previous inex next Calculating Viscous Flow: Velocity Profiles in Rivers an Pipes Michael Fowler, UVa 9/8/1 Introuction In this lecture, we ll erive the velocity istribution for two examples of laminar flow. First we ll consier a wie river, by which we mean wie compare with its epth (which we take to be uniform) an we ignore the more complicate flow pattern near the banks. Our secon example is smooth flow own a circular pipe. For the wie river, the water flow can be thought of as being in horizontal sheets, so all the water at the same epth is moving at the same velocity. As mentione in the last lecture, the flow can be picture as like a pile of printer paper left on a sloping esk: it all slies own, assume the bottom sheet stays stuck to the esk, each other sheet moves ownhill a little faster than the sheet immeiately beneath it. For flow own a circular pipe, the laminar sheets are hollow tubes centere on the line own the mile of the pipe. The fastest flowing flui is right at that central line. For both river an tube flow, the rag force between ajacent small elements of neighboring sheets is given by force per unit area F v z A where now the z-irection means perpenicular to the small element of sheet. A Flowing River: Fining the Velocity Profile For a river flowing steaily own a gentle incline uner gravity, we ll assume all the streamlines point in the same irection, the river is wie an of uniform epth, an the epth is much smaller than the with. This means almost all the flow is well away from the eges (the river banks), so we ll ignore the slowing own there, an just analyze the flow rate per meter of river with, taking it to be uniform across the river. The simplest basic question is: given the slope of the lan an the epth of the river, what is the total flow rate? To answer, we nee to fin the spee of flow v(z) as a function of epth (we know the water in contact with the river be isn t flowing at all), an then a the flow contributions from the ifferent epths (this will be an integral) to fin the total flow. The function v(z) is calle the velocity profile. We ll prove it looks something like this:
Stream velocity v(z) at ifferent epths (For a smoothly flowing river, the ownhill groun slope woul be imperceptible on this scale.) But how o we begin to calculate v(z)? Recall that (in an earlier lecture) to fin how hyrostatic pressure varie with epth, we mentally separate a cyliner of flui from its surrounings, an applie Newton s Laws: it wasn t moving, so we figure its weight ha to be balance by the sum of the pressure forces it experience from the rest of the flui surrouning it. In fact, its weight was balance by the ifference between the pressure unerneath an that on top. Taking a cue from that, here we isolate mentally a thin layer of the river, like one of those sheets of printer paper, lying between height z above the be an z z. This layer is moving, but at a steay spee, so the total force on it will still be zero. Like the whole river, this layer isn t quite horizontal, its weight has a small but nonzero component ragging it ownhill, an this weight component is balance by the ifference between the viscous force from the faster water above an that from slower water below. Bear in min that the iagram below is at a tiny angle to the horizontal: Forwar viscous rag from flui above mgsin ownhill Backwar viscous rag from flui below z + z z Obviously, for the forces to balance, the backwar rag on the thin layer from the slower moving water beneath has to be stronger than the forwar rag from the faster water above, so the rate of change of spee with height above the river be is ecreasing on going up from the be. Let us fin the total force (which must be zero) on one square meter of the thin layer of water between heights z an z z: First, gravity: if the river is flowing ownhill at some small angle, this square meter of the 3 layer (volume z m 1 m z m, ensity ) experiences a gravitational force mg sin gz tugging it ownstream (taking the small angle approximation, sin.)
3 Next, the viscous rag forces: the square meter of layer experiences two viscous forces, one from the slower water below, equal to v z /, tening to slow it own, one from the faster water above it, v z z /, tening to spee it up. Gravity must balance out the ifference between the two viscous forces: g z v z z v z 0 We can alreay see from this equation that, unlike the flui between the plates, v(z) can t possibly be linear in z the equation woul not balance if v / were the same at z an z z! Diviing throughout by an by z, v z z v z z g. Taking now the limit z 0, an recalling the efinition of the ifferential f x f x x f x lim x x 0 x we fin the ifferential equation v z g. The solution of this equation is easy: with C, D constants of integration. g vz z Cz D Remember that the velocity v(z) is zero at the bottom of the river, z = 0, so the constant D must be zero, an can be roppe immeiately. But we re not through we haven t foun C. To o that, we nee to go to the top. Velocity Profile Near the River Surface What happens to the thin layer of river water at the very top the layer in contact with the air? Assuming there is negligible win, there is essentially zero parallel-to-the-surface force from above.
