Author Topic: Basic Rocket Science Q & A  (Read 370086 times)

Offline Proponent

  • Senior Member
  • *****
  • Posts: 5587
  • Liked: 1130
  • Likes Given: 682
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

  • Senior Member
  • *****
  • Posts: 5587
  • Liked: 1130
  • Likes Given: 682
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 Kosmos2001

  • Full Member
  • *
  • Posts: 176
  • CAT
  • Liked: 58
  • Likes Given: 127
Re: Basic Rocket Science Q & A
« Reply #1122 on: 09/20/2018 12:26 PM »
Why some booster nozzles are pointing not vertically but with some angle outside?

Offline Proponent

  • Senior Member
  • *****
  • Posts: 5587
  • Liked: 1130
  • Likes Given: 682
Re: Basic Rocket Science Q & A
« Reply #1123 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.

Online envy887

  • Senior Member
  • *****
  • Posts: 4721
  • Liked: 2624
  • Likes Given: 1410
Re: Basic Rocket Science Q & A
« Reply #1124 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

  • Full Member
  • ***
  • Posts: 367
  • eny, sed, woz, shuga English spelling is messed up
  • bristol, uk
  • Liked: 11
  • Likes Given: 25
Re: Basic Rocket Science Q & A
« Reply #1125 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

  • Senior Member
  • *****
  • Posts: 5587
  • Liked: 1130
  • Likes Given: 682
Re: Basic Rocket Science Q & A
« Reply #1126 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).

Online cppetrie

  • Full Member
  • ****
  • Posts: 620
  • Liked: 352
  • Likes Given: 3
Re: Basic Rocket Science Q & A
« Reply #1127 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.

Online envy887

  • Senior Member
  • *****
  • Posts: 4721
  • Liked: 2624
  • Likes Given: 1410
Re: Basic Rocket Science Q & A
« Reply #1128 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.