Author Topic: Basic Rocket Science Q & A  (Read 281163 times)

Offline Nomadd

  • Senior Member
  • *****
  • Posts: 2607
  • Boca Chica, Texas
  • Liked: 3239
  • Likes Given: 231
Re: Basic Rocket Science Q & A
« Reply #1080 on: 08/07/2016 11:11 PM »
It didn't really. They published it's speed relative to Earth. It was mostly luck that Jupiter and Earth were at a point where they were receding from each other. Add that to Juno's approach speed. If the spacecraft had gotten there at a different time, it's speed relative to Earth could have been much less than 165k. Just figuring the difference between Earth receding or approaching in it's orbit can make a 120,000 mph difference in relative speed.
 I'm not sure what Juno's orbital speed is at closest approach.
« Last Edit: 08/08/2016 02:56 AM by Nomadd »

Offline Burninate

  • Full Member
  • ****
  • Posts: 1129
  • Liked: 344
  • Likes Given: 72
Re: Basic Rocket Science Q & A
« Reply #1081 on: 08/07/2016 11:21 PM »
How did the Juno space probe reach 165,000 MPH?

"The Ferrari hit Arthur at a relative speed of 104mph.  How did Arthur reach 104mph, a speed no human runner has ever attained before?"

Because Arthur was running at 8mph and the Ferrari was speeding at 96mph.  Only Arthur's interaction with the Ferrari propelled him to such a speed.

It's not a perfect analogy because there's no gravity in it, but it's close enough for government work.

When you're talking about the capabilities of a planetary spaceprobe, ask questions like "What was the mission delta V, the sum of the changes in speed from all the burns of its engine?", but that has relatively little to do with current velocity if the mission traverses multiple gravity wells, and current velocity is only even a useful concept relative to some other object or as a time-varying metric at a particular part of an orbit.
« Last Edit: 08/07/2016 11:22 PM by Burninate »

Offline joek

  • Senior Member
  • *****
  • Posts: 2782
  • Liked: 534
  • Likes Given: 328
Re: Basic Rocket Science Q & A
« Reply #1082 on: 08/08/2016 12:14 AM »
I'm not sure what Juno's orbital speed is at closest approach.

At perijove, approximately 130K mph Jupiter-relative and approximately 160K mph Earth-relative.  The difference is due to Earth-Jupiter orbital velocity.

Online eeergo

  • Phystronaut
  • Global Moderator
  • Senior Member
  • *****
  • Posts: 4759
  • Milan, Italy; Spain; Virginia
  • Liked: 434
  • Likes Given: 352
Re: Basic Rocket Science Q & A
« Reply #1083 on: 08/08/2016 01:44 AM »
My references for this issue are jcm's authoritative analysis (http://planet4589.org/space/jsr/back/news.728.txt) and the very complete and didactic report Daniel Marin compiled a couple of weeks ago (http://danielmarin.naukas.com/2016/07/09/cual-ha-sido-la-nave-espacial-mas-rapida/ (in Spanish))

-DaviD-

Offline Nomadd

  • Senior Member
  • *****
  • Posts: 2607
  • Boca Chica, Texas
  • Liked: 3239
  • Likes Given: 231
Re: Basic Rocket Science Q & A
« Reply #1084 on: 08/08/2016 02:52 AM »
I'm not sure what Juno's orbital speed is at closest approach.

At perijove, approximately 130K mph Jupiter-relative and approximately 160K mph Earth-relative.  The difference is due to Earth-Jupiter orbital velocity.
So. the Earth relative speed could have come in at something like 65K to 195K mph depending on orbital positions.
 I think the Jupiter relative speed might have been a more honest one to report.
« Last Edit: 08/08/2016 02:56 AM by Nomadd »

Offline gin455res

  • Full Member
  • ***
  • Posts: 324
  • bristol, uk
  • Liked: 7
  • Likes Given: 18
Re: Basic Rocket Science Q & A
« Reply #1085 on: 09/30/2016 09:21 PM »
I just read that 70% peroxide freezes at -40degC and 100% at about 0.


Has anyone studied the use of subcooled peroxide?
What kind of density increases does 70% peroxide gain from subcooling?


I also read 60% peroxide goes down to -55degC. 


Has any theoretical work been done on finding interesting mixtures of water, peroxide (and perhaps nitrous) as rocket oxidisers?


I imagine dissolving nitrous in 60%(-55degC) peroxide, and seeing what happens to the freezing point. Perhaps, adding nitrous (and further cooling (if possible)), until the mixture doesn't quite have enough energy to decompose.

