Homework 1

Solution. 1

Step. Differential equation from shell momentum balance

For a plane narrow slit, the natural choice is rectangular Cartesian coordinates. Since the fluid

flow is in the z-direction, vx = 0, vy = 0, and only vz exists. Further, vz is independent of z and it is

meaningful to postulate that velocity vz = vz(x) and pressure p = p(z). The only nonvanishing

components of the stress tensor are τxz = τzx, which depend only on x.

Consider now a thin rectangular slab (shell) perpendicular to the x-direction extending a distance

W in the y-direction and a distance L in the z-direction. A 'rate of z-momentum' balance over this

thin shell of thickness ∆x 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(x) is the same at both ends of the slit,

the convective terms cancel out because (ρ vz vz W ∆x)|z = 0 = (ρ vz vz W ∆x)|z = L. Only the

molecular term (L W τxz ) remains to be considered. Generation of z-momentum occurs by the

pressure force (p W ∆x) and gravity force (ρ g W L ∆x). On substituting these contributions into

the z-momentum balance, we get

(L W τxz ) | x − (L W τxz ) | x+∆x+ ( p 0 − p L ) W ∆x + ρ g W L ∆x = 0

(1)

Dividing the above equation by L W ∆x yields

τxz| x+∆x − τxz| x

∆x

=

p 0 − p L + ρ g L

L

(2)

On taking the limit as ∆x → 0, the left-hand side of the above equation is the definition of the

derivative. The right-hand side may be compactly and conveniently written 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 distance upward (in the

direction opposed to gravity) from a reference plane of choice. Since the z-axis points downward

in this problem, h = − z and therefore P = p − ρ g z . Thus, P0 = p0 at z = 0 and PL = pL − ρ g

L at z = L giving p0 − pL + ρ g L = P0 − PL ≡ ∆P. Thus, equation (2) gives

Homework 1

dτxz

dx

=

∆P

L

(3)

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

τxz =

∆P

L

x + C1

(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 rectangular Cartesian coordinates

Step. Shear stress distribution and velocity profile

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

− µ

dvz

dx

=

∆P

L

x + C1

(5)

The above first-order differential equation is simply integrated to obtain the following velocity

profile:

vz = −

∆P

2 µ L

x2 −

C1

µ

x + C2

(6)

The no-slip boundary conditions at the two fixed walls are

BC 1: at x = B, vz = 0 (7)

BC 2: at x = −B, vz = 0 (8)

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Using these, the integration constants may be evaluated as C1 = 0 and C2 = ∆P B2 / (2 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

τxz =

∆P

L

x

(9)

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

the velocity profile as

vz =

∆P B 2

2 µ L

1 −

x

B

2

(10)

It is observed that the velocity distribution for laminar, incompressible flow of a Newtonian fluid

in a plane narrow slit 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 x = 0 (where dvz/dx = 0). Therefore,

vz,max = vz| x = 0 =

∆P B2

2 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 = − B∫ B vz W dx

− B∫ B W dx

=

1

2B

B

∫ −B

vz dx =

∆P B2

3 µ L

=

2

3

vz,max

(12)

Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a

narrow slit is 2/3.

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(iii) The mass rate of flow is obtained by integrating the velocity profile over the cross section of

the slit as follows.

w =

B

∫ −B

ρ vz W dx = ρ W (2 B) vz,avg

(13)

Thus, the mass flow rate is the product of the density ρ, the cross-sectional area (2 B W) and the

average velocity vz,avg. On substituting vz,avg from equation (12), the final expression for the mass

rate of flow is

w =

2 ∆P B3 W ρ

3 µ L

(14)

The flow rate vs. pressure drop (w vs. ∆P) expression above is the slit analog of the Hagen-

Poiseuille equation (originally for circular tubes). It is a result worth noting because it provides

the starting point for creeping flow in many systems (e.g., radial flow between two parallel

circular disks; and flow between two stationary concentric spheres).

Finally, it may be noted that the above analysis is valid when B << W. If the slit thickness B is of

the same order of magnitude as the slit width W, then vz = vz (x, y), i.e., vz is not a function of

only x. If W = 2B, then a solution can be obtained for flow in a square duct.

Homework 1

Solution. 2

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 ∆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 (ρ vz vz 2πr ∆r)|z = 0 = (ρ vz vz 2πr ∆r)|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 ∆r] and gravity force acting on the volume [(ρ g cos β) 2πr ∆r L]. On

substituting these contributions into the z-momentum balance, we get

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

(1)

Dividing the above equation by 2π L ∆r yields

(r τrz ) | r + ∆r − (r τrz ) | r

∆r

=

p 0 − p L + ρ g L cos β

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 + ρ g h , where h is the height (in the

direction opposed to gravity) above some arbitrary preselected datum plane. The advantages of

Homework 1

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 ≡

∆P. Thus, equation (2) gives

d

dr

(r τrz) =

∆P

L

r

(3)

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

τrz =

∆P

2 L

r +

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

Step. Shear stress distribution and velocity profile

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

− µ

dvz

dr

=

∆P

2 L

r +

C1

r

(5)

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

vz = −

∆P

4 µ L

r2 −

C1

µ

ln 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)

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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 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 =

∆P

2 L

r

(9)

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

the velocity profile as

vz =

∆P 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 =

∆P 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

R

∫

vz r dr = ∆P R

2

=

1

vz,max (12)

Homework 1

0∫ R 2 π r dr R

2 0 8 µ L 2

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 ∆P R4 ρ

8 µ 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)

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 ∆P = π R

2 ∆p + π R

2 ρ g L cos β

(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.

