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Successful SpaceX launch



 
 
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  #1  
Old April 8th 16, 11:24 PM posted to sci.space.policy
Jeff Findley[_6_]
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Posts: 2,307
Default Successful SpaceX launch

I watched the Falcon 9/Dragon launch live and it looked successful.
This Dragon is carrying BEAM to ISS, which is a first.

Also, and likely the most historic, the first stage landed successfully
on the barge, which is a first!

Congrats to SpaceX! Here is hoping that they inspect and re-launch that
recovered first stage, which would be another first (re-launch of a
liquid fueled first stage recovered from an orbital launch).


Oh yea, cite:

http://www.theverge.com/2016/4/8/113...uccess-falcon-
9-rocket-barge-at-sea

Why is this important? Because (from the above article):

The Falcon 9 costs $60 million to make and only $200,000 to fuel.

I'm tired of people saying "chemical propulsion is too expensive" when
the fuel is so damn cheap! Fuel costs are *not* the problem.
Skylon/Sabre is a solution to a problem which *does not exist*!

Jeff
--
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These posts do not reflect the opinions of my family, friends,
employer, or any organization that I am a member of.
  #2  
Old April 8th 16, 11:32 PM posted to sci.space.policy
Rick Jones[_6_]
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Posts: 106
Default Successful SpaceX launch

Jeff Findley wrote:
Also, and likely the most historic, the first stage landed successfully
on the barge, which is a first!


I managed to get online to see the youtube-carried technical feed.
How long does the stage just sit there bobbing in the ocean with the
barge waiting for a rogueish wave to come along and upset the apple
cart?

rick
--
web2.0 n, the dot.com reunion tour...
these opinions are mine, all mine; HPE might not want them anyway...
feel free to post, OR email to rick.jones2 in hpe.com but NOT BOTH...
  #4  
Old April 9th 16, 02:40 AM posted to sci.space.policy
Sylvia Else
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Posts: 1,063
Default Successful SpaceX launch

On 9/04/2016 8:24 AM, Jeff Findley wrote:

I'm tired of people saying "chemical propulsion is too expensive" when
the fuel is so damn cheap! Fuel costs are *not* the problem.
Skylon/Sabre is a solution to a problem which *does not exist*!


Skylon/Sabre is not about reducing the amount of fuel consumed.

Sylvia.

  #7  
Old April 10th 16, 10:27 AM posted to sci.space.policy
William Mook[_2_]
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Posts: 3,840
Default Successful SpaceX launch

Alright, I've stated my prejudices against the Skylon's approach to build a LACE propulsion system. SO, let's look at this more closely and see if my prejudice is justified.

A Liquid Air Cycle Engine (LACE) is a type of spacecraft propulsion engine that attempts to increase its efficiency by gathering part of its oxidizer from the atmosphere. A liquid air cycle engine uses liquid hydrogen (LH2) fuel to liquefy the air.

At its atmospheric boiling point, the specific (constant-pressure) heat
capacity of

liquid hydrogen is about 14.4 J/(g K).
liquid oxygen is about 1 J/(g K).
liquid nitrogen is about 1 J/(g K).

The heat of fusion of LH2: 58.5 J/(g K)

The heat of vaporization of LH2 is 452 J/g
The heat of vaporization of LOX is 216 J/g
The heat of vaporization of LN2 is 199 J/g

Boiling point: LH2: 20.28 K
LOX: 90.19 K
LN2: 77.00 K

Now, at room temperature (293 K) a pure oxygen atmosphere will require 203 Joules to reduce a gram of gas to the boiling point of LOX. It will take another 216 J per g to liquefy the oxygen. A total of 419 J per g of oxygen..

Now a solid block of hydrogen at 14 K will absorb 76x14.4 J = 1,094 J rising to 90.19 K, and will also absorb an added 58.5 J along with another 452 J for each gram. A grand total of 1,605 J. So, each gram will have sufficient capacity to absorb energy to liquefy 3.83 grams of oxygen at room temperature.

To obtain 5.5 grams of oxygen for each gram of hydrogen which is the ideal O:F ratio, we must absorb 2,095 J if we are to liquefy all the oxygen. This means we must raise the temperature to room temperature. Of course doing this, allows the oxygen to boil and then to rise in temperature to room temperature, which reduces the energy required to that lost in the process.

So, how is this supposed to work then?

Liquid hydrogen runs through a heat exchanger which cools incoming air. That air eventually liquefies the oxygen, but not the nitrogen which boils at a lower temperature. The cold nitrogen gas is used to chill the air, so that the hydrogen doesn't have to. The purified oxygen, also chills the air, so the hydrogen doesn't have to.