Offline Proponent

  • Senior Member
  • *****
  • Posts: 5162
  • Liked: 786
  • Likes Given: 549
Re: Basic Rocket Science Q & A
« Reply #1086 on: 10/01/2016 09:36 AM »
The characteristics of peroxide-water mixtures have been thoroughly studied (see the Hydrogen Peroxide Handbook produced by Rocketdyne in June 1967 as AFRPL-TR-67-144.

While it's true that, to some extent, diluting 100% peroxide with water lowers its freezing point, adding water lowers the density too.  The result is that, for example, 70% peroxide has a lower density (1.36 kg/m3) at its freezing point (-40 oC) than does 100% peroxide (1.47  kg/m3) at its freezing point (-0.4 oC).  And the water does not contribute to specific impulse, so the impulse density of the more dilute mixtures is lower still.

I don't know about peroxide-nitrous mixtures, but since nitrous generally has lower performance as an oxidizer, I don't have high hopes.
« Last Edit: 10/01/2016 11:01 AM by Proponent »

Online adrianwyard

  • Full Member
  • ****
  • Posts: 955
  • Liked: 172
  • Likes Given: 220
Re: Basic Rocket Science Q & A
« Reply #1087 on: 10/09/2016 12:59 AM »
For long duration missions cryogenic propellants are problematic because they tend to boil off to a gaseous state. To circumvent this we hear of designs that go to the trouble of keeping it cold enough that it stays liquid, or just switch to another propellant. But is there a straightforward reason that it couldn't be allowed to turn to gas, and then recooled and liquified only when needed?

Perhaps the energy and mass of the equipment needed to reliquify is always going to be demonstrably greater than insulation?

Offline TrevorMonty

Re: Basic Rocket Science Q & A
« Reply #1088 on: 10/09/2016 08:16 AM »
For long duration missions cryogenic propellants are problematic because they tend to boil off to a gaseous state. To circumvent this we hear of designs that go to the trouble of keeping it cold enough that it stays liquid, or just switch to another propellant. But is there a straightforward reason that it couldn't be allowed to turn to gas, and then recooled and liquified only when needed?

Perhaps the energy and mass of the equipment needed to reliquify is always going to be demonstrably greater than insulation?
The lastest Lockheed Martin Mars mission idea uses active cryogenic cooling to keep LH/LOX  in liquid state. The fuel is pumped from storage tanks into propulsion stage's tanks a few hours before it is required for a burn.

ULA think they can keep LH/LOX cool for days to weeks even months by using lots of insulation and sun shade. Any H gas boil off can be used to help keep rest of fuel cool. Even though it is in gas state the H is still extremely cold and can be used to cool outer skin of tank. This surplus gas can be used in gas thrusters or burnt in H/O gas thrusters for station keeping.


Offline sdsds

  • Senior Member
  • *****
  • Posts: 5478
  • "With peace and hope for all mankind."
  • Seattle
  • Liked: 577
  • Likes Given: 677
Re: Basic Rocket Science Q & A
« Reply #1089 on: 10/09/2016 08:39 AM »
This surplus gas can be used in gas thrusters or burnt in H/O gas thrusters for station keeping.

Or used in an internal combustion engine to produce mechanical energy, which can be used for a variety of purposes. See the Integrated Vehicle Fluids paper: http://www.ulalaunch.com/uploads/docs/Published_Papers/Supporting_Technologies/Space_Access_Society_2012.pdf
-- sdsds --

Online nicp

  • Member
  • Posts: 28
  • UK
  • Liked: 7
  • Likes Given: 73
Re: Basic Rocket Science Q & A
« Reply #1090 on: 10/12/2016 05:41 PM »
A quick question. Somewhere I got the impression that the oxygen-rich staged combustion cycle is inherently more efficient than a fuel rich staged combustion cycle.
Perhaps that is incorrect - I can't think of a good reason why it would be.

But if it is more efficient, why?

Something is niggling in the back of my brain now about smaller molecules of oxygen vs kerosene, but my last chemistry lesson was in 1984 when I was 18...

Where's my Guinness?

Offline rocx

  • Full Member
  • ***
  • Posts: 384
  • NL
  • Liked: 263
  • Likes Given: 145
Re: Basic Rocket Science Q & A
« Reply #1091 on: 10/13/2016 07:41 AM »
A quick question. Somewhere I got the impression that the oxygen-rich staged combustion cycle is inherently more efficient than a fuel rich staged combustion cycle.
Perhaps that is incorrect - I can't think of a good reason why it would be.

But if it is more efficient, why?

Something is niggling in the back of my brain now about smaller molecules of oxygen vs kerosene, but my last chemistry lesson was in 1984 when I was 18...

