# Definitions of jet thrusts

Discussion in 'Jet Drives' started by Ben Land, Apr 1, 2015.

1. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

dr
This is a pretty omniscient statement and incorrect.
Mass times velocity is MOMENTUM not force.
The change in momentum is Force,
The equation F= ma is the correct formula

Your earlier analogy about throwing a bucket of water off the back of the boat is not quite correct

The topic that we are discussing is THRUST which is a force

If you take your bucket of water and throw it off the back of a boat at 45 mph, it will produce say X amount of thrust.

But what you have missed is that you need to say that if you throw this bucket off the back of the boat at 45 mph and it took you 2 SECONDS TO ACCELERATE the water to 45 mph it will produce X amount of thrust.

If you take the same bucket and throw it off the back of the boat at 45 mph but it takes 1 SECOND TO ACCELERATE the water to 45 mph, it will produce two times X amount of thrust.

Mass times velocity is momentum

Mass times a change in velocity, acceleration, is force, thrust

DR states
Therefore, the THRUST of the propeller decreases at some speed, and the thrust of the propeller approaches zero.

The thrust of a propeller does not approach zero as speed increases, if it did the boat would slow down. The limiting factor is the amount of horsepower that you have to turn the prop that stops the boat from accelerating

2. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

Ben Writes
My guess is that the Gross thust is the power measured coming out of the nozzle, which as you said is not actually the thrust that moves the boat forward. But this is the force that gives the steering force when making a turn

First a necessary definition
Velocity has two components. One is speed, say 45 mph and the other is direction.
It is incorrect to say that the boat has a velocity of 45 mph. It is correct to say that the boat has a speed of 45 mph.
It is correct to say that a boat has a velocity of 45 mph NORTH

Why is this important
If you have the jet stream coming out of the jet at 60 mph, and you change the direction, you have changed the velocity.
A change in velocity is an acceleration. When you accelerate the mass by changing direction ( mass flow rate the same) you get the turning force. The force acts on the side of the nozzle and turns the boat

3. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

Some help here if you guys can
I was going to submit a units equation to clear up a couple of items. I scan the document, throw it on desktop, and have been unable to copy or attach it to this message board.
Thanks

4. Joined: Feb 2010
Posts: 536
Likes: 126, Points: 43, Legacy Rep: 1004
Location: www.boatdesign.net

### Boat Design Net ModeratorModerator

Hi Barry -- is it a pdf or a jpeg? When you click Go Advanced, and then Manage Attachments and select the file to upload, do you get any error message?
If it still won't attach for you, please email the file in question to webmaster@boatdesign.net and include a link to this thread.

5. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

Great, I will try this
Thanks

6. Joined: Oct 2009
Posts: 5,055
Likes: 531, Points: 113, Legacy Rep: 1485
Location: Midcoast Maine

### DCockeySenior Member

Simplified analysis of a jet:

Thurst = Mass flow rate * Velocity difference

Note that "Mass flow rate" has units of mass/time. When multiplied by velocity with units of length/time the result has units of mass*length/time^2 which is the same as the units of force.

7. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

So in summary

If you have a fixed item and accelerate it then the formula is
F=mass times acceleration

by units lbf = [(lbf)(sec)(sec)]/ feet] X [feet/(sec)(sec)]
Throwing the bucket of water off the back of a boat

If you have mass flow rate

F= (mdot) v or
F=mass/sec X Velocity difference ( between inlet and exhaust velocity)

by units lbf= {[(lbf)(sec)(sec)/(feet)]/sec} X [(feet)/(sec)]

but never
F=mass X velocity the units on the right side of the equation do not result in a force in poundf

8. Joined: Feb 2009
Posts: 92
Likes: 0, Points: 0, Legacy Rep: 2
Location: Idaho

### drmiller100Junior Member

You were right, and I was wrong about the equation. It is indeed mass flow rate, and not mass.

using your analogy, if I shorten the gullet of the pump and the bowl I can double my thrust with no change in energy applied. I don't think it works this way.

Mass flow rate doesn't really have a variable for how long it takes to accelerate the water - it only worries the water was indeed accelerated.

Good discussion helps me understand. Thank you.

9. Joined: Oct 2009
Posts: 5,055
Likes: 531, Points: 113, Legacy Rep: 1485
Location: Midcoast Maine

### DCockeySenior Member

If you throw the water off the boat at the same velocity but twice the rate (one bucket every second vs onee bucket every two seconds) the power required (rate of energy) doubles.

One very useful concept in fluid mechanics is the "control volume". Draw a surface around the system and keep track of what goes in and out of the system and the rate of change inside the system. Frequently the analysis then does not require studying the details of what happens inside the control volume.

10. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

Absolutely correct but I think you meant to say the rate of work, which is say horsepower, doubles

11. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

I am not sure what you mean by shorten the gullet of the pump

What I said regarding the bucket example is that if you accelerate the quantity at a higher rate, you get more thrust.
Because in order to accelerate a given mass faster, you have to apply more force. And force through distance with respect to time is horsepower. (foot pounds per second)
So accelerate it twice as fast, then twice the horsepower

12. Joined: Oct 2009
Posts: 5,055
Likes: 531, Points: 113, Legacy Rep: 1485
Location: Midcoast Maine

### DCockeySenior Member

Power is the rate of expenditure of energy. Work is the energy applied to do something. So rate of work is rate of expenditure of energy and is power.

Also "horsepower" is a unit of energy/time.

13. Joined: Feb 2009
Posts: 92
Likes: 0, Points: 0, Legacy Rep: 2
Location: Idaho

### drmiller100Junior Member

I get confused with your use of "accelerate". I think of the pump as a continuous system, not discrete particles, so "accelerate it twice as fast" makes no sense to me. Could you use different terms?

"accelerate the water so it has twice the velocity" works for me. We can assume the direction is out the back of the boat?

If I double the velocity, does it take twice the horsepower,or 4 times the horsepower?

14. Joined: Mar 2002
Posts: 1,539
Likes: 312, Points: 83, Legacy Rep: 158

### BarrySenior Member

There is a distinction between the two
throwing a bucket of water off the back of the boat you would use the F=ma
Fix mass, the bucket starting at zero velocity and you accelerate it to get the force

The second
A given mass flow assume the velocity starts at zero you would use F= mass Flow rate times exit velocity

So for your question if you double the exit velocity you would need theoretically twice the hp if the mass flow rate is the same

15. Joined: Feb 2009
Posts: 92
Likes: 0, Points: 0, Legacy Rep: 2
Location: Idaho

### drmiller100Junior Member

so, I'm still trying to understand this. If the boat is standing still in the water, the intake velocity is "zero". I get that. So let's say for a given setup, the jet stream goes out the back of the boat at 70 mph.

For a new case, for a boat going 40 mph, what is the intake velocity? My understanding is the velocity of the water is "still" 70 mph relative to the boat. Which means it is only going 30 mph backwards relative to the lake.

Does this mean my thrust is now less at 40 mph than it was at zero mph?

To get double the velocity with the same mass flow rate, I'd have to change impellers and nozzle and maybe intake gullet. Is that correct?

I really appreciate your patience. It seems like I'm missing something here as it seems to me your thrust will go down as your speed goes up, leading to a theoretical top speed as a function of jet velocity.

Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.