8.01SC Chapter 28: Fluid Dynamics

Chapter 28 Fluid Dynamics 28.1 Ideal Fluids An ideal fluid is a fluid that is incompressible and has no internal resistance to flow (zero viscosity). In addition, ideal fluid particles undergo no rotation about their center of mass (irrotational). An ideal fluid can flow in a circular pattern, but the individual fluid particles are irrotational. Real fluids exhibit all of these properties to some degree, but we shall often model fluids as ideal in order to approximate the behavior of real fluids. When we do so, one must be extremely cautious in applying results associated with ideal fluids to non-ideal fluids. For a non-ideal fluid, the differential equations describing the motion of the fluid are quite complicated and beyond the scope of this discussion. 28.2 Velocity Vector Field The flow of a fluid like water consists of the movement of individual particles, (water molecules). These particles interact with each other through forces. Because the number of particles is very large applying the laws of motion to each individual particle in the fluid would be an extremely difficult computation problem. To circumvent this problem, describe the state of a moving fluid by specifying the velocity of the fluid at each point in space and at each instant in time. Using Cartesian coordinates, a point in space-time is specified by the ordered triple (x, y, z) for the special location and the variable t to describe the instant in time. The distribution of fluid ! velocities is then given by the vector function v(x, y, z,t) . This represents the velocity of  the fluid at the point (x, y, z) at the instant t . The quantity v(x, y, z,t) is called the velocity vector field. It can be thought of at each instant in time as a collection of vectors, one for each point in space whose direction and magnitude describes the direction and magnitude of the velocity of the fluid at that point (Figure 28.1). This description of the velocity vector field of the fluid refers to fixed points in space and not to moving particles in the fluid. v(x, y, z,t) Velocity vector field at time t 28-1 Figure 28.1: Velocity vector field for fluid flow at time t The three functions vx (x, y, z,t) , v y (x, y, z,t) , and vz (x, y, z,t) to describe the components of the velocity vector field ! v(x, y, z,t) = vx (x, y, z,t) î + v y (x, y, z,t) ĵ+ vz (x, y, z,t)k̂ . (28.2.1) The three component functions are scalar fields. The velocity vector field is in general quite complicated for a three-dimensional time dependent flow. For most flows, the velocity field varies in time. A steady flow is a model in which the velocity field does not change in time,  ∂v (x, y, z,t)  =0 ∂t (steady flow) . (28.2.2) For steady flows the velocity field is independent of time,  v(x, y, z) = vx (x, y, z) î + v y (x, y, z) ĵ+ vz (x, y, z)k̂ (steady flow) . (28.2.3) although the velocities may still vary in space (non-uniform steady flow). . v(A) A . B v(B) . . v(A) v(C) C (a) trajectory of particle that is located at: point A at time t1 ; point B at time t2 ; and point C at time t3 . . v(B) B A (b) trajectory of particle 2 that is located at: point A at time t 2 ; and point B at time t 3 . Figure 28.2: (a) trajectory of particle 1, (b) trajectory of particle 2 Let’s trace the motion of particles in an ideal fluid undergoing steady flow during a succession of intervals of duration Δt . Consider particle 1 located at point A with ! ! coordinates (xA , yA , zA ) . At the instant t1 , particle 1 has velocity v (xA , yA , zA ) = v(A) . During the time [t1 ,t2 ] , where t2 = t1 + Δt1 , the particle moves to point B arriving there at ! ! the instant t2 . At point B , the particle now has velocity v (xB , yB , z B ) = v(B) . During the next interval [t2 ,t3 ] , where t3 = t2 + Δt , particle 1 will move to point C arriving there at ! ! instant t3 , where it has velocity v (xC , yC , zC ) = v(C) . (Figure 28.2(a)). Because the flow 28-2 has been assumed to be steady, at instant t2 , a different particle, particle 2, is now located ! at point A but it has the same velocity v (xA , yA , zA ) as particle 1 had at point A and hence will arrive at point B at the end of the next interval, at the instant t3 (Figure 28.2(b)). In this way every particle that lies on the trajectory that our first particle traces out in time will follow the same trajectory. This trajectory is called a streamline. The particles in the fluid will not have the same velocities at points along a streamline because we have not assumed that the velocity field is uniform. A set of streamlines for an ideal fluid undergoing steady flow in which there are no sources or sinks for the fluid is shown in Figure 28.3. end-cap 2 end-cap 1 Figure 28.3: Set of streamlines for an ideal fluid flow Figure 28.4: Flux Tube associated with set of streaml 28.3 Mass Continuity Equation ! Let v(x, y, z,t) denote the vector field associated with the fluid flow. Let ρ (x, y, z,t) denote the density of the fluid at each point in space and at each instant in time. The density function is an example of a scalar field because there is only one number with appropriate units associated with each point in space at each instant in time. A set of closely separated streamlines that form a flow tube are shown in Figure 28.4. The flow tube has two open surfaces (end-caps 1 and 2) of areas A1 and A2 , respectively, that are perpendicular to the velocity of the fluid. All fluid particles that enter end-cap 1 must follow their respective streamlines, therefore they all leave end-cap 2. If the streamlines that form the tube are sufficiently close together, then the velocity of the fluid in the vicinity of each end-cap surfaces can be assumed to be uniform. 28-3 length dl1 = v1dt end-cap 1 area A1 length dl2 = v2 dt end-cap 2 area A2 Figure 28.5: Mass flow through flux tube Let v1 denote the speed of the fluid near end-cap 1 and v2 denote the speed of the fluid near end-cap 2. Let ρ1 denote the density of the fluid near end-cap 1 and ρ2 denote the density of the fluid near end-cap 2. The amount of mass that enters and leaves the tube in a time interval dt can be calculated as follows (Figure 28.5): consider a small volume of space of cross-sectional area A1 and length dl1 = v1dt near end-cap 1. The mass that enters the tube in time interval dt is dm1 = ρ1dV1 = ρ1 A1dl1 = ρ1 A1v1dt . (28.2.4) In a similar fashion, consider a small volume of space of cross-sectional area A2 and length dl2 = v2 dt near end-cap 2. The mass that leaves the tube in the time interval dt is then dm2 = ρ2 dV2 = ρ2 A2 dl2 = ρ2 A2 v2 dt . (28.2.5) An equal amount of mass that enters end-cap 1 in the time interval dt must leave end-cap 2 in the same time interval, thus dm1 = dm2 . Therefore using Eqs. (28.2.4) and (28.2.5), we have that ρ1 A1v1dt = ρ2 A2 v2 dt . Dividing through by dt implies that ρ1 A1v1 = ρ2 A2v2 (steady flow) . (28.2.6) Eq. (28.2.6) generalizes to any cross sectional area A of the thin tube, where the density is ρ , and the speed is v , (28.2.7) ρ Av = constant (steady flow) . Eq. (28.2.6) is referred to as the mass continuity equation for steady flow. For an incompressible fluid, Eq. (28.2.6) becomes 28-4 A1v1 = A2 v2 (incompressable fluid, steady flow) . (28.2.8) Consider the steady flow of an incompressible fluid with streamlines and closed surface formed by a streamline tube shown in Figure 28.5. According to Eq. (28.2.8), when the spacing of the streamlines increases, the speed of the fluid must decrease. Therefore the speed of the fluid is greater entering end-cap 1 then when it is leaving end-cap 2. When representing fluid flow by streamlines, regions in which the streamlines are widely spaced have lower speeds than regions in which the streamlines are closely spaced. 