rocket efficiency

[n...@null.com]

I'm trying to work out how efficient a mechanical water rocket would be,
and I'm a bit confused about the theory:

Rocket 1 accelerates 1Kg water / Second of exhaust to a velocity of 20M/S
Rocket 2 accelerates 0.5 Kg water / Second to an exhaust velocity of 40M/S

By my reckoning, both these rockets give the same amount of thrust (Mass x
exhaust velocity) but it takes twice as much energy to power rocket 2 as
it does to power rocket 1 (kinetic energy of the exhaust = 0.5MV^2).

I'm sure I must have made a mistake - can anyone point it out to me?

Thanks,

VNE

[PeteAlway]

"n...@null.com" n...@null.com wrote:

Energy efficiency in a rocket is a tough thing to define, because the
performance of a rocket motor/engine is appropriately measured as thrust
integrated over time (average thrust times time) which gives an impulse, or
momentum imparted. Work (kinetic energy imparted) is measured by thrust
integrated over distance (thrust times distance for a constant speed and
thrust, which generally doesn't happen) Because power depends on speed, and a
rocket's speed is generally constantly changing, depending on mass and time of
flight, calculating power and energy is very difficult and not particularly
useful.

At one point, Goddard calculated the energy efficiency of a rocket by determing
the kinetic energy imparted to the exhaust from a rocket held in a fixed mount.
He found that something like 2/3 of the chemical energy in a propellant could
be converted to the kinetic energy of the exhaust gas.

But in general, it is better to measure the effectiveness of a rocket by
determining its exhaust velocity or specific impulse. Specific impulse is the
total impulse (thrust integrated over time, or average thrust times time, the
momentum imparted by the rocket--total impuls is the quantity specified by the
letter in an NAR designation, such as D12-3) divided by the weight of the
propellant consumed. Turns out that these are proportional (except for some
wacky nozzle/external air pressure effects), such that exhaust velocity =
specific impulse x g.

>I'm trying to work out how efficient a mechanical water rocket would be,
>and I'm a bit confused about the theory:
>
>Rocket 1 accelerates 1Kg water / Second of exhaust to a velocity of 20M/S

The exhaust velocity is actually the best measure of efficiency right there.
0r if you prefer, specific impulse = 20 m/s /9.8 m/s^2 = 2.04s

>Rocket 2 accelerates 0.5 Kg water / Second to an exhaust velocity of 40M/S

Again, the exhaust velocity is the best measure of efficiency, though you could
express it as Specific impulse = 4.08 s.

>By my reckoning, both these rockets give the same amount of thrust (Mass x
exhaust velocity)

Mass x exhaust velocity gives kg m/s, units of momentum or impulse, not thrust.
They give the same total impulse. 20 m/s * 1 kg = 20 kg m/s = 20 Newton-s
(=20 N-s) which is the impulse of a full D engine. This is the same for both
rockets you describe. And total impulse is a good measure of total "kick."

>but it takes twice as much energy to power rocket 2 a

>it does to power rocket 1 (kinetic energy of the exhaust = 0.5MV^2).

Yes, you impart twice the energy into the water in rocket two. You pumped more
air pressure into rocket two. You are rewarded with a higher exhaust velocity.
this is good, not because you get more total kick out of the rocket (20 N-s
either way) but because the weight of propellant you had to accelerate is less.
If the weight of propellant is a significant part of the total weight of the
rocket, rocket two will be significantly lighter, and will fly higher.

>I'm sure I must have made a mistake - can anyone point it out to me?

neglecting gravity and drag, rocket performance is described by Tsiolkovsky's
formula. Note that burnout velocity (well, the water isn't burning, but the
veloocity at the end of thrust), Vf is the measure of performance. Ve is
exhaust velocity, and performance improves with that. Mf is final mass, at
burnout--it is the empty or "dead" weight. Mi is the launch mass, including
propellant. ln (lowercase LN) in the natural log function:

Vf = Ve ln (Mi/Mf)

This takes into account the advantages of a higher exhaust velocity, more
propellant (bigger Mi) and less dead weight (smaller Mf)

Notice that energy doesn't really come into play here. Not that energy isn't
exchanged, but that energy calculations aren't really all that useful in this
particular aspect of rocket physics.

