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From: Henry Spencer <>
Subject: Re: Specific impulse
Date: Thu, 22 Aug 1996 15:26:00 GMT

In article <> "david (d.) plumpton" <> writes:
>Can anybody explain to me why specific impulse is measured in seconds?
>Since it is a measure of the efficiency of a fuel, I thought it would
>be something like: meters per second per kilogram.

Specific impulse measures the impulse -- that is, force*time -- obtained
from a specific amount of fuel.  That's why it's called "specific impulse".

Now, N*s/kg, that is (kg*m/s^2)*s/kg, ought to give you m/s, i.e., a
velocity.  And indeed, you will occasionally see m/s (or N*s/kg, the same
thing) used as the unit of specific impulse, especially in European
papers.  If you look at the rocket physics and ignore losses etc., such a
specific impulse is just the velocity of the exhaust jet.  In practice it
is called the "effective exhaust velocity" because it's the exhaust
velocity an ideal rocket would need to have to match the performance of
the real rocket -- it measures overall performance, including the effect
of some losses which make the real rocket not quite as good as its actual
jet velocity would suggest.

So whence comes this business of measuring specific impulse in seconds?
It's because of the long-standing confusion about whether the pound is a
unit of mass or weight.  If you measure both thrust and fuel quantity in
pounds, the pounds cancel top and bottom to give you seconds.  To convert
into a physically meaningful quantity -- the effective exhaust velocity --
you have to multiply by g (Earth's gravitational acceleration) to turn the
fuel units back into mass. 

It's probably too late to fix this, and it is not disastrously harmful.
Specific impulse is useful as a figure of merit regardless of funny units
(there are other engineering figures of merit which use much stranger
units), and converting to effective exhaust velocity is not difficult.
Moreover, the appealing physical significance of effective exhaust
velocity is lost when you start talking about jet engines, because the
definition of effective exhaust velocity implicitly assumes a rocket,
in which the exhaust mass is the fuel only.
 ...the truly fundamental discoveries seldom       |       Henry Spencer
occur where we have decided to look.  --B. Forman  |

From: Henry Spencer <>
Subject: engine thermodynamics (was Re: Aerodesic)
Date: Thu, 20 Jun 1996 05:40:15 GMT

In article <> "Gordon D. Pusch" <> writes:
>...claiming that the total impulse
>I can get from a given quantity of, say, H2 and O2 is bounded from
>above by that obtainable from an ideal H2/O2 rocket-engine ???

Subject to some quibbling about that word "ideal", that's the idea.

>If I were to instead take that same amount of H2 and O2 and use a
>fraction of it to run a fuel cell at stoichiometric ratio (venting 
>the resulting water overboard, say) then use the generated electricity 
>and the excess H2 to run an arcjet, the total impulse will be less
>than the H2/O2 rocket, even though the molecular weight of the 
>exhaust is lower, and the specific impulse of the arcjet is higher ???

The key are those little words "venting the resulting water overboard".
Your propellant consumption is not that of your arcjet, it's that of
your arcjet *plus* *your* *power* *supply*.

Consider the general case:  we expel H2/O2 reaction products at some
low velocity, and tap off some of their energy to power a high-velocity
engine with a much smaller mass flow.  What is the optimal split of the 
energy between the two?

Since kinetic energy is 0.5*m*v^2 and momentum is m*v, the energy needed
to produce a given amount of impulse (neglecting losses) is 0.5*i*v.
Rearranging this, we get i = 2E/v.  That is, for constant energy, you get
maximum impulse by using the *lowest* available exhaust velocity.  (High
exhaust velocities use less mass at the expense of using more energy.)

For our general case, we have i = 2(1-k)E/v + 2kE/rv, where k is the
fraction of the energy used for the high-velocity engine and r is the
ratio of the exhaust velocities (high/low).  Differentiating, we get
di/dk = -2E/v + 2E/rv = 2E(-r + 1)/rv.  Since r>1, this is always
negative, and for k between 0 and 1, you get maximum i for k=0.

In other words, the optimal way to use the H2/O2 energy is to put it all
into the low-velocity H2/O2 reaction-product stream, and none into the 
high-velocity engine.  If you've got to carry the H2/O2 along anyway,
it is the best reaction mass available.  Electric rockets make sense
only if the fuel consumption of the power source is *less* than that of
the rocket engine.
If we feared danger, mankind would never           |       Henry Spencer
go to space.                  --Ellison S. Onizuka |

From: Henry Spencer <>
Subject: Re: Mo) better rockets
Date: Wed, 13 Mar 1996 20:26:49 GMT

In article <4i37ar$> (Trakar) writes:
>I may be (and quite probably am) wrong, but it seems to me that mercury
>vapor at 5000 C would produce more thrust than hydrogen at 5000 C.

More thrust, perhaps (depending on assumptions), but only at the cost of
consuming much more propellant.  For most applications, what you want to
maximize is not the thrust, but the specific impulse -- how much impulse
(thrust*time) you get out of a given mass of propellant.  If you measure
mass in kilograms, as sane people should :-), specific impulse turns out
to be equal to the effective exhaust velocity.  (You will also see it
quoted in "seconds", which is the result of confusing mass and weight. 
Isp(m/s) = Isp(s) * g, where g is 9.81m/s^2, the acceleration of gravity
at Earth's surface.)

To maximize the exhaust velocity obtained from a given temperature, other
things being equal (note that sometimes they aren't!), you want the lowest
possible molecular weight.  Lighter molecules move faster.  Any rocket
which heats its exhaust to a constant temperature, independent of the
exhaust's composition, will perform much better on hydrogen than on
anything else. 
Space will not be opened by always                 |       Henry Spencer
leaving it to another generation.   --Bill Gaubatz |

From: Henry Spencer <>
Subject: Re: Spacecraft design
Date: Sun, 7 Apr 1996 00:50:25 GMT

In article <4jq1he$> (Filip De Vos) writes:
>: Actually, advanced engine concepts -- on paper -- can hit 480s or so with
>: LOX/LH2.  Those engines are not available off the shelf, mind you.
>...Where can I find a Web-resource for theoretical Isp of 
>propellant combinations? I once saw one, but I remember slightly 
>lower Isp for H2-O2.

The theoretical Isp depends heavily on details like expansion ratio and
chamber pressure.  You have to be very cautious in applying theoretical
numbers to real engines -- often the underlying assumptions differ.  Even
just using the theoretical numbers to compare different propellant
combinations is a bit perilous, because sufficiently differing assumptions
can greatly change the comparative picture.  (For example, as I recall,
modern high chamber pressures considerably diminish the advantage of
fluorine over oxygen.)

>...mentions a Rocketdyne engine with a thrust of 1200 kg (I know, the wrong 
>unit)  and an Isp of 476 (supposedly seconds, no unit given...

Sounds like an orbital maneuvering engine with a very high expansion ratio.
There have been a number of those proposed over the years, and a few of
them have even been tested.

>: The RL10 has been fired successfully with liquid fluorine instead of LOX...
>Interesting! What was the Isp achieved and what was the trust-level and 
>chamber-pressure compared with the 'normal' version? 

Unfortunately, I have no details; it was mentioned briefly in a paper on
the engine's history (no reference, either).
Americans proved to be more bureaucratic           |       Henry Spencer
than I ever thought.  --Valery Ryumin, RKK Energia |

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