Accelerating Control Volume - Fluid Momentum Example Problem

Don't use the Tsiolkovsky Rocket Equation! It doesn't account for external forces like gravity or drag. You must use Fluid Momentum to solve accelerating control volume problems.
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Draw your Free Body Diagram, set it equal to your Kinetic Diagram, write your equilibrium equation, and solve. That's every momentum problem in your Fluid Mechanics course. Fluid Momentum problems are Force problems. Fluid Momentum Acceleration problems are probably the hardest type of fluid mechanics momentum problems you'll see, since they may involve calculus. But, the accelerating cv doesn't actually change how you start, only the difficulty of the math at the end.
CHAPTERS
0:00 Tsiolkovsky Rocket Equation
0:56 Free Body Diagram = Kinetic Diagram
2:29 Equilibrium Equation
4:17 Set up the Integrals
6:38 Solve the Integrals
9:56 How to check final answer

[@torusx8564]

Very nice video, that problem wasn't easy ngl. But great job because I did understand everything. I wish you had WAYY more subscribers

[@BrianBernardEngineering]

Good news, is that this is one of the harder momentum problems you are likely to see, so if you can follow this, good chance your homework and tests might be similar or even a step down in difficulty.

[@JacobRen]

Amazing! So surprised to see that only 5 like of this video😅

[@BrianBernardEngineering]

we're up to 6 now, Let's goooooooooo

[@umarshafi5414]

4:08. Since the value of g does not depends upon the change in velocity and change in time, then why are we considering it as rate of change of velocity (or simply time derivative of velocity) ? The value of g varies as we move away from earth, or we can say it varies inversely by square of distance b/w object and radius of earth (i.e g=[G.M]/[r^{2}] or g=[G.M]/[(r+h)^{2}].? Any help whould be greatly appreciated..

[@BrianBernardEngineering]

g is not being considered as rate of change of velocity. If you let the video play a few seconds more, you'll see that the right hand side of the equation has another term in nit. And think of the overall problem, if g = acceleration, then the rocket would be in free fall. Since the rocket is accelerating upwards, not freely downwards, dv/dt will definitely not equal g. However, you are correct that technically g is not constant. This is why I needed to specifically state in my assumptions at the top that g = constant. To realistically model the rocket's motion all the way up to a zero g environment, you would definitely need to have g rewritten as a function of height.

[@aqastiwar]

Much appreciated. I have a question, i am given an assignment in which i am required to find miminimum fuel rate and no time interval is given. What should i do ? I think that i have to derivate the equation and derivative the minima, is it correct ?

[@BrianBernardEngineering]

It may be easier than that. Consider the implications of minimum fuel rate. I can think of 2. First, your fuel tanks would be empty and you only account for the weight of the rocket itself. If you have extra fuel, that requires a higher fuel rate just to lift the fuel itself. Second, your acceleration would be 0, or something incredibly small like .00001, like the rocket is just barely able to overcome gravity. Any faster fuel burn rate than this would cause a faster acceleration. I think you may be able to just do the same FBD = kinetic diagram type analysis, where the force due to burning fuel only has to exactly counter the force of weight of the rocket, and all the other terms are assumed to be zero, in this minimum fuel rate scenario.

[@LeandroOliveira-hs6su]

I firstly aplaud you for the detailed and clarifying explanations in the solution of this classical problem. However, it still bothers me that this solution became a classical one even though it does not consider the relative velocity (Ve - V) when dealing with the net momentum that is crossing the control surface at the rocket nozzle in relation to the reference frame that is moving with the control volume.

[@BrianBernardEngineering]

I think Ve is already a relative velocity in this problem. For a rocket, the particles emitted for thrust would always have their velocity measured relative to the rocket itself, not with respect to ground. Sort of like the experiment where a car drives 1 direction, and you fire a bullet from a gun out of the car shooting backwards. If the speed of the bullet and the speed of the car are the same, then a video camera on the ground would show the bullet just dropping straight down, not moving backwards or forwards at all. Same thing here, the particles exiting the rocket are moving very fast backwards relative to the rocket, but relative to the ground, they might be stationary, or might even still be traveling forwards, just not as fast forward as the ship.

[@LeandroOliveira-hs6su]

@@BrianBernardEngineering My reasoning is the following. If the rocket burn the fuel at a constant rate, then the velocity at which the gases scape the rocket through the nozzle (lets call it Vj) should be constant in regard to a reference frame moving with the rocket. However, the reference frame itself is accelerating with a velocity U = U(t) and the velocity of the gases at the nozzle exit is constant in time. Therefore, the relative velocity is (Vj - U(t)), which is not constant and should be dealt with in the integral of dU.