I'm answering the question of after the Edit. Simple momentum theory:
$$ T = C_T \cdot \rho \cdot A \cdot \Omega^2 \cdot R^2 \tag{Thrust}$$
$$ P = C_P \cdot \rho \cdot A \cdot \Omega^3 \cdot R^3 \tag{Power}$$
$$ C_P = \frac{{C_T}^{3/2}}{\sqrt2} \tag{Ideal Power}$$
$$ FM = \frac{Ideal Power}{InducedPower+ProfilePower} \tag{FigureOfMerit}$$
The Figure Of Merit is a dimensional unit and provides an efficiency measure. It always gives a better result for higher disk loadings, but does provide a correction for real life effects on the ideal power. So for your case, with A = 3.14 m$^2$, $\rho$ = 1.225, T = 981N, the ideal power is:
$$P_{ideal} = 2 \cdot \left(\frac{(T/2)^{3/2}}{\sqrt{2 \cdot \rho \cdot A}}\right) = 2 \cdot \left(\frac{(981/2)^{3/2}}{\sqrt{2 \cdot 1.225 \cdot 3.14}} \right) = 7.8 kW $$
From J. Gordon Leishman, Principles of Helicopter Aerodynamics:

$$C_T = \frac{T}{\rho \cdot A \cdot \Omega^2 \cdot R^2} \tag{CT}$$
Tip speed $V_{tip} = \Omega \cdot R$ should not exceed critical Mach = 0.7 * 340 m/s = 238 m/s. At this tip speed, $C_T$ = 981/(1.225 * 3.14 * 238$^2$) = 0.009 0.0045. Corresponding Figure of Merit = 0.75 0.55. So the power to drive the rotor would be $$7.8 / 0.55 = 14.2 kW$$
The rotor would turn at 238 rad/s = 2,270 rpm, hopefully that is around the rpm for max. torque for your engine. Otherwise you need to gear up/down as required, you have gearboxes anyway.
You would need to add transmission losses for driving the two rotors from your engine. Prouty gives a power loss per stage of 0.0025[max.power + actual power]. Each disk has 2 stages, so a power loss of about 0.005 * 2 * 14.2 = 0.14 kW per disk.
$$Torque = P / \Omega = 14,200 / 238 = 60 Nm$$
EDIT
A numerical error in calculating $C_T$ = 0.0045 not 0.009, so the FM is lower and torque may not be enough. In that case the rotor diameter should be reduced and the rpm increased.