### Author Topic: Basic Rocket Science Q & A  (Read 381335 times)

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #860 on: 12/22/2013 11:31 pm »
Alright, I'm having some trouble calculating the ISP of certain rocket engines.

As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight.

You need to divide the total impulse by the weight of the propellant, not the weight of the entire motor.

2008 N * s / (1.618 kg * 9.8 m/s^2) = 127 s.

Oh, thank you.
Clayton Birchenough
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#### AJA

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##### Re: Basic Rocket Science Q & A
« Reply #861 on: 12/24/2013 10:07 pm »
I've read in quite a few places now, that the Isp of different propellant mixes scales with the density of the fuel. So, for example, RP-1 offers more ISP than methane etc. (All other things being equal.. consider some hypothetical engine that burns Metholox as well as it burns Kerolox).

Why is this? I would've thought that larger molecules would have more degrees of freedom to soak up the thermal energy liberated from combustion, and thereby impart a lesser fraction to translational velocity  - thereby leading to a lower Vexhaust. Same reason why γ (=Cp/Cv) takes on different values in the cases of monoatomic, and polyatomic gases (where Cp = Specific heat at constant pressure, and correspondingly volume)

So what changes in these rocket engines? Like I said before, the Isp seems to be quoted for fuel mixtures, as opposed to specific power plants.

(Yes, I'm assuming there's going to be incomplete combustion, which will bring the energy distribution amongst different modes into play. Otherwise - if an engine burns through the fuel-ox mixture completely, considering the case of Methane vs RP1 - you're only going to have a difference in the percentage of the exhaust that is water and the percentage that is carbon di-oxide. Isn't the momentum carried by these the same?)
« Last Edit: 12/24/2013 10:08 pm by AJA »

#### Proponent

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##### Re: Basic Rocket Science Q & A
« Reply #862 on: 12/31/2013 09:56 pm »
Specific impulse scales with density?  That's a new one on me.  Hydrolox has a notoriously low density yet a very high specific impulse.

Specific impulses are usually quoted for a particular expansion ratio.  In effect, specific impulses given for propellant combinations are actually for particular idealized engines.

More complex molecules (in the exhaust products) will have more non-translational degrees of freedom.  In principle this does make complex exhaust products less attractive, but for a reasonably large expansion ratio, the effect is pretty small.

#### deltaV

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##### Re: Basic Rocket Science Q & A
« Reply #863 on: 01/02/2014 01:11 am »
Here's a high-level view AIUI. (Lots of details omitted.)

For a 100% efficient rocket we have
H = 1/2 m v^2
where H is the heat of combustion of a parcel of mass m of propellants and v is the exhaust velocity. Solving for v we have v = sqrt(2 H/m). One should therefore expect propellants with a higher heat of combustion (per mass) to produce greater specific impulse.

Water has a molar mass of 18 amu and a heat of formation of -242 kJ/mol (as gas). Carbon dioxide has a molar mass of 44 amu and a heat of formation of -394 kJ/mol (as gas). Observe that a carbon dioxide molecule masses over twice water but provides less than twice the heat of formation, so producing H2O releases more heat per kilogram than producing CO2 does. The mainstream hydrocarbon fuels have heat of formation that can be approximated by zero for present purposes so you can think of e.g. a methane atom as just a convenient way to store a carbon atom and 4 hydrogen atoms. To maximize H/m you want to produce as much water as possible. That's why methane has better ISP than RP-1.

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #864 on: 01/12/2014 02:46 pm »
Is there a number that describes the efficiency of stages as a whole? ISP is for propellant combination and engine performance but does not factor in the dry mass of the stage needed to support the propellant and its associated engine...

(Not sure if that made any sense... everything I said sounded weird )
Clayton Birchenough
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#### Proponent

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##### Re: Basic Rocket Science Q & A
« Reply #865 on: 01/12/2014 02:52 pm »
Propellant mass fraction is another relevant number.  In some cases, (initial) thrust-to-weight ratio matters a lot.

