Proposals for air breathing hypersonic craft. III
Robert Clark
propellant because of its low molecular weight, half that of hydrogen
in its more common H2 form. The problem is storing it because it has
the tendency to recombine to the H2 form:
Use of Atomic Fuels for Rocket-Powered Launch Vehicles Analyzed.
http://www.grc.nasa.gov/WWW/RT1998/5000/5830palaszewski.html
High temperatures though are known to create atomic hydrogen, and
scramjets by slowing down the hypersonic air stream create high
temperatures in the thousands of degrees Kelvin.
Page 3 of this report gives a graph showing temperatures reached by
the decelerated air stream at hypersonic speeds:
Advanced propulsion systems.
http://ronney.usc.edu/AME514F06/Lecture13/AME514-F04-lecture13.pdf
The temperature can reach above 5000 K at Mach 15 and nearly 3000 K at
Mach 10.
This page gives a graph showing the rate of dissociation of hydrogen
according to temperature:
The Moller's Atomic Hydrogen Generator tests by JL Naudin.
http://jlnlabs.imars.com/mahg/tests/index.htm
The particular graph is from the work of Nobelist Irving Langmuir. It
shows 9% dissociation at 3000 K, 98% dissociation at 6000 K and 99.99%
at 8000 K. A key fact though is the percent dissociation is dependent
on the total pressure of the hydrogen: at lower pressure the
dissociation is easier and the percentage of atomic hydrogen is higher.
Then to get high percentage of atomic hydrogen at the lower Mach
numbers and lower temperatures we could use hydrogen at lower total
pressure.
To heat the hydrogen however you don't want to mix it with the air as
when you are cumbusting the fuel. This would create a mixture of high
molecular weight, which reduces exhaust velocity. What would be
necessary would be to use efficient heat exchangers to quickly heat the
hydrogen up to the temperature of the decelerated air:
Shooting at the moon.
06 August 1994
NewScientist.com news service
CHARLENE CRABB
"INSTANT HEAT...Hunter does not plan to use electricity to heat the 40
tonnes of hydrogen that will be required for each firing of the JVL.
Instead, the gun will have sophisticated heat exchangers developed at
Brookhaven National Laboratory in New York state that will take just 20
milliseconds to heat this huge quantity of gas to 1200 degree C at a
pressure of only 250 atmospheres. The key to the nearly instantaneous
heating is millimetre-size graphite beads that can absorb the heat
without breaking down. But at these temperatures graphite can react
with hydrogen, so the beads must be covered with thin layers of the
inert compound zirconium oxide. The tiny spheres will be initially
heated by burning methane or propane. The particles will then heat
hydrogen gas as it flows through the bed."
http://www.newscientist.com/article/mg14319373.900.html
Using heat exchangers to heat the hydrogen would also eliminate the
problem of sustaining combustion at the high temperatures produced in
the scramjet engines.
Bob Clark
Robert Clark
> On 23 Jun 2006 09:52:53 -0700, in sci.engr.mech "Robert Clark"
> <rgrego...@yahoo.com> wrote:
>
>
> > High temperatures though are known to create atomic hydrogen, and
> >scramjets by slowing down the hypersonic air stream create high
> >temperatures in the thousands of degrees Kelvin.
> > Page 3 of this report gives a graph showing temperatures reached by
> >the decelerated air stream at hypersonic speeds:
> >
> >Advanced propulsion systems.
> >http://ronney.usc.edu/AME514F06/Lecture13/AME514-F04-lecture13.pdf
> >
> > The temperature can reach above 5000 K at Mach 15 and nearly 3000 K at
> >Mach 10.
>
> when one is trying to talk about simulation flight conditions and a few
> other things related to heat transfer.
>
> > This page gives a graph showing the rate of dissociation of hydrogen
> >according to temperature:
> >
> >The Moller's Atomic Hydrogen Generator tests by JL Naudin.
> >http://jlnlabs.imars.com/mahg/tests/index.htm
>
> and pressure of the constituent, not stagnation temperature. You've made up
> a problem which doesn't exist. Time to go back and review basic
> thermodynamics and gas dynamics.
> --
> Ed Ruf (Use...@EdwardG.Ruf.com)
The purpose of citing the report showing stagnation temperatures was
to give an idea of the high temperatures reached when air is slowed
from hypersonic speed to zero.
