Transport phenomena Solved problems

1 |ABNET MENGESHA DUBE
Transport Phenomena - Fluid Mechanics Theory
Differential Shell Momentum Balance in Rectangular Cartesian Coordinates
Preparation Just what you need to know !
A shell momentum balance is used below to derive a general differential equation that can
be then employed to solve several fluid flow problems in rectangular Cartesian
coordinates. For this purpose, consider an incompressible fluid in laminar flow under the
effects of both pressure and gravity in a system of length L and width W, which is at an
angle β to the vertical. End effects are neglected assuming the dimension of the system in
the x-direction is relatively very small compared to those in the y-direction (W) and the z-
direction (L).
Figure. Differential rectangular slab (shell) of fluid of thickness Δx used in z-momentum
balance for flow in rectangular Cartesian coordinates. The y-axis is pointing outward from
the plane of the computer screen.
Since the fluid flow is in the z-direction, vx = 0, vy = 0, and only vz exists. For small flow
rates, the viscous forces prevent continual acceleration of the fluid. So, 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 Δx 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 system, the convective terms cancel out because (ρ vz vz W Δx)|z = 0 = (ρ vz vz W Δx)|z = L.
Only the molecular term (L W τxz ) remains to be considered, whose 'in' and 'out' directions
are taken in the positive direction of the x-axis. Generation of z-momentum occurs by the
pressure force acting on the surface [p W Δx] and gravity force acting on the volume [(ρ
g cos β) L W Δx].
The different contributions may be listed as follows:
• rate of z-momentum in by viscous transfer across surface at x is (L W τxz )| x
• rate of z-momentum out by viscous transfer across surface at x + Δx is (L W τxz )| x +
Δx
• rate of z-momentum in by overall bulk fluid motion across surface at z = 0 is (ρ
vz vz W Δx )| z = 0
• rate of z-momentum out by overall bulk fluid motion across surface at z = L is (ρ
vz vz W Δx )| z = L
• pressure force acting on surface at z = 0 is p0 W Δx
• pressure force acting on surface at z = L is − pL W Δx
• gravity force acting in z-direction on volume of rectangular slab is (ρ g cos β) L
W Δx
On substituting these contributions into the z-momentum balance, we get
(L W τxz ) | x − (L W τxz ) | x+Δx+ ( p 0 − p L ) W Δx + (ρ g cos β) L W Δx = 0 (1)
Dividing the equation by L W Δx yields
τxz | x+Δx − τxz | x
Δx
=
p 0 − p L + ρ g L cos β
L
(2)
On taking the limit as Δx → 0, the left-hand side of the above equation is exactly the
definition of the 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 distance upward (in the direction opposed to gravity) from a reference plane
of choice. The advantages of using the modified pressure P are that (i) the components of
the gravity vector g need not be calculated; (ii) the solution holds for any flow orientation;

and (iii) the fluid may flow as a result of a pressure difference, gravity or both. 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 βat z = L giving p0 − pL + ρ g
L cos β = P0 − PL ≡ ΔP. Thus, equation (2) yields
dτxz
dx
=
ΔP
L
(3)
The first-order differential equation may be simply integrated to give
τxz =
ΔP
L
x + C1 (4)
Here, C1 is an integration constant, which is determined using an appropriate boundary
condition based on the flow problem. Equation (4) shows that the momentum flux (or
shear stress) distribution is linear in systems in rectangular Cartesian coordinates.
Since equations (3) and (4) have been derived without making any assumption about the
type of fluid, they are applicable to both Newtonian and non-Newtonian fluids. Some of
the axial flow problems in rectangular Cartesian coordinates where these equations may
be used as starting points are given below.
Problem.
Consider a fluid (of density ρ) in incompressible, laminar flow in a plane narrow slit of
length L and width W formed by two flat parallel walls that are a distance 2B apart. End effects
may be neglected because B << W << L. The fluid flows under the influence of both a pressure
difference Δp and gravity.

Figure. Fluid flow in plane narrow slit.
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 viscosity μ).
b) Obtain expressions for the maximum velocity, average velocity and the mass flow rate for slit
flow.
Solution.
Click here for stepwise solution
a)
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 Δx in the fluid is of the form:

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 vzW Δx)|z = 0 = (ρ vz vz W Δx)|z = L. Only the
molecular term (L W τxz ) remains to be considered. Generation of z-momentum occurs by the
pressure force (p W Δx) and gravity force (ρ g W L Δx). On substituting these contributions into
the z-momentum balance, we get
(L W τxz ) | x − (L W τxz ) | x+Δx+ ( p 0 − p L ) W Δx + ρ g W L Δx = 0 (1)
Dividing the above equation by L W Δx yields
τxz| x+Δx − τxz| x
Δx
=
p 0 − p L + ρ g L
L
(2)
On taking the limit as Δx → 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 ≡ ΔP. Thus, equation (2) gives
dτxz
dx
=
ΔP
L
(3)
Equation (3) on integration leads to the following expression for the shear stress distribution:
τxz =
ΔP
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
(click here for detailed discussion).

Step. Shear stress distribution and velocity profile
Substituting Newton's law of viscosity for τxz in equation (4) gives
− μ
dvz
dx
=
ΔP
L
x + C1 (5)
The above first-order differential equation is simply integrated to obtain the following velocity
profile:
vz = −
ΔP
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)
Using these, the integration constants may be evaluated as C1 = 0 and C2 = ΔP 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 =
ΔP
L
x (9)
Further, substitution of the integration constants into equation (6) gives the final expression for
the velocity profile as
vz =
ΔP 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 =
ΔP 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
=
ΔP 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.
(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 ΔP B3
W ρ
3 μ L
(14)
The flow rate vs. pressure drop (w vs. ΔP) 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.

Problem.
Consider a fluid (of density ρ) in incompressible, laminar flow in a plane narrow slit of
length L and width W formed by two flat parallel walls that are a distance 2B apart. End effects
may be neglected because B << W << L. The fluid flows under the influence of a pressure
difference Δp, gravity or both.
Figure. Fluid flow in plane narrow slit.
a) Determine the steady-state velocity distribution for a non-Newtonian fluid that obeys the
power law model (e.g., a polymer liquid).
b) Obtain the mass flow rate for a power law fluid in slit flow.
Solution.
a)
Step. Shear stress distribution
For axial flow in rectangular Cartesian coordinates, the differential equation for the momentum
flux is (click here for derivation)
dτxz
dx
=
ΔP
L
(1)

where P is a modified pressure, which is the sum of both the pressure and gravity terms, i.e.,
ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical.
On integration, this gives the expression for the shear stress τxz for laminar flow in a plane
narrow slit as (click here for derivation)
τxz =
ΔP
L
x (2)
It must be noted that the above momentum flux expression holds for both Newtonian and non-
Newtonian fluids (and does not depend on the type of fluid). Further, τxz = 0 at x = 0 in slit flow
based on symmetry arguments, i.e., the velocity profile is symmetric about the midplane x = 0.
For the region 0 ≤ x ≤ B, the velocity decreases with increasing x and therefore dvz / dx ≤ 0. On
the other hand, for the region −B ≤ x ≤ 0, the velocity decreases with decreasing x and
therefore dvz / dx ≥ 0. Thus, the power law model is given as
τxz = m −
dvz
dx
n
for 0 ≤ x ≤ B (3)
τxz = − m
dvz
dx
n
for −B ≤ x ≤ 0 (4)
Step. Velocity distribution
To obtain the velocity distribution for 0 ≤ x ≤ B, equations (2) and (3) are combined to
eliminate τxz and get
−
dvz
dx
=
ΔP x
m L
1/n
for 0 ≤ x ≤ B (5)
On integrating and using the no-slip boundary condition (vz = 0 at x = B ), we get

vz =
ΔP B
m L
1/n
B
(1/ n) + 1
1 −
x
B
(1/ n) + 1
for 0 ≤ x ≤ B (6)
Similarly, for −B ≤ x ≤ 0, equations (2) and (4) may be combined to eliminate τxz and obtain
dvz
dx
= −
ΔP x
m L
1/n
for −B ≤ x ≤ 0 (7)
Again, integrating and using the no-slip boundary condition (vz = 0 at x = −B) gives
vz =
ΔP B
m L
1/n
B
(1/ n) + 1
1 − −
x
B
(1/ n) + 1
for −B ≤ x ≤ 0 (8)
Equations (6) and (8) may be combined and represented as a single equation for the velocity
profile as follows:
vz =
ΔP B
m L
1/n
B
(1/ n) + 1
1 −
| x |
B
(1/ n) + 1
for | x | ≤ B (9)
Note that the maximum velocity in the slit occurs at the midplane x = 0.
b)
Step. Mass flow rate
Since the velocity profile is symmetric about the midplane x = 0, the mass flow rate may be
obtained by integrating the velocity profile over half the cross section of the slit as shown below.
w = 2
B
∫
0
ρ vz W dx
= 2
ΔP B
mL
1/n
B2
W ρ
(1/ n) + 1
1
∫
0
1 −
x
B
(1/ n) + 1
d
x
B
(10)
w =
2 B2
W ρ
(1/ n) + 2
ΔP B
mL
1/n
(11)

