Problem. Consider a fluid (of constant density ) in incompressible, laminar flow in a tube of circular cross section, inclined at an angle to the vertical. End effects may be neglected because the tube length L is relatively very large compared to the tube radius R. The fluid flows under the influence of both a pressure difference p and gravity. Figure. Fluid flow in circular tube.a) Using a differential shell momentum balance, determine expressions for the steady-state shear stress distribution and the velocity profile for a Newtonian fluid (of constant viscosity ).b) Obtain expressions for the maximum velocity, average velocity and the mass flow rate for pipe flow.c) Find the force exerted by the fluid along the pipe wall.

Solution. a) Flow in pipes occurs in a large variety of situations in the real world and is studied in various engineering disciplines as well as in physics, chemistry, and biology.

For a circular tube, the natural choice is cylindrical coordinates. Since the fluid flow is in the z-direction, vr = 0, v = 0, and only vz exists. Further, vz is independent of z and it is meaningful to postulate that velocity vz = vz(r) and pressure p = p(z). The only non-vanishing components of the stress tensor are rz = zr, which depend only on r. Consider now a thin cylindrical shell perpendicular to the radial direction and of length L. A 'rate of z-momentum' balance over this thin shell of thickness r in the fluid is of the form: Rate of z-momentum In Out + Generation = Accumulation At steady-state, the accumulation term is zero. Momentum can go 'in' and 'out' of the shell by both the convective and molecular mechanisms. Since vz(r) is the same at both ends of the tube, the convective terms cancel out because ( vz vz 2r r)|z = 0 = ( vz vz 2r r)|z = L. Only the molecular term (2r L rz ) remains to be considered, whose 'in' and 'out' directions are taken in the positive direction of the r-axis. Generation of z-momentum occurs by the pressure force acting on the surface [p 2r r] and gravity force acting on the volume [( g cos ) 2r r L]. On substituting these contributions into the z-momentum balance, we get (2r L rz ) | r (2r L rz ) | r + r+ ( p 0 p L ) 2r r + ( g cos ) 2r r L = 0

(1)

Dividing the above equation by 2 L r yields (r rz ) | r + r (r rz ) | r

r = p 0 p L + g L cos

L r

(2)

On taking the limit as r 0, the left-hand side of the above equation is the definition of the first derivative. The right-hand side may be written in a compact and convenient way by introducing the modified pressure P, which is the sum of the pressure and gravitational terms. The general definition of the modified pressure is P = p + g h , where h is the height (in the direction opposed to gravity) above some arbitrary preselected datum plane. The advantages of using the modified pressure P are that (i) the components of the gravity vector g need not be calculated in cylindrical coordinates; (ii) the solution holds for any orientation of the tube axis; and (iii) the effects of both pressure and gravity are in general considered. Here, h is negative since the z-axis points downward, giving h = z cos and therefore P = p g z cos . Thus, P0 = p0 at z = 0 and PL = pL g L cos at z = L giving p0 pL + g L cos = P0 PL P. Thus, equation (2) gives d

dr(r rz) = P

Lr

(3)

Equation (3) on integration leads to the following expression for the shear stress distribution: rz = P

2 Lr + C1

r

(4)

The constant of integration C1 is determined later using boundary conditions. It is worth noting that equations (3) and (4) apply to both Newtonian and non-Newtonian fluids, and provide starting points for many fluid flow problems in cylindrical coordinates (click here for detailed discussion). Substituting Newton's law of viscosity for rz in equation (4) gives dvz

dr= P

2 Lr + C1

r

(5)

The above differential equation is simply integrated to obtain the following velocity profile: vz = P

4 Lr2 C1

ln r + C2

(6)

The integration constants C1 and C2 are evaluated from the following boundary conditions: BC 1: at r = 0, rz and vz are finite(7)

BC 2: at r = R, vz = 0 (8)