The same conclusion can be reached from reentry data:
Atmospheric reentry
"Shock layer gas physics
An approximate rule-of-thumb used by heat shield designers for
estimating peak shock layer temperature is to assume the air
temperature in kelvins to be equal to the entry speed in meters per
second. For example, a spacecraft entering the atmosphere at 7.8 km/s
would experience a peak shock layer temperature of 7800 K. This method
of estimation is a mathematical accident and a consequence of peak heat
flux for terrestrial entry typically occurring around 60 km altitude."
http://en.wikipedia.org/wiki/Reentry
And this page shows an ICBM reentry cone can experience temperatures
of 6000 K at Mach 20 to 25:
Aerothermodynamic Heating
Skin Temperature Data
http://www.aerodyn.org/Atm-flight/table-skin-t.html
The idea to be used here is not to alleviate the high stagnation
temperatures as with scramjets and reentry vehicles but to maintain
them by having a blunt body or sharp pointed body always facing into
the hypersonic air stream so that high temperatures are always
maintained in front of them.
Indeed the terminology "air breather" would be somewhat of a misnomer
here since the air would not be used for combustion as with scramjets
but only to heat the hydrogen to high temperatures and which alone
would be the propellant and exhaust.
A couple of key questions are could the high temperature air heat
enough hydrogen to produce sufficient thrust to counteract the drag
produced by slowing the air down to zero?
Could the heat exchanger be made light enough and operate fast enough
for the amount of hydrogen required to be heated?
Let's do a rough calculation toward the first question. Let's say the
air flow in was at the rate of m given in kg/s and say at a speed of
7000 m/s. Then the force backward would be 7000m.
This page gives the exhaust velocity for a rocket engine according to
temperature:
Rocket engine nozzles.
http://en.wikipedia.org/wiki/Rocket_engine_nozzles
At very high altitudes the exit pressure in the formula is nearly zero
and the velocity equation simplifies to:
V = sqrt[2kRT/((k-1)M)]
If we assume the hydrogen can be heated to 7000 K, then most or all
will dissociate to atomic hydrogen. The molecular weight will be 1 and
monoatomic gases have a ratio of specific heats k of 1.6. So the
velocity will be v = sqrt[2*1.6*8314.5*7000/.6] = 17,618 m/s. This is
about 2.5 times the velocity of the incoming air. So to have net thrust
the mass flow of the hydrogen would have to be at least 1/2.5 times
that of the air flow.
Bob Clark
Robert Clark
use the high temperatures produced to heat hydrogen fuel onboard via
efficient heat exchangers. There would be no combustion involved.
However, it would seem that the best you could do with this would be to
break even: you're assuming the kinetic energy of the air with respect
to the craft can all be converted to heat, and all this heat can be
transferred to the hydrogen then all this heat can be converted back
into kinetic flow for propulsion. The last step of course is where big
losses would occur.
So additionally to this, why not combust the now still air with
further hydrogen? I thought of using the very hot hydrogen already
heated by the heat exchanger but you want to keep the high exhaust
velocity that this achieves, and mixing this with the air reduces this.
So you would use separate hydrogen for this purpose. Now that the air
is still with respect to the craft there is no prolem of the fuel and
air being at different velocities prior to combustion. And the problem
of combusting the fuel with air at temperatures higher than the
dissociation temperature of the combustion products will be eliminated
since the air will now be cooled by the heat exchangers.
Now that the air is still with respect to the craft we can use a
variety of turbojet or rocket engines. The compression ratio will also
be quite high since the air will be stopped from hypersonic speed.
Bob Clark
Paul F. Dietz
> The idea was to bring the hypersonic airstream to a complete stop and
> use the high temperatures produced to heat hydrogen fuel onboard via
> efficient heat exchangers. There would be no combustion involved.
> However, it would seem that the best you could do with this would be to
> break even: you're assuming the kinetic energy of the air with respect
> to the craft can all be converted to heat, and all this heat can be
> transferred to the hydrogen then all this heat can be converted back
> into kinetic flow for propulsion. The last step of course is where big
> losses would occur.
Actually, if there are no losses, this scheme would result in net
thrust, even if no combustion were to occur. You want to expel
the air and the reaction mass at the same speed. What is happening
is that (in the reference frame of the atmosphere) the kinetic
energy of the expelled propellant is being concentrated into
the vehicle.
I've proposed a similar scheme for interstellar vehicles, where
some sort of electromagnetic coupling to the (extremely thin)
interstellar plasma would be used to accelerate onboard reaction
mass backwards. Done properly, this can accelerate a vehicle
that is initially in motion w.r.t. the interstellar medium, and
can do so without violating any conservation laws.
Paul
Robert Clark
Hmm, I keep getting a result that says the kinetic energy out would
be greater than the heat energy input.
Let's say you heat the hydrogen in gas form. You could store the
hydrogen initially as liquid and then use the gaseous hydrogen formed
from the liquid hydrogen that vaporizes when used to cool the external
surfaces and engine nozzles and combustion chambers.