When the mass flow rate w in equation (11) is divided by the density ρ and the cross-sectional
area (2 B W), an expression for the average velocity is obtained.
For n = 1 and m = μ, equation (11) reduces to the Newtonian result, i.e., w = 2 ΔP B3
W ρ / (3 μ
L).
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.
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:
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 vz2πr 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 L
r +
C1
r
(4)
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
− μ
dvz
dr
=
ΔP
2 L
r +
C1
r
(5)
The above differential equation is simply integrated to obtain the following velocity profile:
vz = −
ΔP
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)
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 L
r (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
0∫ R
2 π r dr
=
2
R2
R
∫
0
vz r dr
=
ΔP R2
8 μ L
=
1
2
vz,max (12)
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.
Problem.
Consider a Newtonian liquid (of viscosity μ and density ρ) in laminar flow down an inclined flat
plate of length L and width W. The liquid flows as a falling film with negligible rippling under
the influence of gravity. End effects may be neglected because L and W are large compared to
the film thickness δ.

Figure. Newtonian liquid flow in a falling film.
a) Determine the steady-state velocity distribution.
b) Obtain the mass rate of flow and average velocity in the falling film.
c) What is the force exerted by the liquid on the plate in the flow direction?
d) Derive the velocity distribution and average velocity for the case where x is replaced by a
coordinate x' measured away from the plate (i.e., x' = 0 at the plate and x' = δ at the liquid-gas
interface).
Solution.
a)
dτxz
dx
=
ΔP
L
(1)
where P is a modified pressure, which is the sum of both the pressure and gravity terms, that is,
ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical. Since the
flow is solely under the influence of gravity, Δp = 0 and therefore ΔP/L = ρ g cos β. On
integration,
τxz = ρ g x cos β + C1 (2)

On using the boundary condition at the liquid-gas interface (τxz = 0 at x = 0), the constant of
integration C1 is found to be zero. This gives the expression for the shear stress τxz for laminar
flow in a falling film as
τxz = ρ g x cos β (3)
Newtonian fluids (and does not depend on the type of fluid).
The shear stress for a Newtonian fluid (as per Newton's law of viscosity) is given by
τxz = − μ
dvz
dx
(4)
To obtain the velocity distribution, there are two possible approaches.
In the first approach, equations (3) and (4) are combined to eliminate τxz and get a first-order
differential equation for the velocity as given below.
dvz
dx
= −
ρ g cos β
μ
x (5)
On integrating, vz = − ρ g x2
cos β/(2μ) + C2. On using the no-slip boundary condition at the solid
surface (vz = 0 at x = δ ), C2 = ρ g δ2
cos β/(2μ). Thus, the final expression for the velocity
distribution is parabolic as given below.
vz =
ρ g δ2
cos β
2μ
1 −
x
δ
2
(6)
In the second approach, equations (1) and (4) are directly combined to eliminate τxz and get a
second-order differential equation for the velocity as given below.
− μ
d2
vz
dx2 = ρ g cos β (7)

Integration gives
dvz
dx
= −
ρ g cos β
μ
x + C1 (8)
On integrating again,
vz = −
ρ g x2
cos β
2μ
+ C1 x + C2 (9)
On using the boundary condition at the liquid-gas interface (at x = 0, τxz = 0 ⇒ dvz/dx = 0), the
constant of integration C1 is found to be zero from equation (8). On using the other boundary
condition at the solid surface (at x = δ, vz = 0), equation (9) gives C2 = ρ g δ2
cos β/(2μ). On
substituting C1 and C2 in equation (9), the parabolic velocity profile in equation (6) is again
obtained.
b)
The mass flow rate may be obtained by integrating the velocity profile given by equation (6)
over the film thickness as shown below.
w =
δ
∫
0
ρ vz W dx
=
ρ2
g W δ3
cos β
2μ
1
∫
0
1 −
x
δ
2
d
x
δ
(10)
w =
ρ2
g W δ3
cos β
3μ
(11)
When the mass flow rate w in equation (11) is divided by the density ρ and the film cross-
sectional area (W δ), an expression for the average velocity <vz> is obtained. Also, the maximum
velocity vz, max occurs at the interface x = 0 and is readily obtained from equation (6). Thus,
equations (6) and (11) give
<vz> =
ρ g δ2
cos β
3μ
=
2
3
vz, max (12)

c)
Step. Force on the plate
The z-component of the force of the fluid on the solid surface is given by the shear stress
integrated over the wetted surface area. Using equation (3) gives
Fz = L W τxz|x = δ = ρ g δ L W cos β (13)
As expected, the force exerted by the fluid on the plate is simply the z-component of the weight
of the liquid film.
d)
Step. Choice of coordinates
Consider the coordinate x' starting from the plate (i.e., x' = 0 is the plate surface and x' = δ is the
liquid-gas interface). The form of the differential equation for the momentum flux given by
equation (1) will remain unchanged. Thus, equation (2) for the new choice of coordinates will be
τx'z = ρ g x' cos β + C1 (14)
On using the boundary condition at the liquid-gas interface (τx'z = 0 at x' = δ), the constant of
integration is now found to be given by C1 = − ρ g δ cos β. This gives the shear stress as
τx'z = ρ g (x' − δ) cos β (15)
Note that the momentum flux is in the negative x'-direction. On substituting Newton's law of
viscosity,
dvz
dx'
=
ρ g cos β
μ
(δ − x') (16)
On integrating, vz = (ρg/μ) (δ x' − ½ x'2
) cos β + C2. On using the no-slip boundary condition at
the solid surface (vz = 0 at x' = 0), it is found that C2 = 0. Thus, the final expression for the
velocity distribution is

vz =
ρ g δ2
cos β
μ
x'
δ
−
1
2
x'
δ
2
(17)
The average velocity may be obtained by integrating the velocity profile given in equation (17)
over the film thickness as shown below.
w =
1
δ
δ
∫
0
vz dx'
=
ρ g δ2
cos β
μ
1
∫
0
x'
δ
−
1
2
x'
δ
2
d
x
δ
(18)
Integration of equation (18) yields the same expression for <vz> obtained earlier in equation (12).
It must be noted that the two coordinates x and x' are related as follows: x + x' = δ. By
substituting (x/δ)2
= [1 − (x'/δ)]2
= 1 − 2(x'/δ) + (x'/δ)2
, equation (17) is directly obtained from
equation (6).
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.
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 dzfrom 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 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).
Problem.
A wire - coating die essentially consists of a cylindrical wire of radius κR moving horizontally at
a constant velocity V along the axis of a cylindrical die of radius R. If the pressure in the die is
uniform, then the Newtonian fluid (of constant viscosity μ and constant density ρ) flows through
the narrow annular region solely by the drag due to the motion of the wire (and is referred to as
'axial annular Couette flow'). Neglect end effects and assume an isothermal system.

Figure. Fluid flow in wire coating die.
a) Establish the expression for the steady-state velocity profile in the annular region of the die.
b) Obtain the mass flow rate through the annular die region.
c) Estimate the coating thickness δ some distance downstream of the die exit.
d) Find the force that must be applied per unit length of the wire.
e) Write the force expression in d) as a 'plane slit' formula with a 'curvature correction'.
Solution.
a)
Step. Shear stress and velocity distribution
The fluid flow is in the z-direction shearing constant-r surfaces. Therefore, the velocity
component that exists is vz(r) and the shear stress component is τrz(r).
A momentum balance in cylindrical coordinates (click here for detailed discussion) gives
d
dr
(r τrz) =
ΔP
L
r (1)
Since there is no pressure gradient and gravity does not play any role in the motion of the fluid,
ΔP = 0. Equation (1) with the right-hand side set to zero leads to the following expression for the
shear stress distribution on integration:
τrz =
C1
r
(2)

Substituting Newton's law of viscosity for τrz in equation (2) gives
− μ
dvz
dr
=
C1
r
(3)
The above differential equation is simply integrated to obtain the following velocity profile:
vz = −
C1
μ
ln r + C2 (4)
The integration constants C1 and C2 are evaluated from the following no-slip boundary
conditions:
BC 1: at r = R, vz = 0 (5)
BC 2: at r = κR, vz = V (6)
From BC 1, C2 = (C1/μ) ln R. So, vz = (C1/μ) ln (R/r) and BC 2 yields C1 = − μV/ ln κ. On
substituting the integration constants into equations (2) and (4), the final expressions for the
shear stress and velocity distribution are
τrz =
μV
r ln (1/κ)
(7)
vz
V
=
ln (r/R)
ln κ
(8)
b)

