The Continuity Equation (Fluid Mechanics - Lesson 6)

Поділитися
Вставка
  • Опубліковано 4 лис 2024

КОМЕНТАРІ • 74

  • @MrHEAVY19
    @MrHEAVY19 Рік тому +6

    This 6 minute video just clarified 3 hrs of my professor's lecture

  • @blee52995
    @blee52995 4 роки тому +12

    Jazz major here! Thanks for the sweet vid!

  • @jessicacooper5205
    @jessicacooper5205 6 років тому +21

    Engineering student here, thank you for the video!

  • @SarahCarvela
    @SarahCarvela 3 роки тому +8

    Thank you sir! From a biochem major trying to understand physics.

  • @NK-iy6if
    @NK-iy6if Рік тому +2

    Thank you for the thorough but concise explanation, it was very helpful!

  • @alphatech21
    @alphatech21 23 дні тому

    Thank you very much. It clear and easy to learn than I thought.

  • @MatisCCCC
    @MatisCCCC 3 роки тому +5

    environmental engineering student here, thanks Prof.

  • @ebimeneakeme4163
    @ebimeneakeme4163 3 роки тому +1

    Americans are the BEST
    RESPECT

  • @thefenerbahcesk4156
    @thefenerbahcesk4156 5 років тому +13

    For whatever reason this is important for the MCAT

    • @sagestation7831
      @sagestation7831 4 роки тому +3

      Fluid dynamics, and mechanics are quite possibly the most important physics lessons in medicine lol... Way more important than things like electricity and magnetism, and optics. Not everyone needs MRIs or wears glasses, and any specialist on those machines will learn this stuff to use them. But most people take medications, and have veins...

    • @frozenrats
      @frozenrats 11 місяців тому +1

      Blood flow

    • @thefenerbahcesk4156
      @thefenerbahcesk4156 11 місяців тому

      @@frozenrats I'm in the third year of med school now. This is not important.

    • @kirrans7926
      @kirrans7926 12 днів тому

      i'm in my 5th year. this is important ​@@thefenerbahcesk4156

  • @ehda540
    @ehda540 6 років тому +4

    I might be a 5 years late for this video ,but sir you are fking amazing , teaching is a talent and you got that talent

    • @StrongMed
      @StrongMed  6 років тому +5

      It's never too late for that kind of praise! Thanks! ;)

  • @51bano
    @51bano 10 років тому +4

    continuity equation the first key to solve hydraulic problems.
    Q=V/t
    Q=v.A
    The tow equation the same but the 1st V volume,the 2nd equation v mean velocity
    verify your equation by its units
    Q=cm3/sec. V=cm3 t=sec if you divide =cm3/sec the equation right.
    Q=cm3/sec v=cm/sec A =area=cm2 then{ cm/sec x cm2} =cm3/sec the equation right.

    • @surprisedpikachu3782
      @surprisedpikachu3782 7 років тому +1

      Ibrahim el sayed can anyone tell in the first problem why he hasent converted 2cm into meters ...all other units are in meter

    • @goodman5836
      @goodman5836 7 років тому

      the other radius was also in cm. So the units will basically cancel each other out

  • @sara-hr2fg
    @sara-hr2fg 6 років тому

    2 hours before my test i want to say thank you soooooooooooooo much that helps me alot i understand everything in 6 minutes from you and my teacher who explained it for a week well nothing from him so thank you so much

  • @Startpixie12
    @Startpixie12 Рік тому

    Was it necessary to do unit conversion in the second example because all the units would be ultimately canceled out?

  • @christinadc
    @christinadc 8 років тому +2

    Superb video!! It's so informative now I could explain it easily and no need to study on books. Thanks sir amazing explanation.I wish if my physics teacher would have watched this video before giving that boring lecture. :D

  • @김성원-j8k
    @김성원-j8k Рік тому

    Amazing Video

  • @Startpixie12
    @Startpixie12 Рік тому

    Thanks for a great video! But in the first example, how come he didn’t convert the cm values into m for raidus ?the speed was given in m/s

    • @lumbalumba2436
      @lumbalumba2436 11 місяців тому

      I am pretty sure we still have to convert it. If you try to do it you will still get the same answer.

