DaveM said:
A National LM138 linear regulator will do the job for you. It's currently
stocked by many of the parts houses. Get the datasheet from
www.national.com.
Be sure to mount it on a good heatsink. From your numbers, it will be
dissipating about 50 watts at full load.
http://cache.national.com/ds/LM/LM138.pdf
The T-package (TO-220) has a thermal resistance of 4C/W from junction
to case - which would be a 200C temperature rise at 50W, so you can't
use that.
The K-package (TO-3) is a lot better at 1C/W, but you've got to limit
the junction temperature to less than about 125C (otherwise the LM338
will do it for you) so your total thermal resistance from junction to
ambient has to be less than 2C/W, which isn't easy to achieve - I'd be
looking at a 0.5C/W heatsink, like Farnell's150-019, which costs about
$25, stand 2"" high , 5" wide and 6" long.
Paralleling a couple of linear regulators might save you money and
heat-sink volume, or you might put a beefy power transistor in series
with a low-drop-out linear regulator, mounting it on the same heat-sink
as the linear regulator to exploit the linear regulators thermally
driven current limiter.