Homework 1

Solution. 3

Step. Simplification of continuity equation

In steady laminar flow, the fluid is expected to travel in a circular motion (for small values of Ω).

Only the tangential component of velocity exists. The radial and axial components of velocity

are zero; so, vr = 0 and vz = 0.

For incompressible flow, the continuity equation gives ∇.v = 0.

In cylindrical coordinates,

1

r

∂

∂r

(r vr ) +

1

r

∂vθ

∂θ

+

∂vz

∂z

= 0 ⇒

∂vθ

∂θ

= 0

(1)

So, vθ = vθ (r, z).

Step. Simplification of equation of motion

For a Newtonian fluid, the components of the Navier - Stokes equation may be simplified as

given below.

r - component

: −

ρvθ2

r

= −

∂p

∂r

(2)

θ - component

: 0 =

∂

∂r

1

r

∂

∂r

(r vθ )

+

∂2vθ

∂z2

(3)

z - component

: 0 = −

∂p

∂z

− ρ g

(4)

From symmetry arguments, there is no pressure gradient in the θ-direction; so, p = p(r, z).

Homework 1

b)

Step. Velocity profile

The no-slip boundary conditions at the two disk surfaces are

BC 1 : at z = 0, any r, vθ = 0

(5)

BC 2 : at z = B, any r, vθ = Ω r

(6)

It is possible to make an educated guess for the form of the tangential velocity taking a clue from

BC 2. Thus, it is reasonable to postulate vθ (r, z) = r f(z). Substituting in equation (3) then gives

d2f

dz

2 = 0 ⇒

df

dz

= C1 or f = C1 z + C2

(7)

Equation (5) gives f = 0 at z = 0 or C2 = 0. On the other hand, equation (6) gives f = Ω at z = B or

C1 = Ω/B. On substituting f = Ω z/B, the tangential velocity profile is ultimately obtained as

vθ =

Ω r z

B

(8)

The above result is expected and could have been guessed by considering a fluid contained

between two parallel plates (with the lower plate at z = 0 held stationary and the upper plate at z

= B moving at a constant velocity V). Then, the linear velocity distribution is given by v = V z/B,

which is the velocity profile at any r in the parallel - disk viscometer. On noting that V = Ω r at

the upper plate, equation (8) is obtained.

c)

Step. Determination of torque

From the velocity profile in equation (8), the momentum flux (shear stress) is determined as

τzθ = − µ

∂vθ

∂z

= − µ

Ω r

B

and τrθ = − µ r

∂

∂r

vθ

r

= 0

(9)

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The differential torque dT is rigorously obtained as the cross product of the lever arm r and the

differential force dF exerted on the fluid by a solid surface element. Thus, dT = r x dF where dF

= n.(p δ + τ) dS, r is the vector drawn from the axis of rotation to the element of surface area dS,

n is the unit normal to the surface, p is the pressure, δ is the identity (unit) tensor, and τ is the

viscous stress tensor. For the parallel - disk viscometer,

dT = r x n.τ dS = r δr x (−δz).τzθ δzδθ (2 π r dr)

(10)

The pressure term is neglected in equation (10). Since δz.δz = 1 and δr x δθ = δz, we obtain dT = r

(−τzθ) (2 π r dr) δz. Thus, the z-component of the differential torque is dTz = r (−τzθ) (2 π r dr),

which is simply the product of the lever arm, the momentum flux and the differential surface

area. Integrating over the area of the upper disk after substituting for τzθ from equation (9) yields

Tz |z = B =

R

∫ 0

[ (−τzθ) (2 π r2 dr)]z = B =

2 π µ Ω

B

R

∫ 0

r3 dr =

π µ Ω R4

2 B

(11)

Equation (11) provides the formula for determining the viscosity µ of a Newtonian fluid from

measurements of the torque Tz and angular velocity Ω in a parallel - disk viscometer as µ =

2BTz/(πΩR4).

Note that the shear stress is not uniform [as given by equation 9)] in a parallel - disk viscometer,

whereas it is uniform throughout the gap in a cone - and - plate viscometer.

Homework 1

Solution. 4

Step. Hagen-Poiseuille equation and lubrication approximation

The mass flow rate vs. pressure drop (w vs. ∆P) relationship for a Newtonian fluid in a circular

tube of constant radius R is

w =

π ∆P R 4

ρ

8 µ L

(1)

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

∆P

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

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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

R0

3

−

1

RL

3

(7)

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

w =

π ∆P R0 4

ρ

8 µ L

3 (λ − 1)

1 − λ

− 3

=

π ∆P R0 4

ρ

8 µ L

3 λ 3

1 + λ + λ

2

(8)

Homework 1

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).