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? -- Cheers Clive
OT: Hydraulics query
Started by ●December 5, 2023
Reply by ●December 5, 20232023-12-05
The arsehole Clive Arthur <clive@nowaytoday.co.uk> persisting in being an Off-topic troll... -- Clive Arthur <clive@nowaytoday.co.uk> wrote:> Path: not-for-mail > From: Clive Arthur <clive@nowaytoday.co.uk> > Newsgroups: sci.electronics.design > Subject: OT: Hydraulics query > Date: Tue, 5 Dec 2023 12:00:12 +0000 > Organization: A noiseless patient Spider > Lines: 18 > Message-ID: <ukn3ce$51k5$1@dont-email.me> > Reply-To: clive@nowaytoday.co.uk > MIME-Version: 1.0 > Content-Type: text/plain; charset=UTF-8; format=flowed > Content-Transfer-Encoding: 7bit > Injection-Date: Tue, 5 Dec 2023 12:00:14 -0000 (UTC) > Injection-Info: dont-email.me; posting-host="a3ce545be3b0a467008adadb92affabd"; > logging-data="165509"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX180qkv4nejqBsnbuBdGGSKCNcGuCfoGHW0=" > User-Agent: Mozilla Thunderbird > Cancel-Lock: sha1:dNxnYg/F8D0JdmLBioVi0/AgdXY= > Content-Language: en-GB > X-Received-Bytes: 1732
Reply by ●December 5, 20232023-12-05
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
Reply by ●December 5, 20232023-12-05
The arsehole Fred Bloggs <bloggs.fredbloggs.fred@gmail.com> persisting in being an Off-topic troll... -- Fred Bloggs <bloggs.fredbloggs.fred@gmail.com> wrote:> X-Received: by 2002:a05:620a:81c:b0:77f:8ce:8465 with SMTP id s28-20020a05620a081c00b0077f08ce8465mr20788qks.9.1701788401882; > Tue, 05 Dec 2023 07:00:01 -0800 (PST) > X-Received: by 2002:a9d:740d:0:b0:6cd:ed53:128b with SMTP id > n13-20020a9d740d000000b006cded53128bmr674405otk.2.1701788401584; Tue, 05 Dec > 2023 07:00:01 -0800 (PST) > Path: not-for-mail > Newsgroups: sci.electronics.design > Date: Tue, 5 Dec 2023 07:00:01 -0800 (PST) > In-Reply-To: <ukn3ce$51k5$1@dont-email.me> > Injection-Info: google-groups.googlegroups.com; posting-host=2601:5cc:4701:5250:65ea:7135:bbe2:3e81; > posting-account=iGtwSwoAAABNNwPORfvAs6OM4AR9GRHt > NNTP-Posting-Host: 2601:5cc:4701:5250:65ea:7135:bbe2:3e81 > References: <ukn3ce$51k5$1@dont-email.me> > User-Agent: G2/1.0 > MIME-Version: 1.0 > Message-ID: <cc63c9bc-2598-44ad-8bc9-f49771b5a1cdn@googlegroups.com> > Subject: Re: OT: Hydraulics query > From: Fred Bloggs <bloggs.fredbloggs.fred@gmail.com> > Injection-Date: Tue, 05 Dec 2023 15:00:01 +0000 > Content-Type: text/plain; charset="UTF-8" > Content-Transfer-Encoding: quoted-printable > X-Received-Bytes: 2551
Reply by ●December 5, 20232023-12-05
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 >> CliveIf 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 -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
Reply by ●December 5, 20232023-12-05
possibly related https://youtu.be/ued2cEcfAio?si=ConBsCoSq8pRbjM1 https://youtu.be/z3scTRJCm7w?si=VRTbuXBGiUcRrJch
Reply by ●December 5, 20232023-12-05
On 05/12/2023 17:00, Phil Hobbs wrote:> 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 >Does the fact that in the U-tube, the water's direction is changed by 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 -- Cheers Clive
Reply by ●December 5, 20232023-12-05
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: >>> 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 >> > Does the fact that in the U-tube, the water's direction is changed by > 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. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics
Reply by ●December 5, 20232023-12-05
The idiot Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> persisting in being an Off-topic troll... -- Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> wrote:> Path: not-for-mail > From: Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> > Newsgroups: sci.electronics.design > Subject: Re: OT: Hydraulics query > Date: Tue, 5 Dec 2023 12:00:00 -0500 > Organization: A noiseless patient Spider > Lines: 57 > Message-ID: <5223fc78-187f-9134-84f6-e156a603afe9@electrooptical.net> > References: <ukn3ce$51k5$1@dont-email.me> > <cc63c9bc-2598-44ad-8bc9-f49771b5a1cdn@googlegroups.com> > MIME-Version: 1.0 > Content-Type: text/plain; charset=UTF-8; format=flowed > Content-Transfer-Encoding: 8bit > Injection-Info: dont-email.me; posting-host="ae10fd86b3facb721cc895b62e14bdb8"; > logging-data="257572"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX198DsRz1j8Ppvtp4BSBOvSU" > User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:91.0) Gecko/20100101 > Thunderbird/91.0 > Cancel-Lock: sha1:FRsxzxAFevVbT4jmni5t+TznUQM= > In-Reply-To: <cc63c9bc-2598-44ad-8bc9-f49771b5a1cdn@googlegroups.com> > X-Received-Bytes: 3599
Reply by ●December 5, 20232023-12-05
The idiot Clive Arthur <clive@nowaytoday.co.uk> persisting in being an Off-topic troll... -- Clive Arthur <clive@nowaytoday.co.uk> wrote:> Path: not-for-mail > From: Clive Arthur <clive@nowaytoday.co.uk> > Newsgroups: sci.electronics.design > Subject: Re: OT: Hydraulics query > Date: Tue, 5 Dec 2023 18:29:21 +0000 > Organization: A noiseless patient Spider > Lines: 68 > Message-ID: <uknq63$8oo0$1@dont-email.me> > References: <ukn3ce$51k5$1@dont-email.me> > <cc63c9bc-2598-44ad-8bc9-f49771b5a1cdn@googlegroups.com> > <5223fc78-187f-9134-84f6-e156a603afe9@electrooptical.net> > Reply-To: clive@nowaytoday.co.uk > MIME-Version: 1.0 > Content-Type: text/plain; charset=UTF-8; format=flowed > Content-Transfer-Encoding: 8bit > Injection-Date: Tue, 5 Dec 2023 18:29:23 -0000 (UTC) > Injection-Info: dont-email.me; posting-host="a3ce545be3b0a467008adadb92affabd"; > logging-data="287488"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19Ki2zoCs/RABP7M6+4kMx/03LPFBbAxPs=" > User-Agent: Mozilla Thunderbird > Cancel-Lock: sha1:K0TpXxIC6ORULumeuFgoNWb0K9o= > Content-Language: en-GB > In-Reply-To: <5223fc78-187f-9134-84f6-e156a603afe9@electrooptical.net> > X-Received-Bytes: 3975