The reason why I'm suggesting using the hydrogen in gas form is that
you want to apply the full heating effect from stopping the hypersonic
air to raise the hydrogen to high temperature. However, if you use
liquid hydrogen a large portion of the heat energy just goes to
converting the liquid to gas, the "heat of vaporization".
Say you have 1 kg of hydrogen gas at 1 bar and 300 K. This page gives
properties of hydrogen at various temperatures and pressures:
Hydrogen Properties Package.
http://www.inspi.ufl.edu/data/h_prop_package.html
These are the results at 1 bar and 300 K:
Results
Pressure = 1.000e+00 bar
Temperature = 3.000e+02 K
Enthalpy = 4.199e+03 kJ/kg
Entropy = 6.483e+01 kJ/kg.K
Vel.of sound = 1.310e+03 m/s
Density = 8.080e-02 kg/m**3
Them. cond. = 1.939e-01 W/m.K
Viscosity = 8.948e-06 N.s/m**2
Spec. heat = 1.485e+01 kJ/Kg.K
Gamma = 1.385e+00
The specific heat given is that at constant pressure, Cp. For my
scenario, I'll heat the hydrogen in a constant volume vessel, so I need
the specific heat at constant volume, Cv. Gamma is the ratio of these
so Cv = Cp/gamma = 14,850/1.385 = 10,700 joules per kilo per degree
Kelvin temperature increase.
Let's say you were able to get 10,000,000 joules of heat energy from
stopping the air flow. Then this would raise the temperature of one
kilo of hydrogen at 300 K by 10,000,000/10,700 = 934.5 K, so to 1234.5
K.
This page gives the exhaust velocity for a rocket engine according
to temperature in the combustion chamber:
Rocket engine nozzles.
http://en.wikipedia.org/wiki/Rocket_engine_nozzles
At very high altitudes the exit pressure in the formula is nearly zero
and the velocity equation simplifies to:
V = sqrt[2kRT/((k-1)M)] , k the ratio of specific heats, which is the
same as the gamma from the Hydrogen Properties Package page, R the
ideal gas constant, and M the molecular weight.
We need to use the Hydrogen Properties Package page to find k, or
gamma, at the new temperature 1234.5 K. This requires knowing the
pressure also. The pressure can be found from the ideal gas law: P =
density*R*T/M . The density stays the same at a constant volume so is
.0808 kg/m^3, and P = .0808*8314*1234.5/2 = 414,651 pascals.
Now this can be input to the Hydrogen Properties Package as 4.15 bar
and 1234.5 K. The result is that gamma, the k in the velocity equation,
is 1.364, about the same as in the 1 bar, 300 K case. Then V =
sqrt[2*1.384*8314*1234.5/(.384*2)] = 6,082 m/s.
The kinetic energy of 1 kg of hydrogen at 6,082 m/s is .5*1*6082^2 =
18,500,000 joules, which is more than the 10,000,000 joules of heat
energy input.
The kinetic energy of .41 kg of air at 7000 m/s is about 10,000,000
joules. We are assuming most of this kinetic energy could be converted
to heat energy [though the above calculations make me wonder about the
validity of this too.] The momentum change of stopping this air would
be .41*7000 = 2870 N-s. However, the impulse from the 1 kg of hydrogen
at 6,082 m/s would 6,082 N-s, resulting in a net forward velocity.
Bob Clark
Robert Clark
K so this would need a method to cool the combustion chamber.
In the method or regenerative cooling, liquid hydrogen is used to cool
the nozzles of the rocket engines. In the case of LH/LOX engines the
temperature of the exhaust can be 3000 K. This is higher than the
melting point of steel.
Is the reason why the nozzles do not melt because to melt a material
has to absorb an amount of heat greater than the heat of fusion? And
the regenerative cooling method removes much of the heat delivered to
the nozzle so that the amount of heat required for melting is not
acquired?
Would this work for nozzles or chambers at 7000 K if sufficient liquid
hydrogen was circulated outside or inside the subsurface of the
chambers or nozzles?
Paul F. Dietz
> Hmm, I keep getting a result that says the kinetic energy out would
> be greater than the heat energy input.
Hmm? You must have done something wrong.
Paul
Robert Clark
Even if you used hydrogen gas stored at room temperature and 1 bar and
added no additional heat to it you would get high exhaust velocity by
using a convergent-divergent nozzle when exhausting into a vacuum or at
very high altitude.
From "Rocket engine nozzles",
http://en.wikipedia.org/wiki/Rocket_engine_nozzles, the exhaust
velocity is V = sqrt[2kRT/((k-1)M)] , k the ratio of specific heats, R
the ideal gas constant, M the molecular weight, and T the temperature.