28.4 Bernoulli’s Principle Consider the case of an ideal fluid that undergoes steady flow. Let P(x, y, z,t) denote the pressure scalar field at each point in space and at each instant in time. The equation of state relates pressure, density, and speed of the flow at different points in the fluid. A steady horizontal flow is shown in the overhead view in Figure 28.6. The flow is represented by streamlines along with a flow tube. Consider the motion of a fluid particle along one streamline passing through points A and B in Figure 28.6. The crosssectional area of the flow tube at point A is less than the cross-sectional area of the flow tube at point B . Figure 28.6 Overhead view of steady horizontal flow: in regions where spacing of the streamlines increases, the speed of the fluid must decrease According to Eq. (28.2.8), the particle located at point A has a greater speed than the particle located at point B . Therefore a particle traveling along the streamline from point A to point B must decelerate. Because the streamline is horizontal, the force responsible is due to pressure differences in the fluid. Thus, for this steady horizontal flow in regions of lower speed there must be greater pressure than in regions of higher speed. Now suppose the steady flow of the ideal fluid is not horizontal, with the y -axis representing the vertical direction. A side view of the streamlines and flow tube for this steady flow are shown in Figure 28.7. 28-5 Figure 28.7: Side view of a non-horizontal steady flow Define the system by the mass contained in the flow tube shown in Figure 28.7. The external forces acting on the system are due to the pressure acting at the two ends of the flow tube and the gravitational force. Consider a streamline passing through points 1 and 2 at opposite ends of the flow tube. Assume that the flow tube is narrow enough such that the velocity of the fluid is uniform on the cross-sectional areas of the tube at points 1 and 2 . At point 1 , denote the speed of a fluid particle by v1 , the cross-sectional area by A1 , the fluid pressure by P1 , and the height of the center of the cross-sectional area by y1 . At point 2 , denote the speed of a fluid particle by v2 , the cross-sectional area by A2 , the fluid pressure by P2 , and the height of the center of the cross-sectional area by y2 . At the left end of the flow, tube in a time interval dt , a particle at point 1 travels a distance dl1 = v1dt . Therefore a small volume dV1 = A1dl1 = A1v1dt of fluid is displaced at the right end of the flow tube. In a similar fashion, at particle at point 2 , travels a distance dl2 = v2 dt . Therefore a small volume of fluid dV2 = A2 dl2 = A2 v2 dt is also displaced to the right in the flow tube during the time interval dt . Because the fluid is assumed to be incompressible, by Eq.(28.2.8), these volume elements are equal, dV ≡ dV1 = dV2 . There is a force of magnitude F1 = P1 A1 in the direction of the flow arising from the fluid pressure at the left end of the tube acting on the mass element that enters the tube. The work done displacing the mass element is then dW1 = F1dl1 = P1 A1dl1 = P1dV . (28.3.1) There is also a force of magnitude F2 = P2 A2 in the direction opposing the flow arising from the fluid pressure at the right end of the tube. The work done opposing the displacement of the mass element leaving the tube is then dW1 = − F2 dl2 = − P2 A2 dl2 = − P2 dV . (28.3.2) 28-6 Therefore the external work done by the force associated with the fluid pressure is the sum of the work done at each end of the tube dW ext = dW1 + dW2 = (P1 − P2 )dV . (28.3.3) In a time interval