Author Topic: Basic Rocket Science Q & A  (Read 383730 times)

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #1120 on: 04/14/2018 10:27 pm »
The general thrust equation for rockets is: F=mdot*Ve + (Pe - Pa)*Ae where,
     F=thrust,
     mdot=mass flow rate,
     Ve=velocity of the exhaust at the nozzle exit,
     Pe=exhaust pressure at the nozzle exit,
     Pa=ambient pressure, and
     Ae=area of the nozzle exit.


When considering the Pa term during the low altitude segment of the flight, is that very strictly the general atmospheric pressure at the rocket's altitude?  I believe the forward travel of the rocket and backward travel of the high velocity exhaust gasses creates a localized low pressure at the base of the rocket (this causes the plume recirculation seen on some launches, right?).  Does this area's lowered pressure need to be taken into account for a higher accuracy calculation of the thrust?

I imagine this could come down to a matter of definition.  What really matters is the total force on the rocket, and how that's divided up between thrust and drag may be a little bit arbitrary.  However, I am inclined to use atmospheric pressure in calculating the pressure term in the thrust equation.  That equation is typically derived by noting that if a rocket engine imparts momentum to its exhaust at a rate
The usual way of deriving the thrust equation quoted above is to argue that if the rocket engine imparts momentum at a rate mdot*Ve, then by Newton's third law momentum in the opposite direction is imparted to the engine.  Then the back-pressure term is rather artificially, in my view, tacked on.

However, the equation can also be derived by considering nothing but the pressure forces on the engine.  As one proceeds aft from the forward dome of the combustion chamber, diverging sections (including the forward dome itself) exert forward forward on the engine wherever the internal pressure exceeds ambient, and converging sections exert rearward force.  Where internal pressure is lower than external, diverging sections exert rearward force and converging sections exert forward force.  Add up all of those forces (i.e., integrate from the top of the combustion chamber to the nozzle exit), and you get the usual expression for thrust.

If you look at thrust this way, then it's natural that the general ambient pressure appears in the equation.

EDIT:  "thats" -> "that's"
« Last Edit: 04/15/2018 04:17 pm by Proponent »

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #1121 on: 04/15/2018 08:44 pm »
For a simple illustration of my point, above, imagine a very simple rocket engine in the shape of a cylinder with the top end closed and the bottom end open.  We inject fuel and oxidizer near the top.  They burn, increasing the pressure there.  The net force on the engine is cylinder's cross-sectional area multiplied by the difference in pressures across the top end.

Offline parsec

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Re: Basic Rocket Science Q & A
« Reply #1122 on: 09/13/2018 11:11 am »
I'm working my way through the isentropic flow section of Sutton (9th edition). The equations derived for exhaust velocity and a subsequent plot seem to indicate that specific impulse decreases with increasing k (the ratio of specific heats).

This seems counterintuitive to me. The more degrees of freedom a molecule has, the more k tends towards one, since k is (f/2 +1)/(f/2), where f is the amount of degrees of freedom. However, I would expect that less degrees of freedom result in a higher translational exhaust velocity due to the equipartition of energy between those degrees of freedom.

Wikipedia (I know, not the best source) seems to confirm this qualitative intuition;

https://en.wikipedia.org/wiki/Rocket_engine#Propellant_efficiency

Quote
For a rocket engine to be propellant efficient..... using propellants which are, or decompose to, simple molecules with few degrees of freedom to maximise translational velocity

Sutton's equations and text seem to indicate the contrary. In fact, one of Sutton's equations for the maximum theoretical value for exhaust velocity for infinite expansion produces an infinite velocity (divide by zero) for k=1, i.e. molecules with a large amount of degrees of freedom.

