Rocket efficiency

wrote in message
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I wonder if someone on this list can help me with something that's
confusing me:

In rocketry we want to get the maximum amount of thrust from the minimum
mass of propellant, so we need to accelerate the minimum of propellant to
the maximum velocity possible in order to achieve this aim.

In some other applications (e.g. if the mass of propellant is small
compared to the mass of the rocket - as in some rocket-powered aircraft)
we don't need to worry about using propellant mass so efficiently.

This is what's confusing me:

It should take 200 J to accelerate 1kg of exhaust to 20m/s

It should take 400 J to accelerate 0.5kg of exhaust to 40m/s

Both scenarios provide the same amount of thrust (1 x 20; 0.5 x 40), but

Same amount of thrust, or same amount of impulse (change in momentum)?
Thrust will be related to the time in which the mass is expelled from the
motor. Both scenarios provide the same total impulse, if thats what you
mean.

scenario 1 seems to be more efficient than scenario 2. I don't see where
any energy is lost, so I'm confused as to why that should be.

I think the issue is that momentum (i.e. thrust, impulse, time, all that
stuff) is related to energy, but not the same. For a particular amount of
momentum, say in a moving object, the amount of energy required to acheive
this momentum depends on the mass, and the relationship is not linear. P=mv,
KE=1/2mv^2

Thus for the same momentum at half the mass, v must double, meaning that KE
will be 4 times as high. Momentum and KE are not the same thing. It makes
more sense to talk about rocket motors in terms of momentum (i.e. impulse),
since that tells you what the motor does, and in terms of specific impulse
(which tells you effectively how much 'woosh' you get from a unit amount of
propellant).

I'm sure there must be some flaw in my reasoning but I don't see it. I'd
be very grateful if someone could point it out to me.

Thanks in advance,

vne




Just going back to simple physics:

Kinetic Energy = 1/2mv^2
Momentum = mv
(Impulse = ft = Delta mv)
Average Velocity - Impulse/propellant mass (mv/m)


So double the velocity, you have 4 times the KE, but only twice the momentum
(and thus for the same time, say one second, double the thrust). I was
thinking about the exhaust velocity/propellant mass stuff the other day, but
I seem to remember there being another term relating to pressure and nozzle
area in there.

HTH

--
Niall Oswald
=========
UKRA 1345 L0
EARS 1151
MARS

"Gravity assisted pieces of the rocket raining from the sky should be
avoided. It is also financially undesirable."
-Portland State Aerospace Society

" wrote in message ...
I wonder if someone on this list can help me with something that's
confusing me:

In rocketry we want to get the maximum amount of thrust from the minimum
mass of propellant, so we need to accelerate the minimum of propellant to
the maximum velocity possible in order to achieve this aim.


No, actually we do NOT want to get the maximum amount of thrust from
the minimum mass of propellant. We want to get the maximum exhaust
velocity. We do NOT want to get the maximum exhaust velocity in order
to increase thrust. Getting maximum exhaust velocity is in and of
itself, the aim.

Put very simply:
If your exhaust velocity is Mach 1, then you need a 24:1 fuel to
structure weight ratio in order for a rocket to reach Mach 24, or
17,200 mph...ie: orbital velocity.

If your exhaust velocity is Mach 2, you only need 12:1 fuel to
structure weight ratio. For mach 3 exhaust, you only need 8:1 ratio,
etc.

Rocket 2, in the example you have sited, has an exhaust velocity which
is twice, consequently (in a vacuum), the rocket it is lifting would
end up going twice as fast.

Maximizing thrust is definately NOT the aim. Consider ION propulsion
craft...they have EXTREMILY low thrust...they thrust to weight ratio
is less than 1:1000...that is, the thrust is 1000 times to weak to
lift it off the groun on earth. But Ion propulsion has a specific
impulse of up to 10,000..that means that the exhaust velocity of the
ions is so high that if the spacecraft is 50% fuel by weight, it would
take 10,000 * (1000/1) or 10,000,000 seconds to run out of fuel. TEN
MILLION SECONDS. IE: The rocket can keep on slowly accelerating for
YEARS and DECADES without running out of fuel. A chemical rocket, on
the other hand, runs out of fuel in at most a few minutes. Lets take
a typical Lox/Kerisone engine with a specific impulse of around 300.
The Lox/P4 engine can produce 5gs of thrust! Thats 5:1 thrust ration
as compared to 1:1000 for the ion engine. However...if the
Lox/Kerosine rocket is 50% fuel by weight, same as the Ion drive
rocket I describe above...the fuel will run out in 300*(1/5) = 60
seconds. One minute. Then your out of fuel.

