If you are consuming X watts per cell, then each cell will drain by the same amount of current regardless of whether the cells are in series or parallel (or more likely a combination of the two).
This is supposed to be able to handle "up to" 2-hour flights, and (for jets at least) takeoff power is about 3x cruise power, so about 1.5 watts per watt-hour of battery, or a 1.5C peak discharge rate. What chemistries are you thinking of that cannot handle this?
The full power system for a prop looks like this: you have a battery of a specific voltage, which then runs a motor, which then runs a gearbox, which then turns a prop.
The motor and the gearbox can be considered as one unit - an electric motor has two factors, KV(RPM/volt) and KT(torque/amp). The higher the KV, the lower the KT. A high KV motor spins fast, but draws a lot of current for the same torque - putting it through a reduction gearbox turns it into a low KV, high KT motor. Naturally, low KV motors or (low KV setups) are more efficient because they draw less current for a given torque, and the heating power loss varies with current^2.
The prop needs to spin at certain RPM for max aerodynamic efficiency. Given the slider for motor/gearbox selection between high KV/low KT and the opposites, you generally want to have as high voltage as possible, so that you can run a low KV/high KT setup, which means that for the given torque, the current is minimal. I.e you have a motor spinning really fast, through a large reduction gear, driving a prop at the necessary speed and torque without much load on the motor.
So lets say you determine that you want a certain voltage, which requires a stack of cells in series. The only way to get more capacity is to duplicate that stack and put them in parallel. So your weight becomes quantized by the number of stacks you have in parallel. And the more stacks you have in series, the higher the weight jumps between parallel stacks counts.
Subtracting cells from stacks doesn't work well. Lets say you have a single stack of 10 cells 10s1p. If you do something like 8s1p, you lower the output voltage, which means you need to have slightly higher gear ratio to spin the prop at the same efficient rpm, which means you draw more current, which means the extra capacity in the cells doesn't really matter if you are drawing more current.
If your motors draw W Watts at peak, and you have N cells at V volts then the peak per-cell current will be (approximately -- there are internal losses) W/(NV) regardless of the geometry of your stacks.
If N is small, your designs may be restricted, but I'll hazard a guess that N will be large for a 30 passenger plane capable of 2 hour flights (e.g. a Tesla model 3 is almost 3000 cells; 31s96p).
A plane needs to be able to takeoff in a given runway length, with a given payload, and climb at that power to a safe altitude for an option of an aborted landing. And just Tesla set to a high power mode, if when starting to drive it you do max acceleration for 1-2 minute you drain a significant portion of your battery very quickly.
The overall point that Im trying to make is that slight battery specific energy density improvements don't matter when compared against the power losses which are proportional to square root of the current.
> And just Tesla set to a high power mode, if when starting to drive it you do max acceleration for 1-2 minute you drain a significant portion of your battery very quickly.
An airplane under typical operation has a much narrower ranges of power than a Tesla in high power mode. Takeoff power is a low single-digit multiple of cruise power. For my math earlier I used a 3:1 ratio. A Model S Plaid is more like 50:1, or 30:1 for the "regular" Model S.
> The overall point that Im trying to make is that slight battery specific energy density improvements don't matter when compared against the power losses which are proportional to square root of the current.
You've completely missed making that point, since I still haven't seen an argument so far that the current must be very high.
The only way to reduce current is to put batteries in parallel, which for a given voltage doubles the weight of the battery pack.
You could go with lower voltage, but that means for a given, power, you need more current, which drops your efficiency.