I. T. Sept. 10 meeting presentation slides 1

Bias Magnet for the Booster’s 2-nd Harmonic

Cavity

An attempt to evaluate the scope of workbased of the existing RF design of the cavity

9/10/2015

Geometry and AssumptionsRequired change of Frequency vs Time:Tan, July 17, 2015First question is about a sale of the system

average power consumption must be found

Cavity geometry: G.R., July 2015

Frequency response and average power

Results of modeling by G.R. Combining functions f(t) and f(I), we get the required current shape.

t (ms) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 33Iw (A) 6330 6660 7250 8280 9700 11300 13000 14700 16500 17800 18980 19900 20500 20900 21250 21600 21800 21900Iw_an1 6330 6320.464 6979.824 8125.584 9595.984 11250 12967.34 14648.46 16214.54 17607.5 18790 19745.42 20477.9 21012.3 21394.22 21690 21986.7 22168.28Iw_an2 6330 6578.057 7299.391 8428.959 9869.187 11503.86 13213.47 14889.31 16443.93 17816.99 18976.22 19914.36 20643.61 21188.93 21581.67 21854.33 22036.95 22102.66Sine 6330 6543.278 7169.69 8170.016 9482.316 11026.77 12711.92 14441.74 16123.09 17672.8 19023.96 20130.76 20971.7 21550.64 21895.88 22057.16 22100.96 22102.66

Having the function Iw(t) expressed analytically allows making analytical evaluation of power loss and generate other requirements needed for a power supply design.

Harmonic Representation of the Current Cycle

The second order expansion reproduce the required current shape quite accurately.

Iw(t) = 15728 - 7676 cos(2∙ π∙t/T) - 1523 cos(4∙ π∙t/T)

Coil Design- 0.41” (10.41 mm) rectangular copper wire with 0.229” (5.8 mm) hole for water cooling;- Copper cross-section 80 mm2; cooling hole cross-section 26.5 mm2;- Number of turns – 50 (conservative), 56 max;

- Double-layer winding technique required length of wire Lwire ≈ 80 m- Four reels of wire must be procured to avoid splices (90 feet reel)

- Current range from Imin = 126.6 A to Imax = 440 A- Current shape: I(w) = I0 – I1 cos(2∙ π∙t/T) – I2 cos(4∙ π∙t/T

with I0 = 314.56 A, I1 = 153.52 A and I2 = 30.46 A

- Wire resistance R = 0.019 Ohm

- Power is found by integrating the losses through the cycle Wcycle = 142.5 J;

- f = 15 Hz P = 2140 W

Coil CoolingTwo cooling loops assumed. In the loop formed by the outer part of the coil, the power loss is ~1100 W. Assuming 10 ºC cooling water temperature rise, the required water flow is

Q [l/s] = 0.24 P[kW]/ΔT[C] = 0.0264 [l/s] ∙ 1.58 l/min ~0.4 GPM.Water velocity in the channel with the cross-section S = 26.6 mm2 will be

v[m/s] = 1000 Q[l/s]/S[mm∙ 2] ≈ 1 m/s, which seems OK. This water flow will require the next pressure head:

h = v2 L/2d∙ 1.33

where h is measured in kG/cm2, v is the velocity in m/s, d is the hole diameter in mm, and L is the length of the circuit in meters. For our parameters (L = 39 m)

h = 12 39/10.4 = 3.75 kG/cm∙ 2 or ~53 PSI. Volume of the coil is 10500 cm3 weight is ~100 kG.Power load is 2140/10500 = 0.2 W/cm3.

Flux Return. Inductance. VoltageCentral part of the flux return is made of low-frequency high permeability ferrite; peripheral part of the flux return is assembled using silicon steel plates.

Thickness of the yoke will be optimized to avoid saturation.Shape of the poles will be optimized for better field uniformity in the garnet material

Expected inductance of the coil is ~2.5 mH at the maximum current and ~3.5 mH at the minimum current

Expected resistive voltage is ~9 V;Inductive voltage is ~45 V.

Power Loss in the Garnet MaterialRunning a 2D RF model gives frightening numbers for the power loss: at 125 A f = 75.22 MHz; Q = 3452; ΔU = 20.5 V power scaling factor to 100 kV is 23.8E6Maximum power loss is ~50 W/cm3 Average power must be evaluated to have realistic understanding of the cooling problem

Power loss in the hot spot – 6.1 W/m3

ΔU = 33.27 V power scaling factor to 100 kV is 9.03E6So, the power density at the hot spot at 100 kV is 54.2 MW/m3 or 54.2 W/cm3

 Total dissipated power in the garnet at 75. 225 MHz is 0.0038477*9.03E6 = 34765 WMagnetic portion of the losses is 0.00305*9.03E6 = 27542 WStored energy in the cavity is 2*1.71E-8*9.03E6 = 0.31 JBased on these numbers, quality factor associated with the losses in the garnet QAL800 ≈ 4150

 

Power Loss in the AL-800 Material

I (A) 125 150 175 200 250 300 350 400 450 f (MHz) 75.22 81.37 85.94 89.63 95.26 99.09 101.89 104.09 105.85 k 9.0E6 8.6E6 8.28E6 8.01E6 7.6E6 7.33E6 7.14E6 6.99E6 6.87E6 pmax (W/cm3) 54 10 6.8 5.5 4.7 4.2 3.8 3.5 3.23 pav (W/cm3) 5.6 4.3 3.6 3 2.1 1.6 1.3 1.1 0.98 PGM (W) 27542 19780 15150 11800 7524 5220 3820 2890 2246 PG (W) 34765 26832 22440 19410 12844 10013 7180 6935 6025 Q_calc 2087 2667 3125 3480 4984 6090 8200 8200 9270 Q_glob 4111 5573 6950 8332 11100 13620 15900 18065 20040

