Volume of water moved

Is there a way to determine how much water, in cubic feet, a ship propeller moves in one minute.
Particulars would be a six hundred foot ship, propeller size and pitch typical for type, rpm for 18 knots.
Not interested in slippage, torque, thrust, acceleration, load, etc. etc.
In a perfect world, how many cubic feet of water per minute is being moved.

 

Too hard a question, If you mean water that is somehow perturbed by the action of the propellor, separate from the effect of the hull through the water, I don't like your chances of getting an accurate answer. And I don't see what use the information would be if you could extract it.

 

In a perfect world, it is quite simple.

It is simple the Disc area x the water velocity.

That's it.

But the water velocity, at the prop, is dependent upon many factors owing to the increase in velocity of the from a location fwd of the prop to aft of the prop. This "factor" that alters the flow/velocity is called the axial inflow factor. It is this that controls the efficiency etc too..but you're not interested in that

 

Trick question,

In a perfect world its the ship that moves not the water.
Also in a perfect world the ship will move by the pitch of the prop.

 

Is this an academic question, or a real-life problem? If it is a school/university task, then do the following steps.

In an ideal world, you'd have to know at least the flow speed (corrected for the hull wake influence), the prop diameter (hence, the disc area) and the estimated prop efficiency. With these data, go to the Chapter 11.7.3 (Actuator Disk Theory) of this page: http://mit.edu/16.unified/www/FALL/thermodynamics/notes/node86.html and take note of the following equations:

• 1) Prop thrust: T = massflow * (Ve - Vo)
• 2) Mass flow = rho * Ad * Vd
• 3) Vd = 0.5 * (Vo + Ve)
• 4) Prop efficiency = the last equation in that chapter, too time-coonsuming to write it here.

The symbols used:

• Vo = the flow speed far ahead of the prop
• Ve = flow speed in the propeller's wake
• Ad = prop disc area
• rho = fluid density.

Use the above formulae, make the necessary substitutions and you will arrive to the required (and pretty simple) equation for the mass flow across the prop.

Cheers

 

So you think the only disturbance due to the propellor ("moved" water, as in the original question) is that of the water that actually passes through the propellor ? I don't think so.

 

How about getting to the point?

 

Sounds like a government research project costing several million dollars to show how ships disturb the underwater fauna.

 

Volume of water solved

Thanks for all the replies. First of all this is not a trick question and no barnacles were harmed during the asking of this question.
Actually, your replies led me to the answer I was looking for.
Here goes. Once again a perfect world, 100% efficiency with no other contributing factors.
A ship propeller; fifteen feet in diameter, five foot pitch and turning at 60 rpm. will "screw" its way down a 300' tunnel of water in one minute. So...in order for the ship to move that 300' the prop will have to move, or process, 53,000 cu.ft. of water-the area of a cylinder 300x15.
I know this is absurdly simplistic, but it did spring from a twelth grade public school education and the answer feeds my lust for tinkering.
Now, all I need is a supply of dilithium crystals.

 

Not at all. If you are considering a propeller with no slip, there will be zero water displacement.

 

Gonzo is right. A prop with 100% efficiency would give zero thrust. A zero thrust means that Ve = Vo in the equations from my previous post, and that means that the water flow wouldn't get any backwards acceleration. In other words, a 100% efficient prop would be "invisible" to the waterflow, which would just stream along the hull as if there was no prop at all. Hence, no water is moved by the prop in that particular case.

 

Hi Riobdriver, your original question “

Has headed in 2 different directions.
1. How much water does a 15ft prop with 5ft pitch move at 60rpm after 1 minute.
2. How far will a ship travel with a 15ft prop, 5ft pitch rpm is 60 after 1 minute.

Assuming 100% efficiency

Question 1. This is a pump and will pump/move/process 3.14x7.5x7.5x5x60= 53,000cu.ft
This is the result you came to but it will only apply to water moving in a tube.

Question 2. The water doesn’t move, 100% efficiency will mean the ship moves through the water so the water moved by the prop is zero.
The distance travelled is 5 x 60 = 300ft.


Good luck with the dilithium crystals and of you happen to spot a pair or anti-gravity boots on the way please let me know.

Thanks Don

 

The point is more water is being moved than that which passes through the propellor "disc", the water being accelerated rearwards interacts with, and presumably gives a nett rearward movement to some water that did not travel through that disc. He wants for some mysterious reason to know how much water is moved, I can't see it producing anything better than a migraine finding the right answer.

 

This is fairly simple and Daiquiri got it exactly right. This is right from Newton; F=ma. If you know (or can estimate) the thrust at a given speed and can estimate the efficiency then you know how much water is being accelerated. The Ve (velocity in the race) is related to the inflow velocity is roughly 1/eta.
You can also use the "slip velocity" to get the Ve, but slip has always seemed to me to be less than useful for discussing propeller design.

 

So you're telling me the water being accelerated by the prop does not impart any acceleration to water that did not pass through the prop "disc" ? The seemingly pointless question posed in the opening post was how much water was moved, we know where the force is coming from to accelerate water, but how much water ultimately gets "moved" is a different matter, probably useless to know, but not the same.