Lec 25: Viscosity, Bernoulli
equation
FLUID MECHANICS
1
• For next time:
– Read: § 11-4 to 11-12.
• Outline:
– Viscosity
– Bernoulli equation
– Examples
• Important points:
– Know what a Newtonian fluid is and how
to calculate shear forces
– Understand when you can apply the
Bernoulli equation
– Know how to use different forms of the
Bernoulli equation to solve problems
2
Viscosity
• Consider a stack of copy paper laying on a
flat surface. Push horizontally near the
top and it will resist your push.
F
3
Viscosity
• Think of a fluid as being composed of
layers like the individual sheets of paper.
When one layer moves relative to another,
there is a resisting force.
• This frictional resistance to a shear force
and to flow is called viscosity. It is greater
for oil, for example, than water.
4
Shearing of a solid (a) and
a fluid (b)
The crosshatching represents (a) solid
plates or planes bonded to the solid being
sheared and (b) two parallel plates
bounding the fluid in (b). The fluid might
be a thick oil or glycerin, for example.
5
Shearing of a solid and a fluid
• Within the elastic limit of the solid, the
shear stress  = F/A where A is the area of
the surface in contact with the solid plate.
• However, for the fluid, the top plate does
not stop. It continues to move as time t
goes on and the fluid continues to deform.
6
Shearing of a fluid
• Fluids are broadly classified in terms of
the relation between the shear stress
and the rate of deformation of the fluid.
• Fluids for which the shear stress is
directly proportional to the rate of
deformation are know as Newtonian
fluids.
7
Shearing of a fluid
• Engineering fluids are mostly Newtonian.
Examples are water, refrigerants and
hydrocarbon fluids (e.g., propane).
• Examples of non-Newtonian fluids include
toothpaste, ketchup, and some paints.
8
Shearing of a fluid
• Consider a block or plane sliding at
constant velocity u over a well-oiled
surface under the influence of a
constant force Fx.
• The oil next to the block sticks to the
block and moves at velocity u. The
surface beneath the oil is stationary and
the oil there sticks to that surface and
has velocity zero.
9
Shearing of a fluid
• No-slip boundary condition--The
condition of zero velocity at a boundary is
known in fluid mechanics as the “no-slip”
boundary condition.
10
Shearing of a fluid
11
Shearing of a fluid
• It can be shown that the shear stress  is
given by
du

dy
• The term du/dy is known as the velocity
gradient and as the rate of shear strain.
• The coefficient is the coefficient of
dynamic viscosity, .
12
Coefficient of dynamic viscosity
• Intensive property.
• Dependent upon both temperature and
pressure for a single phase of a pure
substance.
• Pressure dependence is usually weak and
temperature dependence is important.
• Can be found in Figure 9-10--note
conversion factor in caption.
13
14
TEAMPLAY
• Determine the force to slide, at a speed
of 0.5 m/s, two blocks of 1.0 m square
separated by 2 cm with SAE 10W-30 oil
and determine the same force if the
blocks are separated by water. Assume
that the temperature is 40 C.
15
Shearing of a fluid
• And we see that for the simple case of
two plates separated by distance d, one
plate stationary, and the other moving at
constant speed V
du
V
τ μ
μ
dy
d
16
Shearing of a fluid
• Two concentric cylinders can be used as a
viscometer to measure viscosity
For the inner cylinder,
The torque is T=FR,
V=R, and A=2RL. So
at the inner cylinder,
F=2 R2 L/d
17
Fluid Mechanics
• The text obtains the Bernoulli equation
from momentum considerations, as will
most fluid mechanics courses. We will
obtain it from the first law of
thermodyamics.
• Consider the following equation for
steady-state flow:
2
2








dE cv
v
v
e
i


 i h i 
 e h e 
 Q  Wcv  m
 gz i   m
 gz e 
dt
2
2




18
Fluid mechanics
• The result is
2
2








v
v
e
i


 i h i 
 e h e 
0  Q  Wcv  m
 gz i   m
 gz e 
2
2




• or
2
2






v
v
2
1


 h 2  h1 
Q  Wcv  m

 gz 2  z 1 
2
2


19
Fluid mechanics
• On a mass-specific basis

v2 

2
q  w  h 2  h1
2

v1 

2
2
 gz 2  z 1 
• And rearranging the enthalpy terms
q  w  P2 v2  P1v1  (u2
2
2


v2 
v1 
u )

1
2
2
 g z 2  z1 
20
Fluid Mechanics
• With v = constant (incompressible)
P2 v 2  P1v1  v( P2  P1 )
• so
q  ( u 2  u1 )  w  v ( P2
2
2


v2 
v1 
P)

