Current draw calculations for a brush-less motor are way off
I think you have to go back to the physics instead of treating Matlab as some magic black box that spits out the answers.

For a given (say 40V) supply you can work out how the current varies as RPM, ((via the speed constant to give the back EMF, and the resistance to give the current) and hence both electrical power and mechanical power (= electric power minus loss in that resistance). 

Your specific mistake is to calculate Iin as Vin/R. This is the current at speed=0, or the stall current. The motor will take this instantaneously on starting if the full voltage is applied. However as the propellor spins up, the motor generates back EMF (1V for every 100rpm) reducing the voltage across the winding resistance, and therefore reducing the current draw.

Your job is to find the point where the speed is high enough that there is only enough current to supply enough power to push the air, overcome mechanical losses (propellor efficiency) and generate waste heat in the winding resistance.

Now you also need to work from the mechanical side, to find how mechanical power required, thrust, and air velocity vary with propellor speed, given diameter (hence swept area) and pitch (hence nominal airspeed at a given RPM.

This is easiest if you convert everything into SI, so RPM into radians/second etc.

Airspeed * density * swept area gives you mass flow rate (mass/second), 

0.5 * mass flow * airspeed^2 gives you power (via the usual formula for kinetic energy) - that's power (Pout) imparted to the air. 

Then there's a fudge factor for propellor efficiency (Pconst=1.15) = Pmech is Pconst * Pout)

Put these two curves together, and you should find one speed where both expressions yield the same mechanical power. That will be the working speed of the motor.

Now you know the current at that speed from the first curve.

Also work out the thrust at that speed, from Pout/airspeed. Hopefully it'll match the assumption you originally made, of 130N.

I'm not spelling it out in full detail because I think you'll learn better if you have to think about the physics involved.