Is there something I am missing here? Why do molecules with more degrees of freedom seem to produce higher specific impulse values according to Sutton?
« Last Edit: 09/13/2018 11:15 am by parsec »

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #1123 on: 09/13/2018 01:50 pm »
Since the expansion ratio is infinite, the temperature at the nozzle exit is zero, and all of the energy that was in all of those degrees of freedom back in the combustion chamber has been converted into bulk kinetic energy, regardless of how many degrees of freedom exist.  If you hold the exit temperature constant at a relatively high value while varying the ratio of specific heats, then flows with lower ratios of specific heats should do better over some range.  When examining the behavior of an equation as one variable changes, one must ask oneself which of the relevant quantities are also varying and which are (perhaps only implicitly) being held constant.

When thinking about the exhaust-velocity equation, by the way, I find that replacing k Cp/(k - 1) with CpT, where Cp is the mass-specific heat at constant pressure and T, of course, is the absolute temperature, aids my intuition.  The point is that CpT is the enthalpy of an ideal gas.

Offline Kosmos2001

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Re: Basic Rocket Science Q & A
« Reply #1124 on: 09/20/2018 12:26 pm »
Why some booster nozzles are pointing not vertically but with some angle outside?

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #1125 on: 09/20/2018 01:17 pm »
If nozzles point directly backward, then variations in thrust of one engine with respect to another will tend to make the rocket turn a bit.  That  is eliminated if each engine's thrust vector points at the rocket's center of mass.

Offline envy887

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Re: Basic Rocket Science Q & A
« Reply #1126 on: 10/03/2018 08:15 pm »
If nozzles point directly backward, then variations in thrust of one engine with respect to another will tend to make the rocket turn a bit.  That  is eliminated if each engine's thrust vector points at the rocket's center of mass.

Also, if the engines are not symmetric about the center of mass (as in the SSME on the Shuttle), then they have to be pointed through the center of mass to prevent an excess torque - even if the thrust is precisely controlled.

Offline gin455res

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Re: Basic Rocket Science Q & A
« Reply #1127 on: 11/07/2018 09:41 pm »
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.


Lets say the first stage is HTP/kerosene then this has a O/F ratio of about 7:1.  Then if all the fuel was carried in the second stage and it was LOX/Kerosene the ratio would be down to about 2.5:1.


If the first stage + 1st stage fuel is 3ish times bigger than the second
and it has O:F of 126:18 (144).


then the upper-stage -1st stage fuel would be about 48 split about 34:14


The upper stage would need to have tanks for  O:F of 34:32 (14+18)


Would the extra weight of the larger fuel tank kill performance or simplify construction having tanks of 126:34:32 (O:O:F), instead of 126:18:34:14?

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #1128 on: 11/10/2018 02:05 pm »
Gravity losses accumulate at the rate g sin γ, where g is, as you guess, the same g that appears in the rocket equation, and γ is the angle between the flight path and the local horizontal.  Some time ago, I put together a comprehensive look at velocity losses in general, attached to this post (warning: it uses a fair amount of math).

Offline cppetrie

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Re: Basic Rocket Science Q & A
« Reply #1129 on: 11/10/2018 04:33 pm »
This post isnít really meant to answer the gravity losses question but rather summarize my understanding of what gravity losses are. So if Iíve got something wrong please point it out.

Orbit isnít about altitude per se but rather velocity. You need an angular velocity such that the centrifugal force is equal to the pull of gravity to essentially nullify it and enter orbit. Given that we have an atmosphere on earth, achieving the required velocity is easier/possible if your altitude places you outside or at least in the very very thinnest parts of the atmosphere. In order to achieve that altitude, you have to spend some of your fuel countering gravity to increase altitude rather than devoting all the fuel to increasing tangential velocity, thus you lose something to gravity. If you had a perfectly spherical body with no atmosphere you could orbit just off the surface given you have the necessary angular velocity, and orbit could be achieved with almost no gravity losses. Is that at least a decent conceptualization of gravity losses?

If Iím understanding the concept correctly the launch trajectory would affect the amount of gravity losses that occur. A more lofted trajectory would have greater losses versus one that pitches down range earlier in the ascent. But there is a sweet spot between fighting atmospheric drag versus gravity losses that would define the ďidealĒ trajectory. Different trajectories from the ideal might be chosen for various reasons (such as?). What other factors in rocket design and launch trajectories affect gravity losses.