" wrote in message ...
I wonder if someone on this list can help me with something that's
confusing me:

In rocketry we want to get the maximum amount of thrust from the minimum
mass of propellant, so we need to accelerate the minimum of propellant to
the maximum velocity possible in order to achieve this aim.


No, actually we do NOT want to get the maximum amount of thrust from
the minimum mass of propellant. We want to get the maximum exhaust
velocity. We do NOT want to get the maximum exhaust velocity in order
to increase thrust. Getting maximum exhaust velocity is in and of
itself, the aim.

Put very simply:
If your exhaust velocity is Mach 1, then you need a 24:1 fuel to
structure weight ratio in order for a rocket to reach Mach 24, or
17,200 mph...ie: orbital velocity.

If your exhaust velocity is Mach 2, you only need 12:1 fuel to
structure weight ratio. For mach 3 exhaust, you only need 8:1 ratio,
etc.

Rocket 2, in the example you have sited, has an exhaust velocity which
is twice, consequently (in a vacuum), the rocket it is lifting would
end up going twice as fast.

Maximizing thrust is definately NOT the aim. Consider ION propulsion
craft...they have EXTREMILY low thrust...they thrust to weight ratio
is less than 1:1000...that is, the thrust is 1000 times to weak to
lift it off the groun on earth. But Ion propulsion has a specific
impulse of up to 10,000..that means that the exhaust velocity of the
ions is so high that if the spacecraft is 50% fuel by weight, it would
take 10,000 * (1000/1) or 10,000,000 seconds to run out of fuel. TEN
MILLION SECONDS. IE: The rocket can keep on slowly accelerating for
YEARS and DECADES without running out of fuel. A chemical rocket, on
the other hand, runs out of fuel in at most a few minutes. Lets take
a typical Lox/Kerisone engine with a specific impulse of around 300.
The Lox/P4 engine can produce 5gs of thrust! Thats 5:1 thrust ration
as compared to 1:1000 for the ion engine. However...if the
Lox/Kerosine rocket is 50% fuel by weight, same as the Ion drive
rocket I describe above...the fuel will run out in 300*(1/5) = 60
seconds. One minute. Then your out of fuel.

Take ion engines. They use massive amounts of power to produce very low
thrust, but since they do not consume much mass they can keep producing
thrust almost indefinitely, so the total delta-v can be quite high.

The whole mass efficiency/energy efficiency tradeoff can be applied to many
more things than rocket engines. For example airplanes. They work by
accelerating air downwards to produce a force to compensate gravity. If you
want a very energy efficient airplane you need to accelerate a lot of air
downward a little bit. That is why sailplanes have such high wing aspect
ratios: to have as much air as possible to work with.

Or a propeller. All else being equal, a large propeller will be more energy
efficient in producing thrust at low velocities than a small propeller
because it has more air to work with and thus needs to accelerate that air
less.

Hope that helped.


To summarize, maximum thrust is *not* what one is primarily after. In
the vacuum of space, maximum exhaust velocity is the determinant of a
rocket final velocity. In going from ground to orbit, the rocket
*does* have to produce enough thrust to counteract the forces acting
on it; however, the exhaust velocity will *still* be the most
inmportant factor in determining the final velocity of the rocket, as
long as the rocket produces enough thrust to overcome air resistance
and gravity.

The basic rocket equation follows:

V(t) = Vxln(dM/dt) -gt

where V is the velocity of the rocket, Vx the exhaust velocity, dM/dt
the change in the weight of the rocket between liftoff and the current
weight (ie the fuel used), and gt is gravity times the the time passed
since liftoff.

Note: while my basic premise was correct, Last night I also posted
some goblygook for examples. I was going to come on and flame
*myself* this morning for posting nonsense, but I see the moderater
never allowed my post through. I guess I should thank him; now I
won't have to flame myself