We know that F = PcAtCf, and we know that the peak thrust of the L2300 is 661 lbf (from ThrustCurve), so to solve Pc, we need a throat area and a Cf. The first one's easy -- the throat diameter is 0.656", and so the throat area is pi/4*0.656^2 = 0.338 in^2.
For Cf, we need a bunch of stuff (see the Cf equation at
https://nakka-rocketry.net/th_thrst.html). The exit diameter is approximately 1.35" (scaled from the drawing at
https://www.aerotech-rocketry.com/c...sy_dwgs/54mm_kba_ac/54-2550_l2300g-p_assy.pdf ). This gives an expansion area ratio of 1.35^2/0.656^2 = 4.2. From a quick and dirty PEP run for a BaNO3-containing propellant, we get a gamma of 1.2, an exhaust molecular weight of 26, and a chamber temperature of 5000 F. Further, it's reasonable to assume that TRA static tested the L2300 near sea level, so let's guess the external pressure at ~14.7 psi.
We still need the exit pressure and the chamber pressure to solve Cf. Since Pe is a function of Pc (see the isentropic cheat sheet; Pc is the stagnation pressure and Pe is the static pressure at the exit Mach number), and we're actually trying to solve Pc with Cf (which is itself a function of Pc), the solution for Pc is implicit. So, time to pull out Excel and do a couple of goal seeks (or MATLAB and a numerical solution method). The result is around 1250 psi.
Now, to figure out what the new max pressure is, we calculate the Kn difference. Decreasing the nozzle throat size to 0.563 will increase the max Kn by a factor of 1.36. We know that Pc is proportional to Kn^(1/1-n), from the burning rate law. For this propellant, let's guess n at 0.3. This means that the maximum chamber pressure will increase by a factor of (1.36)^(1/(1-0.3)) = 1.55ish. The max Pc we calculated using Excel is ~1250 psi with the 2550 nozzle throat, and so with the 1750 nozzle, the max Pc will be (1250)(1.55) = 1930 psi.
