The parameters are: windspeed is 7 M/s, cart velocity is 9 m/s DDW, both measured wrt the ground which is not moving. Cart weighs 270 Kg and has a 5 m diameter propeller.
Let the wheels generate 1000 Watts as the cart moves over the ground at 9 m/s and the wheel drag is 111.1 N.
Note: All calculations are in the frame of reference where the ground is stationary and has NO energy. The cart, moving at 9 m/s and with a mass of 270 Kg has 10,935 Joules of Kinetic Energy. The 1000 J/s developed at the wheels comes entirely from this KE of the cart being transformed into 1000 J/s of energy at the wheels and then this energy is used to do Work at the propeller to exert a thrust force on the air, accelerating it.
Let the overall efficiency of the cart be 87.5% so that there is 875 J/s of Work being done at the propeller on the air. (In reality such a high overall efficiency is impossible as the propeller taken by itself will not have this high an efficiency, but I am using a best case analysis here)
Thrust = mass x acceleration = л/4•D^2•[v+∆v/2] • ρ•∆v where л/4•D^2•[v+∆v/2] • ρ is the mass of the air moving through the propeller and ∆v is the acceleration of the air by the propeller. With the 5 meter prop, л/4•D^2• ρ works out to 24. V is the velocity of the relative headwind and that we know to be 2 m/s in this case. We don’t yet know what the value of ∆v is and we don’t yet know the value of thrust.
However we do know the power absorbed by the propeller is 875 Watts (875 J/s) and there is another equation that says *Engine power = Thrust•[v+∆v/2]. So we can equate these two relationships to solve for ∆v and then solve for thrust:
л/4•D^2•[v+∆v/2] • ρ•∆v = *Engine power / •[v+∆v/2]
Let ∆v be replaced by x
48x + 12 x^2 = 875 / (2 +x/2)
12x^3 + 96X^2 + 192 x =1750, x ~ 2.99
So ∆v = 2.99 and plugging that into either or both of the above equations gives a value for Thrust = 250 N.
Wheel Drag is 111 N and we can calculate Rolling Resistance from mgCrr using Crr of 0.02 and RR = 53 N. There is also air drag due to the 2 m/s headwind but I am not going to bother calculating it and assume a reasonably low value of 10 N for a Total drag force of 174 N.
At this point the less informed among the cart crowd will say I have just shown the cart works, because the thrust is 250 N and the drag is 174 N, right? No, forces alone are not the whole story as we need to consider energy and power. The thrust force is what would be there IF and only IF, there is sufficient energy in the system to develop and sustain that thrust. Now we will see if that is the case or not:
We know the prop will accelerate the air mass passing through it by ∆v and that is 2.99 m/s (effectively slowing the 7 m/s wind down by that amount)
We can calculate the air mass from the thrust equation: air mass = л/4•D^2•[v+∆v/2] • ρ and we already have all the values to fill in: = 24 (2 +1.5) = 84 Kg.
The initial KE of that mass is 2058 Joules and after being slowed it is 672 Joules so the air has lost 1386 Joules of KE.
In this best case analysis I will allow for ALL of that energy to be transferred to the cart! (In reality there will be air turbulence around the propeller, and such a 100% transfer is absolutely impossible).
Now we must consider the losses: In order for the cart to generate that 1000 Joules of energy at the wheels, it must move a distance of 9 meters over the ground against the drag force that we calculated to be 174 N. 174 N x 9 meters = 1,566 Joules of energy the cart requires in order to do the work at the propeller and overcome all drag at the wheels and due to air resistance.
Now it just a simple matter of seeing if the energy taken from the air is enough to replace the energy taken from the cart’s kinetic energy to do the work at the propeller and overcome all losses. 1566 J/s needed vs 1386 J/s supplied leaves a shortfall 180 J/s.
It should be obvious at this point, to even the most devout member of the cart cult, the cart simply will not work as advertised!