So, by boiling away a gram of solid hydrogen and raising it to room temperature, 2,095 Joules of energy is absorbed in the process. This liquefies 5.5 grams of oxygen starting at room temperature, assuming the nitrogen is recovered.

With air, 78% is nitrogen and 21% oxygen. By liquefying the oxygen in the air, the nitrogen remains gaseous. So, it can flow through a heat exchanger and exit the craft, reducing the work the liquid hydrogen has to do. The lox in a similar fashion can absorb heat to reduce the load on the liquid hydrogen process.

km/sec R..... K...... J/gram grams grams

0.30 500 277.78 403.59 1.01 0.47
0.60 750 416.67 542.48 1.36 0.82
0.90 1000 555.56 681.37 1.70 1.16
1.20 1500 833.33 959.14 2.40 1.86
1.50 2000 1,111.11 1,236.92 3.09 2.55
1.80 3000 1,666.67 1,792.48 4.48 3.94
2.10 4000 2,222.22 2,348.03 5.87 5.33
2.40 5000 2,777.78 2,903.59 7.26 6.72

At 5.5 to 1 and 60 atmosphere pressure, the flame temperature is 3400 K.

I don't see how this can work in practice. At high speeds there isn't enough time for the air to be chilled. If the air is stopped and held until chilled, the drag and temperatures become too high.

Evaporating liquid hydrogen in a heat exchanger could cool ambient air at a certain rate to extract LOX seems doable. Yet, looking at LOX production plants it takes about 800 kWh to produce a ton of LOX from 5 tons of air. That's 2880 Joules per gram. That's the amount of energy absorbed by bringing 1 gram of solid hydrogen up to room temperature.

https://books.google.co.nz/books?id=...%20air&f=false


So, I don't see how it can work realistically.



..
On Sunday, April 10, 2016 at 2:41:49 PM UTC+12, Sylvia Else wrote:
On 10/04/2016 6:34 AM, Jeff Findley wrote:
In article ,
ess says...

On 9/04/2016 8:24 AM, Jeff Findley wrote:

I'm tired of people saying "chemical propulsion is too expensive" when
the fuel is so damn cheap! Fuel costs are *not* the problem.
Skylon/Sabre is a solution to a problem which *does not exist*!

Skylon/Sabre is not about reducing the amount of fuel consumed.


Sorry, I was being sloppy and lumping the mass of the oxidizer in with
the fuel.

How about this. LOX is one of the cheapest fluids used in the aerospace
industry. It's literally made from air. Trying to reduce LOX
consumption is quite counter-intuitive if reducing launch costs is the
goal.

Jeff


The point of Sabre is not to reduce the amount of LOX that is consumed,
the cost of LOX being, as you point out, negligible in this context.

The goal is to build an SSTO vehicle, with the economic advantages that
brings. Sabre is a means to that end, because it reduces the amount of
LOX that has to be *lifted*, and allows atmospheric nitrogen to be used
as reaction mass during the air-breathing phase.

Sylvia.

  #8  
Old April 10th 16, 10:50 AM posted to sci.space.policy
Sylvia Else
external usenet poster
 
Posts: 1,063
Default Successful SpaceX launch

On 10/04/2016 7:27 PM, William Mook wrote:
Alright, I've stated my prejudices against the Skylon's approach to
build a LACE propulsion system. SO, let's look at this more closely
and see if my prejudice is justified.

A Liquid Air Cycle Engine (LACE) is a type of spacecraft propulsion
engine that attempts to increase its efficiency by gathering part of
its oxidizer from the atmosphere. A liquid air cycle engine uses
liquid hydrogen (LH2) fuel to liquefy the air.

At its atmospheric boiling point, the specific (constant-pressure)
heat capacity of

liquid hydrogen is about 14.4 J/(g K). liquid oxygen is about 1 J/(g
K). liquid nitrogen is about 1 J/(g K).

The heat of fusion of LH2: 58.5 J/(g K)

The heat of vaporization of LH2 is 452 J/g The heat of vaporization
of LOX is 216 J/g The heat of vaporization of LN2 is 199 J/g

Boiling point: LH2: 20.28 K LOX: 90.19 K LN2: 77.00 K

Now, at room temperature (293 K) a pure oxygen atmosphere will
require 203 Joules to reduce a gram of gas to the boiling point of
LOX. It will take another 216 J per g to liquefy the oxygen. A
total of 419 J per g of oxygen.