dt , the work done by the gravitational force is equal to dW g = −dm g( y2 − y1 ) = − ρ dVg( y2 − y1 ) . (28.3.4) The assumption that the fluid is ideal means that there are no frictional losses due to viscosity. The change in the potential energy of the mass in the flow tube (the system) is then (28.3.5) dU = −W g = ρ dVg( y2 − y1 ) . At time t , the kinetic energy of the system is the sum of the kinetic energy of the small mass element of volume dV = A1dl1 moving with speed v1 and the rest of the mass in the flow tube. At time t + dt , the kinetic energy of the system is the sum of the kinetic energy of the small mass element of volume dV = A2 dl2 moving with speed v2 and the rest of the mass in the flow tube. The change in the kinetic energy of the system is due to the mass elements at the two ends. Therefore dK = 1 1 1 dm2 v2 2 − dm1v12 = ρ dV (v2 2 − v12 ) . 2 2 2 (28.3.6) The work-energy theorem dW ext = dU + dK for system is then (P1 − P2 )dV = 1 ρ dV (v2 2 − v12 ) + ρ g( y2 − y1 )dV . 2 (28.3.7) Divide each term in Eq. (28.3.7) by the volume dV and rearrange terms, yielding 1 1 P1 + ρ gy1 + ρ v12 = P2 + ρ gy2 + ρ v2 2 . 2 2 Because points 1 and 2 were arbitrarily chosen, drop the subscripts and write Eq. (28.3.8) as 1 P + ρ gy + ρ v 2 = constant (ideal fluid, steady flow) . 2 (28.3.8) (28.3.9) Eq. (28.3.9) is known as Bernoulli’s Equation. 28-7 28.5 Worked Examples: Bernoulli’s Equation Example 28.1 Venturi Meter Figure 28.8 shows a Venturi Meter, a device used to measure the speed of a fluid in a pipe. A fluid of density ρ f is flowing through a pipe. A U-shaped tube partially filled with mercury of density ρ Hg lies underneath the points 1 and 2. Figure 28.8: Venturi Meter The cross-sectional areas of the pipe at points 1 and 2 are A1 and A2 respectively. Determine an expression for the flow speed at the point 1 in terms of the cross-sectional areas A1 and A2 , and the difference in height h of the liquid levels of the two arms of the U-shaped tube. Solution: Figure 28.8: Coordinate system for Venturi tube 28-8 We shall assume that the pressure and speed are constant in the cross-sectional areas A1 and A2 . We also assume the fluid is incompressible so the density ρ f is constant throughout the tube. The two points 1 and 2 lie on the streamline passing through the midpoint of the tube so they are at the same height. Using y1 = y2 in Eq. (28.3.8), the pressure and flow speeds at the two points 1 and 2 are related by 1 1 P1 + ρ f v12 = P2 + ρ f v2 2 . 2 2 (28.3.10) We can rewrite Eq. (28.3.10) as P1 − P2 = 1 ρ f (v2 2 − v12 ) . 2 (28.3.11) Let h1 and h2 denote the heights of the liquid level in the arms of the U-shaped tube directly beneath points 1 and 2 respectively. Pascal’s Law relates the pressure difference between the two arms of the U-shaped tube according to in the left arm of the U-shaped tube according to Pbottom = P1 + ρ f gd1 + ρ Hg gh1 . (28.3.12) In a similar fashion, the pressure at point 2 is given by Pbottom = P2 + ρ f gd2 + ρ Hg gh2 . (28.3.13) Therefore, setting Eq. (28.3.12) equal to Eq. (28.3.13), we determine that the pressure difference on the two sides of the U-shaped tube is P1 − P2 = ρ f g(d2 − d1 ) + ρ Hg g(h2 − h1 ). (28.3.14) From Figure 28.8, d2 + h2 = d1 + h1 , therefore d2 − d1 = h1 − h2 = −h . We can rewrite Eq. (28.3.14) as P1 − P2 = ( ρ Hg − ρ f )gh. (28.3.15) Substituting Eq. (28.3.11) into Eq. (28.3.15) yields 1 ρ (v 2 − v12 ) = ( ρ Hg − ρ f )gh . 2 f 2 (28.3.16) The mass continuity condition (Eq.(28.2.8)) implies that v2 = ( A1 / A2 )v1 and so we can rewrite Eq. (28.3.16) as 28-9 1 ρ f (( A1 / A2 )2 − 1)v12 = ( ρ Hg − ρ f )gh . 