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One complication is that the current thru the rectifier and the resistor isn't a constant DC, but a large pulse of current happening at the top of the incoming AC waveform. This pulse of current is essentially the first filter cap being topped off by the incoming top of the AC waveform, to replenish the voltage charge that was drained off the B+ by the rest of the set since the last replenishment The duty cycle of this pulse is related to the amount of ripple the B+ will have. Not a linear relationship, but still related. A rough approximation: If the duty cycle of the recharge pulse is 5%, that would make for 20 times the normal B+ load current this pulse will peak at. 100ma of B+ load current would make a 2A max pulse recharge current happen thru the rectifier. (You'll find a spec in rectifier tube data sheets "peak current", that's what this spec is for, and why it's possible to kill a rectifier tube with caps that are too big, as bigger caps will make for shorter duty cycles but larger current peaks). The resistor in series with the new silicon diode will only see this recharge current pulse, so when you want to figure out the required resistance you'd need to know that peak current. Problem is that you probably won't know it well enough to make a good calculation, so be prepared to adjust the resistance once you have it in the set. And why you end up with a much lower resistance than what you though you'd need...
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