Now a solid block of hydrogen at 14 K will absorb 76x14.4 J = 1,094 J
rising to 90.19 K, and will also absorb an added 58.5 J along with
another 452 J for each gram. A grand total of 1,605 J. So, each
gram will have sufficient capacity to absorb energy to liquefy 3.83
grams of oxygen at room temperature.

To obtain 5.5 grams of oxygen for each gram of hydrogen which is the
ideal O:F ratio, we must absorb 2,095 J if we are to liquefy all the
oxygen. This means we must raise the temperature to room
temperature. Of course doing this, allows the oxygen to boil and
then to rise in temperature to room temperature, which reduces the
energy required to that lost in the process.

So, how is this supposed to work then?

Liquid hydrogen runs through a heat exchanger which cools incoming
air. That air eventually liquefies the oxygen, but not the nitrogen
which boils at a lower temperature. The cold nitrogen gas is used to
chill the air, so that the hydrogen doesn't have to. The purified
oxygen, also chills the air, so the hydrogen doesn't have to.

So, by boiling away a gram of solid hydrogen and raising it to room
temperature, 2,095 Joules of energy is absorbed in the process. This
liquefies 5.5 grams of oxygen starting at room temperature, assuming
the nitrogen is recovered.

With air, 78% is nitrogen and 21% oxygen. By liquefying the oxygen
in the air, the nitrogen remains gaseous. So, it can flow through a
heat exchanger and exit the craft, reducing the work the liquid
hydrogen has to do. The lox in a similar fashion can absorb heat to
reduce the load on the liquid hydrogen process.

km/sec R..... K...... J/gram grams grams

0.30 500 277.78 403.59 1.01 0.47 0.60 750 416.67 542.48 1.36
0.82 0.90 1000 555.56 681.37 1.70 1.16 1.20 1500 833.33 959.14
2.40 1.86 1.50 2000 1,111.11 1,236.92 3.09 2.55 1.80 3000 1,666.67
1,792.48 4.48 3.94 2.10 4000 2,222.22 2,348.03 5.87 5.33 2.40 5000
2,777.78 2,903.59 7.26 6.72

At 5.5 to 1 and 60 atmosphere pressure, the flame temperature is 3400
K.

I don't see how this can work in practice. At high speeds there
isn't enough time for the air to be chilled. If the air is stopped
and held until chilled, the drag and temperatures become too high.

Evaporating liquid hydrogen in a heat exchanger could cool ambient
air at a certain rate to extract LOX seems doable. Yet, looking at
LOX production plants it takes about 800 kWh to produce a ton of LOX
from 5 tons of air. That's 2880 Joules per gram. That's the amount
of energy absorbed by bringing 1 gram of solid hydrogen up to room
temperature.

https://books.google.co.nz/books?id=...%20air&f=false



So, I don't see how it can work realistically.



. On Sunday, April 10, 2016 at 2:41:49 PM UTC+12, Sylvia Else wrote:
On 10/04/2016 6:34 AM, Jeff Findley wrote:
In article ,
ess says...

On 9/04/2016 8:24 AM, Jeff Findley wrote:

I'm tired of people saying "chemical propulsion is too
expensive" when the fuel is so damn cheap! Fuel costs are
*not* the problem. Skylon/Sabre is a solution to a problem
which *does not exist*!

Skylon/Sabre is not about reducing the amount of fuel
consumed.

Sorry, I was being sloppy and lumping the mass of the oxidizer in
with the fuel.

How about this. LOX is one of the cheapest fluids used in the
aerospace industry. It's literally made from air. Trying to
reduce LOX consumption is quite counter-intuitive if reducing
launch costs is the goal.

Jeff


The point of Sabre is not to reduce the amount of LOX that is
consumed, the cost of LOX being, as you point out, negligible in
this context.

The goal is to build an SSTO vehicle, with the economic advantages
that brings. Sabre is a means to that end, because it reduces the
amount of LOX that has to be *lifted*, and allows atmospheric
nitrogen to be used as reaction mass during the air-breathing
phase.

Sylvia.


Sabre is not a LACE. The air is only cooled to the vapour phase
boundary, not liquified. The cycle would be less efficient if the air
were liquified (even in part).

The system involves cooling the air at the rate it arrives. If this is
not achieved, then the engine won't work, but you can't a priori say
that it cannot be achieved.

Sylvia/
  #9  
Old April 11th 16, 12:15 AM posted to sci.space.policy
William Mook[_2_]
external usenet poster
 
Posts: 3,840
Default Successful SpaceX launch

On Sunday, April 10, 2016 at 9:50:23 PM UTC+12, Sylvia Else wrote:
On 10/04/2016 7:27 PM, William Mook wrote:
Alright, I've stated my prejudices against the Skylon's approach to
build a LACE propulsion system. SO, let's look at this more closely
and see if my prejudice is justified.

