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B Faster velocity of water in tunnel Decrease in pressure?

  1. Jan 15, 2017 #1
    In bernoulli's equation: when water is moving at a faster velocity, why is pressure decreased? I searched the forum and someone said that there is less water touching the surface? I do not conceptually see this.

    I understand for the lift factor with the example of blowing air over the paper:(more spread out air particles as you blow ontop of the paper while the air below is more concentrated)

    Please, need some help! Thanks in advance.
     
  2. jcsd
  3. Jan 15, 2017 #2
    I think you are confused about few things here.

    1) The velocity increases in a funnel due to conservation of mass. Mass conservation means that the mass flow rate in the tube, should be constant at any point inside the tunnel. Mass flow rate can be given as the product of density and volume flow rate. If we consider the fluid to be incompressible or nearly incompressible, i.e. constant density, then the volumetric flow rate must also be constant. Now volumetric flow rate can be given as the product of flow velocity and cross sectional area of the tunnel. When area drops velocity should increase to keep the volume flow rate a constant.
    Pressure is a force divided by the area but that's not a very suitable of looking at this problem. An alternative way of looking at pressure is to see it as mechanical potential (more precisely it is potential energy per unit volume). In this way, we can then think about pressure as the driving force for flow. Systems like to go from high potential to low potential so if the pressure is constant everywhere, there will be no flow in the tunnel. So flow must be from high potential to low potential and so the velocity will point from high pressure to low pressure.
    Mechanical potential energy (pressure) is converted into mechanical kinetic energy (velocity) and so the pressure must decrease as the velocity increases while the liquid flows through the tunnel.
    And you say "...someone said that pressure drops because there's little water touching the surface".Here that person have got himself confused.
    Pressure by itself does not drive flow. Only a gradient in pressure will exist in the tunnel if there is a flow through it. This gradient is what causes the net force on the fluid that drives it through the tunnel but dividing the net force by the area does not give you the pressure .And I think that the person has got himself confused about Pressure and Impulse. Impulse is the force required to completely stop the flow inside the tunnel. Since momentum is conserved in the flow, the force to stop it will be the same but since the tunnel produces a flow with a smaller cross section, the force is applied over a smaller area and thus the "pressure" there is higher.

    I hope you gained a clear idea about the problem.
     
  4. Jan 16, 2017 #3
    Ah, that makes sense, thank you nickyferandezzz! I see there was a gradient from the start and as the flow moves, to maintain this conserved flow through an incompressible fluid, there is a trade off in PE to KE.

    Could you help me with one more problem. In the human body, why might there be less pressure in the capillaries as composed to the artery? Is the mechanism above very similar to this condition?

    My thought experiment is that somehow(not sure conceptually) an increase in resistance causes friction to dissipate kinetic energy a long the walls of the blood vessel as heat and therefore would decrease velocity.(not an ideal system) I can't seem to stretch my mind further than this.

    Or maybe it has something to do with less blood volume in a given space, for example, in the artery there is a certain amount of blood volume that it can hold and causes a hydrostatic pressure, but there are way more capillaries and together there is a greater volume that it can hold and have less pressure?
     