The mass rate of flow is obtained by integrating the velocity profile over the cross section of the
annular region as follows.
w = 2 π ρ
R
∫
κR
rvz dr =
2 π R2
V ρ
ln κ
1
∫
κ
r
R
ln
r
R
d
r
R
(9)
Integration by parts then gives
w =
2 π R2
V ρ
ln κ
1
2
r
R
2
ln
r
R
−
1
4
r
R
2 1
κ
(10)
On evaluation at the limits of integration, the final expression for the mass rate of flow is
w =
π R2
V ρ
2
1 − κ2
ln(1/κ)
− 2 κ2
(11)
c)
Step. Coating thickness
Some distance downstream of the die exit, the Newtonian liquid is transported as a coating of
thickness δ exposed to air. A momentum balance in the 'air region' yields the same shear stress
expression [equation (2)]. The boundary condition at the liquid-gas interface is τrz = 0
at r = κR + δ, which now gives C1 = 0 and τrz = 0 throughout the coating. Furthermore, equations
(4) and (6) are valid and give the velocity profile as vz = V through the entire coating. As the
velocity becomes uniform and identical to that of the wire, the liquid is transported as a rigid
coating. The mass flow rate in the air region is given by
wair = 2πVρ
κR + δ
∫
κR
r dr = πR2
Vρ [(κ + δ/R)2
− κ2
] (12)
Equating the above expression for the flow rate wair in the air region to the flow rate w in the die
region gives

δ
R
= κ2
+
w
πR2
Vρ
1/2
− κ (13)
On substituting w from equation (11), the final expression for the coating thickness is
δ
R
=
1 − κ2
2 ln(1/κ)
1/2
− κ (14)
d)
Step. Force applied to wire
The force applied to the wire is given by the shear stress integrated over the wetted surface area.
Therefore, on using equation (7),
Fz
L
= (2 π κR) τrz| r = κR =
2πμV
ln (1/κ)
(15)
e)
Step. Plane slit formula for force
A very narrow annular region with outer radius R and inner radius (1 − ε)R, where ε is very
small, may be approximated by a plane slit with area = 2π(1 − ½ε)RL ≈ 2πRL and gap = εR.
Then, the velocity profile in the plane slit will be linear and the force applied to the 'plane' wire is
given by
Fplane
2πRL
= μ
V
εR
or
Fplane
L
=
2πμV
ε
(16)
On substituting kappa; = 1 − ε (since R − κR = εR for the thin annular gap) in equation (15), the
force applied to the wire is given by
Fz
L
=
2πμV
− ln (1 − ε)
(17)

The Taylor series expansion given below is next used.
ln (1 − ε) = −ε −
1
2
ε2
−
1
3
ε3
−
1
4
ε4
−
1
5
ε5
− ... (18)
1
− ln (1
− ε)
= 1
ε
1
+
1
2
ε
+
1
3
ε2
+
1
4
ε3
+
1
5
ε4
+
...
−1
= 1
ε
1
−
1
2
ε
−
1
1
2
ε2
−
1
2
4
ε3
−
19
72
0
ε4
−
...
(19
)
Equation (19) may be obtained by using (1 + u)−1
= 1 − u + u2
− u3
+ u4
− ... (which is an infinite
geometric progression with |u| < 1). On substituting in equation (17), the following force
expression is finally obtained as a 'plane slit' formula with a 'curvature correction'.
Fz
L
=
2πμV
ε
1 −
1
2
ε −
1
12
ε2
−
1
24
ε3
−
19
720
ε4
− ... (20)
The above problem solution considers only drag flow in a wire-coating die. The problem is
solved considering both drag and pressure by Malik, R. and Shenoy, U. V. (1991) 'Generalized
annular Couette flow of a power-law fluid', Ind. Eng. Chem. Res. 30(8): 1950 - 1954.
Problem.
Consider an incompressible isothermal fluid in laminar flow between two coaxial cylinders,
whose inner and outer wetted surfaces have radii of kR and R, respectively. The inner and outer
cylinders are rotating at angular velocities Wi and Wo, respectively. End effects may be
neglected.

Figure. Tangential annular flow between two slowly
rotating cylinders.
a) Determine the steady-state velocity distribution in the fluid (for small values of Wi and Wo).
b) Find the torques acting on the two cylinders during the tangential annular flow of a Newtonian
fluid.
Solution.
a)
Step. Simplification of continuity equation
In steady laminar flow, the fluid is expected to travel in a circular motion (for low values
of Wi and Wo). Only the tangential component of velocity exists. The radial and axial
components of velocity are zero; so, vr = 0 and vz = 0.
For an incompressible fluid, the continuity equation gives .v = 0.
In cylindrical coordinates,
(1)
So, vq = vq (r, z).
If end effects are neglected, then vq does not depend on z. Thus, vq = vq (r).

Step. Simplification of equation of motion
There is no pressure gradient in the q-direction. The components of the equation of motion
simplify to
(2)
(3)
(4)
The r-component provides the radial pressure distribution due to centrifugal forces and the z-
component gives the axial pressure distribution due to gravitational forces (the hydrostatic head
effect).
Step. Velocity profile by solving differential equation
The q-component of the equation of motion is integrated to get the velocity profile:
(5)
(6)
The no-slip boundary conditions at the two cylindrical surfaces are
(7)
(8)

The integration constants are thus given by
(9)
On substituting for C1 and C2, the velocity profile is obtained as
(10)
The velocity distribution given by the above expression can be written in the following
alternative form:
(11)
In the above form, the first term on the right-hand-side corresponds to the velocity profile when
the inner cylinder is held stationary and the outer cylinder is rotating with an angular
velocity Wo. This is essentially the configuration of the Couette - Hatschek viscometer for
determining viscosity by measuring the rate of rotation of the outer cylinder under the
application of a given torque (as discussed below).
The second term on the right-hand-side of the above equation corresponds to the velocity profile
when the outer cylinder is fixed and the inner cylinder is rotating with an angular velocity Wi.
This is essentially the configuration of the Stormer viscometer for determining viscosity by
measuring the rate of rotation of the inner cylinder under the application of a given torque (as
discussed next).
b)
Step. Determination of momentum flux
From the velocity profile, the momentum flux (shear stress) is determined as
(12)
Step. Determination of torque
The torque is obtained as the product of the force and the lever arm. The force acting on the inner
cylinder itself is the product of the inward momentum flux and the wetted surface area of the
inner cylinder. Thus,

(13)
Similarly, the torque acting on the outer cylinder is obtained as the product of the outward
momentum flux, the wetted surface area of the outer cylinder, and the lever arm. Thus,
(14)
The tangential annular flow system above provides the basic model for rotational viscometers (to
determine fluid viscosities from measurements of torque and angular velocities) and for some
kinds of friction bearings.
Problem.
An incompressible isothermal liquid in laminar flow is open to the atmosphere at the top.
Determine the shape of the free liquid surface z(r) at steady state (neglecting end effects, if any)
for the following cases:
Case a) Case b)

Case c) Case d)
Figure. Free surface shape of liquid in four cases of tangential flow.
a) when the liquid is in the annular space between two vertical coaxial cylinders, whose inner
and outer wetted surfaces have radii of k R and R, respectively. Here, the inner cylinder is
rotating at a constant angular velocity Wi and the outer cylinder is stationary. Let zR be the liquid
height at the outer cylinder.
b) when a single vertical cylindrical rod of radius Ri is rotating at a constant angular
velocity Wi in a large body of quiescent liquid. Let zRo be the liquid height far away from the
rotating rod.
Hint: Use the result in a).
c) when the liquid is in the annular space between two vertical coaxial cylinders, whose inner
and outer wetted surfaces have radii of k R and R, respectively. Here, the outer cylinder is
rotating at a constant angular velocity Wo and the inner cylinder is stationary. Let zR be the liquid
height at the outer cylinder.
d) when the liquid is in a vertical cylindrical vessel of radius R, which is rotating about its own
axis at a constant angular velocity Wo. Let zR be the liquid height at the vessel wall.
Hint: Use the result in c).
Solution.
(Click link again to hide stepwise solution)

Consider the liquid in the annular space between two vertical coaxial cylinders, whose inner and
outer wetted surfaces have radii of k R and R, respectively. Let the inner and outer cylinders be
rotating at constant angular velocities of Wi andWo, respectively. The first task is to determine
the velocity profile by simplifying the continuity equation and the equation of motion. The next
task is to obtain the free surface shape of the liquid from the velocity profile.
In steady laminar flow, the liquid is expected to travel in a circular motion with only the
tangential component of velocity. The radial and axial components of velocity are zero; so, vr = 0
and vz = 0.
For an incompressible fluid, the continuity equation gives .v = 0. In cylindrical coordinates,
(1)
So, vq = vq (r, z). If end effects are neglected, then vq does not depend on z. Thus, vq = vq (r).
simplify to
(2)
(3)
(4)
As discussed later, the r-component provides the radial pressure distribution due to centrifugal
forces and the z-component gives the axial pressure distribution due to gravitational forces (the
hydrostatic head effect).