2 (28.3.17) We can now solve Eq. (28.3.17) for the speed of the flow at point 1; v1 = 2( ρ Hg − ρ f )gh ρ f (( A1 / A2 )2 − 1) . (28.3.18) Example 28.2 Water Pressure A cylindrical water tower of diameter 3.0 m supplies water to a house. The level of water in the water tower is 35 m above the point where the water enters the house through a pipe that has an inside diameter 5.1cm . The intake pipe delivers water at a maximum rate of 2.0 × 10−3 m 3 ⋅s −1 . The pipe is connected to a narrower pipe leading to the second floor that has an inside diameter 2.5 cm . What is the pressure and speed of the water in the narrower pipe at a point that is a height 5.0 m above the level where the pipe enters the house? Figure 28.9: Example 28.2 (not to scale) Solution: We shall assume that the water is an ideal fluid and that the flow is a steady flow and that the level of water in the water tower is constantly maintained. Let’s choose three points, point 1 at the top of the water in the tower, point 2 where the water just 28-10 enters the house, and point 3 in the narrow pipe at a height h2 = 5.0 m above the level where the pipe enters the house. We begin by applying Bernoulli’s Equation to the flow from the water tower at point 1, to where the water just enters the house at point 2. Bernoulli’s equation (Eq. (28.3.8)) tells us that 1 1 P1 + ρ gy1 + ρ v12 = P2 + ρ gy2 + ρ v2 2 . (28.3.19) 2 2 We assume that the speed of the water at the top of the tower is negligibly small due to the fact that the water level in the tower is maintained at the same height and so we set v1 = 0 . The pressure at point 2 is then 1 P2 = P1 + ρ g( y1 − y2 ) − ρ v2 2 . 2 (28.3.20) In Eq. (28.3.20) we use the value for the density of water ρ = 1.0 × 103 kg ⋅ m −3 , the change in height is ( y1 − y2 ) = 35 m , and the pressure at the top of the water tower is P1 = 1atm . The rate R that the water flows at point 1 satisfies R = A1v1 = π (d1 / 2)2 v1 . Therefore, the speed of the water at point 1 is v1 = R 2.0 × 10−3 m 3 ⋅s −1 = = 2.8 × 10−4 m ⋅s −1 , 2 2 π (d1 / 2) π (1.5 m) (28.3.21) which is negligibly small and so we are justified in setting v1 = 0 . Similarly the speed of the water at point 2 is v2 = R 2.0 × 10−3 m 3 ⋅s −1 = = 1.0 m ⋅s −1 , π (d2 / 2)2 π (2.5 × 10−2 m)2 (28.3.22) We can substitute Eq. (28.3.21) into Eq. (28.3.22), yielding v2 = (d12 / d2 2 )v1 , (28.3.23) a result which we will shortly find useful. Therefore the pressure at point 2 is 1 P2 = 1.01× 105 Pa + (1.0 × 103 kg ⋅ m −3 )(9.8 m ⋅s −2 )(35 m) − (1.0 × 103 kg ⋅ m −3 )(1.0 m ⋅s −1 )2 2 5 5 2 5 P2 = 1.01× 10 Pa + 3.43× 10 Pa − 5.1× 10 Pa = 4.4 × 10 Pa. 28-11 (28.3.24) The dominant contribution is due to the height difference between the top of the water tower and the pipe entering the house. The quantity (1/ 2) ρ v2 2 is called the dynamic pressure due to the fact that the water is moving. The amount of reduction in pressure due to the fact that the water is moving at point 2 is given by 1 2 1 ρ v2 = (1.0 × 103 kg ⋅ m −3 )(1.0 m ⋅s −1 )2 = 5.1× 103 Pa , 2 2 (28.3.25) which is much smaller than the contributions from the other two terms. We now apply Bernoulli’s Equation to the points 2 and 3, 1 1 P2 + ρ v22 + ρ gy2 = P3 + ρ v32 + ρ gy3 . 2 2 (28.3.26) Therefore the pressure at point 3 is 1 P3 = P2 + ρ (v22 − v32 ) + ρ g( y2 − y3 ) . 2 (28.3.27) The change in height y2 − y3 = −5.0 m . The speed of the water at point 3 is v3 = R 2.0 × 10−3 m 3 ⋅s −1 = = 3.9 m ⋅s −1 , π (d3 / 2)2 π (1.27 × 10−2 m)2 (28.3.28) Then the pressure at point 3 is 1 P3 = (4.4 × 105 Pa) + (1.0 × 103 kg ⋅ m −3 )((1.0 m ⋅s −1 )2 − (3.9 m ⋅s −1 )2 ) 2 3 −3 −(1.0 × 10 kg ⋅ m )(9.8 m ⋅s −2 )(5.0 m) . (28.3.29) = (4.4 × 10 Pa) − (7.1× 10 Pa) − 4.9 × 10 Pa 5 3 4 = 3.8 × 105 Pa Because the speed of the water at point 3 is much greater than at point 2, the dynamic pressure contribution at point 3 is much larger than at point 2. 