On Tuesday, December 5, 2023 at 7:00:23 AM UTC-5, Clive Arthur wrote:the tube forward. Centripetal is Newton, differential pressure is Bernoulli. This is my guess...
If I have a U-tube, horizontal and submerged in water, with a propeller
half way along inside one leg, what happens when the prop is rotated?
Obviously, the same amount of water comes out of one leg at the same
velocity as it goes in the other. Because the in and out are separated
in distance, there will be a torque generated. But if we constrain the
tube so it can't rotate - or simply attach a mirror-image tube with
propeller to cancel the torque - will the tube(s) move?
I think so - the water changing direction as it flows around the bend
will generate a force, pushing the tube along in the direction of the U.
It won't matter which way the propeller is turning.
Is that right, or should I stay off the drugs?
The tube restrains the water to turn through the radius with a counter-centripetal for mv^2/r. That will push the tube forward. Then the propeller by the act of moving the water induces a pressure differential from front to back on its blades, pushing
--
Cheers
Clive
On 2023-12-05 10:00, Fred Bloggs wrote:
On Tuesday, December 5, 2023 at 7:00:23 AM UTC-5, Clive Arthur wrote:
If I have a U-tube, horizontal and submerged in water, with a propeller
half way along inside one leg, what happens when the prop is rotated?
Obviously, the same amount of water comes out of one leg at the same
velocity as it goes in the other. Because the in and out are separated
in distance, there will be a torque generated. But if we constrain the
tube so it can't rotate - or simply attach a mirror-image tube with
propeller to cancel the torque - will the tube(s) move?
I think so - the water changing direction as it flows around the bend
will generate a force, pushing the tube along in the direction of the U. >>> It won't matter which way the propeller is turning.
Is that right, or should I stay off the drugs?
The tube restrains the water to turn through the radius with a
counter-centripetal for mv^2/r. That will push the tube forward. Then
the propeller by the act of moving the water induces a pressure
differential from front to back on its blades, pushing the tube
forward. Centripetal is Newton, differential pressure is Bernoulli.
This is my guess...
--
Cheers
Clive
If we dial back the camera a bit, and allow the angle between the tube
axes to be anything, it gets clearer. Force is the time rate of change
of momentum. At the intake end, the force is -m*V, where m is the mass
of water entering per second, and V is the velocity, and at the outlet
end it's +m*V.
If the angle is pi (i.e. the tube is straight) then the answer is obvious--the two contributions add. If not, we have to do vector
addition. With the inlet pointing towards positive X, and the outlet at
some angle theta from there, the total force on the tube is
F = m*V [(1 - cos theta) Xhat - sin theta Yhat).
So when they're pointing the same way, we expect the force to be zero.
Of course this is a zero-order approximation, because the actual motion
of the surrounding water will be a complicated mess, and there's
viscosity and friction and all.
Cheers
Phil Hobbs
On 05/12/2023 17:00, Phil Hobbs wrote:
On 2023-12-05 10:00, Fred Bloggs wrote:Does the fact that in the U-tube, the water's direction is changed by
On Tuesday, December 5, 2023 at 7:00:23 AM UTC-5, Clive Arthur wrote: >>>> If I have a U-tube, horizontal and submerged in water, with a propeller >>>> half way along inside one leg, what happens when the prop is rotated?
Obviously, the same amount of water comes out of one leg at the same
velocity as it goes in the other. Because the in and out are separated >>>> in distance, there will be a torque generated. But if we constrain the >>>> tube so it can't rotate - or simply attach a mirror-image tube with
propeller to cancel the torque - will the tube(s) move?
I think so - the water changing direction as it flows around the bend
will generate a force, pushing the tube along in the direction of the U. >>>> It won't matter which way the propeller is turning.
Is that right, or should I stay off the drugs?
The tube restrains the water to turn through the radius with a
counter-centripetal for mv^2/r. That will push the tube forward. Then
the propeller by the act of moving the water induces a pressure
differential from front to back on its blades, pushing the tube
forward. Centripetal is Newton, differential pressure is Bernoulli.
This is my guess...
--
Cheers
Clive
If we dial back the camera a bit, and allow the angle between the tube
axes to be anything, it gets clearer. Force is the time rate of change
of momentum. At the intake end, the force is -m*V, where m is the mass
of water entering per second, and V is the velocity, and at the outlet
end it's +m*V.
