0
$\begingroup$

I'm trying to understand rockets in a vacuum.

As I understand it the, Tsiolkovsky rocket equation gives me the change in velocity of a rocket + payload after expelling a certain amount of fuel.

But how does acceleration fit into this equation? Acceleration is the change in velocity divided by the change in time. But I have nothing in the Tsiolkovsky equation to measure the time the burn took, only how much delta v it gave me.

Does the Effective Exhaust Velocity tell me how long it took for it to burn that amount of mass? Because this page here http://en.wikipedia.org/wiki/Specific_impulse#Specific_impulse_as_a_speed_.28effective_exhaust_velocity.29 gives me an equation to convert the Effective Exhaust Velocity into Specific Impulse as time.

However, this equation is using the gravity of earth, whereas I want to assume there's no gravitational forces at play. I can't just remove it because then they're the same value.

And if I had the time in here, how does it relate back to the change in mass from the propellant being used?

So at the end of the day, how to you determine:

  1. How much fuel a rocket burns in a second? (is this the effective exhaust velocity, or is there a relationship?)
  2. What's the acceleration of a rocket burning this mass of fuel for a second?

I understand that the changing mass comes into play; but I can't figure it out.

This question is for interests sake; no real reason.

$\endgroup$
0
$\begingroup$
  1. Most rocket variants, other than solid rockets (and even some of those) have throttles or some other means of controlling flow/burn rates. Some also have variable Propellant Nozzles, such that there is no one set burn rate. You could perhaps use 'full throttle' or 'max flow rate' if you like, but you have to make that caveat.

  2. No, burn rate is not equivalent to effective exhaust velocity. You can get a rough idea of typical exhaust velocities if you know what you're burning (Kerosene/LOX, Perchlorate/LOX, Hybrid HTPB/N2O, etc...), the rocket velocity, the atmospheric conditions (I.E. earth altitude, vacuum, etc.) and the flow characteristics of the Propellant Nozzle at those conditions. For a rough estimate of Kerosene/LOX, you could start with 4.4km/s.

The velocity of an exhaust stream after reduction by effects such as friction, non-axially directed flow, and pressure differences between the inside of the rocket and its surroundings. The effective exhaust velocity is one of two factors determining the thrust, or accelerating force, that a rocket can develop, the other factor being the quantity of reaction mass expelled from the rocket in unit time. In most cases, the effective exhaust velocity is close to the actual exhaust velocity.

Effective Exhaust Velocity

  1. For #2 you are looking for Specific Impulse.

Specific impulse (usually abbreviated Isp) is a measure of the efficiency of rocket and jet engines. It represents the force with respect to the amount of propellant used per unit time. If the "amount" of propellant is given in terms of mass (such as in kilograms), then specific impulse has units of velocity. If it is given in terms of weight (such as in kiloponds or newtons), then specific impulse has units of time (seconds). The conversion constant between these two versions is thus essentially "gravity" (more specifically g0). The higher the specific impulse, the lower the propellant flow rate required for a given thrust, and in the case of a rocket the less propellant needed for a given delta-v per the Tsiolkovsky rocket equation.

Wikipedia: Specific Impulse

Now, for what you're after, with a few assumptions (constant acceleration) you can just use the Tsiolkovski + time + Isp of the engine.

Example: I have 4000 kg of a fuel in a 8000 kg fully fueled rocket in space, the fuel is of a certain composition such that through my nozzle design at a vacuum it exits my craft at effectively 4.0km/s ($I_{sp}$), and at full throttle it would burn all it's fuel in 10 seconds. Using the Tsiolkovski:

$ \Delta V = v_e * ln(\frac {m_0} {m_1} ) $

$ \Delta V = 4.0 km/s * ln (\frac {8000} {4000} ) $

$ \Delta V = 4.0 km/s * ln (2) $

$ \Delta V = 4.0 km/s * 0.693... = 2.77 km/s $

This change in velocity was made over 10 seconds, so

$ a = \frac {\Delta V\ km/s} {10\ s} = \frac { 2.77\ km/s } {10\ s} = 0.277\ km/s^2 = 277\ m/s^2 $

