So, I seem to have answered the question myself. Please feel free to critique.
Consider the following, design parameters:
MTOW 700 kg
Rotor Dia 7 m
Chord 0.18 m
Number of blades 3
Tip speed 192 m/s
Zero Lift Drag co. 0.008
To avoid super long calculations, lets employ basic aerodynamic design formulas.
First, what is the induced power for the above mentioned rotor?
Assuming the industry standard starting value of 1.15 for K and standard air density of 1.225 kg/m3 for p, we can find that:
P= (1.15 * (700*9.81)^1.15)/sqrt(2*1.225*38.48)
= 67394 Watt = 67.39 kW
Plus the profile power calculated from the solidity ratio and zero lift drag co.
Solidity ratio = total blade area / disk area = 0.0491
Profile power = 0.125*p*A*(Vtip)^3*sol ratio*zero lift drag co
= 16.39 kW
Now the total power becomes 83.78 kW for a conventional helicopter. Plus 10% for the anti torque rotor the total power = 83.78 * 1.1 = 92.15 kW
For the case of the tandem layout, induced power for each rotor will be calculated assuming a 50:50 sharing of thrust. Profile power remains the same but must be multiplied by 2 due to two rotors.
So, induced power per rotor = Pi = 23.83 kW
Adding profile power brings us to 23.83*2 + 16.39*2 = 80.44 kW
The above is still missing transmission losses and various other sources of parasitic drag but it does give a good view of what to expect. Thank you for the answers though
Last edited by alwynhartman; 17th Oct 2012 at 23:33.