You're killing me Colas. I started to write a reply 4 times last night, and deleted them all; I think you keep posting before you fully grasp what we're saying.
Read, and let soak in, each of the points below:
PRR's calculation applies to class A operation of the output tubes.
The current used in the calculation is
not the current of both tubes, but the
peak current of one-half the output stage (eith the "push" or the "pull" side). You've been confusing that with the idle current of the entire output stage.
Instead,
in class A the
peak current of one side of the output stage is double the idle current (and the negative peak is 0mA). That is where you see double the idle current in the other calculations.
You are right that if we tried to run a pair of EL34's
in class A at 500v B+, that to keep the operation in class A (with a peak current of 70mA, double that side's idle current of 35mA), we would need a 14k impedance at the OT primary.
But as you found, 14kΩ plate to plate is an awkwardly high impedance. So amp designers
do not keep running the output stage in class A while raising the B+ to such voltage levels (really, 400-450v B+ is probably a practical limit).
Instead, they operate the tubes in class AB, use a smaller primary impedance, and have a higher peak current for one half the output stage. Class AB allows this in part because the effective load is not plate-to-plate impedance/2 (as in class A, which is why this number is used in the earlier calculation), but plate-to-plate impedance/4. That's a higher peak current, and in combination with the higher supply voltage results in greater output power.
This is also why class AB has to idle at less than 100% dissipation, because the current peaks are so much greater.
In fact, keep calculating with your 500v, 14k impedance example, and find power output. The EL34 won't be able to pull its plate below 50v, so you get 450v peak voltage swing at best, and
one tube has a 70mA peak current, double its idle of 35mA. 450v*0.07A = 31.5w
peak. 31.5w / 2 = 15.75w RMS
So trying to keep the EL34's class A, while raising B+ above some level, and raising OT primary impedance to maintain class A operation results in
less power output than you'd get with the same tubes in class A at lower B+ voltage and lower primary impedance. You should be able to get 25w with a pair of class A EL34's and properly chosen B+ and OT impedance.
So you see, you're calculation isn't wrong (but you got the right answer for the wrong reasons/assumptions)... it just proves why class A isn't used at high supply voltages. Instead, class AB is used, idling the tubes cooler and getting higher power output power with bigger voltage and current peaks.
here is a chart I have prepared:
PLATE VOLTAGE POWER 70% CURRENT LOAD
You're wrongly applying the approach demonstrated to 70% dissipation, presumably related to class AB. So the whole chart is invalid. Peak current in class AB is never double idle current (cause then it would still be class A), but very often 3-4 times idle current. The power you listed is also idle plate dissipation, not output power found by multiplying peak voltage swing by peak current, and divided by 2 to convert peak watts to RMS watts.
There's actually a similar approach, but with very different assumptions used to determine a useful load and power output in class AB. Review the Radiotron Designer's Handbook 4th Edition, Chapter 13, Sections 3-5. You can download a pdf of it in the Library of Information.
All of this is shown in the chart below for a 6L6, which shows that as OT primary impedance is increased beyond a certain point (generally as a result of increasing supply voltage while trying to maintain class A operation), power output drops.