Brown University Physics 0050/0070Physics Department FLUID FLOW THROUGH A TUBEIn this experiment we will determine how well a physical relationship (also called “law”),namely Poiseulle’s equation, applies.In the supplementary reading material this equation was derived as πpa 4 Q= 8ηlwhere Q is the flow rate in units of volume per second; p is the pressure differencebetween the ends of the tube; a, the tube radius; l , the length of the tube, and η theviscosity of the fluid.We will measure in the experiment Q, as a function of these parameters mentioned above.One parameter will be varied while the others are kept constant, and Q, as a result ofvariation in this parameter, plotted on log-log paper. The reason for choosing log- logpaper is that it brings out a power law dependence, such as Q ∝ a 4 and Q ∝ l −1 muchbetter than linear graphs. Convince yourself that on log-log paper the exponent (4 or -1here) just determines the slope of a straight line, while on linear paper, curved graphswith a continuously changing slope would result. If we write the dependence of Q, on thevarious parameters mentioned in terms of an exponential power, we expect that Q getslarger when, everything else remaining the same, the pressure increases or the radiusincreases; Q gets smaller when the tube length increases or the liquid gets more viscous.One would not expect a significant effect from the material the tube wall is made from,unless it were very rough and prevented a smooth laminar flow to take place.Temperature could enter indirectly through the dependence of the viscosity ontemperature.Collecting all these terms results in a formula of the kind: Kp δ aα Q= γ β η lwhere K is a proportionality constant.We can determine the coefficients α, β, and γ by collecting the fluid coming out of thetube during a given amount of time as a function of that particular parameter. Forexample, to find α, one would measure the amount of water flowing through varioustubes of different diameters, everything else (length, pressure difference) remaining thesame. One then writes the equation: ⎡ Kp δ ⎤ log Q = log ⎢ γ β ⎥ + α log a ⎣η l ⎦This is a straight line of the form: logQ = B +α log a , with a as slope in a logQ versusloga plot. In a similar way one can test for the other parameters and thus check theequation.It should be kept in mind that Poiseuille’s equation holds only for laminar flow. If thefluid moves too fast, smooth shear drag between layers disappears and turbulence sets in.The transition is determined by Reynold’s number, Re. As mentioned in the supplement,
Brown University Physics 0050/0070Physics Department 2v aρ pa 2 a3 p2 ρRe = or if one substitutes = v , Re = . η 8ηl 4η 2 lThis transition occurs when Re equals 2000. PROCEDURE We will use the apparatus shown in the figure. The reservoir has small openings in its side closed by corks. By removing a cork, the water level is set at the height, h, of the resulting opening above the horizontal pipe. The pressure difference between the two ends of the pipe, p, is the height of the water column, thus given by p = ρgh . There is a supply of pipes of various radii and lengths. The only liquid to be used here is water so that the viscosity η, will not be varied. Attach the desired pipe to the reservoir; set the water level to such a height that the Reynold’s number is less than 2000. Adjust the water flow from the faucet, so that the water level in the reservoir remains stable at the desired height. Put the graduated beaker under the opening of the tube and measure the time it takes to collect 500 ml.This will give you the flow rate Q.Measure the flow rate for 4 different radii tubes (keeping the length l and pressure pconstant). Repeat the procedure for 2 different lengths and 3 different pressures. Onecould plot log Q, versus the log of the radius a, the length l and the height h. The slope ofthese lines would give you the various exponents. Instead of calculating the log eachtime, it is much easier to plot the value directly, but on log-log graph paper.Now choose your parameters such that Re >>2000. Measure the flow rate in thisturbulent case and observe the deviation from the lines found earlier.There is a short second experiment designed to test Bernoulli’s equation in a simple case.The apparatus is as shown in the figure. We use the same reservoir as was used earlier.As water flows through the horizontal tube, the water level in the vertical tubes will riseproportionately to the water pressure. Applying Bernoulli’s equation to points l and 2gives: 1 2 1 2p1 + ρv1 = p 2 + ρv 2 2 2p1 = p 0 + ρgh1 , p 2 = p 0 + ρgh2Thus; 1 1 2gh1 + v12 = gh2 + v 2 2 2where h1 and h2 refer to the height of the
