paulmarch wrote:I also think 93143’s comment that “getting energy conservation in an analysis of a thruster of any kind requires one to account for energy transferred to or from what the thruster is pushing on” is spot on.

Thanks for the support...

GIThruster wrote:http://en.wikipedia.org/wiki/Linear-quadratic_regulator

What has dynamic system control got to do with the issue at hand?

GIThruster wrote:If you don't believe me, take the case of an Ion thruster and do the calcs.

All right, you asked for it. Watch this:

Let us posit a thruster with an input power of P, an efficiency of η, and an exhaust velocity of v_exh, on a spacecraft with total initial mass M0 and initial velocity V0.

The thruster is turned on at t = 0 and begins expending propellant at a rate mdot. Without loss of generality, we specify that the thrust is parallel to the initial velocity with respect to the observer, so as to avoid the unnecessary complication of vector math (recall that in accordance with Pythagoras' Theorem, momenta in different axes do not sum linearly, but kinetic energies do).

The static jet power is:

η*P = 0.5*mdot*v_exh^2

Therefore

mdot = 2*η*P/v_exh^2

After Tsiolkovsky,

deltaV = v_exh*ln(M0/(M0-mdot*t))

Kinetic energy E of the vehicle at time t is

E = 0.5*(M0-mdot*t)*(V0+deltaV)^2

and the change in vehicle kinetic energy since t = 0 is

deltaE = 0.5*(M0-mdot*t)*(V0+deltaV)^2 - 0.5*M0*V0^2

=

0.5*M0*V0^2 + M0*V0*deltaV + 0.5*M0*deltaV^2 - 0.5*mdot*t*V0^2 - mdot*t*V0*deltaV - 0.5*mdot*t*deltaV^2

- 0.5*M0*V0^2 = (M0-mdot*t)*V0*deltaV + 0.5*(M0-mdot*t)*deltaV^2 - 0.5*mdot*t*V0^2

This is a quadratic in V0. In order to characterize it, we now take its derivatives. For a given deltaV, the first and second derivatives of deltaE with respect to initial velocity V0 are

ddeltaE/dV0 = (M0-mdot*t)*deltaV - mdot*t*V0

d^2deltaE/dV0^2 = -mdot*t

Note that the second derivative is always negative. This means the vehicle kinetic energy delta as a function of starting velocity, for a given deltaV, is concave downward; that is, it exhibits a maximum. Where? Let the first derivative equal zero:

(M0-mdot*t)*deltaV - mdot*t*V0 = 0

Therefore the maximum occurs at

V0 = (M0-mdot*t)*deltaV/(mdot*t)

The limit of this expression as t -> 0 is

V0 = M0*deltaV/(mdot*t)

which is singular. However, as the numerator and denominator are both zero to the same order, it can be regularized. The limit of deltaV as t -> 0 is

dV/dt*t [@ t=0] = v_exh*mdot*t/(M0-mdot*t) [@ t=0] = v_exh*mdot*t/M0

So the maximum rate of increase of deltaE occurs at

V0 = M0*v_exh*mdot*t/(M0*mdot*t)

V0 = v_exhWhat is the rate of change of E with respect to time at this point?

E = 0.5*(M0-mdot*t)*(V0^2 + 2*V0*deltaV + deltaV^2)

dE/dt = M0*V0*dVdt + M0*deltaV*dVdt - mdot*t*V0*dVdt - mdot*t*deltaV*dVdt - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)

dE/dt[@t=0] =

M0*V0*v_exh*mdot

/(M0-0) + 0 - 0 - 0 - 0.5*mdot*(V0^2 + 0 + 0)

= mdot*V0*v_exh - 0.5*mdot*V0^2

so if V0 = v_exh, as established above, then

dE/dt[@t=0,V=V0=v_exh] = 0.5*mdot*v_exh^2 = η*Pwhich makes sense, as at V = v_exh the exhaust is observed to be stationary, so all the propulsive energy must be going into the vehicle.

Remember, this is a

maximum.

In other words, the energy gained by an accelerating spacecraft cannot exceed the energy provided to its thrusters in any reference frame, assuming the thrusters use internally-carried propellant. This is because the loss of mass due to expenditure of propellant becomes a large kinetic energy sink as the initial velocity increases.

But, you will note, the energy gained by the spacecraft can easily be

less than that expended. Where does the excess go? Into the exhaust, of course.

The rate of outflow of exhaust kinetic energy e at time t is given by

de/dt = 0.5*mdot*(V0 + deltaV - v_exh)^2

= 0.5*mdot*(V0^2 + 2*V0*deltaV - 2*V0*v_exh + deltaV^2 - 2*deltaV*v_exh + v_exh^2)

Earlier, we had

dE/dt = M0*V0*dVdt + M0*deltaV*dVdt - mdot*t*V0*dVdt - mdot*t*deltaV*dVdt - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)

dV/dT = v_exh*mdot/(M0-mdot*t)

which results in

dE/dt =

(M0-mdot*t)*V0*v_exh*mdot

/(M0-mdot*t) +

(M0-mdot*t)*deltaV*v_exh*mdot

/(M0-mdot*t) - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)

= mdot*(V0*v_exh + deltaV*v_exh) - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2)

Sum the two rates of increase of kinetic energy (the vehicle and the exhaust) and we get:

dE/dt + de/dt =

mdot*(V0*v_exh + deltaV*v_exh) - 0.5*mdot*(V0^2 + 2*V0*deltaV + deltaV^2) + 0.5*mdot*(

V0^2 + 2*V0*deltaV - 2*V0*v_exh + deltaV^2 - 2*deltaV*v_exh + v_exh^2)

dE/dt + de/dt = 0.5*mdot*v_exh^2 = η*PThis result holds for all values of V0 and t. In other words,

the kinetic energies gained by an accelerating spacecraft and its exhaust always sum to the value of the total energy productively expended in the operation of the drive. Any inefficiency generally results in the wasted energy appearing as heat in both the exhaust and the structure of the vehicle. It can be shown that thermal energy is invariant under Galilean transformations, which results in conservation being maintained for the full energy budget; this is left as an exercise for the reader.