Then V = sqrt[2*1.385*8314*300/(.385*2)] = 2995.4 m/s.
Bob Clark
Robert Clark
> ...
> At very high altitudes the exit pressure in the formula is nearly zero
> and the velocity equation simplifies to:
>
> V = sqrt[2kRT/((k-1)M)] , k the ratio of specific heats, which is the
> same as the gamma from the Hydrogen Properties Package page, R the
> ideal gas constant, and M the molecular weight.
> We need to use the Hydrogen Properties Package page to find k, or
> gamma, at the new temperature 1234.5 K. This requires knowing the
> pressure also. The pressure can be found from the ideal gas law: P =
> density*R*T/M . The density stays the same at a constant volume so is
> .0808 kg/m^3, and P = .0808*8314*1234.5/2 = 414,651 pascals.
> Now this can be input to the Hydrogen Properties Package as 4.15 bar
> and 1234.5 K. The result is that gamma, the k in the velocity equation,
> is 1.364, about the same as in the 1 bar, 300 K case. Then V =
> sqrt[2*1.384*8314*1234.5/(.384*2)] = 6,082 m/s.
> The kinetic energy of 1 kg of hydrogen at 6,082 m/s is .5*1*6082^2 =
> 18,500,000 joules, which is more than the 10,000,000 joules of heat
> energy input.
> The kinetic energy of .41 kg of air at 7000 m/s is about 10,000,000
> joules. We are assuming most of this kinetic energy could be converted
> to heat energy [though the above calculations make me wonder about the
> validity of this too.] The momentum change of stopping this air would
> be .41*7000 = 2870 N-s. However, the impulse from the 1 kg of hydrogen
> at 6,082 m/s would 6,082 N-s, resulting in a net forward velocity.
>
I plugged in the wrong number for gamma, which doesn't change the
basic conclusion.
At 4.15 bar and temperature 124.5K, gamma is 1.364. This is the number
I should have plugged for the value of k in the velocity equation :
V = sqrt[2*1.364*8314*1234.5/(.364*2)] = 6,201.6 m/s.
This still gives more kinetic energy out than heat energy input and a
net forward velocity for the rocket.
Bob Clark
Robert Clark
I can simplify the question by assuming an ideal gas throughout:
==========================================================
Take 1 kg of hydrogen gas at 1 bar and 300 K. The specific heat of
hydrogen at constant pressure Cp is commonly given as 14,000 J/kg-°K.
For my scenario, I'll heat the hydrogen in a constant volume vessel, so
I need the specific heat at constant volume, Cv. The ratio of Cp to Cv
is k = 1.4 for an ideal diatomic gas, so Cv = 10,000. Let's say you
were able to get 10,000,000 joules of heat energy from stopping the air
flow. Then this would raise the temperature of one kilo of hydrogen at
300 K by 10,000,000/10,000 = 1000 K, so to 1300 K.
I'll use now the equation for the exhaust velocity for a rocket
engine:
Rocket engine nozzles.
http://en.wikipedia.org/wiki/Rocket_engine_nozzles
At very high altitudes the exit pressure in the formula is nearly zero
and the velocity equation simplifies to:
V = sqrt[2kRT/((k-1)M)], k the ratio of specific heats, R the ideal gas
constant, and M the molecular weight, T the temperature.
So V = sqrt[2*1.4*8314*1300/(.4*2)] = 6,150 m/s.
The kinetic energy of 1 kg of hydrogen at 6,150 m/s is .5*1*6,150^2 =
18,900,000 joules, which is more than the 10,000,000 joules of heat
energy input.
The kinetic energy of .41 kg of air at 7000 m/s amounts to 10,000,000
joules. I am assuming most of this kinetic energy could be converted to
heat energy [though the above calculations make me wonder about the
validity of this too.] The momentum change of stopping this air would
be .41*7000 = 2870 N-s. However, the impulse from the 1 kg of hydrogen
at 6,150 m/s would be 6,150 N-s, resulting in a net forward velocity.
==========================================================
From discussions on some physics online forums, I gather what goes
wrong is that the hydrogen stored at 300K already has some energy and
the equation for the exhaust velocity from a rocket nozzle assumes Cp
not Cv in the calculation. When you substract off the kinetic energy
you already can get from the nozzle for the gas at 300K and also divide
by Cp instead of Cv in finding the temperature increase for the
10,000,000 J energy added, you wind up with the same kinetic energy out
as the heat energy input.
However, it *seems* like you should be able to heat the gas at
constant volume and get a higher temperature. I mean the equation for
the velocity should only depend on the temperature on entering the
nozzle. So it should still work with the higher temperature you got by
heating at constant volume just by introducing this gas at the higher
temperature into the nozzle.
Bob Clark