(5)
(6)
The no-slip boundary conditions at the two cylindrical surfaces are
(7)
(8)
(9)
(10)
alternative form:
(11)
the inner cylinder is held stationary and the outer cylinder is rotating with an angular
velocity Wo. The second term on the right-hand-side of the above equation corresponds to the
velocity profile when the outer cylinder is fixed and the inner cylinder is rotating with an angular
velocity Wi.
a)

Step. Free surface shape for rotating inner cylinder and fixed outer cylinder
Now, consider the case where the inner cylinder is rotating at a constant angular velocity Wi and
the outer cylinder is stationary. With Wo = 0, the velocity profile in Eq. 11 is substituted into
the r-component of the equation of motion to get
(12)
Integration gives
(13)
The above expression for the pressure p is next substituted into the z-component of the equation
of motion to obtain
(14)
On substituting the expression for f1 in Eq. 13,
(15)
The constant C3 is determined from the following condition:
(16)
where zR is the liquid height at the outer cylinder and patm denotes the atmospheric pressure.
Subtraction of the above two equations allows for elimination of C3 and the determination of the
pressure distribution in the liquid as
(17)
Since p = patm at the liquid-air interface, the shape of the free liquid surface is obtained as
(18)

b)
Step. Free surface shape for rotating cylinder in an infinite expanse of liquid
The result in a) may be used to derive the free surface shape when a single vertical cylindrical
rod of radius Ri is rotating at a constant angular velocity Wi in a large body of quiescent liquid.
On substituting R = Ro and k = Ri / Ro, Eq. 11 (with Wo = 0) and Eq. 18 yield
(19)
The above equations give the velocity profile and the shape of the free liquid surface for the case
where the inner cylinder of radius Ri is rotating at a constant angular velocity Wi and the outer
cylinder of radius Ro is stationary. Here, zRo be the liquid height far away from the rotating rod.
In the limit as Ro tends to infinity, the velocity profile and free surface shape for a single vertical
rod rotating in a large body of quiescent liquid are obtained as
(20)
c)
Step. Free surface shape for rotating outer cylinder and fixed inner cylinder
Next, consider the case where the outer cylinder is rotating at a constant angular velocity Wo and
the inner cylinder is stationary. With Wi = 0, the velocity profile in Eq. 11 is substituted into
the r-component of the equation of motion to get
(21)
Integration gives
(22)

The above expression for the pressure p is now substituted into the z-component of the equation
of motion to obtain
(23)
(24)
The constant C4 is found from the following condition:
(25)
where zR is the liquid height at the outer cylinder and patm denotes the atmospheric pressure.
Subtraction of the above two equations eliminates C4 and gives the pressure distribution in the
liquid as
(26)
Since p = patm at the liquid-air interface, the shape of the free liquid surface is finally obtained as
(27)
An alternative approach is to obtain the above equation from the result of a) by
substituting r = k R in Eq. 18 to get
(28)
The above expression gives the difference in elevations of the liquid surface at the outer and
inner cylinders. Subtracting Eq. 28 from Eq. 18 yields
(29)

The cylinders must now be swapped, i.e., the rotating cylinder should be made the outer cylinder
rather than the inner one. So, let k R = Ro and R = bRo. On substituting k = 1/b and R = bRo in the
above equation,
(30)
On recognizing that b (which is a dummy parameter) can be simply replaced by k, Ro may be
replaced by R, and Wi byWo, the above equation is identical to Eq. 27.
d)
Step. Free surface shape for cylindrical vessel rotating about its own axis
The result in c) may be used to derive the free surface shape when a liquid is in a vertical
cylindrical vessel of radius R, which is rotating about its own axis at a constant angular
velocity Wo.
In the limit as k tends to zero, Eq. 11 (with Wi = 0) and Eq. 27 yield the velocity profile and free
surface shape as
(31)
Note that as k tends to zero, the first and second terms in square brackets in Eq. 27 also tend to
zero. The velocity profile suggests that each element of the liquid in a rotating cylindrical vessel
moves like an element of a rigid body. The free surface of the rotating liquid in a cylindrical
vessel is a paraboloid of revolution (since Eq. 31 corresponds to a parabola).
Problem.
A plastic resin is in a vertical cylindrical vessel of radius R, which is rotating about its own axis
at a constant angular velocity W.

Figure. Parabolic mirror from free surface shape of rotating liquid.
a) Determine the shape of the free surface z(r) at steady state (neglecting end effects, if any).
Let zR be the liquid height at the vessel wall.
b) Show that the radius of curvature (of the free surface shape) is given by
c) The cylinder is rotated until the resin hardens in order to fabricate the backing for a parabolic
mirror. Determine the angular velocity (in rpm) necessary for a mirror of focal length 163 cm.
Given: The focal length f is one-half the radius of curvature rc at the axis.
Solution.
(Click link again to hide stepwise solution)
a)
In steady laminar flow, the liquid is expected to travel in a circular motion with only the
tangential component of velocity. The radial and axial components of velocity are zero; so, vr = 0
and vz = 0.
For an incompressible fluid, the continuity equation gives .v = 0. In cylindrical coordinates,

(1)
So, vq = vq (r, z). If end effects are neglected, then vq does not depend on z. Thus, vq = vq (r).
simplify to
(2)
(3)
(4)
As discussed later, the r-component provides the radial pressure distribution due to centrifugal
forces and the z-component gives the axial pressure distribution due to gravitational forces (the
hydrostatic head effect).
(5)
(6)
Since the velocity vθ is finite at r = 0, the integration constant C2 must be zero. Because vθ =
ΩR at r = R, C1 = 2Ω. Thus,

vθ = Ω r (7)
The above velocity profile suggests that each element of the liquid in a rotating cylindrical vessel
moves like an element of a rigid body.
Step. Free surface shape for rotating liquid in cylinder
Next, the velocity profile is substituted into the r-component of the equation of motion (Eq. 2) to
get
∂p
∂r
=
ρvθ
2
r
= ρΩ2
r (8)
Integration gives
p = ρΩ2
r2
/2 + f1(z) (9)
The above expression for the pressure p is now substituted into the z-component of the equation
of motion (Eq. 4) to obtain
∂p
∂z
= −ρg ⇒
df1
dz
= −ρg or f1 = −ρgz + C3 (10)
p = ρΩ2
r2
/2 − ρgz + C3 (11)
The constant C3 is found from the following condition:
at r = R, z = zR and p = patm ⇒ patm = ρΩ2
R2
/2 − ρgzR + C3 (12)
where zR is the liquid height at the vessel wall and patm denotes the atmospheric pressure.
Subtraction of the above two equations eliminates C and gives the pressure distribution in the
liquid as
p − patm = ρΩ2
(r2
− R2
)/2 + ρg(zR − z) (13)
Since p = patm at the liquid-air interface, the shape of the free liquid surface is finally obtained as
zR − z = Ω2
(R2
− r2
)/(2g) (14)

The free surface of the rotating liquid in a cylindrical vessel is a paraboloid of revolution (since
Eq. 14 corresponds to a parabola).
b)
Step. General expression for radius of curvature
The circle that is tangent to a given curve at point P, whose center lies on the concave side of the
curve and which has the same curvature as the curve has at P, is referred to as the circle of
curvature. It radius (CP in the figure) defines the radius of curvature at P. The circle of curvature
has its first and second derivatives respectively equal to the first and second derivatives of the
curve itself at P. If the coordinates of the center C and point P are (r1, z1) and (r, z) respectively,
then the radius of curvature CP is given by
rc = [(z1 − z)2
+ (r − r1)2
]1/2
(15)
Now, the radius CP is perpendicular to the tangent to the curve at P. So, the slope of the radius
CP is given by
−
dr
dz
=
z1 − z
r1 − r
(16)
Thus, the first and second derivatives of the curve at point P are
dz
dr
=
r − r1
z1 − z
and
d2
z
dr2 =
1
z1 − z
+
r − r1
(z1 − z)2
dz
dr
=
[1 + (dz / dr)2
]
z1 − z
(17)
On substituting the above derivatives in Eq. 15, the general expression for the radius of curvature
is obtained as
(18)
c)
Step. Angular velocity necessary for parabolic mirror fabrication