I'm not quite the engine expert that some other forum members are, but I can give a first response.
Fuel-rich staged combustion for kerosene does not work because kerosene decomposes and cokes (leaves carbon soot on the engine) at high temperature instead of becoming a hot, clean gas.
For hydrogen engines, fuel-rich staged combustion is actually preferred. See the Space Shuttle Main Engines.
There are other reasons why oxidiser-rich staged combustion is preferred for methalox, but I can't explain it satisfactorily to myself, so I'll leave that to others.
Any day with a rocket landing is a fantastic day.

Offline Proponent

  • Senior Member
  • *****
  • Posts: 5162
  • Liked: 786
  • Likes Given: 549
Re: Basic Rocket Science Q & A
« Reply #1092 on: 10/13/2016 08:34 AM »
To a very rough first approximation, the various gases involved (hydrogen, oxygen, methane) have similar (within a factor of 2) specific heats per molecule.  In other words, it takes a similar amount of energy to raise the temperature of a mole of oxygen molecules from 699 K to 700 K as it does to raise the temperature of a mole of hydrogen molecules from 699 K to 700 K.  That means that the propellant present in the larger quantity (by number, not mass) will tend to provide the most pumping power at a given temperature.  Lox-hydrogen engines run at O/F mass ratios of 5-6, corresponding to number ratios of about 0.35.  Hence, fuel-rich staged combustion offers about three times the power of the oxygen-rich alternative.

When it comes to lox-methane, O/F is usually just slightly fuel-rich, meaning that O/F is a bit over 3 by number.  Hence, there is more pumping power available in the oxygen-rich cycle (though the polyatomic nature of methane raises its molar heat capacity, reducing oxygen's advantage somewhat).
« Last Edit: 10/13/2016 08:35 AM by Proponent »

Offline baldusi

  • Senior Member
  • *****
  • Posts: 7436
  • Buenos Aires, Argentina
  • Liked: 1434
  • Likes Given: 4475
Re: Basic Rocket Science Q & A
« Reply #1093 on: 10/13/2016 10:19 PM »
To a very rough first approximation, the various gases involved (hydrogen, oxygen, methane) have similar (within a factor of 2) specific heats per molecule.  In other words, it takes a similar amount of energy to raise the temperature of a mole of oxygen molecules from 699 K to 700 K as it does to raise the temperature of a mole of hydrogen molecules from 699 K to 700 K.  That means that the propellant present in the larger quantity (by number, not mass) will tend to provide the most pumping power at a given temperature.  Lox-hydrogen engines run at O/F mass ratios of 5-6, corresponding to number ratios of about 0.35.  Hence, fuel-rich staged combustion offers about three times the power of the oxygen-rich alternative.

When it comes to lox-methane, O/F is usually just slightly fuel-rich, meaning that O/F is a bit over 3 by number.  Hence, there is more pumping power available in the oxygen-rich cycle (though the polyatomic nature of methane raises its molar heat capacity, reducing oxygen's advantage somewhat).
If I might add, the original question was ambiguous. Turbine-pumped rocket engines have the turbine power issue and the main combustion chamber issue. And even then, you have gas generator/staged combustion cycles, the expander cycles and the tap-off cycle. And then you have the more complex full flow, dual expander and staged/expander (like the RD-0162).
So, the answer is, it depends. As it was very well explained by Proponent, the turbine power is, when possible, the defining issue on the O/F ratio for turbines. In the RP-1/LOX staged combustion, it is impossible to do fuel rich precombustion. In the LH/LOX the H2 has better performance and on the CH4/LOX the oxidizer-rich case has more power.
Regarding the main combustion, the specific impulse is directly related to average molecular speed, and thus the lighter the molecule, the better the isp. It is obviously also related to combustion temperature and pressure. But that is usually power and material limited, and thus, your only decision is to use oxidizer rich or fuel rich.
In this case, from my limited understanding, given that the molecular weight of the fuel is usually lower than the weight of the oxidizer, using a fuel rich combustion gives improved specific impulse. The oxidizer-rich case would give you better propellant density (again, only if the fuel molecular weight is lighter than the oxidizer's). And in this case specific impulse usually trumps the density considerations.
Please understand that on rocket engines you have to carry both fuel and oxidizer. On combustion engines on Earth, the oxidizer is taken from the atmosphere, and thus, running leaner gives you much better performance, since you are adding a reactant that's "free" in mass terms. But this does not applies to anything that goes to vacuum.

