4.
Experimental Uncertainty
In any experimental (or even computational) study, attention must be paid to the uncertainties involved in
making measurements. Including the uncertainty allows one to judge the validity or accuracy of the
measurements. Uncerta

The surface forces acting on the control surface are:
ji
FS + FB =
dxdydz + fVonL ,i L (1 ) dxdydz + g i L (1 ) dxdydz
x j
(4.48)
Note that the stress terms are the surfaces forces acting on the sides of the CV. The term fVonL,i is the force
per unit ma

5.
Fluid Element Deformations
In order to relate the stresses acting on a fluid element to the other variables in the momentum equations,
we need to determine a constitutive law. The constitutive law should relate the stresses to the rates of
strain (or d

The forces acting on the control volume include both body and surface forces. The body force acting
on the CV in the x-direction, FB,x, can be written as:
FB, x = f B, x ( dxdydz )
(4.24)
where fB,x is the body force per unit mass acting in the x-directio

Example
Consider the flow of a mixture of liquid water and small water vapor bubbles. The bubble diameters are
very small in comparison to the length scales of interest in the flow so that the properties of the mixture can
be considered point functions. F

By utilizing Gauss Theorem (aka the Divergence Theorem), we can convert the area integrals into
volume integrals:
ui ( u dA ) = ( ui u ) dV =
u j ui dV
x j
CS
CV
(
)
CV
CS
ji n j dA
=
CV
ji
x j
dV
Substitute these expressions back into the LME to get:

Expand the left hand side of Eqn. (4.28) and utilize the continuity equation:
u
u
+ i + ui
u j ui = ui
u j + u j i
( ui ) +
t
x j
t
t
x j
x j
(
)
(
u
u
u j + i + u j i
= ui
+
t
t x j
x j
(
)
= 0 (continuity eqn)
=
)
Dui
Dt
Substituting into back int

Example:
The material element shown below has the following stress tensor components:
y
(0, 2, 0) m
T
0 4
3 0 kPa
4 0 5
1
[ ] = 0
x
(1, 0, 0) m
z
a.
b.
c.
d.
(0, 0, 1) m
Find the components of the traction vector, T, on the plane described by the unit

4.
Momentum Equations (aka the LME for a differential CV)
The momentum equations, which are the simply the linear momentum equations for a differential fluid
element or control volume, can be derived several different ways. Three of these methods are give

Another approach to proving symmetry of the stress tensor is to write the AME explicitly for the small
element (again assuming no body couples):
D
ijk r j f k dV + ijk r j lk nl dA =
ijk r j uk dV
(4.19)
Dt
Vsys
Ssys
torque due to
body forces
=Tk
Vsys
t

Cauchys Formula
Cauchys formula is used to determine the traction vector on an arbitrarily oriented surface with an
orientation vector, , given the stress tensor. Consider the small tetrahedral element shown below.
x2
T
The areas of each face are:
dA = dA

To determine the continuity equation for the liquid water phase, consider the control volume drawn below
where the CV surrounds each vapor bubble.
dx
y
x
dy
The rate of change of liquid mass within the control volume is:
L (1 ) dxdydz = (1 ) L dxdydz
t

Dilation (aka Dilatation)
B
u2
dx2 dt
x2
C
B
C
O
A
The rate of dilation can be described by the rate at which the relative
volume of the element increases with time.
1 dV
volumetric dilation rate =
V dt
dx2
dx1
A
u1
dx1dt
x1
The velocity of point A relat

Angular Deformation
u1
dx2 dt
x2
B
The rate of angular deformation in the 1-2 plane can be
described as the average rate at which the sides of the
element approach one another, i.e. the average rate at
which the angles AOA and BOB increase.
C
B
C
d
A
dx2

1.
Symbolic vs. Numeric Calculations
Consider the following example. You need to determine the trajectory of a projectile fired from a cannon.
The projectile has a mass of 10 kg and the cannon is tilted at an angle of 30 from the horizontal. The
initial v

3.
Taylor Series Expansion Approximation
If we know the value of some quantity, y, at some location, x, then how can we determine the value of y at a
nearby location x+dx?
y
y(x+dx)
y(x)
x
x+dx
x
SOLUTION:
We can use a Taylor series expansion for y about

Now lets try working the same problem using symbols rather than numbers. Well plug in the numbers at
the very end of the problem.
Symbolic Solution:
Draw the FBD as before:
y
V
mg
x
Assume this height is negligible.
Follow the same approach as before but

Lastly, if we wanted to determine the angle that will maximum the distance traveled for a given velocity,
we observe from Eqn. (1.8) that we want sin(2) to be as large as possible. Thus, we should tilt our cannon
at an angle of = 45. Substituting this bac

3.
How is the thermodynamic pressure related to the normal stresses? Define the mechanical pressure,
p , as the average of the normal stresses:
( )
p 1 trace ij = 1 ii = 1 ( 11 + 22 + 33 )
3
3
3
For a Newtonian fluid:
p = 1 ( 11 + 22 + 33 )
3
u
u
u
u
= 1

Assumption 3:
There are no preferred directions in the fluid so that the fluid properties are point functions. This is the
condition of isotropy.
The condition of isotropy means that the fluid properties are the same in all directions. Examples of
non-iso

Rigid Body Rotation
u1
dx2 dt
x2
B
The rate at which the fluid element rotates about the 3axis in rigid body motion can be described as the average
rate at which the sides of the element rotate in the same
direction.
C
B
d
dx2
O
C
d A
dx1
A
u2
dx1dt
x1
Th

Now that we have described the deformation rate components (e.g. dilation, angular deformation, and rigid
body rotation; translations are treated separately) of a fluid element, lets combine these into a single tensor
quantity known as the deformation rat

Stress-Strain Rate Relations for a Newtonian Fluid
The following assumptions are based on observation and intuition.
The key assumptions in deriving the stress-strain rate constitutive relations for a Newtonian fluid are:
1. When the fluid is at rest, the

Example
A fluid has a velocity field given by
u = 2 x 3 y + zk
i
j
At the location (x, y, z) = (-2, -1, 2), calculate:
a. the normal and shearing strain rates at the location, and
b. the rotational velocity of the fluid.
SOLUTION:
The strain rate tensor

Symmetry of the Stress Tensor
Consider the Angular Momentum Principal for the small element of material shown below (only the
stresses acting on the positive faces and causing rotations in the x3 axis are shown for clarity). Also note
that no body couples

Stress Sign Convention
The sign conventions for stresses are as follows:
1. A positive face is a face that has a normal vector pointing in a positive direction.
2. Positive stresses on positive faces point in the positive direction.
3. Positive stresses o

Example:
A piston compresses gas in a cylinder by moving at a constant speed, V. The gas density and the piston
length are initially 0 and L0, respectively. Assume that the gas velocity varies linearly from velocity, V, at
the piston face to zero velocity

4.000
3.500
1.000
CL
A
A
2.250
(4X) R.250
(8X) R.125
CL
.250 TYP
.500 TYP
.0625
.250
.0625
SECTION A-A
UNLESS OTHERWISE SPECIFIED:
DIMENSIONS ARE IN INCHES
TOLERANCES:
FRACTIONAL 1/64
ANGULAR: MACH 1 BEND 1
TWO PLACE DECIMAL
0.05
THREE PLACE DECIMAL 0.005