  5. Jan 16, 2017 #4
    You are correct when you say that the mechanism of blood flow in capillaries and arteries is similar to the above scenario.
    But Bernoulli's principle cannot be directly applied to this situation because in order to apply Bernoullis principle, volume flow rate should be constant throughout the circulatory process.But it is not so.The flow varies from arteries to veins and veins to capillaries. We can apply Poiseulle's equation to understand this. According to the Poiseulle's equation,
    Pressure difference is indirectly proportional to the radius of the blood vessel. Now, as we know there are lots of capillaries inside the body.So collectively the total of all radii will be high. As blood should definitely pass through all capillaries the pressure drop across one capillary will be lower to that of an artery. This can be explained as follows too.When you consider a single capillary, its radius is very small. Therefore according to the equation,
    R = resistivity × l/A
    Resistance across a capillary will be higher than that across an artery.Due to the higher resistance, pressure drop will be high ( like when resistance of a certain resistor is high , the voltage drop across it becomes less according to V = IR.) across a capillary.
    Consider the following analogy for better understanding.
    Think of the blood as electric current, it is a viscous fluid. Imagine you are moving through an electrical circuit.
    So as you're going through the circuit , you go through resistive wires the whole time... and at certain points you also hit resistors. There is a drop in voltage equal to V=IR at each resistor right? When you go through a wire, before an after these resistors, there is also a small loss of voltage to the wire itself. Across the whole LENGTH of the thin wire you are slowly losing voltage to the wire itself whether you hit the "resistors" or not. This loss is dependent on the resistivity of the wire which depends on its cross sectional area and the total length you are travelling across. The "resistors" you pass can be viewed as specially coiled wires of specific length and radius... packaged into a "black box" that drops a specific (usually relatively high compared to circuit's wire) "resistance". So really the resistors we spoke of before , are no different than the wire we are travelling through , but they are packaged into a cool black box.

    So how does this relate to blood?

    As you travel through the body's "wires" ( the capillaries), blood loses its "voltage" (pressure) every bit of length you travel just like the charges passing through a circuit lose voltage. The voltage after a specific resistor is the original voltage minus that of the voltage loss over the resistor right? You can never get more voltage in the middle of a circuit unless you have a battery (an electron "pump") which is equivalent to a heart. The voltage of a circuit is analogous to the pressure for a circulatory system.
    The pressure when blood exits the heart is the "battery voltage". The voltage you lose over the wire is the pressure loss over the length of the vein/artery or even a system like all the capillaries ( Here the capillaries can be considered as the resistors while veins and arteries are wires.) So we know that voltage drops more through the resistors than through the conducting wire. Therefore, drop in pressure when you move into a capillary from an artery is lower than the pressure drop across an artery.Voltage along the journey of a circuit always goes from high to low as you go... so does the pressure in the circulatory system, THAT IS WHY THE PRESSURE DROPS. You start at the heart with 140mmhg ( 35 Volts) let's say, you go through 10 meters of artery (the wire) then pass through something like the capillary network (a resistor which is really just a collection of wires)... plot the voltage in your head vs distance traveled. This a conceptual tool that can make blood flow seem more intuitive. Just remember that the resistance of the circulatory system obeys rules similar but not mathematically identical to the resistivity of wire.


    Hope the analogy helps
     
    Last edited: Jan 16, 2017
  6. Jan 16, 2017 #5

    rcgldr

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    The relationship between greater speed and lower pressure is one of coexistence, not cause and effect. A fluid or gas will accelerate from a higher pressure area towards a lower pressure area. Bernoulli's equation relates the change in pressure versus change in speed during the transition. In the case of a narrowing pipe, the mass flow is constant (otherwise mass would be accumulating at some point). Since the mass flow is constant, the speed is greater when the pipe's cross-sectional area is less. The increase in speed coexists a pressure gradient that decreases as the water's speed increases.

    This is the way I've read an explanation for this. Assume that the total mechanical energy per unit volume of water is constant. If the water is stationary, then this mechanical energy is related to the kinetic energy related to the random movement and collisions between the molecules of the water. If the water is flowing then some of that random movement is organized into the net forward component of velocity. In a pipe, the pressure that the water exerts on the pipe is related to the collisions of the water molecules with the pipe. These collisions require a component of velocity perpendicular to the flow. As the flow speed increases, the component of velocity perpendicular to the flow decreases since more of what was random velocity is organized into forward velocity. This reduces the pressure as sensed by the walls of the pipe.
     
    Last edited: Jan 16, 2017
  7. Jan 16, 2017 #6

    boneh3ad

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    Be careful with this. Blowing (with your mouth) over a sheet of paper is not a good example, as what is actually occurring is quite a bit more complicated than just Bernoulli's principle. In fact, Bernoulli's principle is a fairly poor discriptor of what is actually occurring there.