Let z0 be the height of the liquid surface at the centerline of the cylindrical vessel.
Substituting z = z0 at r = 0 in Eq. 14 gives zR − z0 = Ω2
R2
/(2g). Thus, an alternative expression (in
terms of z0) for the shape of the free surface is
z = z0 + Ω2
r2
/(2g) (19)
Differentiating the above expression twice gives
dz
dr
=
Ω2
r
g
and
d2
z
dr2 =
Ω2
g
(20)
On substituting these derivatives into Eq. 18, the radius of curvature of the parabola is given by
rc = [1 + Ω4
r2
/g2
]3/2
(g/Ω2
) (21)
Since the focal length f of the mirror is one-half the radius of curvature rc at the axis,
f = ½ rc|r = 0 = g/(2Ω2
) (22)
Finally, the rotational speed Ω necessary to fabricate a mirror of focal length f is given by
Ω = [g/(2f)]1/2
(23)
Step. Substitution of numerical values
On substituting the focal length f =
163
cm and the gravitational acceleration g = 981 cm/s2
in
the above equation, the rotational speed may be calculated as 1.73 rad/s = 1.73 x 60/(2π) rpm =
16.57 rpm.
The numerical value above for the focal length f in cm may be changed and the rotational speed
recalculated by clicking on the button below.
Problem.
Consider an incompressible isothermal fluid in laminar flow between two concentric spheres,
whose inner and outer wetted surfaces have radii of kR and R, respectively. The inner and outer
spheres are rotating at constant angular velocities Wi and Wo, respectively. The spheres rotate
slowly enough that the creeping flow assumption is valid.

Figure. Flow between two slowly rotating spheres.
a) Determine the steady-state velocity distribution in the fluid (for small values of Wi and Wo).
b) Find the torques on the two spheres required to maintain the flow of a Newtonian fluid.
c) Simplify the velocity and torque expressions for the case of a single solid sphere of
radius Ri rotating slowly at a constant angular velocity Wi in a very large body of quiescent fluid.
Solution.
a)
In steady laminar flow, the fluid is expected to travel in a circular motion (for low values
of Wi and Wo). So, vr = 0 and vq = 0.
For an incompressible fluid, the continuity equation gives .v = 0.
In spherical coordinates,
(1)
So, vf = vf (r, q).

When the flow is very slow, the term rv. v in the equation of motion can be neglected because
it is quadratic in the velocity. This is referred to as the creeping flow assumption. Thus, for
steady creeping flow, the entire left-hand side of the equation of motion is zero and we get
0 = - p + m 2
v + rg = - P + m 2
v (2)
where P is the modified pressure given by P = p + rgh and h is the elevation in the gravitational
field (i.e., the distance upward in the direction opposite to gravity from some chosen reference
plane). In this problem, h = z = r cos q. The above equation is referred to as the Stokes flow
equation or the creeping flow equation of motion. The components of the creeping flow equation
in spherical coordinates simplify to
(3)
(4)
(5)
There is no dependence on the angle f due to symmetry about the vertical axis. The above partial
differential equation must be solved for the velocity distribution vf (r, q).
The no-slip boundary conditions at the two spherical surfaces are
(6)
(7)
From the boundary conditions, the form vf (r, q) = f(r) sin q appears to be an educated guess to
solve the partial differential equation for the velocity profile. On substituting this form in the f-
component of the equation of motion and simplifying, we get

(8)
The above ordinary differential equation may be solved by substituting f(r) = rn
. The resulting
characteristic equation is n2
+ n - 2 = 0, whose solution is n = 1, -2. Thus,
(9)
On imposing the boundary conditions,
(10)
(11)
(12
)
alternative form:
(13)
the inner sphere is held stationary and the outer sphere is rotating with an angular velocity Wo.
The second term on the right-hand-side of the above equation corresponds to the velocity profile
when the outer sphere is fixed and the inner sphere is rotating with an angular velocity Wi.
b)

Step. Determination of momentum flux
From the velocity profile, the momentum flux (shear stress) is determined as
(14)
The torque required to rotate the inner sphere is obtained as the integral over the sphere surface
of the product of the force exerted on the fluid by the solid surface element and the lever
arm kR sin q. The force required on the inner sphere is the product of the outward momentum
flux and the wetted surface area of the inner sphere. Thus, the differential torque on the inner
sphere surface element is
(15)
On integrating from 0 to p, the torque needed on the inner sphere is
(16)
Note that the integral above evaluates to 4/3.
Similarly, the differential torque needed on the outer sphere surface element is obtained as the
product of the inward momentum flux, the wetted surface area of the outer sphere, and the lever
arm. Thus,
(17)
On integrating from 0 to p, the torque needed on the outer sphere is
(18)

c)
Step. Limiting case of single rotating sphere
For the case of a single solid sphere of radius Ri rotating slowly at a constant angular
velocity Wi in a very large body of quiescent fluid, let R = Ro and k = Ri / Ro. Then, on
letting Wo = 0, Eqs. (13) and (16) give
(19)
The above equations hold when the outer sphere of radius Ro is stationary and the inner sphere of
radius Ri is rotating at an angular velocity Wi. In the limit as Ro tends to infinity, the results for a
single rotating sphere in an infinite body of quiescent fluid are obtained as
(20)
Problem.
An isothermal, incompressible fluid of density ρ flows radially outward owing to a pressure
difference between two fixed porous, concentric spherical shells of radii κR and R. Note that the
velocity is not zero at the solid surfaces. Assume negligible end effects and steady laminar flow
in the region κR ≤ r ≤ R.
Figure. Radial flow between two porous concentric
spheres.

a) Simplify the equation of continuity to show that r2
vr = constant.
b) Simplify the equation of motion for a Newtonian fluid of viscosity μ.
c) Obtain the pressure profile P(r) in terms of PR and vR, the pressure and velocity at the sphere
of radius R.
d) Determine the nonzero components of the viscous stress tensor for the Newtonian case.
Solution.
a)
Since the steady laminar flow is directed radially outward, only the radial velocity
component vr exists. The other two components of velocity are zero; so, vθ = 0 and vφ = 0.
For incompressible flow, the continuity equation gives ∇.v = 0.
In spherical coordinates,
1
r2
∂
∂r
(r2
vr ) +
1
r sin θ
∂
∂θ
(vθ sin θ ) +
1
r sin θ
∂vφ
∂φ
= 0 ⇒
∂
∂r
(r2
vr ) = 0 (1)
On integrating the simplified continuity equation, r2
vr = f(θ, φ). There is no dependence
expected on the angles θ and φfrom symmetry arguments. In other words, r2
vr = C, where C is a
constant. This is simply explained from the fact that mass (or volume, if density ρ is constant) is
conserved; so, ρ (4 π r2
vr ) = w is constant and C = w/(4πρ).
b)
The equation of motion is
ρ
Dv
Dt
= −∇p − ∇ . τ + ρ g (2)

The above equation is simply Newton's second law of motion for a fluid element. It states that,
on a per unit volume basis, the mass multiplied by the acceleration is because of three forces,
namely the pressure force, the viscous force, and the gravity force. For an incompressible,
Newtonian fluid, the term −∇ . τ = μ ∇2
v and the equation of motion yields the Navier - Stokes
equation. On noting that r2
vr = constant from the continuity equation, the components of the
equation of motion for steady flow in spherical coordinates may be simplified as given below.
r - component
:
ρ
æ
è
vr
∂vr
∂r
ö
ø
= −
∂p
∂r
− ρ g cos θ (3)
θ - component
:
0 = −
1
r
∂p
∂θ
+ ρ g sin θ (4)
φ - component
:
0 =
∂p
∂φ
(5)
Note that the modified pressure P may be defined as P = p + ρgh, where h is the elevation or the
height above some arbitrary datum plane, to avoid calculating the components of the
gravitational acceleration vector g in spherical coordinates. Here, P = p + ρg r cos θ giving
∂P/∂r = ∂p/∂r + ρg cos θ and ∂P/∂θ = ∂p/∂θ − ρg r sin θ. Then, the components of the equation
of motion in terms of the modified pressure are
r - component
:
ρ
æ
è
vr
∂vr
∂r
ö
ø
= −
∂P
∂r
(6)
θ - component
:
0 =
∂P
∂θ
(7)
φ - component
:
0 =
∂P
∂φ
(8)
Note that the equations (6) - (8) could have been directly obtained from the following form of the
equation of motion:
ρ
Dv
Dt
= −∇P − ∇ . τ (9)

in which the modified pressure P includes both the pressure and gravitational terms.
From equations (7) and (8), P is a function of only r. Substituting P = P(r) and vr = vr(r) in
equation (6) then gives
dP
dr
= − ρ vr
dvr
dr
(10)
c)
Step. Pressure profile
On substituting vr = C / r2
, equation (10) gives
dP
dr
= 2ρ
C2
r5 (11)
Integration gives
P = −
ρ C2
2r4 + C1 (12)
Since the pressure is P = PR and the velocity is vr = vR at r = R, the integration constants may be
evaluated as C1 = PR +ρ C2
/(2R4
) and C = R2
vR . Then, substitution in equation (12) yields the
pressure profile as
P = PR +
ρ C2
2R4
é
ë
1 −
æ
è
R
r
ö
ø
4
ù
û
= PR +
1
2
ρ vR
2 é
ë
1 −
æ
è
R
r
ö
ø
4
ù
û
(13)
d)
Step. Determination of nonzero components of viscous stress tensor
For an incompressible, Newtonian fluid, the viscous stress tensor is given by τ = −μ [ ∇v +
(∇v)t
]. Since the only velocity component that exists is vr(r) = C / r2
, the nonzero components of
the viscous stress tensor are

