How is mass flow rate a force?

  • Thread starter Nic sign
  • Start date
242083
Lets say if I have a pipe with one end closed, and drill a small hole on the side of the pipe near the closed end. Then I use vacuum at the open end to create mass flow exiting the pipe. Where is the opposite force on the pipe?

[crackpot youtube video deleted by moderator @russ_watters ]
 
Last edited by a moderator:

russ_watters

Mentor
18,385
4,636
How is mass flow rate a force?
[title question]
Obviously mass flow rate is not a force. The units don't match.
Lets say if I have a pipe with one end closed, and drill a small hole on the side of the pipe near the closed end. Then I use vacuum at the open end to create mass flow exiting the pipe. Where is the opposite force on the pipe?

[crackpot youtube video deleted by moderator @russ_watters ]
The geometry of what you are saying isn't totally clear to me, so I'll just say the opposing force is the force (/pressure) against the part of the pipe opposite the hole.

Are you the author of the video I deleted?
 
Obviously mass flow rate is not a force. The units don't match.

The geometry of what you are saying isn't totally clear to me, so I'll just say the opposing force is the force (/pressure) against the part of the pipe opposite the hole.

Are you the author of the video I deleted?
So you are saying there is force in the pipe?
 
205
57
So you are saying there is force in the pipe?
The (mass/sec) x velocity is called thrust (in certain circumstances) and is a force. So what exactly are you asking??
 

russ_watters

Mentor
18,385
4,636
So you are saying there is force in the pipe?
In? On? Again, it isn't clear to me what the setup is. I'm thinking I may have misunderstood; if you are drawing a vacuum in the pipe and air is therefore moving in through the little hole and not out, the force is on the outside, not the inside.....but this isn't related to a rocket, so the description and therefore point is confusing. As @hutchphd says, it would be better if you just made your point rather than trying to ask leading or misleading questions.

It's also worth pointing out here that you didn't answer my other question and on a related note, PF is a place for discussing mainstream physics, not personal theories, much less conspiracy theories. If this is a lead-in to a conspiracy theory, it will be locked or deleted.
 
Last edited:

jim hardy

Science Advisor
Gold Member
2018 Award
9,751
4,750
here's an explanation i used some years ago to make it intuitive.

When a user can make Bernoulli and Nasa's formulas give the same answer as this intuitive approach
he understands both the concept and the math.

But he's gotta start simple.



How I wish I were good with graphics.

Here is a simple thought experiment that will make it intuitive.....

Imagine a very un-streamlined rocket out in space. With no atmosphere there it need not be streamlined, does it?
So in your mind make it a perfectly sealed and rigid cube one foot by one foot by one foot. (I use English units because i'm both old and in USA, use metric if you like and make it a 1 meter cube for same logic will apply).

Now internally pressurize your rocket to 1 PSI.
What are forces on each of the six sides? Clearly 144 pounds pushing outward on each side.
Since the cube is rigid, the forces on opposite sides cancel out so there is no net force on the rocket. Up cancels down, left cancels right and forward cancels backward.

Now open a 1 square inch hole(or valve) on any face - i'll pick the bottom.
This is a simple thought experiment so we'll ignore refinements that would be dictated by proper fluid mechanics - entrance losses and vena contracta and all that.

Bottom face of your cube is now only 143 square inches, but top face is still 144.
So forces are no longer balanced. 143 pounds push down against bottom, but 144 still push up against top.
So rocket will accelerate up.

So - a rocket in a vacuum accelerates not because of propellant pushing against air, but because of propellant NOT pushing against anything!

Last statement was shocking enough that I always remember this oversimplification.
Rocket scientists add converging-diverging nozzles to refine the fluid mechanics.
242178



old jim
 

Drakkith

Staff Emeritus
Science Advisor
2018 Award
20,430
4,123
Imagine a single molecule of fuel has just combusted with a single molecule of oxidizer inside the thrust chamber. Let's say that the resulting propellant molecule obtains a great amount of kinetic energy directed upwards towards the top of the thrust chamber. This molecule moves upwards, impacts the wall at the top of the chamber, and bounces off towards the opening in the chamber leading to the nozzle. In order to bounce off the wall of the chamber the molecule requires a downward acceleration, which is provided by the repulsive force from the wall. In return, the molecule itself exerts a repulsive force on the wall of the chamber.

