Deriving the Equation for Average Water Mass Flow Rate in Water Rocket Launches

In summary, the equation to calculate the average mass flow rate of a water rocket is derived from the application of Bernoulli's equation, which relates pressure, height, and velocity in a fluid flow. The equation is ṁ = A * Cd * √(2ρΔP), where A is the area of the nozzle, Cd is the discharge coefficient, ρ is the density of the fluid, and ΔP is the change in pressure acting on the water. The coefficient Cd is typically less than 1 and takes into account the nozzle shape and other factors. The equation helps to determine the mass flow rate of water for a given rocket design, which can be used to predict the range of the rocket.
  • #1
PhysicsIBStudent
6
0
Hey, I'm a student who has to complete a written assignment for physics. I chose to launch water rockets and investigate the effect of volume on the range. I am trying to also come up with a mathematical model however I am struggling to understand an equation

I found the following equation to calculate the average mass flow rate:

ṁ = A * Cd * √(2ρΔP)
(https://www.palmbeachschools.org/sc/wpcontent/uploads/sites/85/2016/05/Bottle_Rocket_Calculations_Simplified.pdf)

Where:
ṁ is the mass flow rate
A is the area of the nozzle
Cd is the coefficient of drag
p is the density
and ΔP is the average Pressure acting on the water

Does anyone know where this equation is derived from cause I can't find it anywhere?

Thanks in advance for all replies.
 
Physics news on Phys.org
  • #2
This equation starts out from ##\dot{m}=\rho v A##, where v is the exit velocity. Are you familiar with that equation, which is the same as the volumetric flow rate times the density? If the exit velocity were determined by the Bernoulli equation, then it would satisfy:
$$\frac{1}{2}\rho v^2=\Delta P$$##C_d## is a nozzle discharge factor to correct for nozzle shape and non-Bernoulli deviation. It is typically on the order of unity.
 
  • #3
Chestermiller said:
This equation starts out from ##\dot{m}=\rho v A##, where v is the exit velocity. Are you familiar with that equation, which is the same as the volumetric flow rate times the density? If the exit velocity were determined by the Bernoulli equation, then it would satisfy:
$$\frac{1}{2}\rho v^2=\Delta P$$##C_d## is a nozzle discharge factor to correct for nozzle shape and non-Bernoulli deviation. It is typically on the order of unity.

Thanks so much, I now understand. I watched some videos and found this helpful diagram:
bernoul.gif

But I don't see how to get the average pressure(ΔP) from Bernoulli's?
 

Attachments

  • bernoul.gif
    bernoul.gif
    17.7 KB · Views: 1,404
  • #4
PhysicsIBStudent said:
Thanks so much, I now understand. I watched some videos and found this helpful diagram:
View attachment 229734
But I don't see how to get the average pressure(ΔP) from Bernoulli's?
I don’t understand your question.
 
  • #5
Chestermiller said:
I don’t understand your question.
How do I get from this:

upload_2018-8-24_1-57-25.png


to this

upload_2018-8-24_1-57-9.png
 

Attachments

  • upload_2018-8-24_1-56-26.png
    upload_2018-8-24_1-56-26.png
    11.6 KB · Views: 450
  • upload_2018-8-24_1-57-9.png
    upload_2018-8-24_1-57-9.png
    1.5 KB · Views: 1,103
  • upload_2018-8-24_1-57-25.png
    upload_2018-8-24_1-57-25.png
    16.6 KB · Views: 1,079
  • #7
PhysicsIBStudent said:
ṁ = A * Cd * √(2ρΔP)
The equation makes sense except for the Cd term. In many rocketry equations there is a drag term for the air drag on the rocket, but that would not be relevant here since we are concerned only with the speed of the exhaust, not the acceleration of the rocket. So presumably it represents the nozzle drag on the water flow. But in that case the greater the drag the lower the flow rate, which is not what the equation says.
I would have expected ṁ = A * √(2ρΔP/Cd) (though not necessarily the factor 2).
 
  • #8
haruspex said:
The equation makes sense except for the Cd term. In many rocketry equations there is a drag term for the air drag on the rocket, but that would not be relevant here since we are concerned only with the speed of the exhaust, not the acceleration of the rocket. So presumably it represents the nozzle drag on the water flow. But in that case the greater the drag the lower the flow rate, which is not what the equation says.
I would have expected ṁ = A * √(2ρΔP/Cd) (though not necessarily the factor 2).
Cd is the discharge coefficient, related to the nozzle shape (Google discharge coefficient). Typically, Cd is < 1, so it does result in reduced discharge mass flow rate.
 
  • #9
Chestermiller said:
Cd is the discharge coefficient, related to the nozzle shape (Google discharge coefficient). Typically, Cd is < 1, so it does result in reduced discharge mass flow rate.
Ok, thanks.
 
  • #10
Hi guys, pleas can someone run me through exactly how the equation is derived? Up until you end up at
ṁ = A * Cd * √(2ρΔP)
 
  • #11
kelv_01 said:
Hi guys, pleas can someone run me through exactly how the equation is derived? Up until you end up at
ṁ = A * Cd * √(2ρΔP)
Can you answer any of my questions in post #6?
 
  • #12
haruspex said:
Can you answer any of my questions in post #6?
No, unfortunately I cannot. Can you answer mine?
 
  • #13
kelv_01 said:
No, unfortunately I cannot.
Then I'll give you some hints.
We are applying Bernoulli’s equation to the water stream as it goes from a point just inside the rocket to a point just outside.
The first term on each side of the first equation refers to the pressure at a point in the stream flow. What does ##\Delta P## mean in the other equation?
The middle term on each side of the first equation refers to a height at a point in the stream flow. In the actual question, what is the change in height?
The last term each side refers to a velocity. What do we know about the velocities of the water flow at the two points (relative to the rocket)?
 

1. What is average water mass flow rate?

Average water mass flow rate is the amount of water that passes through a certain point in a given period of time. It is usually measured in volume per time, such as liters per second or cubic meters per hour.

2. How is average water mass flow rate calculated?

The average water mass flow rate is calculated by dividing the total volume of water that passes through a point by the total amount of time it takes for the water to pass through. For example, if 100 liters of water passes through a point in 10 seconds, the average water mass flow rate would be 100 liters per 10 seconds, or 10 liters per second.

3. What factors can affect average water mass flow rate?

There are several factors that can affect the average water mass flow rate, including the size and shape of the pipe or channel through which the water is flowing, the viscosity of the water, and any obstructions or changes in elevation along the flow path.

4. How is average water mass flow rate important in scientific research?

Average water mass flow rate is an important measurement in many scientific fields, including hydrology, environmental science, and engineering. It can help researchers understand the movement and distribution of water in natural systems, as well as in man-made systems such as pipelines and irrigation channels.

5. How can average water mass flow rate be measured?

Average water mass flow rate can be measured using various methods, including flow meters, pressure sensors, and velocity sensors. These devices can be installed in pipes or channels to directly measure the flow of water, or they can be used in conjunction with calculations based on the dimensions and properties of the flow path.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
1K
Replies
18
Views
889
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • General Engineering
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top