Simply using the generic drag equation will get you within the ballpark required for FlightGear.
$$D = C_D \cdot \frac{1}{2} \rho V^2 \cdot A$$ with $C_D$ the flat plate drag coefficient and A being a reference area of your cowl flaps. The data for the correct Reynolds number is best used, with $$Re = \frac{\rho V \bar{c}}{\mu}$$ with $\bar{c}$ = mean aerodynamic chord and $\mu$ = kinematic viscosity of air. The Re uses the wing geometry to determine if the flow is laminar or turbulent, which has a large influence on the drag calculation.
So calculating additional drag for air density $\rho$ of 1.1 kg/m$^3$ (function of altitude) and airspeed of 150 m/s, we would get: $D = C_D \cdot ½ \cdot 1.1 \cdot 150^2 \cdot A = 12,375 \cdot C_D \cdot A [N] $. Now the only thing we need to do is look up the $C_D$ and determine reference area A in [$m^2$], and this is where things get a bit more challenging.
Note that we're calculating additional drag. When the flap is fully retracted and aligned with the cowling, additional drag is zero. Pull the flap open and extra drag is created - it will be less than bluff body drag, since the cowl flap opens to expel exhaust gases which prevents some of the pressure drag from occurring.
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To keep things simple, we can use bluff body aerodynamics with the front projected area as our reference A. For instance an airfoil $C_D$ of 0.045, with projected area of 0.002 * sin$\alpha$ [m$^2$]. $\alpha$ is the deflection angle of the flap.
The exhaust gas expelled behind the cowling flap prevents much of the pressure drag from happening, like in an airfoil where the flow gently foils back into the afterbody. Without the exhaust gas the $C_D$ would be higher, more that of a bullet. How much more depends on a whole lot of exhaust gas pressure parameters. Values of between 0.05 and 0.3 seem could be taken.