RocketMan on 11/6/2010 at 18:42
Hey all. I'm writing rocket simulation software and I want to add a special optimization module that requires an energy analysis of thrust vs drag on a rocket. I'm having a problem with my understanding of the formulas:
Volumetric flow of a fluid is cross sectional area X velocity Q = VA (m^3 / sec). Multiply this with density (kg/m^3) and the "m^3" terms cancel leaving kg/sec, which is a mass flow rate for the fluid so now we have mdot = ro x V x A. Now from here I have a problem. If I integrate the function with respect to V, I get the well known drag formula F = 1/2 x ro x A x V^2 but if I simply multiply mdot by another V term I get kg/sec x m/sec which is the same as kg x m/sec^2.....the latter term otherwise known as acceleration. This means we have kg (Mass) x m/sec^2 (accel) = ma = F but the euation is now ro x A x V^2, same as the integrated version but without the 1/2. Both are expressed in units of force...but this can't be because they're different equations.
That's my first problem. My second problem is that actually I'm interested in fluid ENERGY, not force. Now if energy is work and work is F x s (where s is displacement) and power is work/time which is F x s / t where (s/t) is none other than velocity, then this implies that whatever the true force function is (lets go with the well known 1/2 x ro x A x V^2) then fluid power should be 1/2 x ro x A x V^3. This is pure speculation though and I can't find any reference to this at all.
Bottom line is, I need to know what the equation is for fluid power or work is because (and this is key) I need 2 equations: one representing rocket power (derived from the rocket propellant's total impulse) and one representing the work or power of the air incident on the rocket which robs the rocket of it's power. BOTH equations need to have m and V terms in them so that I can perform a curve analysis to find the optimal intersection point where the losses are minimized. This can only be done if my equations depend on the same variables and if they are both energy equations (comparing apples to apples). So far the rocket part is easy. Impulse is known which gives an m dV relationship that is fixed...I can calculate V and thus the kinetic energy of the rocket. I just need the corresponding drag equation to figure out what the air's stopping energy is.
Comments?
Kolya on 11/6/2010 at 22:11
I'd love to say it's not rocket science, but it all sounds Greek to me.
So here's a video of the Greek rocket war. I think you'll like it R-Man.
<embed src="http://www.youtube.com/v/_PijfPZx88I&hd=1&fs=1&showinfo=0&rel=0" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"/>
suliman on 11/6/2010 at 23:43
Now, I don't know much about fluid dynamics, but what are you trying to find by multiplying mdot by V? a=dV/dt, so dV=a*dt, and you use that to integrate mdot*a dt with respect to time and(assuming the acceleration is constant in time?) you get the equation ma=F=1/2 r0AV^2. This may be a fluid dynamics thing, but I really don't see the logic in multiplying both sides by V. It may give you a correct expression(are you sure you're not assuming V to be a constant in time when you're doing it, though?), but it won't give you the one you're looking for.
Second problem, power is the time derivative of energy, so I don't think you can just divide by t. If F is dependant on time(and it most likely is, because it depends on V), you should take a time derivative of F*s.
I'm in no way an authority in this stuff, and It's also 3am and some parts of my brain are probably already asleep, so make sure I'm not horribly wrong here before you do any serious rocket stuff:)
RocketMan on 12/6/2010 at 00:24
I think you found my calculus error on the first part....I hadn't considered the time-domain nature of the equation, which is pretty careless of me. However I think I failed to mention that big A is area and small a is not explicitly part of the equation but comes from the addition of the seemingly arbitrary V. You have m/s from the V and you have /s that's mixed in with the equation you're multiplying the V with and together they become m/s/s or m/s^2 which is little a. You might be on to something though regarding the integration in the time domain so I'll have to check that.
As far as the energy equation however, dividing work by t should be ok since taking the time derivative of s gives ds/dt (which is V) but in this case s/t is just the same exact thing (no calculus needed). In physics it is already established that any linearly moving mass under the influence of a force is being driven with power P=F dV. The question is, can it be applied to my Drag equation as I've outlined? I have a feeling I've missed something. As you pointed out, my derivation of the drag equation was logically flawed....maybe I'm going about the power equation the wrong way too.
suliman on 12/6/2010 at 00:41
Well, isn't the P=FdV thing also assuming that F is a constant in time? According to the Drag equation, F goes like V squared, and V is a*t. If F is a constant in time then d/dt (F*s) really is F*ds/dt, but if F changes in time then you have to take that into account when deriving and do P(t)=d/dt[F(t)*s(t)]=Fdot*s(t)+F(t)*sdot. At least, that's the way I'd do it.
RocketMan on 12/6/2010 at 01:58
Once again, you're right. Problem here is that I can't solve for F empirically in the time domain. My rocket equation has so many interdependencies that I've resorted to numerical methods to calculate all of the performance characteristics during flight. This module I'm writing now is supposed to be a footnote to more than 15 pages of math contained in a java class. The idea is not to get an exact solution that can be expressed as a time dependent function but simply to compare 2 simple relations and determine the intersection point....IF I were able to express F properly in terms of t, I don't think both of us together would be able to find the derivative/integral without a computer.
The scope of my efforts is essentially this: I know total impulse for my rocket which is I = m dV = constant so for every conceivable value of m there is a corresponding theoretical maximum delta V achievable by the rocket. This basic relation generates all my coordinate pairs...for every m there is a known V. Now, I want to find the highest ratio of rocket energy : energy losses which is basically rocket kinetic energy compared to fluid kinetic energy subtracted during flight. The flight is time dependent but I'm only examining a snapshot of the flight because doing anything more makes this far too complicated and it's not necessary to come up with an optimal mass that's usable. So I pick an operating point which makes sense and I won't go into detail on that since it requires substantial background in rocketry. Then I plug in all the parameters from that operating point that are not m or V and then loop through all values of m and V until the energy ratio is maximized in favour of the rocket. The rocket Ek is 1/2 mV^2...beautiful since it contains m and V. The fluid loss equation, which is what this thread is all about, will probably not contain a mass term but that's ok as long as it has a V term (which is implicitly linked to m). Just need to find "an" energy formula that is simple enough and is correct.
If F = 1/2 ro x A x V^2 is force, then I can't think of an easy way to transform this into energy but lets think of this a different way. Instead of focussing on the fluid lets focus on the force imparted by the fluid as a mere reaction force on the rocket. If we solve the F equation without worrying about energy we know what the reaction force is and if we use the rocket's current velocity (negative this time) and do F x -V[rocket] would that not be the power lost by the rocket due to the drag force?
@Kolya
WTF??! Firing homemade rockets at a church at night while there's a mass inside?....and this is a festival? Oh bless you Kolya for making my day a happy one :D
suliman on 15/6/2010 at 07:38
..okay, here's an idea. The speed of fluid going through a certain part of the tube (dx) should be constant at any given time. You find V(x) (probably have to solve a differential equation for that), and assuming there's no flow in the radial direction you integrate over the length of the tube l, multiply that by the amount of time you need, square it and multiply it by m/2 when m is the amount of fuel your tube can hold at a given moment. Seems logical enough, the only problem is the differential equations.