I hope this is of some use to you

Peter Alway

Saturn Press
PO Box 3709
Ann Arbor, MI 48106-3709
http://members.aol.com/satrnpress/saturn.htm
Free scale data at:
http://yellowjacketsystems.com/alway/Default.htm

[Jerry Irvine]

In article <20040708170633...@mb-m02.aol.com>,
pete...@aol.com (PeteAlway) wrote:

Propellant specific impulse
System specific impulse
Ballistic Coefficient

> "n...@null.com" n...@null.com wrote:
>
> Energy efficiency in a rocket is a tough thing to define,

For some I suppose.

--
Jerry Irvine, Box 1242, Claremont, California 91711 USA
Opinion, the whole thing. <mail to:01ro...@gte.net>
Please bring common sense back to rocketry administration.
Produce then publish. http://www.usrockets.com
My articles valuable? Donate http://tinyurl.com/2hmgv

[Doug Sams]

PeteAlway wrote a great post.

Peter,

You've got me thinking about something that's been kicking around
in the back of my head since I BAR'd.

If Work=INT(F*ds), then what am I missing in this scenario?

A constant force is applied to an object at rest for a fixed duration
accelerating that object, and moving it some distance. W=F*d1

That same force is applied for the same duration, this time to
an object already moving with some (non-zero) initial velocity.
Again some acceleration is imparted, but more importantly,
a different distance is traversed since the object was already
moving. W=F*d2

IOW, in each case, the same impulse was applied, but a different
amount of work was done (or was it?). I can see where the same
change in momentum occurred, but using W=F*d seems to indicate
more work was done to the moving object than to the object initially
at rest.

What's wrong here? What am I missing? Where did I blow it? Why
did I skip so many sessions of physics in college?

Thanks.

Doug

--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

[Jerry Irvine]

In article
<f73942f3642a37282f0...@mygate.mailgate.org>,
"Doug Sams" <doug_m...@yahoo.com> wrote:

> PeteAlway wrote a great post.
>
> Peter,
>
> You've got me thinking about something that's been kicking around
> in the back of my head since I BAR'd.
>
> If Work=INT(F*ds), then what am I missing in this scenario?

Air Drag.

Much higher with velocity.

>
> A constant force is applied to an object at rest for a fixed duration
> accelerating that object, and moving it some distance. W=F*d1
>
> That same force is applied for the same duration, this time to
> an object already moving with some (non-zero) initial velocity.
> Again some acceleration is imparted, but more importantly,
> a different distance is traversed since the object was already
> moving. W=F*d2
>
> IOW, in each case, the same impulse was applied, but a different
> amount of work was done (or was it?). I can see where the same
> change in momentum occurred, but using W=F*d seems to indicate
> more work was done to the moving object than to the object initially
> at rest.
>
> What's wrong here? What am I missing? Where did I blow it? Why
> did I skip so many sessions of physics in college?
>
> Thanks.
>
> Doug

--

[Doug Sams]

JI wrote:

> Air Drag.
>
> Much higher with velocity.

Meant to add: Neglecting drag (assume in space).

[n...@null.com]

On Thu, 08 Jul 2004 21:06:33 +0000, PeteAlway wrote:

> "n...@null.com" n...@null.com wrote:
>
> Energy efficiency in a rocket is a tough thing to define, because the
> performance of a rocket motor/engine is appropriately measured as thrust
> integrated over time (average thrust times time) which gives an impulse, or
> momentum imparted. Work (kinetic energy imparted) is measured by thrust
> integrated over distance (thrust times distance for a constant speed and
> thrust, which generally doesn't happen) Because power depends on speed, and a
> rocket's speed is generally constantly changing, depending on mass and time of
> flight, calculating power and energy is very difficult and not particularly
> useful.

