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

AI Thread Summary
The discussion focuses on deriving the equation for average mass flow rate in water rocket launches, specifically ṁ = A * Cd * √(2ρΔP). The equation is based on the mass flow rate formula, where exit velocity is determined by Bernoulli's principle, linking pressure difference (ΔP) to velocity. Participants clarify that Cd is a discharge coefficient accounting for nozzle shape and drag, which can affect flow rate. There is confusion about how to calculate average pressure from Bernoulli's equation and the relationship between drag and flow rate. Overall, the conversation emphasizes understanding the physics behind water rocket propulsion and the mathematical modeling involved.
PhysicsIBStudent
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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.
 
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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.
 
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?
 

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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.
 
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
 

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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).
 
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.
 
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)?
 
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