#### deltaV

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##### Re: Basic Rocket Science Q & A
« Reply #866 on: 01/12/2014 03:19 pm »
Is there a number that describes the efficiency of stages as a whole? ISP is for propellant combination and engine performance but does not factor in the dry mass of the stage needed to support the propellant and its associated engine...

(Not sure if that made any sense... everything I said sounded weird )

One figure I like (which is non-standard as far as I know) is the ISP of theoretical stage that has the same wet mass, zero dry mass, and provides the same delta-vee as the real stage we're evaluating. Call this concept the equivalent specific impulse. Here are the formulae:
deltaV = I * ln((M_w+M_p)/(M_d+M_p))
I_equiv = deltaV / ln((M_w+M_p)/(M_p))
where I is specific impulse (in velocity units), M_w is the wet mass of the stage, M_d is the dry mass of the stage, M_p is the mass of the payload (including any later stages), deltaV is the delta vee of the stage, and I_equiv is the equivalent specific impulse described above. The formulae for deltaV is only valid for a single stage but the definition of I_equiv is meaningful both for individual stages and for a multi-stage rocket as a whole.

One nice thing about this equivalent specific impulse is that the equivalent specific impulse of a multi-stage rocket is the ln(mass ratio) weighted average of the equivalent specific impulses of its constituent stages. For example a two-stage rocket with equivalent specific impulses of 350 s and 370s has an overall equivalent specific impulse somewhere between 350 and 370s. This two-stage rocket is better (in terms of mass ratios) than an alternative single stage rocket design with equivalent specific impulse 340s.

One annoying thing about this concept is the equivalent specific impulse of a stage depends on the mass of its payload.
« Last Edit: 01/12/2014 03:25 pm by deltaV »

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #867 on: 01/12/2014 05:40 pm »
Thank you both Proponent and DeltaV.

Would total impulse divided by mass of the stage be a a good evaluation of the performance of the stage?
Clayton Birchenough
Astro. Engineer and Computational Mathematics @ ERAU

#### M129K

##### Re: Basic Rocket Science Q & A
« Reply #868 on: 01/15/2014 07:38 pm »
How can you calculate the ∆V needed to escape Earth and get on a Mars transfer if using low-thrust propulsion? I recently asked this somewhere else, and why it was useful for something like reaching GEO, I still don't know how to calculate escape velocity ∆V.

#### mheney

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##### Re: Basic Rocket Science Q & A
« Reply #869 on: 01/15/2014 08:33 pm »
Escape velocity is sqrt(2) times your circular orbital velocity - which makes your delta V (sqrt(2) -1) times orbital velocity.  That holds if you only burn at or near perigee.  (you keep the perigee, and raise apogee to infinity.)  If you do continuous burn, then you're essentially trying to raise both apogee and perigee to infinity.  That sees like you'd be doubling the total dV needed that way...

#### M129K

##### Re: Basic Rocket Science Q & A
« Reply #870 on: 01/16/2014 07:06 am »
Escape velocity is sqrt(2) times your circular orbital velocity - which makes your delta V (sqrt(2) -1) times orbital velocity.  That holds if you only burn at or near perigee.  (you keep the perigee, and raise apogee to infinity.)  If you do continuous burn, then you're essentially trying to raise both apogee and perigee to infinity.  That sees like you'd be doubling the total dV needed that way...

So... Just double the delta V from 3.22 to 6.44 km/s to escape?

#### Proponent

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##### Re: Basic Rocket Science Q & A
« Reply #871 on: 01/16/2014 11:57 am »
Escape velocity is sqrt(2) times your circular orbital velocity - which makes your delta V (sqrt(2) -1) times orbital velocity.  That holds if you only burn at or near perigee.  (you keep the perigee, and raise apogee to infinity.)  If you do continuous burn, then you're essentially trying to raise both apogee and perigee to infinity.  That sees like you'd be doubling the total dV needed that way...