28.6 Laminar and Turbulent Flow 28-12 28.6.1 Introduction During the flow of a fluid, different layers of the fluid may be flowing at different speeds relative to each other, one layer sliding over another layer. For example consider a fluid flowing in a long cylindrical pipe. For slow velocities, the fluid particles move along lines parallel to the wall. Far from the entrance of the pipe, the flow is steady (fully developed). This steady flow is called laminar flow. The fluid at the wall of the pipe is at rest with respect to the pipe. This is referred to as the no-slip condition and is experimentally holds for all points in which a fluid is in contact with a wall. The speed of the fluid increases towards the interior of the pipe reaching a maximum, vmax , at the center. The velocity profile across a cross section of the pipe exhibiting fully developed flow is shown in Figure 28.10. This parabolic velocity profile has a non-zero velocity gradient that is normal to the flow. vmax Figure 28.10 Steady laminar flow in a pipe with a non-zero velocity gradient 28.6.2 Viscosity Due to the cylindrical geometry of the pipe, cylindrical layers of fluid are sliding with respect to one another resulting in tangential forces between layers. The tangential force per area is called a shear stress. The viscosity of a fluid is a measure of the resistance to this sliding motion of one layer of the fluid with respect to another layer. A perfect fluid has no tangential forces between layers. A fluid is called Newtonian if the shear forces per unit area are proportional to the velocity gradient. For a Newtonian fluid undergoing laminar flow in the cylindrical pipe, the shear stress, σ S , is given by σS =η dv , dr (28.3.30) where η is the constant of proportionality and is called the absolute viscosity, r is the radial distance form the central axis of the pipe, and dv / dr is the velocity gradient normal to the flow. The SI units for viscosity are poise = 10−1 Pa ⋅s . Some typical values for viscosity for fluids at specified temperatures are given in Table 1. Table 1: Coefficients of absolute viscosity 28-13 fluid oil Water at 0o Coefficient of absolute viscosity η 1− 10 poise Water at 100o 0.28 × 10−2 poise Air at 20o 1.81× 10−4 poise 1.79 × 10−2 poise At a certain flow rate, this resistance suddenly increases and the fluid particles no longer follow straight lines but appear to move randomly although the average motion is still along the axis of the pipe. This type of flow is called turbulent flow. Osbourne Reynolds was the first to experimentally measure these two types of flow. He was able to characterize the transition between these two types of flow by a parameter called the Reynolds number that depends on the average velocity of the fluid in the pipe, the diameter, and the viscosity of the fluid. The transition point between flows corresponds to a value of the Reynolds number that is associated with a sudden increase in the friction between layers of the fluid. Much after Reynolds initial observations, it was experimentally noted that a small disturbance in the laminar flow could rapidly grow and produce turbulent flow. Example 28.3 Couette Flow Consider the flow of a Newtonian fluid between two very long parallel plates, each plate of width w , length s , and separated by a distance d . The upper plate moves with a constant relative speed v0 with respect to the lower plate, (Figure 28.11). v0 d F(x) x F0 l Figure 28.11 Laminar flow between two plates moving with relative speed v0 Choose a reference frame in which the lower plate, located on the plane at x = 0 , is at rest. Choose a volume element of length l and cross sectional area A , with one side in contact with the plate at rest, and the other side located a distance x from the lower plate. The velocity gradient in the direction normal to the flow is dv / dx . The shear force on