$\endgroup$
  • $\begingroup$ thanks a heap, that answers my question completely. And thank you for the additional info on Rockets; I didn't realise there were throttles. $\endgroup$ – NeomerArcana Jan 13 '15 at 21:19
  • $\begingroup$ Well I doubt 'throttle' would be the appropriate name for them (they don't 'have' throttles), but yes they have mechanisms by which they can throttle (verb) their thrust. For example, liquid bi-propellant uses pumps to mix fuel with liquid oxygen, and by varying the flow rate they regulate thrust - the SSME is engaged at low thrust during solid rocket boosting for example. In hybrid rockets, they can regulate gas flow rates, whereas some solid propellant rockets (like boosters) have no method of regulating thrust or reignition - they're full blast until out of fuel. $\endgroup$ – Ehryk Jan 13 '15 at 22:22
  • $\begingroup$ I caveat the above comment to say that there are likely counterexamples to all of those, methods by which solid rockets can be throttled, bi-propellant setups that cannot be, etc. Estes model rocket engines are examples of full thrust 'til empty. $\endgroup$ – Ehryk Jan 13 '15 at 22:24
  • $\begingroup$ A good read to get a better understanding of this is xkcd's rocket golf: what-if.xkcd.com/85 . What's important to note is that the only way to change your velocity (accelerate) is to throw/push/explode/golf something the opposite way. The speed at which it leaves you is $I_{sp}$, quantity of mass you hit/move/throw/burn per second is your flow rate - in my example above, 4,000kg of fuel was used in 10s, so 400kg/s was the burn rate. $\endgroup$ – Ehryk Jan 13 '15 at 22:28
  • $\begingroup$ Once again thanks for your edit with the additional equations. Just a question though, it's seems unintuitive that the delta v goes down if my fuel accounts for more of the total mass? I mean, if more of my total mass is fuel, wouldn't burning it away result in a higher delta V rather than a lower one? (or did I misinterpret your "4000kg of fuel in an 8000kg fully fueled rocked"?) $\endgroup$ – NeomerArcana Jan 14 '15 at 10:17
0
$\begingroup$

Although a previous answer has been selected I think the asker needs a bit more info to understand what sort of questions you should be asking about a rocket. In particular, the Tsiolkovsky rocket equation is only ever used to compute a required mass ratio or delta-V. It's essentially a staging tool; given some total delta-V what are the number of burns and stages I need at what MR? What number of burns or stages would best fit the problem? Is this possible given constraints on payload and inert masses? It should not be used for any sort of real flight dynamics including computing vehicle acceleration, which in an of itself is not typically something people worry too much about. There are some simple models for flight dynamics to consider things like g-t losses or basic aerodynamic effects, but when baselining a rocket you would incorporate all of those into a rough efficiency--say you will have a 10-20% penalty in propellant mass.

The basic rocket relationships which hold regardless of the actual model of the rocket flow are:

(1) Tsiolkovsky equation: $MR = e^{\Delta V/c}$

where $c$ is the effective exhaust velocity, viz. the velocity of the propellant gases after pressure losses have been taken into account. This is used for staging, delta-V, and mass ratio calculations.

(2) Basic Thrust Equation: $F = \dot{m}c$.

This equation is used to determine what the thrust the rocket will produce is given $c$, since most ways of analyzing a rocket return $c$ rather than $F$ (thrust). Thrust is more important to consider in whether a rocket is capable of executing a certain maneuver rather than whether the rocket itself is a good design. High thrust buy low Isp is almost always worse than slightly lower thrust at higher Isp.

(3) Constant Burn Rate Assumption: $m_b = \dot{m}\Delta t$

where $m_b$ is the mass of the burned propellant. This is less of a fundamental result than an assumption. We assume a constant burn rate most of the time, since this is easy to work with from an engineering perspective.

You mention specific impulse, which is formally defined as thrust per unit weight of mass expended, but as you can see from (2) this isn't actually a different parameter from the effective exhaust velocity, e.g.

(4) $I_{sp} = c/g$

Where $g= 9.81$ m/s$^2$, viz. the acceleration near earth, and is only in there because of the definition of $I_{sp}$.

Contrary to the previous answer it is not enough to know simply what your propellants are to determine $c$ or $I_{sp}$--then rocketry would be stupid easy! In general you need the type of rocket (e.g. solid, liquid monopropellant, liquid bipropellant, etc.), type of propellants, mixture ratio of the propellants, nozzle geometry (especially the expansion and contraction ratios), and thermochemical states of the propellants in the combustion chamber (which leads into engine cycles, injector theory, chemical kinetics of the propellant flow and a whole host of other topics). This doesn't even begin to touch on loss mechanisms for a non-ideal rocket, most of which affect $c$ and $I_{sp}$ or whether or not real rocket could work for a given design (e.g. cooling methods, structural integrity, combustion stability). A good place to start if you want to do a first order estimate of $c$ is by selecting some propellants, running a thermochemical code, like NASA's CEA code, and use the results in an isentropic analysis (see a book like Sutton, Rocket Propulsion Elements for more detail). This gives you first order estimates of $c$, and even $\dot{m}$ if you learn how to use $c^*$ velocities correctly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.