A Liquid Air Cycle Engine (LACE) is a type of spacecraft propulsion
engine that attempts to increase its efficiency by gathering part of
its oxidizer from the atmosphere. A liquid air cycle engine uses
liquid hydrogen (LH2) fuel to liquefy the air.

At its atmospheric boiling point, the specific (constant-pressure)
heat capacity of

liquid hydrogen is about 14.4 J/(g K). liquid oxygen is about 1 J/(g
K). liquid nitrogen is about 1 J/(g K).

The heat of fusion of LH2: 58.5 J/(g K)

The heat of vaporization of LH2 is 452 J/g The heat of vaporization
of LOX is 216 J/g The heat of vaporization of LN2 is 199 J/g

Boiling point: LH2: 20.28 K LOX: 90.19 K LN2: 77.00 K

Now, at room temperature (293 K) a pure oxygen atmosphere will
require 203 Joules to reduce a gram of gas to the boiling point of
LOX. It will take another 216 J per g to liquefy the oxygen. A
total of 419 J per g of oxygen.

Now a solid block of hydrogen at 14 K will absorb 76x14.4 J = 1,094 J
rising to 90.19 K, and will also absorb an added 58.5 J along with
another 452 J for each gram. A grand total of 1,605 J. So, each
gram will have sufficient capacity to absorb energy to liquefy 3.83
grams of oxygen at room temperature.

To obtain 5.5 grams of oxygen for each gram of hydrogen which is the
ideal O:F ratio, we must absorb 2,095 J if we are to liquefy all the
oxygen. This means we must raise the temperature to room
temperature. Of course doing this, allows the oxygen to boil and
then to rise in temperature to room temperature, which reduces the
energy required to that lost in the process.

So, how is this supposed to work then?

Liquid hydrogen runs through a heat exchanger which cools incoming
air. That air eventually liquefies the oxygen, but not the nitrogen
which boils at a lower temperature. The cold nitrogen gas is used to
chill the air, so that the hydrogen doesn't have to. The purified
oxygen, also chills the air, so the hydrogen doesn't have to.

So, by boiling away a gram of solid hydrogen and raising it to room
temperature, 2,095 Joules of energy is absorbed in the process. This
liquefies 5.5 grams of oxygen starting at room temperature, assuming
the nitrogen is recovered.

With air, 78% is nitrogen and 21% oxygen. By liquefying the oxygen
in the air, the nitrogen remains gaseous. So, it can flow through a
heat exchanger and exit the craft, reducing the work the liquid
hydrogen has to do. The lox in a similar fashion can absorb heat to
reduce the load on the liquid hydrogen process.

km/sec R..... K...... J/gram grams grams

0.30 500 277.78 403.59 1.01 0.47 0.60 750 416.67 542.48 1.36
0.82 0.90 1000 555.56 681.37 1.70 1.16 1.20 1500 833.33 959.14
2.40 1.86 1.50 2000 1,111.11 1,236.92 3.09 2.55 1.80 3000 1,666.67
1,792.48 4.48 3.94 2.10 4000 2,222.22 2,348.03 5.87 5.33 2.40 5000
2,777.78 2,903.59 7.26 6.72

At 5.5 to 1 and 60 atmosphere pressure, the flame temperature is 3400
K.

I don't see how this can work in practice. At high speeds there
isn't enough time for the air to be chilled. If the air is stopped
and held until chilled, the drag and temperatures become too high.

Evaporating liquid hydrogen in a heat exchanger could cool ambient
air at a certain rate to extract LOX seems doable. Yet, looking at
LOX production plants it takes about 800 kWh to produce a ton of LOX
from 5 tons of air. That's 2880 Joules per gram. That's the amount
of energy absorbed by bringing 1 gram of solid hydrogen up to room
temperature.

https://books.google.co.nz/books?id=...%20air&f=false



So, I don't see how it can work realistically.



. On Sunday, April 10, 2016 at 2:41:49 PM UTC+12, Sylvia Else wrote:
On 10/04/2016 6:34 AM, Jeff Findley wrote:
In article ,
ess says...

On 9/04/2016 8:24 AM, Jeff Findley wrote:

I'm tired of people saying "chemical propulsion is too
expensive" when the fuel is so damn cheap! Fuel costs are
*not* the problem. Skylon/Sabre is a solution to a problem
which *does not exist*!

Skylon/Sabre is not about reducing the amount of fuel
consumed.

