OceanGen Motor Parameters Derived from Initial Lab Tests
P. McGill
v 1.2.0
26Aug10

We have a generator that produces an unknown voltage Vg and contains an unknown resistance Rg (we'll ignore inductive effects for this analysis). Connected to this is a known load resistance RL with a measured load voltage VL. The actual generator is three-phase AC feeding a rectifier, but we can represent it as a single-output DC generator.

Levant Power made some measurements on their test rig:
At 8 cm/s, we saw 25 V across a 50 ohm load and 22 V across a 10 ohm load
At 20 cm/s, we saw 65 V across a 50 ohm load and 55 V across a 10 ohm load

We know that the generator voltage is equal to the load voltage divided by the load resistance (i.e. the system current) times the sum of the two resistances:

(Subscript[V, g]) = (Subscript[V, L])((Subscript[R, g])+(Subscript[R, L]))/(Subscript[R, L])

With the test data we can solve for Vg and RL at each speed

eqn1=Vg==25(Rg+50)/50

eqn2=Vg==22(Rg+10)/10

sol1=Solve[{eqn1,eqn2},{Vg,Rg}]//N

eqn3=Vg==65(Rg+50)/50

eqn4=Vg==55(Rg+10)/10

sol2=Solve[{eqn3,eqn4},{Vg,Rg}]//N

The two generator resistance values vary quite a bit, but these lab measurements are probably pretty rough. Levant can probably tell us exactly what Rg is, but this measurement includes the clip leads used during testing and losses in the rectifier.

Now find the generator voltage constant in volts/(meter/ second) at each speed. It shouldn't change much.

Vg/0.08/.sol1

Vg/0.20/.sol2

From this we can see that at 2 meter/second, we would produce at least 650 volts, and the generator resistance would drop this about 2 volts for every amp of current we draw. Based on these measurements, let's assume a reasonable value for the generator voltage constant and generator winding resistance, and set the velocity to 2 m/s:

Kv=330

Rg=2

V=2

Given these values, we can find the value of load resistance we need to obtain a desired damping force.

We know the power into the system is the damping force times the velocity:

(Subscript[P, in])=(Subscript[F, d])×V

and the power out of the system can be found from the output current and load voltage:

(Subscript[P, out])=((Subscript[I, L])^2)×(Subscript[R, L])

but the load current is determined by the generator voltage and the circuit resistance:

(Subscript[I, L])=((V×(Subscript[K, v]))/((Subscript[R, g])+(Subscript[R, L])))

	(Subscript[P, out])=(((((V×(Subscript[K, v]))/((Subscript[R, g])+(Subscript[R, L])))))^2)×(Subscript[R, L])

Now we can define the efficiency of the system:

E=((Subscript[P, out])/(Subscript[P, in]))=(((((((V×(Subscript[K, v]))/((Subscript[R, g])+(Subscript[R, L])))))^2)×(Subscript[R, L]))/((Subscript[F, d])×V))

We can rearrange this to solve for the damping force:

(Subscript[F, d])=(((((((V×(Subscript[K, v]))/((Subscript[R, g])+(Subscript[R, L])))))^2)×(Subscript[R, L]))/(E×V))

Unfortunately, we don't know the efficiency of the system yet (but we expect have some measurements soon). This is important because if the system is 100% efficient then all of the mechanical input power goes to electrical power at the output, and we can completely control the damping force with the load resistor. On the other hand, if the system is 0% efficient, then all of the mechanical input power is wasted and the load resistor has absolutely no effect on the damping force.

Let's define the damping force as a function of both the load resistor and the system efficiency:

Fd[Rl_,E_]:=Rl(V Kv/(Rg+Rl))^2/(E V)

Now we can plot the function for a range of load resistances at four discrete efficiencies of 30%, 40%, 50%, and 60%:

Plot[Evaluate[Table[Fd[r,n],{n,0.3,0.6,0.1}]],{r,5,50},
{PlotLabel->"Damping Force vs. Load Resistance at 2 m/s",
Frame->True,
FrameLabel->{"Load Resistance (ohms)","Damping Force (newtons)"},
PlotStyle->Thick,
GridLines->Automatic}]

The top, blue curve is at 60% efficiency, and the bottom, green curve is at 30% efficiency. We can see that if we desire a damping force at 2 m/s of 30 kN (i.e. a damping constant of 15 kN/m/s), and we are operating at 50% efficiency (violet line), then we need a load resistor of about 14 ohms. If our efficiency is only 30% (green line), then we'll need a load resistor of about 8 ohms.

Of course the efficiency will certainly change with velocity and load, so our measured curves will vary from these, but this analysis let's us choose a reasonable starting point for our testing.