Unsteady flow of water being discharged from a tank

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Main Question or Discussion Point

I am currently modeling the altitude that a water rocket can attain. In order to do this i am first modeling the the thrust phase.

Due to the water mass decreasing with time (the same goes for the pressure - it decreases over time as the water is expelled out of the nozzle).

This is an unsteady drainage problem in which the bottle is filled with water and pressurized to around 6 Bar. I cannot use Bernoulli's equation as this only describes a steady flow process. The same applies for Torricelli's v = sqrt(2gh).

Does anyone know how i go about modeling this unsteady flow problem?
 

Answers and Replies

  • #2
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You'll need some approximations. Neglecting losses in the system and using energy conservation to find the instantaneous exit velocity should give a reasonable approximation. Then simulate it step by step in time.
 
  • #3
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Can you tell us more about the dimensions of the bottle and the fractional fill of water and gas?
 
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The bottle is a standard 2L coke bottle (as restricted to in my mini-thesis project). The diameter is constantly changing, but for the sake of easy explaining i can use MatLab later to iterate each step at a different height. I am not sure of the ratio of water to air yet, as i was first wanting to create the model and then use an optimization approach to determine what the best ratio would be in terms of thrust.

I saw an answer you left 2 years ago (URL below) where you replied to someone about an unsteady tank draining problem. I think the only difference there is obviously the area changing and the pressurized air.
https://physics.stackexchange.com/questions/260829/tank-draining-derivation-is-it-correct-to-use-bernoulli-and-continuity-equation/423973#423973

Any help will be greatly appreciated and reduce a lot of stress.
 
  • #5
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Can you tell us more about the dimensions of the bottle and the fractional fill of water and gas?
I commented instead of replying, so just making sure you get notified.
 
  • #6
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Since you have gauge pressures on the order of bars, you don't have to account for hydrostatic pressure variations. So Bernoulli gives you $$v=\sqrt{\frac{2p}{\rho}}$$where p is the gauge pressure. Bernoulli can be used directly because the transient variation in velocity at a given location in the bottle is going to be negligible compared to the acceleration of the fluid in approaching the bottle opening (which is much smaller in cross section than the main bottle diameter). So the flow approaching the opening can be considered quasi static. Anyway, this is the approximation I would use, at least to get started.
 
  • #7
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Since you have gauge pressures on the order of bars, you don't have to account for hydrostatic pressure variations. So Bernoulli gives you $$v=\sqrt{\frac{2p}{\rho}}$$where p is the gauge pressure. Bernoulli can be used directly because the transient variation in velocity at a given location in the bottle is going to be negligible compared to the acceleration of the fluid in approaching the bottle opening (which is much smaller in cross section than the main bottle diameter). So the flow approaching the opening can be considered quasi static. Anyway, this is the approximation I would use, at least to get started.
I agree completely that the variation in velocity at a point in the bottle is considered negligible compared to the acceleration of the fluid approaching the nozzle. This is in fact what I did in my previous model. However, when I took this to my Professor, he did not appreciate any "assumptions and short cuts". This is most likely a case of 'killing an ant with a bazooka' - yet i have no choice but to model it considering a fully transient flow.

Currently I am working with what you wrote out with the link above (the unsteady tank drainage problem) and trying to add to equation 8 the rate at which work from pressure is being done on the water. Would this be correct in the difference between that problem and mine? Then I will just run the process at certain y-intervals with the change in area.

Sorry for the inconvenience of not being able to use a quasi-static approach. The accuracy of transient flow for this approach does not seem worth the complexity of modeling it as transient, but sadly it must be done.
 
  • #8
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If your professor doesn't want you to use any approximations and shortcuts, then you better follow the motion of each and every molecule, and include the gravitational pull of Jupiter (to say nothing of viscous effects).

If you want to include the transience in your problem, for a cylinder like a 2L bottle, you can write $$A\frac{dp}{dz}=\rho A \frac{dv}{dt}$$ for the differential force balance on the portion of liquid up to the convergence (where v is the lower axial velocity in the larger area region). So, $$p_{hi}-p_{conv}=\rho h\frac{dv}{dt}=\rho h\frac{d^2h}{dt^2}$$where h is the distance from the gas surface to the convergence. From the convergence on, $$p_{conv}=\frac{1}{2}\rho (V^2-v^2)$$where V is the velocity out the exit hole: ##V=vA/A_x=\frac{A}{A_x}\frac{dh}{dt}##.

This neglects the transient in the converging section.

If your professor wants it more accurate than this, I suggest using computational fluid dynamics.
 
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  • #9
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If your professor doesn't want you to use any approximations and shortcuts, then you better follow the motion of each and every molecule, and include the gravitational pull of Jupiter (to say nothing of viscous effects).

If you want to include the transience in your problem, for a cylinder like a 2L bottle, you can write $$A\frac{dp}{dz}=\rho A \frac{dv}{dt}$$ for the differential force balance on the portion of liquid up to the convergence (where v is the lower axial velocity in the larger area region). So, $$p_{hi}-p_{conv}=\rho h\frac{dv}{dt}=\rho h\frac{d^2h}{dt^2}$$where h is the distance from the gas surface to the convergence. From the convergence on, $$p_{conv}=\frac{1}{2}\rho (V^2-v^2)$$where V is the velocity out the exit hole: ##V=vA/A_x=\frac{A}{A_x}\frac{dh}{dt}##.

This neglects the transient in the converging section.

If your professor wants it more accurate than this, I suggest using computational fluid dynamics.
Thank you so much for your help, it is seriously appreciated.

I only have one more question and then I should be sorted. I am currently setting up a transient energy balance on the entire inner bottle control volume. So far i have:
The rate at which gravitational work is being done on the water + the rate at which the work from pressure is being done on the water = the rate of change of kinetic energy of water + the rate of kinetic energy leaving the bottle.

Firstly, is the above statement accurate? If so, for the change in pressure you wrote above (considering the water until it reaches convergence), how would i write it in terms of energy?

Sadly, I am not allowed to model it in CFD. It would have been really interesting to compare to my theoretical model.

Again, thank you for your help. You will undoubtedly be acknowledged in my Thesis.
 
  • #10
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Thank you so much for your help, it is seriously appreciated.

I only have one more question and then I should be sorted. I am currently setting up a transient energy balance on the entire inner bottle control volume. So far i have:
The rate at which gravitational work is being done on the water + the rate at which the work from pressure is being done on the water = the rate of change of kinetic energy of water + the rate of kinetic energy leaving the bottle.

Firstly, is the above statement accurate? If so, for the change in pressure you wrote above (considering the water until it reaches convergence), how would i write it in terms of energy?

Sadly, I am not allowed to model it in CFD. It would have been really interesting to compare to my theoretical model.

Again, thank you for your help. You will undoubtedly be acknowledged in my Thesis.
I think your energy balance looks OK. I think it should lead to basically the same equation I presented. In my judgment, there is no need to consider gravitational work because of the high pressures involved, which would be sufficient to support a column of water 60 m high.
 

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