    Be careful with your language here, as you aren't exactly correct. You neglect the possibility that a system already has kinetic energy initially. A flow can move from a low pressure region into a high pressure region if it already has enough kinetic energy to overcome that pressure gradient (i.e. enough kinetic energy to be converted back into potential energy in the form of static pressure).

    Again, this is not true. A pressure gradient is what provides the force for a fluid flow, so this is what provides acceleration, not velocity. The velocity can still be nonzero without a pressure gradient. It can't accelerate.

    Your intuition on dissipation of kinetic energy serves you well. The heart essentially must provide enough pressure to push blood through all your blood vessels in your body and back to itself in order to keep blood flowing. The force it is fighting is essentially viscous drag against the walls of your blood vessels. So, if the heart provides a certain total pressure to the outgoing blood, the flow will slowly lose energy (total pressure) as it passes through your system due to viscous dissipation. So, the total pressure is higher at the exit of the heart than it is later in your arteries, which is higher than in the smaller arteries that branch off, which is higher than the capillaries that follow, which is then higher than the veins that follow that and so on.

    This holds (approximately) when you are laying flat. If you are standing up, then you also have to factor in gravity, which will add or subtract from the total pressure depending on the situation. It also gets more complicated if you are moving and you have muscles that are squeezing blood vessels and things like that. Still, the above model is one that is good just to get the general idea.

    I assure you that the total volume flow rate through the body is constant (though the volume flow rate through a single capillary is certainly a lot smaller than a single given vein). Any network of blood vessels branching off of the same larger vessel will, combined, have the same volume flow rate as that larger vessel (not accounting for the transport of nutrients, oxygen, etc. across the walls into surrounding tissue, which should be a relatively small effect. If this wasn't true, blood would pool somewhere in your body and you'd be in a world of hurt.

    Bernoulli's equation does have some important limitations, however, and in this case, the most important is that it assumes the flow is inviscid, which is not a good model for the circulatory system. Not only is viscosity important to blood flow, but it is also a complicated non-Newtonian fluid, so modeling it is even more difficult than just assuming it is something like water.

    Poiseuille's equation can do an okay job of handling blood flow, but remember that blood is a non-Newtonian fluid, so you either have to use an effective viscosity model in place of ##\mu## or else live with the inaccuracies that come along with assuming the flow to be Newtonian when it is not. Also consider that capillaries can be so small that only single blood cells can pass through at times, so in those cases, the flow isn't even really single-phase, and more properly looks like a multiphase flow of what is essentially water and large, flexible particles. It's a really complicated problem.

    Even assuming you can live with the limitations I discussed above, this is poor logic. Poiseuille's equation does not deal with the total of all radii in a system. It deals with the pressure drop through individual tubes. As such, you can't just say that ##r## is very large because there are lots of capillaries. Poiseuille's equation is as follows:

    [tex]\Delta p = \dfrac{128\mu L Q}{\pi d^4}[/tex]

    So, the pressure drop is inversely proportional to ##d^4##. With a capillary, ##d## is absolutely tiny, so the pressure drop per length is going to be very, very high. Of course, the pressure drop per length is also going to be proportional to volumetric flow rate, which is also quite a bit smaller than in arteries. Since ##Q## is proportional to ##v## and ##d^2##, that means that the bottom term is still likely to dominate and the pressure drop per length through a capillary is going to be high. You even state this right after the above quoted text and essentially contradict yourself.

    There is also a question of length. If arteries had appreciably greater length than capillaries then they could make up ground. I don't know the relative total lengths of the two, though.

    However, none of this addresses the OP's original question, which pertained to the pressure in capillaries, not the pressure drop. His/her original intuition about viscous dissipation was the correct one there.
     
  8. Jan 16, 2017 #7
    If there's no pressure gradient across a tube, how will the flow inside be maintained?
     
  9. Jan 16, 2017 #8
    What you have said is correct but, here it is not the case. It is true that if a certain flow has enough kinetic energy to overcome the pressure gradient, it will move from lower pressure region to higher pressure region. Then its mechanical kinetic energy will be converting to potential energy. So then can you explain how pressure drops while velocity increases? Here, the fact that I have exaggerated is that systems LIKE to move from high potential to low potential. So velocity can increase while pressure decreases.I have not said anything that is contradictory to what you have said above. Although the what you say is true, can you please explain how it relates to this situation?
     
