Step 1: make sure the LC res freq is well below 300 or 360 Hz ( else you will cook the L & C ).
Step 2: you generally want to minimise the L size and compensate with larger electrolytics as the caps are cheaper than a large L and you will need some pretty large electrolytics anyway to handle the ripple current.
Step 3: the exact calculation depends on the nominal firing angle for your 135VDC and 100A, which is based on the nominal transformer output, but let us suppose we allow for 10% rms volt ripple out of the SCR's and we want < 0.5% volt ripple ( of 135VDC ) on the output caps at full load - the calculations are pretty simple , we need to go from 13.5Vrms ripple to 675mV rms ripple with 100A in the L.
Step 4: Let us choose an Fo of 20Hz for the filter, i.e. 20.2.pi = 1/( sqrt (L.C) ), or 15,790 = 1/ (L.C)
Step 5: At 360Hz rectifier ripple we need Xc / ( Xc + X L ) = 0.675 / 13.5 = 0.05, i.e. a 20th, Xc = 1/2.pi.F.C X L = 2.pi.F.L
Step 5a: using simple algebra we can solve for L and C using the information in 4 & 5 above
Step 6: more empirically we choose a small-ish value for L, say 150uH ( 100ADC remember )now X L = 0.339 ohms ( 360 Hz )
we need Xc to be 1/20th of this or 16.9 milli ohm approx, this is 26,000uF for C, this will give the required attenuation.
Step 7: Fo for the LC pair is now 1/(2.pi.sqrt(LC)) = 80 Hz, which is a bit high, to get to 20Hz say, C needs to be 420,000uF
For the filter to work the ESR of the electro's needs to be well below the Xc ( 16.9 milli-ohm ) - so you may well need up to 9 x 47,000uF at 200V to get near this figure, else the ESR will be the main ripple voltage constituent.
Step 8: If we assume all the ripple volts are now dropped across the L ( as the C is large and low ESR ) the ripple current in the choke is Vac/X L = 13.5 / 0.339 = 40 amps, all of this ripple will be in the caps - so if we have 9 x 47,000uF they each need to be able to handle 40/9 = 4.5 amps @ 360Hz continuously - this is the major design criteria for the electro's.
looking at a typical cap from element 14, we see 200V 2200uF = 1.98amps and 200v 10,000uF screw terminal at 19.9A 18 milli-ohms
So 2 x Vishay 101/102 PHR-ST 10,000uF 200V gives 40 amp ripple current at 9 milli-ohms total
the combined volts from the Xc & ESR = SQRT (884mV^2 + 360mV^2 ) = 954mV - a bit higher than the 675mV we were hoping for - but never mind,
the overall attenuation is ~ 0.954 / 13.5 = 0.0707 ( -23 dB ) and 954mV / 135VDC = 0.7% ripple rms is inside the spec you wanted.
the 10000uF caps will dissipate 7.2 watts each , for 2 only - which is getting on the warm side ( so 4 might be better 1.8 watts each )
[ 4 x 10,000uF will give lower ripple, and a filter res freq of 65Hz, still a bit on the high side - but live-able ]
Happy designing...