Since the thermal conductivity of Pb changes dramatically with temperature the simple analytical solution for constant k is not applicable over most of the volume. However, this rapid change occurs outside the beam area so the heat equation becomes homogeneous( the external power density g = P/(pa*a*t) = 0.). We know the power in, the beam power, and we know in equilibrium that the power out at the cooled edges( r=b) must be the same.
Suppose the volume in red is the beam. This goes to a radius a. At radius b the temperature is fixed at Tb. From the BNL tables of the integral of thermal conductivity of lead we can calculate the temperature Ta at radius a. Within the region r<a the thermal conductivity of lead is practically constant at k=0.35 W/(cm.K). For Ta > 300K we can write
If the beam power is uniformly distributed across this region then the equation for the temperature distribution for 0<r<a is
At r = 0 this gives a fixed rise in temperature of P/(4ptk)
a=2mm, P=60W, t=0.05cm
effect of different boundary temperatures and central temperature for uniform beam power
|Tb (K)||b (mm)||Ta (K) at 2mm||T at center (K)|
Thus, the best case is if we could bring the cooling to within 4 mm
of the center (494K). The more realistic condition of 6K cooling with a
4mm radius for the cooling line is too close to the melting point for Pb
to be practical (573K).