τrr = −2 μ
dvr
dr
=
4 μ C
r 3 and τθθ = τφφ = −2 μ
vr
r
= −
2 μ C
r 3 (14)
Note that the shear stresses are all zero. The normal stresses in equation (10) are partly
compressive and partly tensile. Not only is the φ-component of [∇ . τ] directly zero, but the r-
component and the θ-components are zero too as evaluated below.
[∇ . τ]r =
1
r 2
d
dr
(r 2
τrr) −
τθθ + τφφ
r
= −
4 μ C
r 4 +
4 μ C
r 4 = 0 (15)
[∇ . τ]θ =
1
r sin θ
∂
∂θ
(τθθ sin θ) −
τφφ cot θ
r
= −
2 μ C cot θ
r 4 +
2 μ C cot θ
r 4 = 0 (16)
Since [∇ . τ] = 0, viscous forces can be neglected in this flow and the Euler equation for inviscid
fluids may be directly used.
Problem.
An isothermal, incompressible fluid of density ρ flows radially outward owing to a pressure
difference between two fixed porous, coaxial cylindrical shells of radii κR and R. Note that the
velocity is not zero at the solid surfaces. Assume long cylinders with negligible end effects and
steady laminar flow in the region κR ≤ r ≤ R.
Figure. Radial flow between two porous coaxial
cylinders (top view).
a) Simplify the equation of continuity to show that r vr = constant.
b) Simplify the equation of motion for a Newtonian fluid of viscosity μ.

c) Obtain the pressure profile P(r) in terms of PR and vR, the pressure and velocity at the cylinder
of radius R.
d) Determine the nonzero components of the viscous stress tensor for the Newtonian case.
Solution.
a)
component vr exists. The tangential and axial components of velocity are zero; so, vθ = 0 and vz =
0.
1
r
∂
∂r
(r vr ) +
1
r
∂vθ
∂θ
+
∂vz
∂z
= 0 ⇒
∂
∂r
(r vr ) = 0 (1)
On integrating the simplified continuity equation, r vr = f(θ, z). Since the solution is expected to
be symmetric about the z-axis, there is no dependence on the angle θ. Furthermore, the cylinders
are long with negligible end effects (considered unbounded in z-direction); so, there is no z-
dependence. In other words, r vr = C, where C is a constant. This is simply explained from the
fact that mass (or volume, if density ρ is constant) is conserved; so, ρ (2 π r vr L) = w is constant
and C= w/(2πLρ).
b)
The equation of motion is
ρ
Dv
Dt
= −∇P − ∇ . τ (2)

in which P includes both the pressure and gravitational terms. Note that the modified
pressure P is defined as P = p + ρgh, where h is the elevation or the height above some arbitrary
datum plane. Then, the formulation is valid for any orientation of the cylinder axis. For an
incompressible, Newtonian fluid, the term −∇ . τ = μ ∇2
v and the equation of motion yields the
Navier - Stokes equation. On noting that r vr = constant from the continuity equation, the
components of the equation of motion for steady flow in cylindrical coordinates may be
simplified as given below.
r - component
:
ρ
æ
è
vr
∂vr
∂r
ö
ø
= −
∂P
∂r
(3)
θ - component
:
0 =
∂P
∂θ
(4)
z - component
:
0 =
∂P
∂z
(5)
From equations (4) and (5), P is a function of only r. Substituting P = P(r) and vr = vr(r) in
equation (3) then gives
dP
dr
= − ρ vr
dvr
dr
(6)
c)
Step. Pressure profile
On substituting vr = C / r, equation (6) gives
dP
dr
= ρ
C2
r3 (7)
Integration gives
P = −
ρ C2
2r2 + C1 (8)

Since the pressure is P = PR and the velocity is vr = vR at r = R, the integration constants may be
evaluated as C1 = PR +ρ C2
/(2R2
) and C = R vR . Then, substitution in equation (8) yields the
pressure profile as
P = PR +
ρ C2
2R2
é
ë
1 −
æ
è
R
r
ö
ø
2
ù
û
= PR +
1
2
ρ vR
2 é
ë
1 −
æ
è
R
r
ö
ø
2
ù
û
(9)
d)
Step. Determination of nonzero components of viscous stress tensor
For an incompressible, Newtonian fluid, the viscous stress tensor is given by τ = −μ [ ∇v +
(∇v)t
]. Since the only velocity component that exists is vr(r) = C / r, the nonzero components of
the viscous stress tensor are
τrr = −2 μ
dvr
dr
=
2 μ C
r 2 and τθθ = −2 μ
vr
r
= −
2 μ C
r 2 (10)
Note that the shear stresses are all zero. The normal stresses in equation (10) are equal in
magnitude and opposite in sign (partly compressive and partly tensile). Not only are the θ-
component and z-component of [∇ . τ] zero, but the r-component is zero too because
[∇ . τ]r =
1
r
d
dr
(r τrr) −
τθθ
r
= −
2 μ C
r 3 +
2 μ C
r 3 = 0 (11)
Since [∇ . τ] = 0, viscous forces can be neglected in this flow and the Euler equation for inviscid
fluids may be directly used.
Problem.
Steady, laminar flow occurs in the space between two fixed parallel, circular disks separated by a
small gap 2b. The fluid flows radially outward owing to a pressure difference (P1 − P2) between
the inner and outer radii r1 and r2, respectively. Neglect end effects and consider the
region r1 ≤ r ≤ r2 only. Such a flow occurs when a lubricant flows in certain lubrication systems.
Figure. Radial flow between two parallel
disks.

a) Simplify the equation of continuity to show that r vr = f, where f is a function of only z.
b) Simplify the equation of motion for incompressible flow of a Newtonian fluid of
viscosity μ and density ρ.
c) Obtain the velocity profile assuming creeping flow.
d) Sketch the velocity profile vr (r, z) and the pressure profile P(r).
e) Determine an expression for the mass flow rate by integrating the velocity profile.
f) Derive the mass flow rate expression in e) using an alternative short-cut method by adapting
the plane narrow slit solution.
Solution.
a)
component vr exists. The tangential and axial components of velocity are zero; so, vθ = 0 and vz =
0.
1
r
∂
∂r
(r vr ) +
1
r
∂vθ
∂θ
+
∂vz
∂z
= 0 ⇒
∂
∂r
(r vr ) = 0 (1)
On integrating the simplified continuity equation, r vr = f(θ, z). Since the solution is expected to
be symmetric about the z-axis, there is no dependence on the angle θ. Thus, f is a function
of z only and not of r or θ. In other words, r vr = f(z). This is simply explained from the fact that
mass (or volume, if density ρ is constant) is conserved; so, ρ (2 π r vr dz) = dw is constant (at a
given z) and is independent of r.
b)

For a Newtonian fluid, the Navier - Stokes equation is
ρ
Dv
Dt
= −∇P + μ∇2
v (2)
in which P includes both the pressure and gravitational terms. On noting that vr = vr(r, z), its
components for steady flow in cylindrical coordinates may be simplified as given below.
r - component
:
ρ
æ
è
vr
∂vr
∂r
ö
ø
= −
∂P
∂r
+ μ
é
ë
∂
∂r
æ
è
1
r
∂
∂r
(r vr )
ö
ø
+
∂2
vr
∂z2
ù
û
(3)
θ - component
:
0 =
∂P
∂θ
(4)
z - component
:
0 =
∂P
∂z
(5)
Recall that r vr = f(z) from the continuity equation. Substituting vr = f / r and P = P(r) in equation
(3) then gives
− ρ
f 2
r3 = −
dP
dr
+
μ
r
d2
f
dz2 (6)
c)
Step. Velocity profile
Equation (6) has no solution unless the nonlinear term (that is, the f 2
term on the left-hand side)
is neglected. Under this 'creeping flow' assumption, equation (6) may be written as
r
dP
dr
= μ
d2
f
dz2 (7)
The left-hand side of equation (7) is a function of r only, whereas the right-hand side is a
function of z only. This is only possible if each side equals a constant (say, C0). Integration with