Actually, what I'm considering is a water-rocket powered glider that would
operate at a roughly constant airspeed. I was wondering whether the energy
in my compressed air would be put to better use powering a propellor or
pushing water out of the back. The thrust doesn't have to last very long -
just enough to get some altitude.

Perhaps I should have said this before - I figured that practicalities
don't always help in understanding the theory, but perhaps I was wrong.

>>I'm trying to work out how efficient a mechanical water rocket would be,
>>and I'm a bit confused about the theory:
>>
>>Rocket 1 accelerates 1Kg water / Second of exhaust to a velocity of 20M/S
>
> The exhaust velocity is actually the best measure of efficiency right there.
> 0r if you prefer, specific impulse = 20 m/s /9.8 m/s^2 = 2.04s
>
>>Rocket 2 accelerates 0.5 Kg water / Second to an exhaust velocity of 40M/S
>
> Again, the exhaust velocity is the best measure of efficiency, though you could
> express it as Specific impulse = 4.08 s.
>
>>By my reckoning, both these rockets give the same amount of thrust (Mass x
> exhaust velocity)
>
> Mass x exhaust velocity gives kg m/s, units of momentum or impulse, not thrust.

I gave Mass / second - which I think converts my units into thrust?

> They give the same total impulse. 20 m/s * 1 kg = 20 kg m/s = 20 Newton-s
> (=20 N-s) which is the impulse of a full D engine. This is the same for both
> rockets you describe. And total impulse is a good measure of total "kick."
>
>>but it takes twice as much energy to power rocket 2 a
>>it does to power rocket 1 (kinetic energy of the exhaust = 0.5MV^2).
>
> Yes, you impart twice the energy into the water in rocket two. You pumped more
> air pressure into rocket two. You are rewarded with a higher exhaust velocity.
> this is good, not because you get more total kick out of the rocket (20 N-s
> either way) but because the weight of propellant you had to accelerate is less.
> If the weight of propellant is a significant part of the total weight of the
> rocket, rocket two will be significantly lighter, and will fly higher.

Perhaps another application where this could make a difference would be
something like an ion-drive, where the power is fixed (by the size of
the solar-cells / reactor...) but the propellant mass can be much lower
compared to the mass of the spacecraft.

So as the specific-impulse of a rocket engine increases, its energy
efficiency really does decrease? I never realised this before.

>>I'm sure I must have made a mistake - can anyone point it out to me?
>
> neglecting gravity and drag, rocket performance is described by
> Tsiolkovsky's formula. Note that burnout velocity (well, the water
> isn't burning, but the veloocity at the end of thrust), Vf is the
> measure of performance. Ve is exhaust velocity, and performance
> improves with that. Mf is final mass, at burnout--it is the empty or
> "dead" weight. Mi is the launch mass, including propellant. ln
> (lowercase LN) in the natural log function:
>
> Vf = Ve ln (Mi/Mf)
>
> This takes into account the advantages of a higher exhaust velocity,
> more propellant (bigger Mi) and less dead weight (smaller Mf)
>
> Notice that energy doesn't really come into play here. Not that energy
> isn't exchanged, but that energy calculations aren't really all that
> useful in this particular aspect of rocket physics.
>
> I hope this is of some use to you

It's helped clear a lot up in my mind.

Thanks,

VNE

[Alan Jones]

You can't realy compare work done with momentum. Rather, the work
done is equal to the change in energy (kinetic plus potential),
neglecting drag and other things not modelled. It is useful for
things like piston launchers, and not so useful for rocket flight.

Alan

[Alan Jones]

Rocket #2 is much better since it has less mass tham Rocket #1.
Since you initaly want infinite thrust, Rocket #2 thrust is closer to
optimal.