The delta-V for a very-low-thrust trajectory between two circular orbits is simply the difference in the orbital speeds of the two orbits.  A "circular orbit at infinity" has an orbital speed of zero.  If you're in a parking at, say, 7 km/s, then the low-thrust delta-V to escape is 7 km/s.  This of course ignores the perturbing influences of the moon and sun, but I think it's a pretty good first approximation.

To get to Mars, of course, you need to do more than just escape: you need a positive asymptotic speed.  It's not obvious to me that there's a good simple approximation available here.  If the acceleration is high enough that the delta-V for a Hohmann-like transfer from Earth's orbit to Mars's can be delivered in a couple of months, then I suppose you could just add the delta-V, recognizing that it will be a bit of an underestimate.  If, on the other hand, thrust is so low that several orbits of the sun will be completed on the way to Mars, then low-thrust approximation works well, it's just that this time the relevant circular speeds are those of Earth and Mars about the sun.  Most likely, though, neither of these limiting cases applies.  At least they give you upper and lower bounds on the delta-V required.
« Last Edit: 01/16/2014 12:38 pm by Proponent »

#### mheney

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##### Re: Basic Rocket Science Q & A
« Reply #872 on: 01/16/2014 02:40 pm »
It seems , though, that you only need to fire your engine near perigee.  The target orbit really isn't a circular orbit at infinity - it's a parabola (Actually, a hyperbola, so you've got some added heliocentric velocity once you escape.)

It seems that you'd be "wasting" any delta-v on the 'upper' half of the orbit, which only serves to raise the perigee.
If you set up your orbit so that perigee is at 6 pm solar time (over the terminator, "behind" the Earth looking down from the North), you'd want to do your burn for some fraction of the orbit around perigee.  The effect would be for the apogee to climb out ahead/away from the earth along Earth's orbit, while the perigee remained (roughly) fixed.  (Note that if you wanted to drop into the lower solar system, you'd reverse this, and have your perigee at 6am solar, with the apogee climbing out behind the Earth...)

The closer you get to instantaneous perigee burns, the closer your total delta-V gets to the theoretical minimum - which is sqrt(2) -1 times your orbital velocity.   For a 7 km/s LEO, that'd be about 2.9 km/s delta-V to escape - much less than the 7km/sec for a constant burn.  You'd be trading reaction mass for time-to-escape.

#### Proponent

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##### Re: Basic Rocket Science Q & A
« Reply #873 on: 01/16/2014 04:38 pm »
To minimize delta-V, one would could indeed burn only near perigee and accept much longer trips.  With electric propulsion offering exhaust velocities in the neighborhood of 20-30 km/s, however, I would think that a mission planner would typically prefer a higher delta-V.  The high exhaust velocity limits the additional propellant mass needed.

#### Hop_David

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##### Re: Basic Rocket Science Q & A
« Reply #874 on: 01/18/2014 04:23 pm »
The delta-V for a very-low-thrust trajectory between two circular orbits is simply the difference in the orbital speeds of the two orbits.  A "circular orbit at infinity" has an orbital speed of zero.  If you're in a parking at, say, 7 km/s, then the low-thrust delta-V to escape is 7 km/s.  ...
To get to Mars, of course, you need to do more than just escape: you need a positive asymptotic speed.  It's not obvious to me that there's a good simple approximation available here.  If the acceleration is high enough that the delta-V for a Hohmann-like transfer from Earth's orbit to Mars's can be delivered in a couple of months, then I suppose you could just add the delta-V, recognizing that it will be a bit of an underestimate.  If, on the other hand, thrust is so low that several orbits of the sun will be completed on the way to Mars, then low-thrust approximation works well, it's just that this time the relevant circular speeds are those of Earth and Mars about the sun.  Most likely, though, neither of these limiting cases applies.  At least they give you upper and lower bounds on the delta-V required.

Indeed. Low earth orbit has a big angular velocity, about 4 degrees a minute (360 degrees/90 minutes).

