Rigorous analysis thrust/propellant power, arbitrary frame.
Posted: Thu Jan 17, 2013 4:56 pm
Take a test mass, M kg, travelling at velocity Q. It begins ejecting mass at the rate of m kg/s of material at Vm/s at t=0. Now consider the energy required for this ejection rate in a time sample of Δt at t=0.
Impulse Equations (Force = rate of change of momentum):
Force on ejectant = d(m[Q+V])/dt = +V.m.Δt
Force on test mass = d((Q-(Vm/M))xM)/dt = -V.m.Δt
Conservation of momentum:
V.mΔt = -M.ΔQ
(note; ΔQ/Δt =~dQ/dt=-V.m/M, therefore Force (on test mass) = -mV (unsurprisingly, same result as above))
OK! So now do the maths!...
A: Power, by rate of change of kinetic energy, at t=0;
Test Mass---
ΔKE = ½ M Q² - ½ M (Q-ΔQ)² = +M.Q.ΔQ - ½ M ΔQ²
lim ΔQ->0; ΔQ²=0
So ΔKE = M.Q.ΔQ = -M.Q.mVΔt/M = -Q.V.m.Δt
Propellant---
ΔKE = ½ {(Q+V)² – Q²} m Δt = Q.V.m.Δt + ½ V².m.Δt
TOTAL ΔKE = -Q.V.m.Δt + Q.V.m.Δt + ½ V².m.Δt = ½ V².m.Δt
NOTE: TOTAL ΔKE independent of Q
B: Alternative calculation of Power, by rate of expended work, at t=0;
Test Mass---
Rate of Work = -V.m.(Q- ΔQ).Δt = -Q.V.m.Δt + Q.V.m.Δt.ΔQ
lim Δt->0; ΔQ.Δt=0
So Work rate = -Q.V.m.Δt
Propellant---
***CAUTION!!.. distance travelled by a mass accelerating to V in time Δt is V.Δt/2, because this is the average distance travelled for the linear acceleration (this is a legitimate assumption on a timescale of lim Δt->0)***
Rate of Work = V.m.(Q+V/2).Δt = Q.V.m.Δt + ½ .V².m.Δt
TOTAL rate of work = -Q.V.m.Δt + Q.V.m.Δt + ½ V².m.Δt = ½ V².m.Δt
TWO DIFFERENT APPROACHES; BY RATE OF WORK DONE AND BY ΔKE.
SAME ANSWER.
BOTH ANSWERS INDEPENDENT OF Q.
There is NO 'over-unity' violation if the maths is done correctly for an accelerated mass using a propellant.
Note that the ±Q.V.m.Δt term in each of the propellant and mass cancel out. This will grow as Q grows, but the result remains frame-independent because the sum of the terms is zero. Always. And forever. Because it is the same, but opposite, term in both the propellant and thrusted mass.
It only becomes an 'un-physical' violation if there is no propellant to counterbalance the Q.V.m.Δt term.
Impulse Equations (Force = rate of change of momentum):
Force on ejectant = d(m[Q+V])/dt = +V.m.Δt
Force on test mass = d((Q-(Vm/M))xM)/dt = -V.m.Δt
Conservation of momentum:
V.mΔt = -M.ΔQ
(note; ΔQ/Δt =~dQ/dt=-V.m/M, therefore Force (on test mass) = -mV (unsurprisingly, same result as above))
OK! So now do the maths!...
A: Power, by rate of change of kinetic energy, at t=0;
Test Mass---
ΔKE = ½ M Q² - ½ M (Q-ΔQ)² = +M.Q.ΔQ - ½ M ΔQ²
lim ΔQ->0; ΔQ²=0
So ΔKE = M.Q.ΔQ = -M.Q.mVΔt/M = -Q.V.m.Δt
Propellant---
ΔKE = ½ {(Q+V)² – Q²} m Δt = Q.V.m.Δt + ½ V².m.Δt
TOTAL ΔKE = -Q.V.m.Δt + Q.V.m.Δt + ½ V².m.Δt = ½ V².m.Δt
NOTE: TOTAL ΔKE independent of Q
B: Alternative calculation of Power, by rate of expended work, at t=0;
Test Mass---
Rate of Work = -V.m.(Q- ΔQ).Δt = -Q.V.m.Δt + Q.V.m.Δt.ΔQ
lim Δt->0; ΔQ.Δt=0
So Work rate = -Q.V.m.Δt
Propellant---
***CAUTION!!.. distance travelled by a mass accelerating to V in time Δt is V.Δt/2, because this is the average distance travelled for the linear acceleration (this is a legitimate assumption on a timescale of lim Δt->0)***
Rate of Work = V.m.(Q+V/2).Δt = Q.V.m.Δt + ½ .V².m.Δt
TOTAL rate of work = -Q.V.m.Δt + Q.V.m.Δt + ½ V².m.Δt = ½ V².m.Δt
TWO DIFFERENT APPROACHES; BY RATE OF WORK DONE AND BY ΔKE.
SAME ANSWER.
BOTH ANSWERS INDEPENDENT OF Q.
There is NO 'over-unity' violation if the maths is done correctly for an accelerated mass using a propellant.
Note that the ±Q.V.m.Δt term in each of the propellant and mass cancel out. This will grow as Q grows, but the result remains frame-independent because the sum of the terms is zero. Always. And forever. Because it is the same, but opposite, term in both the propellant and thrusted mass.
It only becomes an 'un-physical' violation if there is no propellant to counterbalance the Q.V.m.Δt term.