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Foil target heating by the beam

Foil target heating by the beam


The maximum current the target can take is limited by the beam heating of the target. The magnetization of a material drops with temperature, going to zero at the Curie's point (940C° for supermendur).

Heating by a Beam Pulse

A short beam pulse (0.5 ps at CEBAF) heats the area it overlaps with. The heat flow is negligible during this period. The temperature change can be calculated as:
ΔT=dNe/ds·α/CP , where The effect depends neither on the target thickness nor on the target angle with respect to the beam. The beam profile is presented by a Gaussian, as
d²N/dx/dy=Ne/(2·π·rb²)·exp(-(x²+y²)/2/rb²).
The number of electrons in the pulse is Ne = Ibeam/1.6·10-19/F, where:
At CEBAF: We obtain:
ΔTmax=0.013 K / pulse     for rb=30 µm and at x=y=0
This effect of one pulse is negligible.

Heating by a continuous beam

The beam gradually heats up the foil, a thermal wave is propagated across the foil surface, until heat dissipation compensates the beam heating.

Here we assume that a thin foil is normal to Z and the problem is 2-dimensional. The heat equation, taking into account the heat conduction, radiation, and the beam heating, looks like:
∂T/∂t = ∇²T·κ/(ρCP) - 2·σε((T+To)4-To4)/(ΔzρCP) + Bfluxα/CP ,
where

We solved this equation numerically, on a grid with a variable cell size.

Heating during a short time period

Let us consider a foil at a room temperature To. At the moment t=0 the beam starts. For a certain time period the heat wave does not reach yet the foil boundaries. During this period the heating does not depend on the boundary conditions or the size of the foil (if it is large enough). The effect of a 50 μA beam of different spatial distributions is shown on this plot. Radiation starts to play a role when a large foil area is heated. After 1 second the width of the heated spot is about 1 cm. A fast raster can considerably reduce the heating for the first millisecond of the beam pulse. In this case a simple 1-dimensional raster was simulated - a circle of a 1 mm diameter.

Running with a pulsed beam - calculation for PREX

At CEBAF, one can use a pulsed beam with a pulse length of 0.1-1ms and a repetition rate of 30-120 Hz. This provides a low average beam current, while having the "regular" current in the pulse (i.e. the current used in the experiment). Let us consider using a foil perpendicular to the beam, magnetized in a high field of 4 T (a clone of Hall C polarimeter).

The maximum temperature of the foil during the first beam pulse is shown on this plot. With a 1.4×1.4mm² fast raster the maximum foil temperature increases by about
ΔT(t=1ms)max=12K;

The average beam current is <Ibeam>=1.5 μA. The effect of this current is estimated analytically, assuming an axial symmetry:

Neglecting the radiation, the final temperature at the foil center is:
ΔT(r=0,t=∞)=<Ne>·α·ρ/(2π·κ)·[0.5+log(Rfoil/RR)] = 12K

In case of no raster one has to use the Gaussian shape of the beam spot profile:

Neglecting the radiation, the final temperature at the foil center is:
ΔT(r=0,t=∞)=<Ne>·α·ρ/(2π·κ)·∫0R [dr/r·(1-exp(-r²/2/rb²))] = 22K

With the raster, the total temperature rise of about

ΔTmax=25K;
should cause the foil depolarization of less than 0.25%.

With a 12 μm thick target at 20° to the beam, the coincidence Moller counting rate at 0.850 GeV was about 40 kHz at Ibeam = 0.3 μA. The LG single arm counting rates were ∼ 1 MHz, while the aperture counters rates were ∼ 2 MHz. The projected rates will be larger by a factor
(50μA/0.3μA)·(1μm/12μm·sin(20°))=4.75
We can reduce the single rates by about a factor of 2 with a collimator, which would reduce the coincidence rate by a factor of ∼3. Another factor of 2.8 should be gained by upgrading the aperture counters and a part of the electronics used.

About first 150μs of the pulse should not be used for the measurement because of various transitional processes. The time needed for a 1% absolute statistical error measurement will be (the foil polarization is 0.08 and the analyzing power is 0.76):
t = 1/(0.01·0.08·0.76)²/(40·10³Hz·4.75/3.*0.85ms*30Hz) = 28 min


Last updated: 24 April 2006
E-Mail: Eugene Chudakov gen@jlab.org