Presentation on theme: "Practical magnet design 1.Iron saturates, when. Fields higher than 2.2 T with super conducting magnets high I Free currents (coils) magnetization 2.For."— Presentation transcript:
Fields higher than 2.2 T with super conducting magnets high I Free currents (coils) magnetization 2.For a given field we need a certain current (Ampereturns) Can be approximated Final values from computer codes (FEMM, POISSON, TOSCA, etc.
Current in the coils h e.g. B = 2 T, h = 20 cm, NI = 320000 At
Coil parameters NI = (NI) needed, but N = ?, I = ? i.e. which conductor? Dissipated power (heat) P = I 2 R Must be removed from the system Air cooling Indirect cooling Direct water cooling Hollow conductor Relatively ineffective
Coil cooling The dissipated electric power in converted into heat (water temperature). The temperature rise rise between input and output is where T = temperature rise [K] I = electric current [A] R = coil resistance [ ] q = water flow [l/min]
Usually Often the requirement is The resistivity of copper So: The smaller the T, the larger q is needed
For a large q a large pressure difference (pressure gradient) is needed Semi empirically where P = pressure difference [bar] q = water flow [l/min] a = flow cross section [mm 2 ] l = flow length [m]
Note! So, the desired pressure increases very rapidly with electric current Usually If the desired pressure becomes too high the cooling circuit can be divided into several parallel circuits (pressure decreases as 1/n 2 )
T q T q I2RI2R n parallel cooling circuits (electrically in series)
Division into parallel cooling circuits is done e.g. with a double pancake structure “wind” starting from the outer radius into the inner radius, come back along the next layer These double pancakes are connected in series electrically and in parallel Note: The conductivity of the cooling water must be lowion exchanged water, closed cooling circuit, heat exchangers
Example: K130 cyclotron 380 mm < 400 mm parallel 20 layers (axially) 10 turns (double conductor) per coil
Desired NI = 400.000 At In one conductor pair 400.000 At/(2x200 t) = 1000 A In one conductor: 500 A Conductor pair is connected in parallel also electrically Length of a double pancake Conductor parameters d = 17 mm = 7 mm R = 1 mm A = 250 mm 2
Conductor resistance (double pancake/ one conductor) for one conductor Dissipated power: For a hot coil P nominal = 130 kW
Cooling We allow T = 20 o C Hence the cooling water flow: for one conductor/dpc Required pressure drop
Both coils consist of 10 double pancakes (dpc) Each dpc has two conductors in parallel (both electrically and for cooling) 40 pieces of 180 m long cnductors Total flow = 40 x 2.5 l/min = 100 l/min So: the cooling system must be capable of producing about 6.5 bar pressure difference at a 100 l/min flow rate in order to keep the temperature increase below 20 o C. Note: the cooling system has to cool other devices as well!
The conductor was of type What would happen, if it was replaced with a conductor where A cu and A water are the same as well as the conductor outer dimensions?
The flow rate in the cooling hole would be doubled and the cross section would be doubled Change in the pressure drop Would be beneficial to use one hole with the same total cross section But: winding would be difficult r
Alternatively we could use a square conductor with the same cross section Number of turns would increase and the conductor length as well Higher P Also the resistance of one dpc would increase More flow Higher pressure
Let us calculate this as an example: Conductor dimensions = In one coil 400.000 At/2 = 200.000 At Keep the current density unaltered I = 2 x 500 A = 1000 A N = 200 Full double pancakes N = (2 x 7 ) x 15 = 210 I = 952 A i.e. 7 double pancakes (15 turns)
The conductor is insulated diameter 26 mm r= 390 mm Power dissipation for a double pancake Total power
Water flow rate Pressure drop Starts to on the limit! In both cases Power supply: 1000 A/140 V
Power supply Current/Voltage These have also some limitations Current must be taken into account when dimensioning the transport cables from the PS to the magnet The dimensioning ( ) depends naturally upon the heat transfer properties of the surroundings Typically power dissipation (heat) can be < 10 W/m e.g. A cu = 400 mm 2 and I = 425 AP/L = 7.8 W/m
Note: For I = 1000 A, A cu = 1600 mm 2 (7.8 W/m) 4 cm In practice the feeding cable is divided into several thinner cables More cooling surface/unit power loss More flexible (mounting)
For example: K130 cyclotron I nominal = 1000 A, I max = 1170 A 4 x 150 mm 2 cable Maximum current 320 A/cable (depends on the assembly – cooling circumstances) Note also: For high currents the transfer losses (Voltage) may be significant in a long transfer cable The Power Supply sees both the load and the transfer line
RECTANGULAR WITH A ROUND HOLE Dimensions in mm NO.OD.ID.x/yRKg/m 83294x321/0.50.80.070 83304.3x42.50.9/0.750.80.100 83564.67x4.192.411.13/0.890.760.130 83575.08x4.772.641.22/1.0650.760.163 82905.8x22.214.171.124/0.60.80.143 83276x42.51.75/0.750.80.170 83006x531.5/110.200 83146x126.96.36.199/10.80.200 83706.1x43.21.45/0.40.60.143 83427x532/10.60.250 81157x641.5/110.256 68617x188.8.131.52/1.4510.304 83647x8.541.5/2.2510.412 85217.2x184.108.40.206/0.8510.218 68337.5x532.25/10.50.270 82807.5/7x63.31.98/1.3510.310 68647.5x63.52/1.2510.308 82607.5x75.21.15/0.91/0.250.275 83317.5x841.75/210.420 68797.6x7.23.52.05/1.8510.395 81607.8x220.127.116.11/1.870.50.389 83288x63.22.4/18.104.22.1680 81038x63.22.4/1.410.350 82878x642/110.309 83998x22.214.171.124/1.110.329 Important note: choose the hollow conductor dimensions from the manufacturer’s stock list – otherwise you may be surprised by the price!