Alan

[n...@null.com]

I know that's normally the case, but for me I'm not so sure - I'm
considering a rocket-powered glider where the lift is generated
aerodynamically.

The amount of power I have is fixed (the amount of air in a CO2 cylinder).
I want to use it as efficiently as possible, and I don't need to worry as
much as usual about the mass of the propellant (water).

VNE

[Kenneth C. McGoffin]

pete...@aol.com (PeteAlway) wrote in message news:<20040708170633...@mb-m02.aol.com>...
> "n...@null.com" n...@null.com wrote:
>
> snipped

>
> At one point, Goddard calculated the energy efficiency of a rocket by determing
> the kinetic energy imparted to the exhaust from a rocket held in a fixed mount.
> He found that something like 2/3 of the chemical energy in a propellant could
> be converted to the kinetic energy of the exhaust gas.

The maximum theoretical amount of kinetic energy transfer from the
propellant to the spent (single stage)rocket is 64.761% at a mass
ratio of 4.921553.

Interesting but largely useless tidbit of rocket info.

What is useful is that the fraction of this maximum energy transfer
increases rapidly from a mass ratio of one but decreases slowly at
higher mass ratios. What this means for hobby rocketry is that almost
all hobby rockets operate at very low energy efficiencies. A mass
ratio of 1.2 only uses about 17% of that 64.761%---approximately 10%
efficiency. And then drag losses eat up most of that. And this all
doesn't even include the losses converting combustion heat energy into
kinetic energy of the exhaust.

It's very easy for the energy efficiency of a hobby rocket measured as
gravitational potential energy at apogee divided by energy of
combustion of the the propellant to be low single digit percentages.

One advantage of water rockets is that they can operate at mass ratios
that are much more energy efficient.
+McG+

[Kenneth C. McGoffin]

"n...@null.com" <n...@null.com> wrote in message news:<pan.2004.07.08....@null.com>...

No mistake. But the mathematics of optimizing water rockets gets very
difficult--systems of simultaneous differential equations. Most folks
use simple, approximate formulas that work "good 'nuff." These are
available on some water rocket web sites. I got partway through doing
the math the right way before I burned out on the idea. Way too much
work for something that wouldn't make me money! But it was
interesting--I actually understand the basic internal interplay of
energy and momentum in the things. What people forget about water
rockets is that the derivative of momentum wrt time has *two* terms.
For "regular" rockets you can ignore one, but not for water rockets.

See my reply to Peter.
+McG+

[n...@null.com]

Thanks,

It's all becoming much clearer now.

VNE

[Gary]

The "distance" used in the work formula given applies to the
displacement of the object caused by the applied force, not the total
distance moved for an object whose initial velocity is not zero.

This is a classic work-energy situation. The work done will be equal to
the change in energy of the system (kinetic energy for this scenario),
which will, indeed, be equal to the work done on the object if it had
started at rest.

KEfinal = KEinitial + W

--
Gary Bolles
NAR 82636

summum jus, summa injuria est

To contact me; bollesg at comcast dot net
http://home.comcast.net/~bollesg/rockets/rockets.html

[Jerry Irvine]

In article <f130bbe4.04070...@posting.google.com>,

This should be in the anti-terrorist FAQ.

[Larry Curcio]

> > The maximum theoretical amount of kinetic energy transfer from the
> > propellant to the spent (single stage)rocket is 64.761% at a mass
> > ratio of 4.921553.
> >

Warning: Incoming from left field!

Interesting. I just got this very result in water rocket calculations.
The problem was to adjust the density of the working fluid
(traditionally water) to maximize the cutoff velocity. In
outer space, this would be

Vcutoff = Ve*ln[(Mprop + Mdead)/Mdead]

where
Ve is effective exhaust velocity,
Mp is propellant (working fluid) mass
aMdead is dead mass (structural mass)

The kinetic energy of the exhaust gas is then

E = .5*Mp*Ve^2

Thus

Ve = Sqrt[2E/Mp]

Substituting into the top equation, and letting
Mp = Fract * MDead

you discover the maximum cutoff velocity at
constant energy comes at Fract = 3.92...,
corresponding to an optimal mass ratio of
4.92...