4 So the balance of forces equation for the top layer is just gz vz 0. We can take this top layer to be as thin as we like, so let s look what happens in the limit of extreme thinness, z 0. The term gzthen goes to zero, so the other term must as well. Since is constant, this means v z 0 at the surface z h. So the velocity profile function v z has zero slope at the river surface. With this new information, we can finally fix the arbitrary integration constant C. Now the velocity profile so v z gives an h 0 g h C. g vz z Cz, v z g z C, Putting this value for C into v(z) we have the final result: g vz z h z. This velocity profile v(z) is half the top part of a parabola:
5 Total River Flow Knowing the velocity profile v(z) enables us to compute the total flow of water in the river. As explaine earlier, we re assuming a wie river having uniform epth, ignoring the slowown near the eges of the river, taking the same v(z) all the way across. We ll calculate the flow across one meter of with of the river, so the total flow is our result multiplie by the river s with. The flow contribution from a single layer of thickness z at height z is vz z cubic meters per secon across one meter of with. The total flow is the sum over all layers. In the limit of many infinitely thin layers, that is, z 0, the sum becomes an integral, an the total flow rate h h I v( z) ( g / ) z( h z) ( g / 3 ) h 0 0 3 in cubic meters per secon per meter of with of the river. It is worth thinking about what this result means physically. The interesting part is that the flow is proportional to h 3, where h is the epth of the river. So, if there s a storm an the river is twice as eep as normal, an flowing steaily, the flow rate will be eight times normal. Exercise: plot on a graph the velocity profiles for two rivers, one of epth h an one h, having the same values of, g, an. What is the ratio of the surface velocities of the two rivers? Suppose that one meter below the surface of one of the rivers, the water is flowing 0.5 m.sec -1 slower than it is flowing at the surface. Woul that also be true of the other river? Flow own a Circular Tube (Poiseuille Flow) The flow rate for smooth flow through a pipe of circular cross-section can be foun by essentially the same metho. (This was the flow pattern analyze by Poiseuille an use by him to confirm Newton s postulate of flui flow behavior being governe by a coefficient of viscosity.) In the pipe, the flow is fastest in the mile, an the water in contact with the pipe wall (like that at the river be) oesn t flow at all. The river s flow pattern was most naturally analyze by thinking of flat layers of water, all the water in one layer having the same spee. What woul be the corresponing picture for flow own a pipe? Here all the flui at the same istance from the center moves own the pipe at the same spee instea of flat layers of flui, we have concentric hollow cyliners of flui, one insie the next, with a tiny ro of the fastest flui right at the center. This is again laminar flow, even though this time the sheets are rolle into tubes. The blue circular area on the cross-section of the pipe shown below represents one of these cyliners of flui all the flui between r an rrfrom the central line. Each of these hollow cyliners of water is pushe along the pipe by the pressure ifference between the ens of the pipe. Each feels viscous forces from its two neighboring cyliners: the next bigger one, which surrouns it, tening to slow it own, but the next smaller one (insie it) tening to spee it up. Writing own the ifferential equation is a little more tricky that for the
6 river, because we must take into account that the two surfaces of the hollow cyliner (insie an outsie) have ifferent areas, rl r r L. It turns out that the velocity profile is an again parabolic: the etails are given below. a Velocity profile for laminar flow own a circular pipe Circular Pipe Flow: Mathematical Details Suppose the pipe has raius a, length L an pressure rop P, pressure rop per meter P/ L. Let us focus on the flui in the cyliner between r an rrfrom the line own the mile, an we ll take the cyliner to have unit length, for convenience. The pressure force maintaining the flui motion is the ifference between pressure x area for the two ens of this one meter long hollow cyliner: P net pressure force r r. L (We re assuming r r, since we ll be taking the r 0 limit, so the en area r r. The equality becomes exact in the limit.) This force exactly balances the ifference between the outer surface viscous rag force from the slower surrouning flui an the inner viscous force from the central faster-moving flui, very similar to the situation in the previous analysis of river flow. Using F / A v z /, an remembering that the inner an outer surfaces of the cyliner have slightly ifferent areas, that P is positive, but v / r is negative, the force equation is: Rearranging, P r r r r v r r r v r. L r r
7 v r r v r r r r P r r r L r v r r r in the limit r 0, remembering the efinition of the ifferential (see the similar analysis above for the river). This can now be integrate to give v P r r C r L where C is a constant of integration. Diviing both sies by r an integrating again P r vr C ln r D. L 4 The constant C must be zero, since physically the flui velocity is finite at r = 0. The constant D is etermine by the requirement that the flui spee is zero where the flui is in contact with the tube, at r = a. The flui velocity is therefore v r P L a r To fin the total flow rate I own the pipe, we integrate over the flow in each hollow cyliner of water: in cubic meters per secon. a a 3 P a r r P I rv rr r a L 4 8 L 0 0 Notice the flow rate goes as the fourth power of the raius, so oubling the raius results in a sixteen-fol increase in flow. That is why narrowing of arteries is so serious. 4. 4 previous inex next Thanks to Lina Fahlberg-Stojanovska for spotting a sign error in the earlier version.