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?
P=(K(T)^1.5)/sqrt(2pA)
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
= (0.125*1.225*38.48*(192)^3*0.0491*0.008)/1000
= 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 October 2012 at 23:33.