Sorry, I was being sloppy and lumping the mass of the oxidizer in
with the fuel.

How about this. LOX is one of the cheapest fluids used in the
aerospace industry. It's literally made from air. Trying to
reduce LOX consumption is quite counter-intuitive if reducing
launch costs is the goal.

Jeff


The point of Sabre is not to reduce the amount of LOX that is
consumed, the cost of LOX being, as you point out, negligible in
this context.

The goal is to build an SSTO vehicle, with the economic advantages
that brings. Sabre is a means to that end, because it reduces the
amount of LOX that has to be *lifted*, and allows atmospheric
nitrogen to be used as reaction mass during the air-breathing
phase.

Sylvia.


Sabre is not a LACE. The air is only cooled to the vapour phase
boundary, not liquified. The cycle would be less efficient if the air
were liquified (even in part).

The system involves cooling the air at the rate it arrives. If this is
not achieved, then the engine won't work, but you can't a priori say
that it cannot be achieved.

Sylvia/


Here's what I got from online sources;

The RB545 is covered by the official secrets act, so not much can be said about it. This is a barrier right at the outset.

The SABRE design is neither a conventional rocket engine nor jet engine, but a hybrid that uses air from the environment at low speeds/altitudes, and stored liquid oxygen (LOX) at higher altitude.

The SABRE engine "relies on a heat exchanger capable of cooling incoming air to -150 °C (-238 °F), to provide oxygen for mixing with hydrogen and provide jet thrust during atmospheric flight before switching to tanked liquid oxygen when in space.

At the front of the engine, a simple translating axisymmetric shock cone inlet slows the air to subsonic speeds using two shock reflections.

Part of the air then passes through a precooler into the central core, with the remainder passing directly through a ring of bypass ramjets.

The central core of SABRE behind the precooler uses a turbo-compressor run off the same gaseous helium loop Brayton cycle which compresses the air and feeds it into four high pressure combined cycle rocket engine combustion chambers.

The oxygen is also fed to the combustion unit, using a turbopump.

***

So, I don't get how you end up with oxygen without liquefying it. I do get that using liquid hydrogen's larger specific heat its possible to cool air, and in a sense compress it unconventionally. This seems to me critical. Since it appears nowhere in any of the literature available, and it is a fundamental feature, that suggests to me one of the essential features hidden by the official secrets act around the RB545.

Now, this sets me to wondering if I can use a cold plate as a sort of compressor (forget separation which as you point out takes a helluva lot of energy) for an external combustion setup, that rides the shock wave produced by exploding hydrogen rich exhaust brought to rest relative to the moving air, that has been compressed by cooling the air moving along the surface of the ship.

* * *

In 2011, hardware testing of the heat exchanger technology "crucial to [the] hybrid air- and liquid oxygen-breathing [SABRE] rocket motor" was completed, demonstrating that the technology is viable. The tests validated that the heat exchanger could perform as needed for the engine to obtain adequate oxygen from the atmosphere to support the low-altitude, high-performance operation.

* * *

Not clear how this helps with high altitude high speed operation! lol.

The equation of state; at constant pressure, says;

PV=nRT -- V = nRT/P --- V ~ T * constant

So, if you go from say 273 K to 93 K you reduce volume by 34% or increase density by 2.94x Nearly 3x! Which is useful. Oxygen and Nitrogen have slightly different ratios of specific heat and gas constant, so rapid chilling of air very likely does cause an increase of oxygen near the chilled surface, though the major effect would be the increase in gas density with lower temperature. Injecting hydrogen into this increased density mixture, could produce a useful propulsive effect, without the need for ramjet or compressor blades. Of course, ram pressure caused by discontinuities in a 'propulsive surface' might be exploited.

1 kg of hydrogen contains 141.8 MJ of energy. Combining 1 kg of hydrogen with 8 kg of oxygen releases 141.8 MJ of energy which when expanded at 78% efficiency produces an exhaust speed of 4.38 km/sec.

Now, running the exhaust hydrogen rich means we accelerate hydrogen to the exhaust speed.

H2 O2 Fraction Ve (m/sec)

1.0 0.45 0.056 2,587
1.0 0.40 0.050 2,482
1.0 0.35 0.044 2,364
1.0 0.30 0.038 2,230
1.0 0.25 0.031 2,076
1.0 0.20 0.025 1,895
1.0 0.15 0.019 1,677
1.0 0.10 0.013 1,400
1.0 0.05 0.006 1,013

So, as the vehicle accelerates the LOX/LH2 rocket increases its oxygen fraction, and its exhaust velocity, so that the water vapour and excess hydrogen are standing still relative to the air that is moving past the aircraft. The mixture is then detonated to produce a shock wave, which interacts with the tail of the aircraft.