(It turns out that when thrust to weight ratios are
very high, and altitudes at cutoff are short, then
cutoff velocities on earth are very near outer
space cutoff velocities. All this holds true for
water rockets.)

[Jerry Irvine]

In article <f%uHc.32796$6e7....@nwrddc03.gnilink.net>,
"Larry Curcio" <lcu...@verizon.net> wrote:

> > > The maximum theoretical amount of kinetic energy transfer from the
> > > propellant to the spent (single stage)rocket is 64.761% at a mass
> > > ratio of 4.921553.
> > >
>
>
> Warning: Incoming from left field!
>
> Interesting. I just got this very result in water rocket calculations.
> The problem was to adjust the density of the working fluid
> (traditionally water)

Beer works better.

How do you model the liberation of the dissolved gases?

Jerry

> to maximize the cutoff velocity. In
> outer space, this would be
>
> Vcutoff = Ve*ln[(Mprop + Mdead)/Mdead]
>
> where
> Ve is effective exhaust velocity,
> Mp is propellant (working fluid) mass
> aMdead is dead mass (structural mass)
>
> The kinetic energy of the exhaust gas is then
>
> E = .5*Mp*Ve^2
>
> Thus
>
> Ve = Sqrt[2E/Mp]
>
> Substituting into the top equation, and letting
> Mp = Fract * MDead
>
> you discover the maximum cutoff velocity at
> constant energy comes at Fract = 3.92...,
> corresponding to an optimal mass ratio of
> 4.92...
>
> (It turns out that when thrust to weight ratios are
> very high, and altitudes at cutoff are short, then
> cutoff velocities on earth are very near outer
> space cutoff velocities. All this holds true for
> water rockets.)
>
>

--

[shockwaveriderz]

What would happen if you used a piston launcher with a water rocket? Would
it do the same as it does for regular rockets?
shockie B)

"Alan Jones" <ala...@nospam.mchsi.com> wrote in message
news:vn4se0l6i9o0dlfdu...@4ax.com...

[Alan Jones]

Then you will get much better performance using the CO2 to drive a
propeller. But you did spawn an interesting discussion.

Alan

[PeteAlway]

>You've got me thinking about something that's been kicking around
>in the back of my head since I BAR'd.

>If Work=INT(F*ds), then what am I missing in this scenario?

>A constant force is applied to an object at rest for a fixed duration
>accelerating that object, and moving it some distance. W=F*d1
>
>That same force is applied for the same duration, this time to
>an object already moving with some (non-zero) initial velocity.
>Again some acceleration is imparted, but more importantly,
a different distance is traversed since the object was already
>moving. W=F*d2
>
>IOW, in each case, the same impulse was applied, but a different
>amount of work was done (or was it?). I can see where the same
>change in momentum occurred, but using W=F*d seems to indicate
>more work was done to the moving object than to the object initially
>at rest.
>
>What's wrong here? What am I missing? Where did I blow it? Why
>did I skip so many sessions of physics in college?

Kinetic energy is imparted to the exhaust as well as to the rocket. For the
rocket starting at zero velocity, and, say, moving very slowly compared to the
exhaust for the duration of the interval in question, the exhaust is moving
very fast, and the rocket is moving very slowly, os the exhaust gets almost all
of the kinetic energy.

When the rocket is moving at the exhaust velocity (very near the exhaust
velocity for the time interval in question), the exhaust ends up effectively at
a standstill, and so the rocket itself gets almost all the kinetic energy.

When the rocket is going much faster than the exhaust velocity, the exhaust is
actually travelling more slowly than it was when it was in the rocket, and so
the rocket actually "steals" the kinetic energy from the exhaust.

