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Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a circular tube

Problem.

Consider a fluid (of constant density ρ) in incompressible, laminar flow in a tube of circular cross section, inclined at an angle β 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 Δp 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 μ).

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.

Click here for stepwise 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.

Step. Differential equation from shell momentum balance

For a circular tube, the natural choice is cylindrical coordinates. Since the fluid flow is in the z-direction, vr = 0, vθ = 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 nonvanishing components of the stress tensor are τrz = τzr, 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 Δr 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 (ρ vz vz 2πr Δr)|z = 0 = (ρ vz vz 2πr Δr)|z = L. Only the molecular term (2πr L τrz ) 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 2πr Δr] and gravity force acting on the volume [(ρ g cos β) 2πr Δr L]. On substituting these contributions into the z-momentum balance, we get

(2πr L τrz ) | r − (2πr L τrz ) | r + Δr+ ( p 0 − p L ) 2πr Δr + (ρ g cos β) 2πr Δr L = 0 (1)

Dividing the above equation by 2π L Δr yields

(r τrz ) | r + Δr − (r τrz ) | r

Δr  =  

p 0 − p L + ρ g L cos β

L

  r   (2)

On taking the limit as Δr → 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 + ρ 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 β and therefore P  =  p − ρ g z cos β. Thus, P0  =  p0 at z = 0 and PL  =  pL − ρ g L cos β at z = L giving p0 − pL + ρ g L cos β  =  P0 − PL  ≡  ΔP. Thus, equation (2) gives

d

dr(r τrz)  =  

ΔP

L r (3)

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

τrz  =   ΔP

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).

Step. Shear stress distribution and velocity profile

Substituting Newton's law of viscosity for τrz in equation (4) gives

− μdvz

dr = 

ΔP

2 Lr +

C1

r(5)

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

vz   =  − ΔP

4 μ Lr2  − 

C1

μln r  +  C2 (6)

The integration constants C1 and C2 are evaluated from the following boundary conditions:

BC 1:  at r = 0,     τrz 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 = ΔP R2 / (4 μ 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

τrz =   ΔP

2 Lr (9)

Further, substitution of the integration constants into equation (6) gives the final expression for the velocity profile as

vz =   ΔP R 2

4 μ 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)

Step. Maximum velocity, average velocity and mass flow rate

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   =   ΔP R2

4 μ L(11)

(ii) The average velocity is obtained by dividing the volumetric flow rate by the cross-sectional area as shown below.

vz,avg =       0∫  R vz 2 π r dr   =   2

R2

  R∫ 0

  vz r dr

 =  ΔP R2

8 μ L

 =  1

2

vz,max (12)

    0∫  R 2 π r dr

Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a circular tube is ½.

(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

 ρ vz 2 π r dr  =   ρ π R2 vz,avg (13)

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

w   =   π ΔP R4 ρ

8 μ L(14)

The flow rate vs. pressure drop (w vs. ΔP) 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)

Step. Force along tube wall

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 π R L) τrz| r = R   =  π R2 ΔP =  π R2 Δp + π R2 ρ g L cos β (15)

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

Related Problems in Transport Phenomena - Fluid Mechanics :

Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a slightly tapered tube- Determination of mass flow rate for Newtonian fluid in slightly tapered tube of changing radius rather than constant radius

Transport Phenomena - Fluid Mechanics Problem : Power law fluid flow in circular tube- Determination of shear stress, velocity profile and mass flow rate for power law fluid rather than Newtonian fluid

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Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a slightly tapered tubehttp://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid203.html?0

Problem.

A fluid (of constant density ρ) is in incompressible, laminar flow through a tube of length L. The radius of the tube of circular cross section changes linearly from R0 at the tube entrance (z = 0) to a slightly smaller value RL at the tube exit (z = L).

      Figure. Fluid flow in a slightly tapered tube.

Using the lubrication approximation, determine the mass flow rate vs. pressure drop (w vs. ΔP) relationship for a Newtonian fluid (of constant viscosity μ).

Solution.

Click here for stepwise solution

Step. Hagen-Poiseuille equation and lubrication approximation

The mass flow rate vs. pressure drop (w vs. ΔP) relationship for a Newtonian fluid in a circular tube of constant radius R is (click here for derivation)

w   =  π ΔP R 4 ρ

8 μ L(1)

The above equation, which is the famous Hagen-Poiseuille equation, may be re-arranged as

ΔP

L  =  

8 μ w

ρ π

1

R 4(2)

For the tapered tube, note that the mass flow rate w does not change with axial distance z. If the above equation is assumed to be approximately valid for a differential length dz of the tube whose radius R is slowly changing with axial distance z, then it may be re-written as

−  dP

dz  =  

8 μ w

ρ π  

1

[R(z)] 4(3)

The approximation used above where a flow between non-parallel surfaces is treated locally as a flow between parallel surfaces is commonly called the lubrication approximation because it is often employed in the theory of lubrication. The lubrication approximation, simply speaking, is a local application of a one-dimensional solution and therefore may be referred to as a quasi-one-dimensional approach.

Equation (3) may be integrated to obtain the pressure drop across the tube on substituting the taper function R(z), which is determined next.

Step. Taper function

As the tube radius R varies linearly from R0 at the tube entrance (z = 0) to RL at the tube exit (z = L), the taper function may be expressed as R(z) = R0 + (RL − R0) z / L. On differentiating with respect to z, we get

dR

dz  =  

RL − R0

L(4)

Step. Pressure P as a function of radius R

Equation (3) is readily integrated with respect to radius R rather than axial distance z. Using equation (4) to eliminate dz from equation (3) yields

(− dP )   =  8 μ w

ρ π  

L

RL − R0

  dR

R 4(5)

Integrating the above equation between z = 0 and z = L, we get

PL

∫P0

(−dP )   =  8 μ w

ρ π  

L

RL − R0

  

RL

∫R0

dR

R 4(6)

P0 − PL

L  =  

8 μ w

ρ π  

1

3 (RL − R0 )   

1

R03

− 1

RL3

(7)

Equation (7) may be re-arranged into the following standard form in terms of mass flow rate:

w   =  π ΔP R0

4 ρ

8 μ L  

3 (λ − 1)

1 − λ− 3

  =  

π ΔP R0 4 ρ

8 μ L  

3 λ 3

1 + λ + λ 2

(8)

where the taper ratio λ ≡ RL / R0. The term in square brackets on the right-hand side of the above equation may be viewed as a taper correction to equation (1).

Related Problems in Transport Phenomena - Fluid Mechanics :

Transport Phenomena - Fluid Mechanics Problem : Newtonian fluid flow in a circular tube- Determination of shear stress, velocity profile and mass flow rate for tube of constant radius rather than changing radius

Transport Phenomena - Fluid Mechanics Problem : Power law fluid flow in a slightly tapered tube- Determination of mass flow rate in slightly tapered tube for power law fluid rather than Newtonian fluid

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