  • @sarangabbasi2560
    @sarangabbasi2560 5 років тому +2

    powerful lecture............

  • @firman.akhmad
    @firman.akhmad 2 роки тому

    How to understanding this case, when we have reduced cross section (A) in pipeline (caused by valve half closing) the time to fill up in a tank is different when I fully open the valve and half open, does this mean flow (Q) in faucet valve end is less than the source? Kindly explain, thanks

  • @elifugur5935
    @elifugur5935 7 років тому +2

    quick question.. So since the blood vessels are considered to be a closed circuit, the flow rate should be constant at every point in the circulatory system. The law of continuity basically states this. But when you consider poiseuille's law, the flow rate is proportional to r^4. So if we were to have constriction of the blood vessels, then the flow rate would decrease. But then as stated earlier shouldn't flow rate be maintained constant throughout the body regardless of the radius of the vessels? I would appreciate it greatly if someone can help me out. Thanks

    • @StrongMed
      @StrongMed  7 років тому +5

      I wouldn't say flow is constant, since that implies sameness over time. Instead, at any one point in time, total flow should be the same everywhere in the CV system. For example, flow through the aortic valve approximately equals flow through the pulmonary valve, and approximately equals flow through the entire body's capillary bed. It's approximately, because there are moment to moment changes in the capacity of certain vascular beds - most prominently in the lungs with the respiratory cycle, but also places like the legs with abrupt changes in position.
      While arteries can constrict or dilate in response to various factors, capillaries don't. In addition, when there are changes in peripheral resistance due to changes in artery radius, there may also be changes to cardiac output. For example, giving a pure vasoconstrictor like phenylepherine to a patient usually decreases cardiac output due to increased afterload (i.e. increased work and O2 demand for the myocardium). And this is made that much more complicated because some compounds such as catecholamines can influence between artery radius (and thus resistance) as well as having a direct effect on myocardial contractility.
      So overall, depending on what caused it, changes in artery radius can have variable effects on both the velocity of blood flow through the arteries and that through the capillaries, but the degree of change in velocity in those two areas are not the same.

    • @elifugur5935
      @elifugur5935 7 років тому +1

      Ahhhh ok, I think I finally understand how flow rate can change. basically the amount of vol passing through any pt in time must equal to the amount of vol pass through all other points, i.e flow rates in capillaries will be smaller relative to the artery, but the summation of the flow rates of the capillaries would more or less be equal to the artery that was preceding them. Thank you sooo much for the clarification!!

  • @darkook8472
    @darkook8472 3 роки тому

    thank you
    a petroleum engineering student here 🙋🏻‍♂️

  • @christianarrizon9796
    @christianarrizon9796 2 роки тому

    So basically what ur saying is that any amount that enters in one end the same amount comes out in the other ????

    • @StrongMed
      @StrongMed  2 роки тому

      Yes, that's correct, but it requires the simplifying assumption that the fluid is non-compressible (which is an accurate approximation for everyday applications).

    • @christianarrizon9796
      @christianarrizon9796 2 роки тому

      @@StrongMed so if it enters in the form of ice n it melts inside would it change???

    • @StrongMed
      @StrongMed  2 роки тому +1

      @@christianarrizon9796 Ice isn't a fluid. The physics of fluid mechanics don't apply. Ice also takes up more space (i.e. has a lower density) than water. I would guess that slush (i.e. a slurry mixture of part ice and part water) violates the non-compressible assumption (i.e. increasing pressure would convert some ice to water, thus taking up less room), so the continuity equation wouldn't perfectly apply then either.

  • @zaamk6728
    @zaamk6728 8 років тому +2

    thank you for such an amazing video.

  • @emkalada
    @emkalada 11 років тому

    Great video; I was wondering if you could clarify something for me: I'm confused about this equation (a1v1=a2v2) and this equation: flow = change in pressure/R. it seems like from the first equation, velocity 1 is proportional to 1/r^2 but in the second equation, flow (or velocity?) is directly proportional to r^4.