respect to r from the inner radius r1 to the outer radius r2then gives P2 − P1 = C0 ln (r2/r1). On
replacing C0 in terms of f,
0 = (P1 − P2) +
æ
è
μ ln
r2
r1
ö
ø
d2
f
dz2 (8)
The above equation may be integrated twice with respect to z as follows.
df
dz
=
− ΔP
μ ln (r2/r1)
z + C1 (9)
f =
− ΔP
2 μ ln (r2/r1)
z2
+ C1 z + C2 ⇒ vr =
− ΔP
2 μ r ln (r2/r1)
z2
+ C1
z
r
+
C2
r
(10)
Here, ΔP ≡ P1 − P2. Equation (10) is valid in the region r1 ≤ r ≤ r2 and −b ≤ z ≤ b.
Imposing the no-slip boundary conditions at the two stationary disk surfaces (vz = 0 at z = +b and
any r) gives C1 = 0 and C2 = ΔP b2
/ [2 μ ln (r2/r1)]. On substituting the integration constants in
equation (10), the velocity profile is ultimately obtained as
vr =
ΔP b2
2 μ r ln (r2/r1)
é
ë
1 −
æ
è
z
b
ö
ø
2
ù
û
(11)
d)
Step. Sketch of velocity profile and pressure profile
The velocity profile from equation (11) is observed to be parabolic for each value
of r with vr,max = ΔP b2
/ [2 μ r ln (r2/r1)]. The maximum velocity at z = 0 is thus inversely
proportional to r. In general, it is observed from equation (11) that vr itself is inversely
proportional to r. Sketches of vr(z) for different values of r and vr(r) for different values of |z|
may be plotted.
The pressure profile obtained by integrating the left-hand side of equation (7) is (P − P2) /
(P1 − P2) = [ln(r/r2)] / [ln(r1/r2)]. A sketch of P(r) may be plotted which holds for all z.
e)
Step. Mass flow rate by integrating velocity profile

The mass flow rate w is rigorously obtained by integrating the velocity profile using w =
∫ n . ρv dS, where n is the unit normal to the element of surface area dS and v is the fluid velocity
vector. For the radial flow between parallel disks, n = δr, v = vrδr , and dS = 2πr dz. Then,
substituting the velocity profile from equation (11) and integrating gives
w =
b
∫
−b
ρ vr (2 π r) dz =
π ΔP b2
ρ
μ ln (r2/r1)
é
ë
z −
z3
3b2
ù
û
b
−b
=
4π ΔP b3
ρ
3μ ln (r2/r1)
(12)
f)
Step. Mass flow rate using short-cut method by adapting narrow slit solution
The plane narrow slit solution may be applied locally by recognizing that at all points between
the disks the flow resembles the flow between parallel plates provided vr is small (that is, the
creeping flow is valid).
The mass flow rate for a Newtonian fluid in a plane narrow slit of width W, length L and
thickness 2B is given by w = 2 ΔPB3
W ρ / (3 μ L) (click here for derivation). In this expression,
(ΔP/L) is replaced by (− dP/dr), B by b, and W by 2πr. Note that mass is conserved; so, w is
constant. Then, integrating from r1 to r2 gives the same mass flow rate expression [equation (12)]
as shown below.
w
r2
∫
r1
dr
r
=
4π b3
ρ
3μ
P2
∫
P1
(−dP) ⇒ w =
4π (P1 − P2) b3
ρ
3μ ln (r2/r1)
(13)
This alternative short-cut method for determining the mass flow rate starting from the narrow slit
solution is very powerful because the approach may be used for non-Newtonian fluids where
analytical solutions are difficult to obtain.
Problem.
A parallel - disk viscometer consists of two circular disks of radius R separated by a small
gap B (with R >> B). A fluid of constant density ρ, whose viscosity μ is to be measured, is placed
in the gap between the disks. The lower disk at z = 0 is fixed. The torque Tz necessary to rotate
the upper disk (at z = B) with a constant angular velocity Ω is measured. The task here is to
deduce a working equation for the viscosity when the angular velocity Ω is small (creeping
flow).

Figure. Parallel - disk viscometer.
a) Simplify the equations of continuity and motion to describe the flow in the parallel - disk
viscometer.
b) Obtain the tangential velocity profile after writing down appropriate boundary conditions.
c) Derive 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. Neglect the pressure term.
Solution.
a)
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.
1
r
∂
∂r
(r vr ) +
1
r
∂vθ
∂θ
+
∂vz
∂z
= 0 ⇒
∂vθ
∂θ
= 0 (1)
So, vθ = vθ (r, z).
For a Newtonian fluid, the components of the Navier - Stokes equation may be simplified as
given below.

r - component
:
−
ρvθ
2
r
= −
∂p
∂r
(2)
θ - component
:
0 =
∂
∂r
æ
è
1
r
∂
∂r
(r vθ )
ö
ø
+
∂2
vθ
∂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).
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
d2
f
dz2 = 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)
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
æ
è
vθ
r
ö
ø
= 0 (9)
The differential torque dT is rigorously obtained as the cross product of the lever arm r and the
differential force dFexerted 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.
Problem.

Consider a liquid (of density ρ) in laminar flow down an inclined flat plate of length L and
width W. The fluid flows as a falling film with negligible rippling under the influence of gravity.
End effects may be neglected because L and W are large compared to the film thickness δ.
Figure. Fluid flow in a falling film.
power law model (e.g., a polymer liquid). Reduce the result to the Newtonian case.
b) Obtain the mass flow rate for a power law fluid. Simplify for a Newtonian fluid.
c) What is the force exerted by the fluid on the plate in the flow direction?
Solution.
a)
dτxz
dx
=
ΔP
L
(1)
where P is a modified pressure, which is the sum of both the pressure and gravity terms, that is,
ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical. Since the
flow is solely under the influence of gravity, Δp = 0 and therefore ΔP/L = ρ g cos β. On
integration,

τxz = ρ g x cos β + C1 (2)
On using the boundary condition at the gas-liquid interface (τxz = 0 at x = 0), the constant of
integration C1 is found to be zero. This gives the expression for the shear stress τxz for laminar
flow in a falling film as
τxz = ρ g x cos β (3)
In the falling film, the velocity decreases with increasing x and therefore dvz / dx ≤ 0. Thus, the
power law model is given as
τxz = m
æ
è
−
dvz
dx
ö
ø
n
(4)
To obtain the velocity distribution, equations (3) and (4) are combined to eliminate τxz and get
−
dvz
dx
=
æ
è
ρ g cos β
m
ö
ø
1/n
x1/n
(5)
On integrating, vz = − (ρ g cos β/m)1/n
x1/n + 1
/(1/n + 1) + C2. On using the no-slip boundary
condition at the solid surface (vz = 0 at x = δ ), C2 = (ρ g cos β/m)1/n
δ1/n + 1
/(1/n + 1). Thus, the
final expression for the velocity distribution is
vz =
æ
è
ρ g δ cos β
m
ö
ø
1/n
δ
(1/ n) + 1
é
ë
1 −
æ
è
x
δ
ö
ø
(1/ n) + 1
ù
û
(6a)
Equation (6a) when simplified for a Newtonian fluid (n = 1 and m = μ) gives the following
parabolic velocity profile.

vz =
ρ g δ2
cos β
2μ
é
ë
1 −
æ
è
x
δ
ö
ø
2
ù
û
(6b)
b)
The mass flow rate may be obtained by integrating the velocity profile over the film thickness as
shown below.
w =
δ
∫
0
ρ vz W dx
=
æ
è
ρ g δ cos β
m
ö
ø
1/n
δ2
W ρ
(1/ n) + 1
1
∫
0
é
ë
1 −
æ
è
x
δ
ö
ø
(1/ n) + 1
ù
û
d
æ
è
x
δ
ö
ø
(7)
w =
δ2
W ρ
(1/ n) + 2
æ
è
ρ g δ cos β
m
ö
ø
1/n
(8a)
Equation (8a) for a Newtonian fluid (n = 1 and m = μ) gives
w =
ρ2
g W δ3
cos β
3μ
(8b)
When the mass flow rate w in equation (8) is divided by the density ρ and the film cross-
sectional area (W δ), an expression for the average velocity <vz> is obtained. Also, the maximum
velocity vz, max occurs at the interface x = 0 and is readily obtained from equation (6). Thus,
equations (6a) and (8a) give
<vz> =
δ
(1/ n) + 2
æ
è
ρ g δ cos β
m
ö
ø
1/n
=
(1/ n) + 1
(1/ n) + 2
vz, max (9a)
For n = 1 and m = μ, equation (9a) reduces to the Newtonian result given below.