To enjoy an Oberth benefit, we'd need to do the acceleration in the perigee's neighborhood, let's say over 60 degrees or about 15 minutes. If the ion rocket accelerates at 100 micro-gees, 100 micro-gees * 15 minutes = ~ 1 meter/second. A neglible burn.

Instead of an impulsive burn at perigee sending the rocket to escape, the ship's path would be a gradual spiral out of earth's gravity well. From LEO to escape would take about 7 km/s as you say. Although the delta V budget is more than twice that of an impulsive burn, ion exhaust speeds can be 4 or 5 times that of chemical. So the exponent in the rocket equation (delta V/V exhaust) is smaller in spite of the extra delta V.

However earth's orbit about the sun is a much more leisurely degree per day. (360 degrees/365 days). In this case we'd remain in a 60 degree neighborhood of perihelion for two months. 100 micro-gees * 2 months = 5 km/s.

For heliocentric orbits, low thrust, high ISP engines might be able to enjoy some Oberth benefit. Once out of planetary wells, it seems to me we can have our ion cake and eat it too.
« Last Edit: 01/18/2014 04:31 pm by Hop_David »

#### sdsds

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##### Re: Basic Rocket Science Q & A
« Reply #875 on: 01/18/2014 10:45 pm »
However earth's orbit about the sun is a much more leisurely degree per day. (360 degrees/365 days). In this case [of a spacecraft in heliocentric orbit] we'd remain in a 60 degree neighborhood of perihelion for two months. 100 micro-gees * 2 months = 5 km/s.

For heliocentric orbits, low thrust, high ISP engines might be able to enjoy some Oberth benefit. Once out of planetary wells, it seems to me we can have our ion cake and eat it too.

That's very well expressed. Thanks! And just to check the reverse works too? I.e. a spacecraft that has just managed to escape from Mars (but is still in essentially the same orbit around the Sun as Mars) has plenty of time at aphelion to perform an ion "burn" that drops its perihelion down to intersect Earth?
-- sdsds --

#### Avron

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##### Re: Basic Rocket Science Q & A
« Reply #876 on: 01/19/2014 02:36 pm »
In a LOX/H2 engine, do the two fluids enter the combustion chamber via the injector plate as gasses or as liquids?

#### aga

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##### Re: Basic Rocket Science Q & A
« Reply #877 on: 01/19/2014 03:35 pm »
In a LOX/H2 engine, do the two fluids enter the combustion chamber via the injector plate as gasses or as liquids?

in rl-10 they are gases, acc. to this post by ranulfc

http://forum.nasaspaceflight.com/index.php?topic=30547.msg1125002#msg1125002
Quote
...even though at the same time everyone also understood that the RL-10 used "gas/gas" in operation and why....
42

#### Avron

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##### Re: Basic Rocket Science Q & A
« Reply #878 on: 01/19/2014 04:27 pm »
In a LOX/H2 engine, do the two fluids enter the combustion chamber via the injector plate as gasses or as liquids?

in rl-10 they are gases, acc. to this post by ranulfc

http://forum.nasaspaceflight.com/index.php?topic=30547.msg1125002#msg1125002
Quote
...even though at the same time everyone also understood that the RL-10 used "gas/gas" in operation and why....

I got this "In main combustion chambers oxygen is injected in its liquid state, whereas
the fuel - used for regeneratively cooling the combustor walls - is injected in the gaseous state.
"

ntrs.nasa.gov/archive/nasa/casi.ntrs.../20020045359_2002079435.pdf‎

#### baldusi

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##### Re: Basic Rocket Science Q & A
« Reply #879 on: 01/19/2014 06:27 pm »
Depends on engine cycle. Normally  pressure fed, gas generator and bleed expander means liquid for both, stage combustion and closed expander means gas for the preburner rich or expanded part (Hydrogen, normally for hydrolox), and both gas.
Now, is is a bit simplified, because usually the temperature and pressure is above the critical point. So the actual state might not be clear cut.