That's kind of an ad-hoc explatation, as work and energy just aren't the best
tools to understand reaction propulsion. The work-energy equations work out
and energy is conserved, but I think it's better for understanding for your
head to be in the land of impulse and momentum.

[PeteAlway]

"n...@null.com" n...@null.com wrote:

>On Thu, 08 Jul 2004 21:06:33 +0000, PeteAlway wrote:

>> "n...@null.com" n...@null.com wrote:

>Actually, what I'm considering is a water-rocket powered glider that would
>operate at a roughly constant airspeed. I was wondering whether the energy
>in my compressed air would be put to better use powering a propellor or
>pushing water out of the back. The thrust doesn't have to last very long -
>just enough to get some altitude.

>Perhaps I should have said this before - I figured that practicalities
>don't always help in understanding the theory, but perhaps I was >wrong.

I am blitheringly ignorant in the ways of propellers. But I can't help but
notice that there are more prop-driven planes than rocket powered planes in the
world. I think there must be a reason.

>> Mass x exhaust velocity gives kg m/s, units of momentum or impulse, not
thrust.
>
>I gave Mass / second - which I think converts my units into thrust?

Kg/s is a mass flow rate, not a thrust. Thrust is a force, in units of kg
m/s^2 (mass times acceleration). Mass flow rate times velocity gives units of
thrust. (I'm 90% sure that mass flow rate times velocity is thrust, and 100%
sure it is proportional to thrust. Pardon my rust on this issue)

>Perhaps another application where this could make a difference would be
>something like an ion-drive, where the power is fixed (by the size of
>the solar-cells / reactor...) but the propellant mass can be much lower
>compared to the mass of the spacecraft.

Ion drive gives a much higher exhaust velocity, giving a much higher specific
impulse.

>So as the specific-impulse of a rocket engine increases, its energy
>efficiency really does decrease? I never realised this before.

I don't think I said that anywhere. You can increase specific impulse by using
a propellant with more energy to begin with (liquid oxygen vs gunpowder, or in
this case, high-pressure air and water vs. low-pressure air and water) or you
can increase specific impulse by increasing the energy efficiency of the engine
(de laval nozzle vs a simple hole).

A liquid hydrogen rocket burning with a low energy efficiency (Kinetic energy
of exhaust in static test / chemical energy input) could outperform a gunpowder
rocket with a higher energy efficiency.

[Larry Curcio]

Here are results for the original scenarios
at different dead mass values

Dead Mass (g) 50
100 150 200 250
Cutoff Vel (m/s) (1Kg, 20m/s) 60.89045 47.95791 40.73764
35.83519 32.18876
Cutoff Vel (m/s) (.5KG, 40 m/s) 95.91581 71.67038 58.65348
50.11052 43.94449

300
29.32674
39.23317

Optimized results using
Cd = 0.65 and pressure = 160 psi (Achievable)
Water Temp = 30 deg C

1 Liter:
Water Mass = 403 grams
Dead Mass = 145 grams

2 Liter:
Water Mass = 800 grams
Dead Mass = 289 grams

Altitude for both is about 185 feet
- not including the air burst after the water is run out
About 200 feet in all from direct observation

More is possible if a long copper tube
is used in the launcher. Has the same
effect as a piston launcher.

Regards

[Alan Jones]

On Fri, 9 Jul 2004 14:40:30 -0400, "shockwaveriderz"
<shockwa...@hotmail.com> wrote:

>What would happen if you used a piston launcher with a water rocket? Would
>it do the same as it does for regular rockets?
>shockie B)

I don't know what the rules are for water rocket contests, but you
would want a separate charge of compressed air to activate the piston.