So, you use pure rocket to accelerate from zero to Mach 3, and then from Mach 3 to Mach 9 you use external combustion scramjet and the cold plate compression innovation to assist with combustion, and then switch back to pure rocket after that.

With oblique body HST to provide lift at low speeds, and then changing obliquity of the craft relative to direction of motion, as speed increases, we eliminate air drag and gravity losses of a ballistic ascent. Reducing ideal delta vee from 9.2 km/sec to 8.0 km/sec.

So, 0 to 1.0 km/sec with a 4.6 km/sec exhaust speed requires;

u = 1 - 1/exp(1/4.6) = 0.196

With 0.166 oxygen and 0.030 hydrogen

From 1.0 km/sec to 2.6 km/sec

u = 1 - 1/exp(1.6/4.6) = 0.294

With 0.249 oxygen and 0.045 hydrogen and 0.012 coming from on board oxygen and 0.237 coming from atmospheric oxygen.

From 2.6 km/sec to 8.0 km/sec

u = 1 - 1/exp(5.4/4.6) = 0.691

Where 0.585 is oxygen and 0.106 is hydrogen.

So, in the last phase 1-0.691 = 0.309 is the inert weight. So, 1 kg of inert weight requires 2.236 kg of propellant. Which is 1.893 kg of oxygen and 0.343 kg hydrogen.

Now the second middle phase fraction starts with 3.236 kg per kg of inert weight and requires 0.294 propellant fraction. That means 1-0.294 = 0.706 is inert fraction (which includes propellant of the last phase). So, dividing 3.236 kg by 0.706 obtains 4.584 kg at this phase. Multiplying this figure by the fractions obtained 0.012 * 4.584 = 0.055 kg of oxygen, and 0.045 * 4.584 = 0.206 kg of hydrogen.

1 kg inert
1.893 kg oxygen + 0.343 kg hydrogen - orbital phase -- 3.236 kg overall
0.055 kg oxygen +0.206 kg hydrogen - atmospheric boost -- 3.497 kg overall

Now the first phase fraction starts with 3.497 kg of inert weight and requires 0.196 propellant fraction. This means that 1-0.196 = 0.804 is inert fraction (which includes proellant of the remaining phases). So dividing 3..497 by 0.804 obtains 4.350 kg. Multiplying this figure by fractions obtained for the first phase; 0.166 x 4.350 = 0.722 kg oxygen and 0.030 x 4.350 = 0.131 kg hydrogen.

0.722 kg oxygen + 0.131 kg hydrogen - take off run --- 4.350 kg overall

2.670 kg oxygen + 0.680 kg hydrogen + 1.000 kg inert -- 4.350 kg overall

With a 12.5% structure fraction, this is

0.545 kg structure, 0.455 kg payload, 2.670 kg oxygen, 0.680 kg hydrogen

So, for each ton of payload we have 9.560 tons of take off weight. 1.197 tons of structure. 5.868 tons of oxygen and 1.495 tons of hydrogen.

A two stage to orbit conventional rocket using the same propellant, must attain 9.2 km/sec - and so each stage must attain 4.6 km/sec. So, we have with a 4.6 km/sec exhaust velocity;

u = 1 - 1/exp(4.6/4.6) = 0.6321

propellant fraction. With a 6.25% structure fraction we have

1 - 0.6321 - 0.0625 = 0.3054

payload fraction. So a kg of payload, real payload in this case means a stage weight of 3.275 in the upper stage. This is; 3.275 * 0.6321 = 2.071 propellant, and 0.204 kg of structure and 1 kg of payload.

The booster stage is 3.275 bigger than the upper stage. So, take off weight is 10.726 kg for each kg on orbit. That's 6.781 kg of propellant and 0.670 kg structure.

So, we have a 10.726 ton take off weight for every 1.000 ton of payload. We have 0.204 tons of upper stage structure and 0.670 tons of first stage structure, a total structure of 0.874 tons per ton of payload.

We are carrying a lot more propellant. But we have for each ton of payload 0.874 tons of structure for the two stage rocket and 1.197 tons of structure for the air breathing stage. This is 1.34 times heavier per unit payload, and likely costs at least 1.34 times as much to build and likely 2.70x as much to maintain.

The magnus effect can be used to skip a thermally protected cylinder off the upper atmosphere, to allow it to circle the Earth and land at the launch center, in a manner similar to that proposed by Sanger's Silbervogel. This may be applied to any stage travelling faster than 2.3 km/sec. Since the first stage has an ideal delta vee of 4.6 km/sec and after gravity drag is travelling faster than 3.3 km/sec, this is easily achieved.