From BC 1 (which states that the momentum flux and velocity at the tube axis cannot be infinite), C1 = 0. From BC 2 (which is the no-slip condition at the fixed tube wall), C2 = P R2 / (4 L). On substituting C1 = 0 in equation (4), the final expression for the shear stress (or momentum flux) distribution is found to be linear as given by rz = P

2 Lr

(9)

Further, substitution of the integration constants into equation (6) gives the final expression for the velocity profile as vz = P R 2

4 L 1 r

R 2

(10)

It is observed that the velocity distribution for laminar, incompressible flow of a Newtonian fluid in a long circular tube is parabolic. b) From the velocity profile, various useful quantities may be derived. (i) The maximum velocity occurs at r = 0 (where dvz/dr = 0). Therefore, vz,max = vz| r = 0 = P R2

4 L

(11)

(ii) The average velocity is obtained by dividing the volumetric flow rate by the cross-sectional area as shown below.vz,avg = 0R vz 2 r dr

0R 2 r dr = 2

R2 R 0 vz r dr= P R2

8 L= 1

2vz,max

(12)

Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a circular tube is . (iii) The mass rate of flow is obtained by integrating the velocity profile over the cross section of the circular tube as follows.w = R 0 vz 2 r dr = R2 vz,avg

(13)

Thus, the mass flow rate is the product of the density , the cross-sectional area ( R2) and the average velocity vz,avg. On substituting vz,avg from equation (12), the final expression for the mass rate of flow isw = P R4

8 L

(14)

The flow rate vs. pressure drop (w vs. P) expression above is well-known as the Hagen-Poiseuille equation. It is a result worth noting because it provides the starting point for flow in many systems (e.g., flow in slightly tapered tubes). c) The z-component of the force, Fz, exerted by the fluid on the tube wall is given by the shear stress integrated over the wetted surface area. Therefore, on using equation (9), Fz = (2 R L) rz| r = R = R2 P = R2 p + R2 g L cos

(15)

where the pressure difference p = p0 pL. The above equation simply states that the viscous force is balanced by the net pressure force and the gravity force.

Fully Developed Laminar Flow in a Pipe Considering a fully developed laminar flow in a pipe it is possible to derive an expression for the velocity profile and then calculate useful practical results. This derivation can be carried out in a number of ways - (1) by a Control Volume Analysis, (2) from Navier-Stokes Equation or (3) by Dimensional Analysis.Volumetric Flow Rate A quantity of interest in pipe flows is the volumetric flow rate, which is obtained by integrating the velocity profile. Considering a disc of thickness dr at a radius r then

integrating

leading to

If an average velocity, V, is defined such that it can be verified that

The volumetric flow rate written in terms of pressure gradient becomes,

Correction for non-horizontal pipes If the pipe considered in the previous analysis was not horizontal, then gravity effects should be included when calculating velocity and volumetric flow rates. Referring to Fig.8, if the flow is inclined at an angle to the horizontal, the pressure difference term needs to be modified.

Figure 8: Flow through an inclined pipe. Accordingly, the force balance becomes

The pressure difference term in other equations needs to be replaced as well. Accordingly,

and

Energy Considerations, Friction factor

Figure 9: Energy balance for a pipe flow.Again considering the pipe flow as in the previous section, an energy analysis can be carried out. With reference to Fig. 9, for a mass balance,

Since the flow is incompressible and the pipe cross-section area is constant,

Now applying the energy equation for a steady flow,

Note that every term in the above equation has the dimension of length. As a fully developed flow is being considered , then 1 = 2 . There are no external features between (1) and (2) which could add or remove energy. As a result, loss of head is given by,

A force balance in the x-direction on a control volume gives,

Note that,

Dividing by gives,

An inspection of these equations shows thator Again this result is valid for both laminar and turbulent flows.Dimensional AnalysisThrough a dimensional analysis it is possible to derive an expression for "loss of head" in a pipe flow. Assuming that pressure drop is proportional to pipe length it can be shown that

Thus pressure drop now becomes,

with

The non-dimensional quantity f is referred to as "Darcy's Friction Factor". For a laminar flow it is given by It is possible to show that