1
2
2
 gz 2  z 1 
21
Fluid Mechanics
• The term (u2 –u1) will come up later as a
‘head loss’ term, usually treated with
experimental data.
• It represents losses due to friction as the
fluid flows.
• Often ‘frictionless’, adiabatic flow is
assumed and (u2 –u1) as well as q
disappear.
22
Fluid Mechanics
• The work term w would normally be work
done on the fluid by a compressor, fan, or
pump or done by the fluid in a turbine.
• For example, for frictionless flow in the
absence of kinetic energy or potential
energy changes:
 w  v( P2  P1 )
23
Fluid Mechanics
• If work is done on the fluid by a pump,
the work w will be negative, and P2 will be
greater than P1
 w  v( P2  P1 )
• If work is done by the fluid, as it passes
through a liquid turbine, for example, then
P2 will be less than P1 because w is
positive.
24
Fluid Mechanics
• Return to the complete equation and think
of the case of the pump. The work term
can be that for the pump or fan to
overcome friction in a piping or duct
system.
• For now let us assume the flow is
frictionless and set w = 0.

v2 
P)
2
v ( P2
1
2

v1 

2
2
 gz 2  z 1   0
25
Fluid Mechanics
• In the world of fluid mechanics, somewhat
differently than for thermodynamics,
density is used more often than specific
volume.
• We are considering incompressible fluids,
so
1
v

26
Fluid Mechanics

( P2  P1 ) v 2 
v1 


 gz 2  z1   0

2
2
2
2
•or
P1 v1 
P2 v 2 

 gz 1 

 gz 2  cons tan t

2

2
2
2
•These are forms of the Bernoulli
equation
27
Fluid Mechanics
P v 

 gz  cons tan t
• Bernoulli equation

2
2
• Each of the terms has units of energy per
unit mass. For adiabatic, no work
interactions, incompressible,
frictionless, and steady flow, the
Bernoulli equation says the energy content
of the fluid [along a streamline] is a
constant.
28
Fluid Mechanics
• The p/ term is just the pv term, the old
flow work or flow energy term.
• Energy can be traded between the flow
energy (p/), kinetic energy (V2/2), and
potential energy (gz), but the total energy
of the flow will not change [along a
streamline]. (Remember it is frictionless,
adiabatic, steady, and no work is being
done on it.)
29
Fluid Mechanics
• The Bernoulli equation can also be
expressed in terms of pressures:
1
2
P  v   gz  cons tan t
2
30
Fluid Mechanics
1
2
P  v   gz  cons tan t
2
•P is called the static pressure and
would be measured as shown in a fluid
flow:
Flow direction
31
Fluid Mechanics
• The second term
1
V 2 is the velocity
2
or dynamic pressure.
• For flows where the elevation z is
approximately constant,
1
2
P  ρv   Static Pressure  Dynamic Pressure 
2
Stagnation Pressure  Pstagnation
32
Fluid mechanics
• The second term, and thus the velocity,
can be obtained from a measurement of
the static pressure and the stagnation
pressure as shown.
1
2
• Thus ρV  Pstagnation  Pstatic
2
Static pressure
Stagnation
pressure
Flow direction
33
TEAMPLAY
• A point in a flow where a fluid comes to
rest (V=0) is known as a “stagnation
point.” The pressure there is the
stagnation pressure.
• Use the Bernoulli equation and predict the
stagnation pressure on the leading edge of
a sailplane wing soaring at 40 mph at an
altitude of 2,000 ft where the static
pressure is 13.7 psia and the temperature
is 60°F.
34
Fluid Mechanics
• Pitot-static tube--a device somewhat
similar to the previous one used in
measuring the velocity of aircraft.
• Its operation is based on the Bernoulli
equation and the velocity in the equation
is the velocity of the aircraft.
35
Fluid mechanics
• The third term in the previous form of the
equation is known as the elevation
pressure.
1
P   v 2   gz  constant
2
36
Fluid mechanics
• It is also possible to write the equation in
terms of elevation--often called “head”.
v 2
P

 z  constant
g
2g
37
Fluid Mechanics--example
problem courtesy of Dr. Dennis
O’Neal
• A tank of water has a small nozzle at its
base as shown. Find the velocity in
ft/sec and the volumetric flow rate in
ft3/sec from the nozzle.
38
Fluid Mechanics
• Assume the jet is cylindrical and that the
pressure is atmospheric as soon as it
leaves the nozzle. Apply Bernoulli’s
equation between a point 1 on the surface
of the tank and a point 2 at the nozzle
exit:
P1 v1 
P2 v 2 

 z1 

 z2
g
2g
g
2g
2
2
39
Fluid Mechanics
• The pressure at point 1 and at point 2 is
just that of the atmosphere, so P1 = P2.
• At point 1 the height is z1 = H and at point
2 the height is z2 = 0. If the size of the
top of the tank is very large compared to
the outlet area, then (V1)2 is significantly
less than (V2)2 .
40
Fluid Mechanics
• The Bernoulli Equation becomes
P1
P2 v 2 
0H 

0
g
g
2g
2
• but the terms P1/g and P2/g are equal
and cancel because the pressures are the
same, and what is left is
V2  2gH
41
Fluid Mechanics
• And so the exit velocity is
V2  2(32.2 ft / sec )(16 ft )  32.1 ft / sec
2
• the discharge flow rate is
2


1 2
1  4 in 
ft
Q  A2V2   d V2   
32.1
 2.8

4
4  12 in 
sec

ft 
ft 3
sec
42