Where I was really going with that query was ... what happens ... in SE as opposed to PP. ...
In SE, your only option for output stage class of operation is Class A. Because if your "single end" ever turns off, you hard-clip the output to the speaker.
Where I was really going with that query was how does what happens with a 5K primary relate to what happens with a 1K7 primary? ...
In SE, you can get an operating output stage with willy-nilly selection of B+ and load impedance. But you picked an expensive output tube type, and decided to use a pair of them... If you want to get your money's worth, then you shoot for something approaching maximum clean output power.
If the tube never turns off (never hits, or just touches, 0 current), then for zero distortion it will swing from idle current to 2x idle, and down to zero current, then return to the idle current. Also for zero distortion, the plate voltage change that happen when current moves from idle to 2x idle will be exactly the same as the plate voltage change when current swings in the opposite direction.
For maximum output power, you will also idle the tubes at 100% dissipation, because anything less will simply shift the loadline downward and give less output power.
To make all this work, you have to balance the OT primary Z, B+ voltage, PT current capability and tube characteristics. You've narrowed the playing field for us by selecting an OT (and its primary impedance) before any other consideration.
You have a 1.7kΩ OT to be used with a pair of tubes; that will look like a 3.4kΩ load to a single tube if we plot loadlines on a single tube's plate curves (which is what Jazbo did). Real-deal KT88's have an
absolute maximum plate dissipation of 42w; Jazbo used the design maximum of 35w, which also allows wiggle room if you use 6550's or tubes labeled "KT88" but not living up to all its specs.
35w / 410v = ~85mA idle per tube.
If the tube swings upward to double-idle, the change in plate current is the extra 85mA (peak current of 170mA per tube). Ohm's Law time: what voltage swing is required for an increase of 85mA in a 3.4kΩ load impedance? 85mA * 3.4kΩ = 289v peak. Can the tube & B+ support this? 410vdc - 289v = 121v left across the tube, so plenty even in pentode mode.
You'd plot the loadline to help ensure the swing towards 0 current is a similar length on paper as the swing towards double-current. If they are exactly the same, then you have zero distortion. With real tubes, one will be a bit shorter than the other, and how much different they are tells the amount of one type of distortion.
What if you wanted to use a 5kΩ load, as you alluded? Well, you now know that for class A, SE, you need to swing from idle to double-idle current and your selected B+ (and tube dissipation limit) have pegged that idle current at about 85mA. How much voltage swing is needed to push an extra 85mA through 10kΩ? Why 10k? Because I'm guessing you're talking about a 5kΩ primary impedance for the OT, but you have 2 tubes in parallel so plotting loadlines for a single tube requires treating it as a 10kΩ load. 85mA * 10kΩ = 850v peak. Your B+ is only 410v in Jazbo's loadline plot, and you have to leave 40-80v across the tube in addition to plate voltage swing for most output pentodes/beam power tubes (because they can't pull their plate to 0v).
So you'd need over 900v B+... The result is you'd have to raise B+ voltage to get the needed plate voltage swing + operating plate voltage. You'll also need to drop your idle current to hit dissipation limit with the new B+, and re-figure. But that will be easy because you've exceeded the plate voltage rating of the tube. Obviously, such a high load impedance simply won't allow these tubes to pass enough current.
Starting to see the interplay now?