    • @StrongMed
      @StrongMed  11 років тому +2

      That's a great question. You are correct that from the continuity eq (a1v1=a2v2), velocity 1 is proportional to 1/r^2. With the 2nd eq (flow = pressure gradient / resistance), you are also correct that flow (but not velocity) is proportional to r^4, (as a consequence of Poiseuille's Law), but (and here's the catch) only if the pressure gradient remains constant.
      Consider a straight tube of radius r transmitting some fluid with velocity v, and the flow Q is being driven by a pressure gradient, which we'll call P. What happens if we double the radius of the tube? It's actually impossible to say without more information about the source of the pressure gradient. For example, if it's a water pump designed to pump water at a specific set pressure, the doubling of the radius (which decreases the resistance by a factor of 16) would result in flow being increased by a factor of 16 also. By increasing the flow by factor of 16, and increasing the cross sectional area by a factor of 4 (2^2), the velocity through the tube must increase by a factor of 4 also.
      However, if the pump is designed to pump water at a specific flow rate, the doubling of the radius (again, which decreases the resistance by a factor of 16)would result in the pressure gradient dropping by a factor of 16. If the flow remains constant with the radius doubling (and cross sectional area increaing by factor of 4), the velocity through the tube decreases by a factor of 4.
      It all depends on what is ultimately pushing (or pulling) the fluid. The cardiovascular system acts more like the former situation. The primary relevant feedback mechanisms are baroreceptors, which act to keep the blood pressure (i.e. pressure gradient) constant in response to decreases in peripheral vascular resistance (i.e. blood vessel radius). However, there are secondary mechanisms such as circulating hormones called ANP and BNP which can indirectly detect decreased cardiac output (i.e. flow) and dilate the arteries in response (i.e. increase blood vessel radius), the effect of which is to increase flow at the expense of decreasing the pressure gradient.
      I know that was a bit long-winded, but I hope it helps. It's a little difficult to answer math-related questions in a dialog box.

    • @emkalada
      @emkalada 11 років тому

      that makes sense; so if talking about constant flow, use a1v1=a2v2 and if talking about constant pressure, then q is proportional to r^4; do you know why they use the A1V1=A2V2 to explain that since the capillaries have the largest surface area but then use Q=r^4 when talking about atherosclerosis? I get that in th former situation, as you said because the heart is seen as pressure pump so pressure is constant so use A1v2=a2v2; but what about the second situation?

  • @KevinSmall
    @KevinSmall 11 років тому +2

    Very good medical application...thank you

  • @j-bear7
    @j-bear7 4 роки тому

    thank you for easy and straight forward explanation

  • @anilsharma-ev2my
    @anilsharma-ev2my 3 роки тому

    What is maximum limits
    of siphoning at all under NTP AND STP
    Please make a video over it 🕉🕉🕉🕉

  • @nowthenad3286
    @nowthenad3286 5 років тому

    If I have a tap being supplied with 1/2" pipe and I want to attach a hosepipe to that tap, will I obtain a greater flow rate at the end of a 30m hose pipe if the pipe is 3/4" instead of 1/2"?

    • @not_thareesh
      @not_thareesh 4 роки тому

      No, you wouldn't be getting a greater flow rate. You have to reduce the radius of the pipe. He has explained it clearly on 4:08.
      Pi * r1^2 * V1 = Pi * r2^2 * V2
      r1^2 * V1 = r2^2 * V2
      (1/2)^2 * V1 = (3/4)^2 * V2
      1/4 * V1 = 9/16 * V2
      16 * V1 = 36 * V2
      V1/V2 = 36/16
      On simplifying, you get
      V1 = 9 m/s
      V2 = 4 m/s
      Velocity and Pressure are inversely proportional to the area of the cross-section of the body through which a fluid is flowing.
      So, providing a hose of a greater radius wouldn't increase the velocity of which it flows.

    • @lafdasurmemes1832
      @lafdasurmemes1832 3 роки тому

      Flow is constant at both the sections however velocity increases or decreases depending upon whether the area reduces or increases respectively.