  10. Jan 16, 2017 #9

    rcgldr

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    This describes a flow in an adverse pressure gradient. Think of what happens when air or water flows from a smaller diameter pipe into a larger diameter pipe, the speed decreases and the pressure increases. You could also consider the case of impact pressure into a pitot tube. The air moving with respect to the pitot tube is slowed down to zero speed within the tube, the pressure sensed inside the tube is greater than the pressure of the moving air.

    http://en.wikipedia.org/wiki/Impact_pressure
     
  11. Jan 16, 2017 #10

    boneh3ad

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    This statement contradicts itself. I assure you what I said is both correct and is the case here.

    Simple, this is a velocity increase, i.e. the flow is accelerating. Acceleration requires a force. That force is a pressure difference. If the velocity was constant but nonzero, then the flow would have zero acceleration (and zero pressure gradient in the inviscid case) but there would still be flow. If the pressure was increasing, then the velocity would be decreasing, i.e. a deceleration.

    No, they like to move when they are already moving and they like to stay still when they start out still. They like to accelerate from high potential to low potential.

    Yes you have. See the bolded sections of the following quote:

    In order:
    • If pressure is constant everywhere, there will be no acceleration in the tunnel. Flow can still occur.
    • Flow need not move from high potential to low potential. However, acceleration must point that direction.
    • Again, velocity can appose a pressure gradient. Acceleration cannot.

    In the inviscid case (which is assumed in any situation involving Bernoulli's equation), then no pressure gradient means constant flow. If viscosity is accounted for, then you do require a pressure gradient to maintain constant flow.
     
  12. Jan 16, 2017 #11
    Dear boneh3ad,

    Would an explanation for flow without any pressure gradients i.e. constant velocity be due to say inertia? I can't seem to see the fluid move in any other way.

    Lastly, at the molecular and atomic level, how would dissipation of energy translate into a loss of pressure exerted against the vascular walls? Is this in relation to acceleration(force=ma; P=F/Area)
    So when there is a pressure gradient, there is acceleration, and as the blood travels down to the capillaries it loses kinetic energy and this may decelerate the particles. How would this decelerating the particles contribute to the decrease in the Force that is orientated perpendicular to the flow.



    Thank you for your help!
     
  13. Jan 16, 2017 #12

    boneh3ad

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    Sure. Fluids are just like any other matter. If they are moving and don't have a force opposing their motion, they will keep moving. The only issue is that they are much more convenient to treat as a continuously flowing medium rather than rigid bodies, so the equations take on a different form. They are still based on the same physical principles, though.

    Typically, this energy is dissipated as heat. It's not typically a huge amount of heat, but heat nonetheless.

    A pressure gradient perpendicular to a flow won't speed it up or slow it down, necessarily. It will tend to bend the flow. For example, if you have a vortex, the pressure will decrease as you approach the center, corresponding to an inward force. The speed of the flow doesn't change, but the direction does. This is directly analogous to a centripetal force.
     
  14. Jan 16, 2017 #13
    I apologize boneh3ad, I did not explain my last question properly as I was having trouble even thinking about this problem.

    I was trying to integrate what rc had mentioned on how ,"when blood moves faster, there is less of a perpendicular force direction, therefore there is less pressure on the vascular walls in that scenario." I am trying to wrap my mind around the relationship between a decrease in kinetic energy and pressure at the level of particles and their effect on the wall? As we go from the artery to the capillaries, how is this decrease in kinetic energy(due to increase in resistance) causing the blood to hit the walls less, and therefore less overall pressure?

    Maybe there are other components that I am neglecting to add into this system?

    Thank you again!
     
  15. Jan 16, 2017 #14
    Can you please kindly explain further the above situation you are referring to?
    Thank you.
     
  16. Jan 17, 2017 #15

    boneh3ad

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    The statement you typed literally said two different things that were contradictory.
     
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