Offline FishInferno

  • Member
  • Posts: 95
  • Liked: 39
  • Likes Given: 83
Re: Basic Rocket Science Q & A
« Reply #1094 on: 04/30/2017 02:36 AM »
Why is an overexpanded nozzle less efficient than an ambient nozzle?  According to the attached diagram, the nozzle is "full" all the way to the end with both an ambient and overexpanded nozzle (not grossly overexpanded, the bottom one, I know that has other issues).  How is the rocket affected by the exhaust "collapsing" after it has left the engine?

I understand that an underexpanded nozzle is less efficient because it fails to capture the maximum possible energy from the expanding exhaust, but just to be clear, it is still producing the same amount of thrust that it does at ambient, right?  Because both nozzles are "full".
Comparing SpaceX and SLS is like comparing paying people to plant fruit trees with merely digging holes and filling them.  - Robotbeat

Offline Damon Hill

  • Veteran
  • Full Member
  • ****
  • Posts: 548
  • Auburn, WA
  • Liked: 71
  • Likes Given: 121
Re: Basic Rocket Science Q & A
« Reply #1095 on: 04/30/2017 02:44 AM »
One problem with over-expansion is the turbulent flow as it detaches from the nozzle wall; this can tear up an engine and cause guidance problems--the flow will be asymmetrical.

Offline FishInferno

  • Member
  • Posts: 95
  • Liked: 39
  • Likes Given: 83
Re: Basic Rocket Science Q & A
« Reply #1096 on: 04/30/2017 03:06 AM »
One problem with over-expansion is the turbulent flow as it detaches from the nozzle wall; this can tear up an engine and cause guidance problems--the flow will be asymmetrical.

I thought that was only with grossly over-expanded when it becomes detached from the nozzle like in the bottom picture.  Aren't most rockets designed with engines that are slightly over-expanded at sea level in order to maintain efficiency at a higher altitude? It can't become too unstable at that amount of over-expansion if it is usable on rockets.
Comparing SpaceX and SLS is like comparing paying people to plant fruit trees with merely digging holes and filling them.  - Robotbeat

Offline Jim

  • Night Gator
  • Senior Member
  • *****
  • Posts: 31343
  • Cape Canaveral Spaceport
  • Liked: 9624
  • Likes Given: 299
Re: Basic Rocket Science Q & A
« Reply #1097 on: 04/30/2017 03:12 AM »
Aren't most rockets designed with engines that are slightly over-expanded at sea level

under-expanded

Offline dimovski

  • Member
  • Posts: 1
  • Zagreb
  • Liked: 0
  • Likes Given: 0
Re: Basic Rocket Science Q & A
« Reply #1098 on: 04/30/2017 10:37 AM »
Hello everybody,

I've been lurking around in this forum for quite a while and I sincerely hope that I am not repeating an already asked question or posting in the wrong topic.

Intrigued by the LANTR-concept, I tried to replicate the different Isps for different mixture ratios by using the formula Ve^2=2k/(k-1)*R*T/M, which I've found here and simplified a bit by removing the part with different pressures, as that factor would most likely be rather close to 1.

However, I always get results that are way off, e.g., if I go for pure hydrogen & T=2900, according to the 1st link
I should get an Isp of around 940, so a Ve of around 9000. But what I actually get is: Ve=sqrt(2*1.666/0.666*8.314*2900/1)=347.311m/s, which is off by a factor of 30(!).

Could anybody tell me what exactly I'm missing here?

Thanks in advance and sorry for posting a potentially dumb question.

Offline deruch

  • Full Member
  • ****
  • Posts: 1576
  • California
  • Liked: 1186
  • Likes Given: 1694
Re: Basic Rocket Science Q & A
« Reply #1099 on: 05/01/2017 01:34 AM »
Why is an overexpanded nozzle less efficient than an ambient nozzle?  According to the attached diagram, the nozzle is "full" all the way to the end with both an ambient and overexpanded nozzle (not grossly overexpanded, the bottom one, I know that has other issues).  How is the rocket affected by the exhaust "collapsing" after it has left the engine?

I understand that an underexpanded nozzle is less efficient because it fails to capture the maximum possible energy from the expanding exhaust, but just to be clear, it is still producing the same amount of thrust that it does at ambient, right?  Because both nozzles are "full".
The general thrust equation for rockets is: F=mdot*Ve + (Pe - P0)*Ae where,
     F=thrust,
     mdot=mass flow rate,
     Ve=velocity of the exhaust at the nozzle exit,
     Pe=exhaust pressure at the nozzle exit,
     P0=ambient pressure, and
     Ae=area of the nozzle exit.

So, if the exhaust pressure is below the ambient pressure, you end up with a negative term in your thrust equation.  i.e. (Pe - P0)*Ae < 0.  And ergo, a reduction in thrust. 
Shouldn't reality posts be in "Advanced concepts"?  --Nomadd