If the angle is pi (i.e. the tube is straight) then the answer is
obvious--the two contributions add. If not, we have to do vector
addition. With the inlet pointing towards positive X, and the outlet at
some angle theta from there, the total force on the tube is
F = m*V [(1 - cos theta) Xhat - sin theta Yhat).
So when they're pointing the same way, we expect the force to be zero.
Of course this is a zero-order approximation, because the actual motion
of the surrounding water will be a complicated mess, and there's
viscosity and friction and all.
Cheers
Phil Hobbs
180'. That change in direction of a moving mass must surely result in a force?
Isn't it a bit like blowing yourself along on a skateboard using an
umbrella held in front and a leaf-blower blowing forwards into the
umbrella? eg...
https://www.youtube.com/shorts/1CXB7_gm8I0
Clive Arthur <clive@nowaytoday.co.uk> wrote:
On 05/12/2023 17:00, Phil Hobbs wrote:
On 2023-12-05 10:00, Fred Bloggs wrote:Does the fact that in the U-tube, the water's direction is changed by
On Tuesday, December 5, 2023 at 7:00:23 AM UTC-5, Clive Arthur wrote: >>>>> If I have a U-tube, horizontal and submerged in water, with a propeller >>>>> half way along inside one leg, what happens when the prop is rotated? >>>>>
Obviously, the same amount of water comes out of one leg at the same >>>>> velocity as it goes in the other. Because the in and out are separated >>>>> in distance, there will be a torque generated. But if we constrain the >>>>> tube so it can't rotate - or simply attach a mirror-image tube with
propeller to cancel the torque - will the tube(s) move?
I think so - the water changing direction as it flows around the bend >>>>> will generate a force, pushing the tube along in the direction of the U. >>>>> It won't matter which way the propeller is turning.
Is that right, or should I stay off the drugs?
The tube restrains the water to turn through the radius with a
counter-centripetal for mv^2/r. That will push the tube forward. Then
the propeller by the act of moving the water induces a pressure
differential from front to back on its blades, pushing the tube
forward. Centripetal is Newton, differential pressure is Bernoulli.
This is my guess...
--
Cheers
Clive
If we dial back the camera a bit, and allow the angle between the tube
axes to be anything, it gets clearer. Force is the time rate of change >>> of momentum. At the intake end, the force is -m*V, where m is the mass >>> of water entering per second, and V is the velocity, and at the outlet
end it's +m*V.
If the angle is pi (i.e. the tube is straight) then the answer is
obvious--the two contributions add. If not, we have to do vector
addition. With the inlet pointing towards positive X, and the outlet at
some angle theta from there, the total force on the tube is
F = m*V [(1 - cos theta) Xhat - sin theta Yhat).
So when they're pointing the same way, we expect the force to be zero.
Of course this is a zero-order approximation, because the actual motion
of the surrounding water will be a complicated mess, and there's
viscosity and friction and all.
Cheers
Phil Hobbs
180'. That change in direction of a moving mass must surely result in a
force?
Isn't it a bit like blowing yourself along on a skateboard using an
umbrella held in front and a leaf-blower blowing forwards into the
umbrella? eg...
https://www.youtube.com/shorts/1CXB7_gm8I0
Sure. That force is internal to the device, though—the net change in momentum equals minus that in the fluid outside.
If I have a U-tube, horizontal and submerged in water, with a propeller
half way along inside one leg, what happens when the prop is rotated?
Obviously, the same amount of water comes out of one leg at the same
velocity as it goes in the other. Because the in and out are separated
in distance, there will be a torque generated. But if we constrain the
tube so it can't rotate - or simply attach a mirror-image tube with
propeller to cancel the torque - will the tube(s) move?
I think so - the water changing direction as it flows around the bend
will generate a force, pushing the tube along in the direction of the U.
It won't matter which way the propeller is turning.
Am 05.12.2023 um 13:00 schrieb Clive Arthur:
If I have a U-tube, horizontal and submerged in water, with aI agree, that the water changing direction will generate a force.
propeller half way along inside one leg, what happens when the prop is
rotated?
Obviously, the same amount of water comes out of one leg at the same
velocity as it goes in the other. Because the in and out are
separated in distance, there will be a torque generated. But if we
constrain the tube so it can't rotate - or simply attach a
mirror-image tube with propeller to cancel the torque - will the
tube(s) move?
I think so - the water changing direction as it flows around the bend
will generate a force, pushing the tube along in the direction of the
U. It won't matter which way the propeller is turning.
But doesn't the propeller, which is somehow fixed to the tube, create a
force in the opposite direction?
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