I appreciate the feedback and the tremendous resource NSF is for us amateur armchair rocketeers and space nerds.

Offline envy887

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Re: Basic Rocket Science Q & A
« Reply #1130 on: 11/12/2018 02:24 pm »
This post isnít really meant to answer the gravity losses question but rather summarize my understanding of what gravity losses are. So if Iíve got something wrong please point it out.

Orbit isnít about altitude per se but rather velocity. You need an angular velocity such that the centrifugal force is equal to the pull of gravity to essentially nullify it and enter orbit. Given that we have an atmosphere on earth, achieving the required velocity is easier/possible if your altitude places you outside or at least in the very very thinnest parts of the atmosphere. In order to achieve that altitude, you have to spend some of your fuel countering gravity to increase altitude rather than devoting all the fuel to increasing tangential velocity, thus you lose something to gravity. If you had a perfectly spherical body with no atmosphere you could orbit just off the surface given you have the necessary angular velocity, and orbit could be achieved with almost no gravity losses. Is that at least a decent conceptualization of gravity losses?

If Iím understanding the concept correctly the launch trajectory would affect the amount of gravity losses that occur. A more lofted trajectory would have greater losses versus one that pitches down range earlier in the ascent. But there is a sweet spot between fighting atmospheric drag versus gravity losses that would define the ďidealĒ trajectory. Different trajectories from the ideal might be chosen for various reasons (such as?). What other factors in rocket design and launch trajectories affect gravity losses.

I appreciate the feedback and the tremendous resource NSF is for us amateur armchair rocketeers and space nerds.

Even without an atmosphere it would still have gravity losses, due to the need to angle some part of the thrust down to avoid hitting the ground while accelerating up to orbital velocity.

The only way to not have gravity loss is to instantly reach orbital velocity, or to be physically supported (gun barrel, maglev, launch loop, etc) while accelerating. Or to launch horizontally from e.g. a tower with a high enough elevation that the rocket can thrust only tangentially and still reach orbital velocity before hitting the ground.

Offline Tuna-Fish

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Re: Basic Rocket Science Q & A
« Reply #1131 on: 12/03/2018 05:58 pm »
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

Offline gin455res

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Re: Basic Rocket Science Q & A
« Reply #1132 on: 12/03/2018 06:19 pm »
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

The point was to remove the fuel tank from the first stage. So only 3 tanks need to be manufactured not 4. And also the tanks would be more similar in size. Perhaps, this would mean all three tanks could be made the same thickness. One long cylindrical oxidiser tank on the first stage, and 2 smaller more spherical-ish tanks of oxidiser and fuel on the upper-stage, perhaps with common bulkheads.

Offline John-H

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Re: Basic Rocket Science Q & A
« Reply #1133 on: 12/03/2018 10:36 pm »
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

Didn't Atlas essentially do just that?  The first stage was just booster engines, the next stage had the tanks. Of course, the tanks were the lightest possible.

John

Offline meberbs

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Re: Basic Rocket Science Q & A
« Reply #1134 on: 12/04/2018 12:59 am »
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

The point was to remove the fuel tank from the first stage. So only 3 tanks need to be manufactured not 4. And also the tanks would be more similar in size. Perhaps, this would mean all three tanks could be made the same thickness. One long cylindrical oxidiser tank on the first stage, and 2 smaller more spherical-ish tanks of oxidiser and fuel on the upper-stage, perhaps with common bulkheads.
Basic problem is that starting with any typical 2 stage design, there is something like a 10% penalty to payload capacity for extra dry mass on the first stage. For any final stage, the dry mass penalty to payload is 100% by definition. Even ignoring the complexity that a fluid flow across stage separation entails, that means you are guaranteed to lose performance unless the weight added to the upper stage tank is less than 10% of the mass of the tank that was removed from the first stage. This is unlikely to ever work out.