This mutual repulsive force accelerates the propellant molecule downward and the rocket upward. Repeat this process about a gazillion times per second and you have a simplified view of how a rocket engine works to accelerate a rocket in the vacuum of space.

Lets say if I have a pipe with one end closed, and drill a small hole on the side of the pipe near the closed end. Then I use vacuum at the open end to create mass flow exiting the pipe. Where is the opposite force on the pipe?
At the top of the inside of the pipe. If you add up the force from the molecules that bounce off of the sides you'll find that they sum to zero (since force is a vector). But if you sum up the forces exerted on all the remaining sides (which in your example is just the top surface) you'll find that the sum is non-zero. This non-zero force is what accelerates the pipe.

In fact, it doesn't matter what the shape of the container is. If you integrate the force across the entire inside surface of a closed volume of gas, you'll find that it is always zero. But as soon as you open a hole in one side the total net force no longer zero.
 

Astronuc

Staff Emeritus
Science Advisor
18,491
1,609
Force is related to the change in momentum of a mass per unit time; F = dp/dt. Force is also the product of pressure (or pressure drop) and the area. So flow from a rocket nozzle or pipe produces a net force by virtue of the flow and pressure drop at the exit.

See - https://www.physicsforums.com/threads/why-does-f-dp-dt.100862/
 
here's an explanation i used some years ago to make it intuitive.

When a user can make Bernoulli and Nasa's formulas give the same answer as this intuitive approach
he understands both the concept and the math.

But he's gotta start simple.



So when wind blows on the back of your head why don’t you feel the air in front of you push off your face as it moves away from you.?



View attachment 242178


old jim
 
Imagine a single molecule of fuel has just combusted with a single molecule of oxidizer inside the thrust chamber. Let's say that the resulting propellant molecule obtains a great amount of kinetic energy directed upwards towards the top of the thrust chamber. This molecule moves upwards, impacts the wall at the top of the chamber, and bounces off towards the opening in the chamber leading to the nozzle. In order to bounce off the wall of the chamber the molecule requires a downward acceleration, which is provided by the repulsive force from the wall. In return, the molecule itself exerts a repulsive force on the wall of the chamber.

This mutual repulsive force accelerates the propellant molecule downward and the rocket upward. Repeat this process about a gazillion times per second and you have a simplified view of how a rocket engine works to accelerate a rocket in the vacuum of space.



At the top of the inside of the pipe. If you add up the force from the molecules that bounce off of the sides you'll find that they sum to zero (since force is a vector). But if you sum up the forces exerted on all the remaining sides (which in your example is just the top surface) you'll find that the sum is non-zero. This non-zero force is what accelerates the pipe.

In fact, it doesn't matter what the shape of the container is. If you integrate the force across the entire inside surface of a closed volume of gas, you'll find that it is always zero. But as soon as you open a hole in one side the total net force no longer zero.
So why isn’t there force in the opposite direction of mass flow exiting the pipe? One side of the pipe is open so you’re supposed to have a net force
 
Force is related to the change in momentum of a mass per unit time; F = dp/dt. Force is also the product of pressure (or pressure drop) and the area. So flow from a rocket nozzle or pipe produces a net force by virtue of the flow and pressure drop at the exit.

See - https://www.physicsforums.com/threads/why-does-f-dp-dt.100862/
So why is there no opposite force in the scale when the container is lifted off using vacuum? Around 3:45 of vid
<Video removed. Final warning: do not post that video again.>
 
Last edited by a moderator:

DrClaude

Mentor
6,809
2,926
We do not debunk pseudoscience at PF. That video is full of wrong. The person who made it doesn't understand Newton's third law at all.
 

russ_watters

Mentor
18,385
4,636
Let's just close the thread here. The questions are full of wrong too and I don't think it's going to become productive.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top