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2.016 Hydrodynamics
Reading #2
2.016 Hydrodynamics
Prof. A.H. Techet
Pressure effects
Fluid forces can arise due to flow stresses (pressure and viscous shear), gravity forces,
fluid acceleration, or other body forces.
For now, let us consider a fluid in static
equilibrium – with no velocity gradients (and thus, no viscous stresses).
Forces on fluid in static equilibrium are due only to:
1. Pressure acting on the fluid volume
2. Gravity acting on the mass of the fluid
3. Other external body forces (e.g. electromagnetic)
Using standard conventions, we consider pressure to be positive for compression.
Recall
that we said pressure is isotropic: it acts in all directions equally, and it always acts
normal to surfaces.
Let us prove that it acts in all directions equally by considering a
triangular volume of fluid with height,
dz
, length,
dx,
and unit width,
b
, into the page:
Figure 2.1: Elemental fluid volume
version 4.0
updated 8/30/2005
1
©
2005 A. Techet
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Reading #2
The
fluid element cannot support shear while at rest
(by our definition of a fluid).
Thus
the sum of the forces on the triangle, in the
x
and
z
directions, MUST equal zero:
z
∆−
p
b
(
∆
s
)
s
i
n
θ
=
0
(2.1)
∑
F
=
p
b
()
x
x
n
∑
F
=
p
z
(
b
∆
x
)
−
p
n
(
b
∆
s
θ
−
1
ρ
(
b
∆
x
∆
z
)
=
0
(2.2)
z
2
By geometry,
∆
z
= ∆
s
θ
and
∆
x
= ∆
s
θ
, such that
∑
F
=
p
x
−
p
n
=
0
(2.3)
z
and
∑
F
=
p
z
−
p
n
−
ρ
∆
z
=
0 .
(2.4)
z
2
Taking the limit as
∆
x,
∆
z goes to zero (i.e. the triangle goes to a point), we see that
p
=
p
=
p
=
p
.
(2.5)
x
z
n
Since
θ
is arbitrary, pressure at a point in a fluid is independent of orientation and is thus
isotropic.
Pressure (or any stress for that matter) causes NO net force on a fluid element
unless it varies spatially!
version 4.0
updated 8/30/2005
2
©
2.016 Hydrodynamics
Reading #2
Let us find how spatial pressure variation causes net force by considering a small fluid
element:
Figure 2.2: Fluid Element Volume (z is positive upwards by convention).
Let us assume that the only forces are due to gravity and pressure gradients. Let the
pressure on the bottom equal a reference pressure
p
bot
=
p
o
. Then, the pressure on the top
may be found using a Taylor’s series expansion:
=
(
δ
z
)
+
1
d
2
p
(
z
)
2
+
...
(2.6)
p
top
p
0
+
dp
dz
2!
dz
2
0
0
Taking the limit as
δ
z goes to zero we are able to ignore the higher order terms and keep
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.016 taught by Professor Alexandratechet during the Fall '05 term at MIT.
 Fall '05
 AlexandraTechet
 Shear, Stress

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