Bernoulli's Equation and water rockets

In summary, the conversation is about the exhaust velocity of water rockets and how it is affected by the amount of water used. The participants discuss different approaches to measuring the exhaust velocity and the use of Bernoulli's equation in the calculation. They also consider the role of mass in the equation and the importance of accounting for the expansion of gas during the expulsion of water.
  • #1
MigMRF
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So I'm playing around with some water rockets and I'm trying to figure out how fast the exhaust velocity of the water is. I've had an experimental approach using high fps camera to record and analyse (using tracker) the exhaust velocity. I'm using a 0,5 l soda bottle with 0,085 L ; 0,135 ; L 0,185 L and 0,235 L water. So far I've found, that more water means less exhaust velocity (the pressure is exactly the same at 1,8 bar). This makes sense, because the acceleration of the water comes from the energy of the pressurized air.
However I've found, that a lot of people use the Bernoullis equation (rewritten) to calculate the exhaust velocity with the formula:
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However there is no mass involved here?
 
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  • #2
The denominator is density. Mass is 'involved' there.
 
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  • #3
Dullard said:
The denominator is density. Mass is 'involved' there.
Indeed, but then it doesn't matter if there is 50 ml of water or 450 ml of water. This makes no sense to me tbh.
 
  • #4
It doesn't matter, for the velocity calculation.
I suspect that the reason that more water results in lower velocity is because there is less gas. The pressure decays as the water is expelled - that rate of reduction is higher if there is less gas (initially). The initial velocities should be the same.
 
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  • #5
Ahh, thanks! That makes perfectly sense!
I was wondering if there is a good model for pressure as a function of time?
 
  • #6
MigMRF said:
Indeed, but then it doesn't matter if there is 50 ml of water or 450 ml of water. This makes no sense to me tbh.
If you have density and you want mass, you just multiply by volume. If you have velocity and you want volumetric flow rate, multiply by cross sectional area. Like any equation/model, Bernoulli's equation is adaptable, you just have to ask it the question you want it to answer!
 
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  • #7
MigMRF said:
Indeed, but then it doesn't matter if there is 50 ml of water or 450 ml of water. This makes no sense to me tbh.

If you have nine times as much volume you also have nine times as much mass. Think of it this way. When 450 mL of water is expelled you can imagine breaking that stream of ejected fluid into nine equal 50 mL segments. The last segment, for example, is ejected at the same speed as the first segment, provided the pressure difference is the same.
 
  • #8
Mister T said:
provided the pressure difference is the same.
Yes, but this is not the situation for a 1/2 liter bottle initially filled halfway with liquid. The gas pressure (driving the flow out the nozzle) decreases as the liquid is discharged and the gas expands.

You can use Bernoulli to determine the initial nozzle velocity (pretty much pressure head becomes velocity head) but to solve for the time-dependent velocity you have to also model the expansion of the gas.
 
  • #9
For the gas expansion, I'd just model it as adiabatic expansion, so for air, p*v^1.4 = constant. You can get dv/dt from the exhaust velocity and nozzle diameter, so from that, you should be able to figure out the pressure (and velocity) at all points during the expulsion of the water.
 
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  • #10
cjl said:
For the gas expansion, I'd just model it as adiabatic expansion, so for air, p*v^1.4 = constant. You can get dv/dt from the exhaust velocity and nozzle diameter, so from that, you should be able to figure out the pressure (and velocity) at all points during the expulsion of the water.
Yeah, that might actually be a solution. Just to make sure, when you write dv/dt are you saying v as velocity or do you mean V as in volume. It would make more sense if the change in volume with respect to time is related to the mass flow rate
 
  • #11
gmax137 said:
Yes, but this is not the situation for a 1/2 liter bottle initially filled halfway with liquid.

You introduced the equation ##v_{ex}=\sqrt{\frac{2(p_{in}-p_{out})}{\rho_w}}## into this discussion, and wanted to understand the role played by mass. I tried to explain that to you.

The fact that that ##p_{in}## is not constant in your particular application of that equation has no bearing on the explanation. In other words, try to understand the role played by mass when ##p_{in}## is constant so that you can then understand the more complicated situation when it's not constant.
 
  • #12
MigMRF said:
Yeah, that might actually be a solution. Just to make sure, when you write dv/dt are you saying v as velocity or do you mean V as in volume. It would make more sense if the change in volume with respect to time is related to the mass flow rate
I meant volume, but I can see how that notation was unclear. You can also of course obtain velocity, but you also have to account for the fact that the mass is non-constant.
 
  • #13
Mister T said:
You introduced the equation ##v_{ex}=\sqrt{\frac{2(p_{in}-p_{out})}{\rho_w}}## into this discussion, and wanted to understand the role played by mass. I tried to explain that to you.

The fact that that ##p_{in}## is not constant in your particular application of that equation has no bearing on the explanation. In other words, try to understand the role played by mass when ##p_{in}## is constant so that you can then understand the more complicated situation when it's not constant.
I think you have me confused with the OP.
 

1. What is Bernoulli's Equation and how does it relate to water rockets?

Bernoulli's Equation is a fundamental principle in fluid dynamics that describes the relationship between pressure, velocity, and elevation in a fluid. It states that as the velocity of a fluid increases, the pressure decreases, and vice versa. This equation is important in understanding the physics behind water rockets, as it explains how the rocket is able to launch into the air by using pressurized water and air to create thrust.

2. How does the design of a water rocket impact its flight?

The design of a water rocket can greatly impact its flight. Factors such as the shape and size of the bottle, the amount of water and air inside, and the placement of fins and a nose cone can all affect the rocket's stability, speed, and trajectory. A well-designed water rocket will have a streamlined shape, a balanced weight distribution, and proper fins and nose cone for stability and control.

3. How does air pressure affect the launch of a water rocket?

Air pressure plays a crucial role in the launch of a water rocket. As the air inside the rocket is compressed, it creates a higher pressure than the surrounding air. When the rocket is released, this high-pressure air rushes out of the nozzle, creating thrust that propels the rocket into the air. The higher the air pressure inside the rocket, the more forceful the launch will be.

4. Can Bernoulli's Equation be used to predict the flight of a water rocket?

Bernoulli's Equation can be used to make predictions about the flight of a water rocket, but it is not the only factor that affects the rocket's trajectory. Other variables such as wind, air resistance, and the shape of the rocket can also impact its flight. However, understanding Bernoulli's Equation can help in designing and optimizing a water rocket for better performance.

5. How can we use Bernoulli's Equation to improve the efficiency of a water rocket?

By applying Bernoulli's Equation, we can make adjustments to the design of a water rocket to improve its efficiency. For example, increasing the velocity of the air and water inside the rocket by using a longer nozzle or a smaller opening at the top can increase the thrust and therefore the efficiency of the rocket. Additionally, reducing air resistance by making the rocket more streamlined can also improve its efficiency.

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