Table below reflects results of 2D modeling of the cavity with different current settings in the coil

Comparison with the 3D modeling by G.R. Maximum and average power density

Power loss in AL800 material (magnetic and total)

Average Power Loss in the AL-800 Material

Let’s find the average energy deposited in the garnet block during one cycle. Current change is in accordance with the harmonic representation (as found earlier):

Iw(t) = 15728 - 7676 cos(2∙ π∙t/T) - 1523 cos(4∙ π∙t/T)RF voltage power drops linearly from 100 kV at t = 0 to 0 at t = 23 ms.

U = U0 (1 – t/τ) when t < τ and 0 when t ≥ τ∙

This means that the power scaling factor also changes with time; additional, time dependent coefficient for the power scaling factor is

λ(t) = (1-t/τ)2

Procedure is detailed in the next sheet

Average Power Loss in the AL-800 Material

The procedure is as following:

• Use the expression for the current change:

Iw(t) = 15600 - 7676 cos(2∙ π∙t/T) - 1523 cos(4∙ π∙t/T).

• Use expression for the frequency as a function of current:

f(I) = 22.5 + 12.7 I – 0.868 I∙ ∙ 2 + 0.03 I∙ 3 – 4.1 10∙ -4 I∙ 4 (I is in kilo-Amperes)

• Use expression for the power loss vs frequency (P in W, f in MHz):

P(f) = 215894 – 3447.8*f + 13.831*f2

• Apply additional scaling factor and get adjusted (for the accelerating voltage) power loss.

• Integrate the losses over time 209.2 J per cycle 3140 W in the garnet (0.51 W/cm3 average power density)

Maximum Power Loss Density in the AL-800 MaterialIw (kA) 6.25 6.5 7.0 7.75 8.75 10.0 12.5 15 17.5 20.0 22.5 f (MHz) 75.22 76.62 79.15 82.38 85.94 89.63 95.26 99.09 101.89 104.09 105.85 k 9.03E6 8.94E6 8.76E6 8.53E6 8.28E6 8.01E6 7.6E6 7.33E6 7.14E6 6.99E6 6.87E6 P128 (W/cm3) 51.5 32.2 15.8 8.5 5.63 4.24 3.23 2.57 2.32 2.10 1.89 Rmax (mm) 128 128 127 124 120 108 107 106 105 105 105 ΔR1/2 (mm) 12.5 15 25 50 Wide Linear Linear Linear Linear Linear Linear Zmax (mm) 110 110 110 110 0, 110 0, 110 110 110 110 110 110 ΔZ1/2 (mm) 0.5 1 3 25 Uniform Linear Linear Linear Linear Linear Linear PG (W) 34785 32811 29521 27375 22440 17868 12854 10017 8200 6933 6024 Q_glob 4112 4417 5005 5853 6950 8332 11103 13619 15917 18065 20039

Procedure:1. Use the following expression for the current change:

Iw(t) = 15600 - 7676 cos(2∙ π∙t/T) - 1523 cos(4∙ π∙t/T).2. Use expression for the frequency as a function of current:

f(I) = 22.5 + 12.7 I – 0.868 I∙ ∙ 2 + 0.03 I∙ 3 – 4.1 10∙ -4 I∙ 4 (I is in kilo-Amperes)3. Use expression for the power loss density vs frequency (p in kW/cm3, f in MHz):

p(f) = 1460/(f-70)2 + 0.09*(f-90) (at R = 128 mm, Z = 110) 4. Apply additional scaling factor for the accelerating voltage drop in time.5. Integrate the power loss density over time in the hot spot R = 128 mm, Z = 110 mm The accumulated heat per one cycle is 0.16 J/cm3 per cycle.

Average power density in the hot spot is 2.4 W/cm3.

Temperature rise at the hot spotIs the found maximum power loss density dangerous?

Let’s make simple evaluation. Let’s consider a plate with thickness t. Let’s have coordinate x in the direction perpendicular to the plate with x = 0 at the surface where we will find the temperature rise relative to the other surface of the plate. Power deposition density is p(x). Thermal conductivity is k. Let’s assume

p(x) = p0*exp(-x/d)Power flux in the direction towards x = 0 plane isQ(x) = ∫p(x)dx with the limits from x = t to x = 0

Q(x) = -p0·d·[exp(-x/d) - exp(-t/d)] = k·dT/dx dT/dx = -p0·d/k·[exp(-x/d) - exp(-t/d)]

We assume here that the temperature derivative at x = t is zero ΔT(x) = d2·p0/k·[exp(-x/d) - exp(-t/d)] + p0·d/k·exp(-t/d)·(x-t)

or ΔT(x) = d·p0/k·{[d·exp(-x/d) – (d – x + t)·exp(-t/d)]}

For x = 0 with t = 2 cm and d = 10 cm, and with k = 0.035 W/(m-K) and p0 = 2.4 W/m3 we have:

ΔT(0) = 685.7*(10 – 12*0.82) ≈ 120 K

I = 125 A I = 150 A I = 175 A I = 200 A

I = 250 A I = 300 A I = 350 A I = 450 A