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) = 273K
 

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)
4.2	4	221	494
4.2	6	366	639
4.2	7	527	800
4.2	8	600	873
4.2	10	721	994
6	4	300	573
6	6	443	716
6	8	676	949
6	10	797	1070

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).