Brown University Physics 0050/0070Physics Departmentwater level in the vertical tubes above the center of the main tube.To obtain the velocities we note that the flow rate is given by Q = Av , where A is thecross-section of the tube.Bernoulli’s equation gives; 2 2 1⎡Q⎤ 1⎡Q ⎤ gh1 + ⎢ ⎥ = gh2 + ⎢ ⎥ 2 ⎣ A1 ⎦ 2 ⎣ A2 ⎦Set the water level in the reservoir to about 12-cm. Measure h1, h2 and the flow rate Q.With these values calculate the left and right hand side of the above equation. Howclosely are they equal?Bernoulli’s equation is not applicable for turbulent flow. To see this qualitatively, use atube with an abrupt transition between narrow and wide diameter tubes. Thisdiscontinuity introduces turbulence. Calculate now both sides of the equation and seehow close to equal they are.Questions:1) What is the most likely cause for differences between Poiseuille’s theoretical formulaand your experiments?2) What is the most likely cause in your experiment for deviations from Bernoulli’sequation? FLUID MECHANICS SUPPLEMENTIn Chapter 14 of the book by Bueche, you find various equations of hydrodynamics,notably the equation of continuity and Bernoulli’s equation.We want here to discuss the flow of water through a cylindrical tube, which is notdiscussed in the book.Along the walls of the cylinder the fluid does not flow: it is stationary. There is a gradualincrease of the fluid velocity, from zero at the wall to maximum at the center. The friction dvforce, F, that resists the velocity, is given by F = −ηA where η = viscosity, A, the dy dvcontact area and the velocity gradient, that is, the change of the velocity in going from dythe wall (v = 0) towards the center (vmax) . The center fluid drags the next layer, whichslips somewhat behind etc., until the fluid layer at the wall, which does not move at all.This type of fluid flow is called laminar flow.This friction is the cause of a pressure drop along a cylindrical tube. Poiseuille derived aformula that relates the quantity of liquid, Q, (the volume ) flowing per second through a
Brown University Physics 0050/0070Physics Departmentcylindrical tube of radius a and length l, with the pressure difference, p, between bothends. This formula is: πpa 4 Q= (1) 8ηlThe constant η is the viscosity defined earlier. We will derive this equation. Let usconsider a cylinder with radius r, where r < a. At the entrance surface is a force +πr2p1,and at the exit surface —πr2p2. The force at the cylindrical surface, the friction force, is dvgiven by − η 2πrl according to the definition for η. Since the liquid does not drexperience acceleration but flows with constant velocity, the sum of these three forces dvmust equal zero. Since is negative if we define r in the usual way as the distance from drthe center, we obtain: dv πr 2 p1 − πr 2 p 2 + 2πrηl =0 (2) dr p − p2This yields dv = − 1 rdr and after integration: 2ηl − p(r 2 + c) v= (3) 4ηl 1where c represents the integration coefficient stemming from ∫ rdr = r 2 + c . We know 2that v = 0 at the wall, that is, when r = a. p1—p2 has been written as p, for short.Application of the boundary condition gives us c and results in: p(a 2 − r 2 ) v= (4) 4ηlThis velocity profile is sketched in the figure. The total volume of liquid, Q, flowingthrough a cross section per second can be obtained from: p (a 2 − r 2 )2πrdr a Q=∫ (5) 0 4ηlThis is namely the sum of the fluid flowing through rings bounded by radius r and r + dr.One obtains, using: a 1 a 1 πpa 4 ∫ 0 rdr = a 2 and ∫ r 3 dr = a 4 ; Q = 2 0 4 8ηlWe will now discuss a few applications of this formula. The average velocity of theliquid, v can be obtained from: Q pa 2 =v = πa 2 8ηlReynold made a study of the applicability of Poiseuille’s equation and concluded thatabove a certain v laminar flow breaks up and turbulence sets in. This happens when: 2v r ρ = Reynold’s number Re is > 2000. η
Brown University Physics 0050/0070Physics DepartmentIn turbulent flow there vortex formation, the flow is irregular and noisy. Blood flowthrough the narrow capillaries is laminar as one verifies easily. However, through theaorta, with v = 0.6m/sec, r = 0.012 m, η= 0.003 kg/m sec and a densityρ=1 gram/cm3 = 1000kg/m3, one calculates Re = 4800. So blood flow through the aorta isturbulent. There are several other important medical implications of equation 1.If one requires rapid blood transfusion, increasing the height of the bottle with blood by afactor of two increases the flow by that same factor, but increasing the diameter of thetube by two increases the flow by a factor of sixteen.In the case of a shock victim, the person’s body temperature often drops. This results in asignificant increase of the viscosity and thus a decrease in blood flow. To keep the bodytemperature up, covering of the victim with a blanket and if appropriate giving warmdrinks helps.Finally, it should be remarked that our simple model could only approximate blood flow.Veins are not solid tubes, but are elastic, expanding somewhat with increased pressure.Also blood is not as simple as water in its viscosity. For example, in narrow capillariesthe blood cells line up, reducing the viscosity.