Aluminum Converter

Calculating the temperature of a 1/2 mil Aluminum converter with energy deposited from a 44 MeV electron beam.

Calculating number of particles per second

We have electron beam of:

Frequency: $f=300Hz$

Peak current: $I=50 Amps$

Pulse width: $∆t= 50 ps = 5*10^{-11} seconds$

By $Q=It$, we have $N*e=f*I*∆t$

Where $N$ is the number of electrons that hit the target per second, $e$ is electron charge and $f$, $I$ and $∆t$ are given above.

$N = \frac {f I ∆t} {e} = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 4.6875*10^{12}$

So, we have around $4.6875*10^{12}$ electrons per second or $15.625*10^9$ electrons per pulse.

Calculating the stopping power due to collision of one 44 MeV electron in Aluminum

From NIST ([1] see link here) the stopping power for one electron with energy of 44 MeV in Aluminum is $1.78 MeV cm^2/g$.

$1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm$

The effective length of 1/2 mil Al:

$(2.70 \frac {g}{cm^3})(0.00127 cm) = 0.003429 \frac {g}{cm^2}$

The total stopping power due to collisions on Al per incident electron:

$(1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 \frac {MeV}{electron}$

The energy deposited per pulse:

$(0.0061 \frac {MeV}{electron})(15.625*10^9 \frac {electrons}{pulse}) = 95.3125*10^6 \frac {MeV}{pulse}$

The energy deposited per second:

$(95.3125*10^6 \frac {MeV}{pulse})(300 \frac {pulses}{second}) = 28.6*10^9 \frac {MeV}{second}$

Calculating the temperature increase

The power deposited in 1/2 mil Al is:

$P = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV}) = 4.575*10^{-3} W$

Stefan-Boltzmann Law (Wien Approximation) says $P = (0.924)(Area)(\sigma)(T^4)$

Solving for Temperature and taking into account the two sides of the converter we get:

$T^4 = \frac {P}{(0.924)(2A)(\sigma)}$

where $\sigma$ is the Stefan-Boltzmann constant, $\sigma = 5.67*10^{-8} \frac {W}{m^2 K^4}$. Assume a beam spot diameter on the converter surface of 5mm, or an area of $A = 19.62 mm^2 = 19.62*10^{-6} m^2$.

Plugging in the numbers we see that the temperature will increase $217.2 K$. Now, adding in the temperature of the converter at room temperature we get :

$T = 300 + 217.2 = 517.2 K$

The melting temperature of Aluminum is $933.5 K$.

Conclusion

An Aluminum converter that is 1/2 mil thick being struck by a 44 MeV electron beam with a 50 picosecond pulse width, 300 Hz rep rate, and 50 Amp peak current is found to be safe from melting.

Go Back