 107 A 111-1 the equivalent losses averaged over the facing surfaces would be 4 x 10-5 W m-2. This is barely perceptible. High critical current hard superconducting edges can considerably reduce the edge loss contribution of a sheet conductor.

The asperity calculation of $8 can be used to estimate losses on roughened surfaces. Consider parallel ridges 1 pm high spaced 10 pm apart. Let 105 A m-1 be the critical current of the asperities and consider the mean surface current density to be 104Am-l. The loss in one asperity at 50 Hz is, from equation (32), 4.5 x 10-7 W m-I. Thus the average loss over the surface is 0.045 W m-2. The data have been selected to give a loss level commonly experienced with a view to bringing out the possible significance of surface roughness.

For sharp-edged conductors the loss at low currents depends on the fourth power of the current whereas losses on a smooth surface depend only on the third power.

In the case of the strip this difference arises from neglect of current penetration into the broad sides of the strip. One might argue that the interesting case is where the central part of the strip carries lossless surface currents and only field penetration at the edges is an important source of loss. There is a touch of physical realism but it is not the model we set out to study.

Some correction to the formulae to account for penetration on the broad face may be administered by adding in the losses that would be calculated assuming the current is uniform across the strip and allowing it to have finite thickness. Considering the isolated strip ($5) from this point of view Hancox (1966) gives a loss supposing uniform current :

Materials carrying alternating currents are often tested as strips back to back.

where c is the thickness and 2a the width of the strip. we may find the thickness to be important in determining losses when

Comparing equations (34) and (17),

T C F<- 2a' (344

10. Conclusions The hysteresis losses in isolated wires of hard superconductor carrying alternating

currents have been calculated. A number of different-shaped cross sections have been considered but it appears that shape is not a great influence. An elliptical wire has the same loss as a round wire of equal current-carrying capacity: a thin strip has lower losses.

Losses at the edges of thin strips have also been investigated. The various cases considered give losses at low currents proportional to the fourth power of the current.

Acknowledgments I am grateful to Mr. H. 0. Lorch who first introduced me to the problem of the isolated

strip, to Messrs. R. Hancox, B. J. Maddock and M. T. Taylor for discussions of some aspects of the work, to Mr. D. C . Heap for checking through the mathematics and to one of the referees for helpful suggestions about presentation. The work reported in this paper was carried out at the Central Electricity Research Laboratories and is published by permission of the Central Electricity Generating Board.


References BETH, R. A., 1967, J. appl. Phys., 38, 4689-92. DUNN, W. I., and HLAWICZKA, P., 1968, J. Phys. D : Appl. Phys., 1, 1469-76. GROVER, F. W., 1962, Inductance Calculations (New York : Dover). GRAY, A., 1967, Absolute Measurements in Electricity and Magnetism (New York : Dover). HANCOX, R., 1966, Proc. I.E.E., 113, 1221-8. KOBER, H., 1957, A Dictionary of Conformal Transformations (New York : Dover). LONDON, H., 1963, Phys. Lett., 6, 162-5. SWANK, G. W., 1968, J. math. Phys., 9, 1308-12.

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