<vz> =
ρ g δ2
cos β
3μ
=
2
3
vz, max (9b)
c)
Step. Force on the plate
The z-component of the force of the fluid on the solid surface is given by the shear stress
integrated over the wetted surface area. Using equation (3) gives
Fz = L W τxz|x = δ = ρ g δ L W cos β (10)
As expected, the force exerted by the fluid on the plate is simply the z-component of the weight
of the liquid film.
Problem.
Consider a fluid (of constant density ρ) in incompressible, laminar flow in a long circular tube of
radius R and length L. 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 a pressure difference
Δp, gravity or both.
power law model (e.g., a polymeric liquid).
b) Obtain the mass flow rate for a power law fluid in tube flow.
Solution.

a)
For axial flow in cylindrical coordinates, the differential equation for the momentum flux is
(click here for derivation)
d
dr
(r τrz) =
ΔP
L
r (1)
ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the tube axis with the vertical.
On integration (with the condition that τrz must be finite at r = 0), this gives the expression for the
shear stress τrz for steady laminar flow in a circular tube as (click here for derivation)
τrz =
ΔP
2 L
r (2)
The velocity vz decreases with increasing r and therefore dvz / dr ≤ 0. Thus, the power law model
is given as
τrz = m
æ
è
−
dvz
dr
ö
ø
n
(3)
Equations (2) and (3) are combined to eliminate τrz and get

−
dvz
dr
=
æ
è
ΔP r
2 m L
ö
ø
1/n
(4)
On integrating and using the no-slip boundary condition (vz = 0 at r = R ), we get the velocity
profile as
vz =
æ
è
ΔP R
2 m L
ö
ø
1/n
R
(1/ n) + 1
é
ë
1 −
æ
è
r
R
ö
ø
(1/ n) + 1
ù
û
(5)
Note that the maximum velocity occurs at the tube centerline (r = 0).
b)
Step. Mass flow rate : Hagen-Poiseuille equation analog
The mass flow rate may be obtained by integrating the velocity distribution over the tube cross
section as shown below.
w =
R
∫
0
ρ vz 2 π r dr
(6)
w = 2 π
æ
è
ΔP R
2 m L
ö
ø
1/n
R3
ρ
(1/ n) + 1
1
∫
0
é
ë
1 −
æ
è
r
R
ö
ø
(1/ n) + 1
ù
û
æ
è
r
R
ö
ø
d
æ
è
r
R
ö
ø
(7)
w =
π R3
ρ
(1/ n) + 3
æ
è
ΔP R
2 m L
ö
ø
1/n
(8)
When the mass flow rate w in equation (8) is divided by the density ρ and the cross-sectional
area (π R2
), an expression for the average velocity is obtained.
For n = 1 and m = μ, equation (8) reduces to the Hagen-Poiseuille equation for Newtonian fluids,
i.e., w = π ΔP R4
ρ / (8 μ L).

Problem.
Consider a fluid (of density ρ) in incompressible, laminar flow in a long circular tube of
radius R and length L. End effects may be neglected because the tube length L is relatively large
compared to the tube radius R. The fluid flows under the influence of a pressure difference Δp,
gravity or both.
a) Determine the steady-state velocity distribution for a non-Newtonian fluid that is described by
the Bingham model.
b) Obtain the mass rate of flow for a Bingham fluid in a circular tube.
Solution.
a)
For axial flow in cylindrical coordinates, the differential equation for the momentum flux is
(click here for derivation)
d
dr
(r τrz) =
ΔP
L
r (1)
ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the tube axis with the vertical.

On integration (with the condition that τrz must be finite at r = 0), this gives the expression for the
shear stress τrz for steady laminar flow in a circular tube as (click here for derivation)
τrz =
ΔP
2 L
r (2)
Step. Bingham model
For viscoplastic materials (e.g., thick suspensions and pastes), there is no flow until a critical
stress (called the yield stressτ0) is reached. To describe such a material that exhibits a yield
stress, the simplest model is the Bingham model given below.
η → ∞ or
dvz
dr
= 0 if |τrz| ≤ τ0 (3a)
η = μ0 +
τ0
+ dvz /dr
or τrz = − μ0
dvz
dr
+ τ0 if |τrz| ≥ τ0 (3b)
Here, η is the non-Newtonian viscosity and μ0 is a Bingham model parameter with units of
viscosity. In equation (3b), the positive sign is used with τ0 and the negative sign
with dvz/dr when τrz is positive. On the other hand, the negative sign is used with τ0 and the
positive sign with dvz/dr when τrz is negative.
Step. Inner plug-flow region
Let τrz = τ0 at r = r0. Then, from equation (2), r0 is given by τ0 = ΔP r0 /(2L).
Let the inner region (r ≤ r0) where the shear stress is less than the yield stress (τrz ≤ τ0) be denoted
by subscript i. Sinceτrz is finite and τrz = − η (dvz /dr), the Bingham model [as per equation (3a)]
gives (dvzi /dr) = 0. Integration gives vzi = C1(constant velocity), which implies the fluid is in
plug flow in the inner region.
Let the outer region be denoted by subscript o. For the region r0 ≤ r ≤ R, the velocity decreases
with increasing r and therefore dvz / dr ≤ 0 and τrz ≥ τ0. Thus, the Bingham model according to
equation (3b) is given by

τrz = − μ0
dvzo
dr
+ τ0 or η = μ0 +
τ0
− dvzo /dr
for r0 ≤ r ≤ R (4)
To obtain the velocity profile for r0 ≤ r ≤ R, equations (2) and (4) may be combined to
eliminate τrz and get
dvzo
dr
= −
ΔP
2μ0L
r +
τ0
μ0
for r0 ≤ r ≤ R (5)
Integration gives vzo = − ΔP r2
/(4μ0L) + τ0 r/μ0 + C2. Imposing the boundary condition that vzo =
0 at r = R then yields C2= ΔP R2
/(4μ0L) − τ0 R/μ0. Thus, the velocity distribution in the outer
region is
vzo =
ΔP R2
4 μ0 L
é
ë
1 −
æ
è
r
R
ö
ø
2
ù
û
−
τ0 R
μ0
æ
è
1 −
r
R
ö
ø
for r0 ≤ r ≤ R (6)
For the inner plug-flow region, it was earlier found that vzi = C1. To determine C1, it may be
noted that vzi = vzo at r = r0. Then, equation (6) along with τ0 = ΔP r0/(2L) may be simplified to
give C1 = ΔP R2
/(4μ0L) (1 − r0/R)2
. Thus, the velocity of the plug flow in the inner region is
vzi =
ΔP R2
4 μ0 L
æ
è
1 −
r0
R
ö
ø
2
for r ≤ r0 (7)
Here, r0 = 2 L τ0 / ΔP is the radius of the plug-flow region in the central part of the tube. The
velocity profile is parabolic in the outer region as given by equation (6) and is flat in the inner
region as given by equation (7).
b)
Step. General expression for mass flow rate
The mass flow rate may be obtained by integrating the velocity profile over the cross section of
the circular tube as shown below.
w = = 2 π ρ
R
∫
0
vz r dr
(8a)

Rather than insert two separate expressions from equations (6) and (7) for vz and integrate in two
regions, it is easier to integrate by parts.
w = 2πρ
é
ë
vz
r2
2
|
R
0
−
1
2
R
∫
0
r2
æ
è
dvz
dr
ö
ø
dr
ù
û
= πρ
R
∫
0
r2
æ
è
−
dvz
dr
ö
ø
dr (8b)
The first term in the square brackets above is zero at both limits on using the no-slip boundary
condition (vz = 0 at r = R) at the upper limit. From equation (2), r/R = τrz /τR where τR = ΔP
R/(2L) is the wall shear stress. Thus, a general expression for the mass rate of flow in a circular
tube is
w =
πR3
ρ
τR
3
τR
∫
0
τrz
2
æ
è
−
dvz
dr
ö
ø
dτrz (8c)
Step. Buckingham - Reiner equation for mass flow rate of Bingham fluid
The lower limit of integration is reset to τ0 as per equation (3a). Then, substituting equation (4)
and integrating yields
w =
πR3
ρ
μ0τR
3
τR
∫
τ0
τrz
2
(τrz − τ0) dτrz =
πR3
ρ
μ0τR
3
é
ë
1
4
(τR
4
− τ0
4
) −
1
3
(τ0τR
3
− τ0
4
)
ù
û
(8d)
The final expression for the mass flow rate is
w =
π ΔP R4
ρ
8 μ0 L
é
ë
1 −
4
3
τ0
τR
+
1
3
æ
è
τ0
τR
ö
ø
4
ù
û
(8e)
The above expression is known as the Buckingham - Reiner equation. Note that τR = ΔP R/(2L)
is the wall shear stress and τ0 is the yield stress. Since no flow occurs below the yield stress (that
is, when τR ≤ τ0), the above expression is valid only for τR > τ0.
For τ0 = 0 and μ0 = μ, the Bingham model simplifies to the Newtonian model and equation (8e)
reduces to the Hagen - Poiseuille equation, i.e., w = π ΔP R4
ρ / (8 μ L).

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