Finally, dividing 10.726 - 1 by three obtains 3.242 with 0.202 of that structure, and 3.04 of that propellant, we obtain a rocket that consists of three similar components that are made more cheaply with common tooling throughout.

1 ton payload, 0.202 tons structure in each of the three flight elements, and 3.04 tons of propellant in each of the three elements. This puts up 1..14 tons for the same take off weight when operated as two stages. This is 9.408 tons take off weight for each ton put into orbit when we divide 10.726 by 1.14.

This is the most practical solution near term.

1 ton - 9.56 tons take off weight - air breathing.
1 ton 9.41 tons take off weight - three element booster

I don't know, I just don't see the advantage of a practical air breather, seet as that might be.



  #10  
Old April 11th 16, 02:09 AM posted to sci.space.policy
Sylvia Else
external usenet poster
 
Posts: 1,063
Default Successful SpaceX launch

On 11/04/2016 9:15 AM, William Mook wrote:
On Sunday, April 10, 2016 at 9:50:23 PM UTC+12, Sylvia Else wrote:
On 10/04/2016 7:27 PM, William Mook wrote:
Alright, I've stated my prejudices against the Skylon's approach
to build a LACE propulsion system. SO, let's look at this more
closely and see if my prejudice is justified.

A Liquid Air Cycle Engine (LACE) is a type of spacecraft
propulsion engine that attempts to increase its efficiency by
gathering part of its oxidizer from the atmosphere. A liquid air
cycle engine uses liquid hydrogen (LH2) fuel to liquefy the air.

At its atmospheric boiling point, the specific
(constant-pressure) heat capacity of

liquid hydrogen is about 14.4 J/(g K). liquid oxygen is about 1
J/(g K). liquid nitrogen is about 1 J/(g K).

The heat of fusion of LH2: 58.5 J/(g K)

The heat of vaporization of LH2 is 452 J/g The heat of
vaporization of LOX is 216 J/g The heat of vaporization of LN2 is
199 J/g

Boiling point: LH2: 20.28 K LOX: 90.19 K LN2: 77.00 K

Now, at room temperature (293 K) a pure oxygen atmosphere will
require 203 Joules to reduce a gram of gas to the boiling point
of LOX. It will take another 216 J per g to liquefy the oxygen.
A total of 419 J per g of oxygen.

Now a solid block of hydrogen at 14 K will absorb 76x14.4 J =
1,094 J rising to 90.19 K, and will also absorb an added 58.5 J
along with another 452 J for each gram. A grand total of 1,605
J. So, each gram will have sufficient capacity to absorb energy
to liquefy 3.83 grams of oxygen at room temperature.

To obtain 5.5 grams of oxygen for each gram of hydrogen which is
the ideal O:F ratio, we must absorb 2,095 J if we are to liquefy
all the oxygen. This means we must raise the temperature to
room temperature. Of course doing this, allows the oxygen to
boil and then to rise in temperature to room temperature, which
reduces the energy required to that lost in the process.

So, how is this supposed to work then?

Liquid hydrogen runs through a heat exchanger which cools
incoming air. That air eventually liquefies the oxygen, but not
the nitrogen which boils at a lower temperature. The cold
nitrogen gas is used to chill the air, so that the hydrogen
doesn't have to. The purified oxygen, also chills the air, so
the hydrogen doesn't have to.

So, by boiling away a gram of solid hydrogen and raising it to
room temperature, 2,095 Joules of energy is absorbed in the
process. This liquefies 5.5 grams of oxygen starting at room
temperature, assuming the nitrogen is recovered.

With air, 78% is nitrogen and 21% oxygen. By liquefying the
oxygen in the air, the nitrogen remains gaseous. So, it can flow
through a heat exchanger and exit the craft, reducing the work
the liquid hydrogen has to do. The lox in a similar fashion can
absorb heat to reduce the load on the liquid hydrogen process.

km/sec R..... K...... J/gram grams grams

0.30 500 277.78 403.59 1.01 0.47 0.60 750 416.67 542.48
1.36 0.82 0.90 1000 555.56 681.37 1.70 1.16 1.20 1500 833.33
959.14 2.40 1.86 1.50 2000 1,111.11 1,236.92 3.09 2.55 1.80
3000 1,666.67 1,792.48 4.48 3.94 2.10 4000 2,222.22 2,348.03
5.87 5.33 2.40 5000 2,777.78 2,903.59 7.26 6.72

At 5.5 to 1 and 60 atmosphere pressure, the flame temperature is
3400 K.