[Alan Jones]

Cool! Now how much horsepower does a water rocket generate? ;)

[RayDunakin]

<< Cool! Now how much horsepower does a water rocket generate? ;) >>

That depends on whether it's analog or digital. ;)

[Larry Curcio]

I have this problem. My ISP isn't displaying all the
messages that I see in Google, and I cannot apply
to someone who lists hs address as n...@null.com.
Please bear with me. I'm trying to reply to the
last message by the above poster.

The reason the two scenarios are not equivalent is
that, during operation, the unused propellant (in this
case water) must be accelerated. Thus, when the
water is half expended, the other half is moving at the
same speed as the rest of the rocket. That's not a big
deal if the propellant mass is small compared to the
rocket mass, but in this case, even the unused water
at the halfway mark may be several times the dead
mass. That's a lot of kinetic energy put into the
propellant that you don't get back when you expend it.
There is also the potential energy of lifting that
propellant.

(Actually, you can get some of the kinetic energy back
by retaining some of the the mass and allowing its
associated energy to work against drag. If you were
very clever, you could expend it after it did this.)

This is not just a problem with water rockets. It's the
reason that low Ve solid propellants are used primarily
in boosters. Certainly, if you have an upper stage with
a given impulse, you want a lightwaeight propellant
with a high Ve, rather than a heavy propellant
with a low Ve. After all, you have to lift that propellant
before you burn it. Well... it dosn't really matter if the
propellant is in a segregated stage.

Hope that helps.

-Larry Curcio

[Kenneth C. McGoffin]

Alan Jones <ala...@nospam.mchsi.com> wrote in message news:<thmue09dp35q2ssul...@4ax.com>...

> On 09 Jul 2004 22:16:35 GMT, pete...@aol.com (PeteAlway) wrote:

> Cool! Now how much horsepower does a water rocket generate? ;)

The same formula that applies to all rockets:

HP = (Ve * F) / 375 Ve in mph, F in pounds

Working through the equations gives me

HP = 2.49*At*P^1.5

where Ax is the throat area in square inches and P is the
(relative)pressure in atmospheres. So at 100 psi it's about 44 HP per
square inch of throat area.

You had to ask. ;-)
+McG+

[Jerry Irvine]

In article <f130bbe4.04071...@posting.google.com>,

kmc...@yahoo.com (Kenneth C. McGoffin) wrote:

I am not sure which is more amusing to me, water rocket boosted darts,
or water rocket horsepower calculatons.

:)

Jerry

[Alan Jones]

On Mon, 19 Jul 2004 06:48:29 -0700, Jerry Irvine <01ro...@gte.net>
wrote:

>In article <f130bbe4.04071...@posting.google.com>,
> kmc...@yahoo.com (Kenneth C. McGoffin) wrote:
>
>> Alan Jones <ala...@nospam.mchsi.com> wrote in message
>> news:<thmue09dp35q2ssul...@4ax.com>...
>> > On 09 Jul 2004 22:16:35 GMT, pete...@aol.com (PeteAlway) wrote:
>>
>> > Cool! Now how much horsepower does a water rocket generate? ;)
>>
>> The same formula that applies to all rockets:
>>
>> HP = (Ve * F) / 375 Ve in mph, F in pounds
>>
>> Working through the equations gives me
>>
>> HP = 2.49*At*P^1.5
>>
>> where Ax is the throat area in square inches and P is the
>> (relative)pressure in atmospheres. So at 100 psi it's about 44 HP per
>> square inch of throat area.
>>
>> You had to ask. ;-)
>> +McG+
>
>I am not sure which is more amusing to me, water rocket boosted darts,
>or water rocket horsepower calculatons.
>
>:)
>
>Jerry

Water rockets have very low Isp, so staging and boosted darts are
relatively more important. Rocket horsepower is just silly, but both
can be amusing. But ya know, the development of the V2 was eased
because lightweight turbo pumps for fire engines were available more
or less of the shelf, and turbo pumps require horsepower to produce
forceful streams of water. Now if you use your PAD to drive a high
pressure turbo pump in your water rocket...

Alan