The Atlas example mentioned above is a stage and a half design, so there was no real stage separation, just dropping some of the engines mid flight, but keeping the full tanks all the way to orbit. This required extremely lightweight tanks due to the near SSTO design.

Offline Slarty1080

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Re: Basic Rocket Science Q & A
« Reply #1135 on: 12/22/2018 08:55 am »
I have a question about gravity loss. Am I correct to say that for orbital vehicles there is no gravity loss? If so presumably it follows that the number of engines on an orbital stage should be minimised (excepting the need for redundnacy and thrttling issues)?
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Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #1136 on: 12/22/2018 10:42 am »
Gravity losses accrue at a rate equal to the local acceleration of gravity multiplied by the sine of the angle between the flight path and the local horizontal.  In practice, then, gravity losses are large during the early phases of launch to orbit, when the flight path is nearly vertical, and smaller layer on. It's even possible to have "gravity gains" when a launch vehicle overshoots the target altitude and gains speed as it descends. Gravity losses still exist in orbit, but because they're small, optimal thrust-to-weight ratios are small, as you surmise. E.g., the top stage of an orbital launch vehicle might have a T/W around 0.5.

Suppose you're in a circular orbit and you want to raise the apogee.  An instantaneous delta-V in the direction of motion will raise the apogee without an losses. In reality, though, no delta-V can be instantaneous.  Altitude will tend to increase during the burn, resulting in some gravity loss.  By directing the thrust a little bit inward, you could maintain a constant altitude while accelerating.  That would eliminate gravity loss but introduce steering loss (thrust not aligned with velocity vector).

Very-low-thrust engines, e.g., ion engines, tend to produce substantial losses (though it's a moot point whether the losses are gravity or steering).  The delta-V needed for a very-low-thrust transfer between two circular orbits is just the difference in orbital speeds of the orbits, which is in general quite a bit more than the delta-V needed for a pure Hohmann transfer.
« Last Edit: 12/22/2018 04:05 pm by Proponent »

Offline nicp

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Re: Basic Rocket Science Q & A
« Reply #1137 on: 01/03/2019 12:12 am »
The Titan 1 rocket used an LR-87 engine running kerosene (well RP-1 I guess) and Lox.
Different variants of the engine ran LH2/LOX and (in the Titan 2) Aerozine 50/N2O4.

Ok, these were different engine variants. But _how_ different? And how much work went into redesigning the engine to run on Aerozine 50/N204 vs kerolox?

I can imagine the liquids have different viscosities, densities, etc, so the turbopump will differ - is this the big problem?
Are the thrust chamber or engine bells different? If so I suspect less so...

My question sort of arises from seeing a nice photograph of a Proton and thinking, 'Could you switch this to kerolox?' (EDIT: I'll add - yes I know you now need an ignition source).

Probably a non-starter for good reasons, but Titan (sort of) did it. The other way.

« Last Edit: 01/03/2019 12:15 am by nicp »
For Vectron!

Offline rakaydos

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Re: Basic Rocket Science Q & A
« Reply #1138 on: 04/03/2019 01:18 am »
Hopefully this is the right section. I've encountered this term on the forum, but I never took chemestry in high school. Anyone care to explain?

Online ncb1397

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Re: Basic Rocket Science Q & A
« Reply #1139 on: 04/03/2019 01:25 am »
It is the point on a pressure vs temperature graph where a material can be liquid, solid or gaseous. Sort of how New Mexico, Utah, Colorado and Arizona all converge at one point. At this point, with small adjustments to pressure or temperature, you can change the phase of a substance between all three just as you can change between several states with minor adjustments to latitude and longitude where the four aforementioned states meet.

Wikipedia has a better explanation than what I can come up with:
Quote
In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.[1] It is that temperature and pressure at which the sublimation curve, fusion curve and the vaporisation curve meet. For example, the triple point of mercury occurs at a temperature of −38.83440 įC and a pressure of 0.2 mPa.
https://en.wikipedia.org/wiki/Triple_point



« Last Edit: 04/03/2019 01:29 am by ncb1397 »