I don't see how this can work in practice. At high speeds there
isn't enough time for the air to be chilled. If the air is
stopped and held until chilled, the drag and temperatures become
too high.

Evaporating liquid hydrogen in a heat exchanger could cool
ambient air at a certain rate to extract LOX seems doable. Yet,
looking at LOX production plants it takes about 800 kWh to
produce a ton of LOX from 5 tons of air. That's 2880 Joules per
gram. That's the amount of energy absorbed by bringing 1 gram of
solid hydrogen up to room temperature.

https://books.google.co.nz/books?id=...%20air&f=false





So, I don't see how it can work realistically.



. On Sunday, April 10, 2016 at 2:41:49 PM UTC+12, Sylvia Else
wrote:
On 10/04/2016 6:34 AM, Jeff Findley wrote:
In article ,
ess says...

On 9/04/2016 8:24 AM, Jeff Findley wrote:

I'm tired of people saying "chemical propulsion is too
expensive" when the fuel is so damn cheap! Fuel costs
are *not* the problem. Skylon/Sabre is a solution to a
problem which *does not exist*!

Skylon/Sabre is not about reducing the amount of fuel
consumed.

Sorry, I was being sloppy and lumping the mass of the
oxidizer in with the fuel.

How about this. LOX is one of the cheapest fluids used in
the aerospace industry. It's literally made from air.
Trying to reduce LOX consumption is quite counter-intuitive
if reducing launch costs is the goal.

Jeff


The point of Sabre is not to reduce the amount of LOX that is
consumed, the cost of LOX being, as you point out, negligible
in this context.

The goal is to build an SSTO vehicle, with the economic
advantages that brings. Sabre is a means to that end, because
it reduces the amount of LOX that has to be *lifted*, and
allows atmospheric nitrogen to be used as reaction mass during
the air-breathing phase.

Sylvia.


Sabre is not a LACE. The air is only cooled to the vapour phase
boundary, not liquified. The cycle would be less efficient if the
air were liquified (even in part).

The system involves cooling the air at the rate it arrives. If this
is not achieved, then the engine won't work, but you can't a priori
say that it cannot be achieved.

Sylvia/


Here's what I got from online sources;

The RB545 is covered by the official secrets act, so not much can be
said about it. This is a barrier right at the outset.

The SABRE design is neither a conventional rocket engine nor jet
engine, but a hybrid that uses air from the environment at low
speeds/altitudes, and stored liquid oxygen (LOX) at higher altitude.

The SABRE engine "relies on a heat exchanger capable of cooling
incoming air to -150 °C (-238 °F), to provide oxygen for mixing with
hydrogen and provide jet thrust during atmospheric flight before
switching to tanked liquid oxygen when in space.

At the front of the engine, a simple translating axisymmetric shock
cone inlet slows the air to subsonic speeds using two shock
reflections.

Part of the air then passes through a precooler into the central
core, with the remainder passing directly through a ring of bypass
ramjets.

The central core of SABRE behind the precooler uses a
turbo-compressor run off the same gaseous helium loop Brayton cycle
which compresses the air and feeds it into four high pressure
combined cycle rocket engine combustion chambers.

The oxygen is also fed to the combustion unit, using a turbopump.

***

So, I don't get how you end up with oxygen without liquefying it. I
do get that using liquid hydrogen's larger specific heat its possible
to cool air, and in a sense compress it unconventionally. This seems
to me critical. Since it appears nowhere in any of the literature
available, and it is a fundamental feature, that suggests to me one
of the essential features hidden by the official secrets act around
the RB545.


There's quite a lot of information available in the technical documents
section of the Reaction Engines web site.

http://www.reactionengines.co.uk/tech_docs.html

Importantly, there is no separation of oxygen from the air during the
air breathing ascent. The cooled compressed air is fed into to the
rocket engines.

In 2011, hardware testing of the heat exchanger technology "crucial
to [the] hybrid air- and liquid oxygen-breathing [SABRE] rocket
motor" was completed, demonstrating that the technology is viable.
The tests validated that the heat exchanger could perform as needed
for the engine to obtain adequate oxygen from the atmosphere to
support the low-altitude, high-performance operation.

* * *

Not clear how this helps with high altitude high speed operation!
lol.


Regardless of the speed through the air, the inlet reduces the air speed
to subsonic.

Beyond mach 5.5, the craft operates as a pure rocket.

The rest of your analysis appears to relate to something rather
different from the sabre engine and skylon.

Sylvia.
 




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