the volume element due to the fluid above the element is given by F(x) = η A dv . dx (28.3.31) 28-14 The shear force is balanced by the shear force F0 of the lower plate on the element, such that F(x) = F0 . Hence F0 = η A dv . dx (28.3.32) The velocity of the fluid at the lower plate is zero. The integral version of this differential equation is then v′= v( x ) x′= x 1 (28.3.33) F dx′ = ∫ dv′ . η A x′=∫ 0 0 v′= 0 Integration yields F0 x = v(x) . ηA (28.3.34) The velocity of the fluid at the upper plate is v0 , therefore the constant shear stress is given by F0 ηv0 = , (28.3.35) d A hence the velocity profile is v v(x) = 0 x . (28.3.36) d This type of flow is known as Couette flow. Example 28.4 Laminar flow in a cylindrical pipe. Let’s consider a long cylindrical pipe of radius r0 in which the fluid undergoes laminar flow with each fluid particle moves in a line parallel to the pipe axis. Choose a cylindrical volume element of length dl and radius r , centered along the pipe axis as shown in Figure 28.12. There is a pressure drop dp < 0 over the length of the volume element resulting in forces on each end cap. Denote the force on the left end cap by FL = p / A and the force on the right end cap by FR = ( p + dp) / A on the right end cap, where A = π r 2 is the cross sectional area of the end cap. 28-15 dl r0 r FL = p / A FR = ( p + dp) / A cylindrical volume element Figure 28.12 Volume element for steady laminar flow in a pipe The forces on the volume element sum to zero and are due to the pressure difference and the shear stress; hence FL − FR + σ S 2π rdl = 0 . (28.3.37) Using our Newtonian model for the fluid (Eq. (28.3.30) and expressing the force in terms of pressure, Eq. (28.3.37) becomes dp dv r= . (28.3.38) 2η dl dr Eq. (28.3.38) can be integrated by the method of separation of variables with boundary conditions v(r = 0) = vmax and v(r = r0 ) = 0 . (Recall that for laminar flow of a Newtonian fluid the velocity of a fluid is always zero at the surface of a solid.) dp 2η dl r′=r0 ∫ r′ dr′ = r′=r v′(r=r0 )=0 ∫ dv′ . (28.3.39) v′=v(r ) Integration then yields v(r) = − dp (r0 2 − r 2 ) . 4η dl (28.3.40) Recall that the pressure drop dp < 0 . The maximum velocity at the center is then vmax = v(r = 0) = − dp 2 r . 4η dl 0 (28.3.41) To determine the flow rate through the pipe, choose a ring of radius r and thickness r , oriented normal to the flow. The flow through the ring is then v(r)2π rdr = − dpπ 2 2 (r − r )rdr . 2η dl 0 (28.3.42) Integrating over the cross sectional area of the pipe yields 28-16 Q= r=r0 ∫ v(r)2π r dr r=0 dpπ Q=− 2η dl r=r0 dpπ 2 2 2 4 2 ∫r=0 (r0 − r )r dr = − 2ηdl (r0 r / 2 − r / 4) r=r0 r=0 πr 4 = 0 dp 8η dl (28.3.43) The average velocity is then vave = Q dp 2 =− r 2 8η dl 0 π r0 (28.3.44) Notice that the pressure difference and the volume flow rate are related by dp = 8η dl Q π r0 4 (28.3.45) which is equal to one half the maximum velocity at the center of the pipe. We can rewrite Eq. (28.3.45) in terms of the average velocity as dp = 8η dl 64η dl Q= v 2 4 π r0 vave 2d 2 ave (28.3.46) where d = 2r0 is the diameter of the pipe. For a pipe of length l and pressure difference Δp , the head loss in a pipe is defined as the ratio Δp vave 2 l 64 , hf = = ρ g ( ρ vave d / η ) 2g d (28.3.47) where we have extended Eq. (28.3.46) for the entire length of the pipe. Head loss is also written in terms of a loss coefficient k according to vave 2 hf = k , 2g (28.3.48) For a long straight cylindrical pipe, the loss coefficient can be written in terms of a factor f times an equivalent length of the pipe k= f l . d (28.3.49) The factor f can be determined by comparing Eqs. (28.3.47)-(28.3.49) yielding 28-17 f = 64 64 , = ( ρ vave d / η ) Re (28.3.50) where Re is the Reynolds number and is given by Re = ρ vave d / η . (28.3.51) 28-18