  • @soorajnair2849
    @soorajnair2849 5 років тому

    In your calculations velocity in meter per sec and radius in cm.why don't you convert radius cm in to meters. While calculations all units should be same ryt?

    • @not_thareesh
      @not_thareesh 4 роки тому

      He had already changed it. Please check the video once more.
      He has changed 1.5 cm to 1.5 X 10^-2 which is basically 1.5/100

  • @viviankamanga1650
    @viviankamanga1650 5 місяців тому

    The velocity for the first example was 0.16m/s

  • @daniellecoutre8878
    @daniellecoutre8878 3 роки тому

    so the volume of section 1 equals the volume of section 2??

  • @nichithcn7119
    @nichithcn7119 6 років тому +5

    I came here as a 11th standard student preparing for entrance, XD, so I stopped at 4:14

    • @rachitjoshi6931
      @rachitjoshi6931 5 років тому +1

      Nichith CN VEVO same here lmao, paused the video when he said "medical point of view"

  • @Lanicatt
    @Lanicatt 3 роки тому +1

    Med student here thank uu

  • @mtumiran2899
    @mtumiran2899 3 роки тому

    the real bang jago

  • @MWang-ne9ze
    @MWang-ne9ze 5 років тому

    Very easy to understand, thanks

  • @draxlom6467
    @draxlom6467 7 років тому +3

    love you strong medicine

  • @ayishanuhu4955
    @ayishanuhu4955 4 роки тому

    Thank you

  • @tomsebastian2021
    @tomsebastian2021 3 роки тому

    11th student is here gogogogo

  • @maily3033
    @maily3033 9 років тому

    you didn't convert cm into m

    • @StrongMed
      @StrongMed  9 років тому +1

      Mai Ly No need to. As long as the radius is given on both sides of the equation in one set of units, and the velocity is given (or determined) on both sides of the equation in one set of units, the units for radius and velocity don't necessarily need to be consistent with one another.

  • @ezlearnpoint2357
    @ezlearnpoint2357 4 роки тому

    Please tell me about software used

  • @wardeggerrobertmarius144
    @wardeggerrobertmarius144 2 місяці тому

    Continuity theory is just a theory... But it's good enough for most practical applications.
    In reality pressure drops with the increase of the distance from pressure source. And flow drops with the drop of pressure.
    😂 And to marke it even more complicated inside a gravitational field we have gravitational pressure... Pressure is higher closer to the gravitational center.
    So... If you stand on your head blood pressure will be higher on your brains that on your feet.😂😂😂 Lesson from an engenier for medics😂😂😂😂😂😂😂 Don't put a patient with anevrism to stand on his/her head 😂😂😂😂 based on Pascal's laws and Bernoulli's equation 😂😂😂😂

  • @landegre
    @landegre 4 роки тому

    12th grader here
    Thank you alot

  • @jayampower2519
    @jayampower2519 5 років тому

    you are so 👍👍🙌

  • @RayRay-yt5pe
    @RayRay-yt5pe 4 роки тому

    What the hell! So the concept is this simple.

  • @mamoonaashfaq6311
    @mamoonaashfaq6311 6 років тому

    👍👍👍

  • @UmangSharma-d1c
    @UmangSharma-d1c 8 місяців тому

    😊

  • @ak123cric
    @ak123cric 5 років тому

    This is not the continuity equation.....

    • @StrongMed
      @StrongMed  5 років тому

      There are numerous different formulations of "continuity equations". This (A1v1=A2v2) is the one most commonly cited in medical school for very simple medical applications. (I admit that the title of the video is a misnomer - saying "THE continuity equation" implies there is only one...)

    • @OnlyPainAllDayEveryDay00
      @OnlyPainAllDayEveryDay00 3 роки тому

      @@StrongMed you did a good job. This is a great summary for many who don’t need the partial differential equation derivation of the continuity equation

  • @rashidarif8383
    @rashidarif8383 5 років тому

    wow interesting

  • @SuperMossyG
    @SuperMossyG 6 років тому +1

    Dr. Moore such horrible directions