ENES100_0702
Prof. R. J. Phaneuf
Fall 2002
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Fluids and Pressure
Your project involves converting human power into the motion of water through a
difference in height. Water of course is a liquid, with no definite shape, although it does
have a well defined density (1.00 g/cm3). It acts very nearly as an incompressible fluid.
In a liquid, like water, the molecules are close enough to interact constantly. Their nearneighbor spacing is determined by the minimum in the intermolecular potential, and is
close to what they would have in the solid phase, approximately 0.3 nm. Nevertheless
they are free to move about each other, and this is what allows them to conform to the
shape of the bottom of their container. In fact, they are in constant relative motion,
effectively colliding with their nearest neighbors. Each collision results in a change in
the momentum of a molecule. Consider a molecule colliding with the wall of the
container. After the collision, its momentum has changed:
 




p  p f  pi  mv f  mvi  mv
(1)
We can relate this change in momentum to a force which the wall exerts on the molecule
during the collision, using Newton’s second law:




v p
Fwallmol  ma  m

t t
(2)
But since the force is internal to the system consisting of the wall + molecule, Newton’s
third law tells us that the molecule must exert an equal and opposite force on the wall:



p
Fwallmol   Fmolwall  
t
(3).
The force applied by the molecule on the wall is an impulse-it only exists for the short
collision time t . Nevertheless, the collisions occur very frequently, due to the huge
number of molecules, approximately 1015 molecules per cm2 of wall area, and there
results an almost constant force, distributed over the area of contact between the wall and
fluid. It makes sense to talk about this distributed force per unit area as a pressure.
P F t/A
(4),
where F t denotes the time average value over the area considered of the force imparted
by each of the molecules which collides with the wall. The direction of the average force
is normal to the surface, and acts outward. You can also imagine that if an internal wall
were put into the fluid, that equal and opposite pressures would act on the two sides of it.
Furthermore, the orientation of the wall is not important-the motion of the fluid particles
is independent of orientation, so that the pressure acting on a small part of is independent
of direction.
This immediately allows us to determine what the pressure must be at any depth beneath
the top surface of a liquid in equilibrium. For a wall oriented horizontally the upward
pressure must be equal to the weight of the fluid above it so that the overall forces are
equal and opposite, and so that the wall remains (on average) at rest. The weight of the
column of fluid above the wall is given by:
W  Mg  Ahg
(5)
where M is the mass of the column of liquid  is its mass density, A is the area of the
column, and h is the height of the column, measured above the wall. The corresponding
pressure is given by:
P  W / A  gh ,
(6)
i.e., the pressure is proportional to the depth beneath the surface of the liquid.
Similar considerations apply within a gas, such as the atmosphere, although the density
varies with height. It is still true that the pressure at a given altitude is equal to the weight
of the column of air extending above a surface, divided by the surface area. At sea level,
atmospheric pressure is
Patm  14.7lb / in 2  1.01x105 N / m 2  1atm
(7)
You can immediately see that if a liquid is in a container at sea level and exposed to
atmosphere, the overall weight of atmosphere plus liquid should be considered, and the
true or absolute pressure will be the sum of that given by equations (6) and (7). Since
many pressure gauges use atmospheric pressure as a reference, it is common to talk about
the gauge pressure, which is the pressure given by equation (6), Pgauge  gh .
Pabsolute  Pgauge  Patm
(8)
Fluids confined to a vessel
We know consider a fluid enclosed in a vessel. Pascal’s principle states that an external
pressure applied to a fluid is transmitted to every portion of the fluid and the walls of the
vessel. We can use this principle to understand how simple pressure gauges work.
Consider a U-shaped tube, called a manometer, containing a liquid, say water. If both
sides of the tube are open to atmospheric pressure than the height of the water in each of
the two arms must be equal, as the pressure exerted on each of the two surfaces is 1
atmosphere. Now let’s say that one arm is connected to a pump, which can produce a
pressure different from atmospheric, as we’ll discuss later. If the water initially moves,
but then comes to rest such that the level in the opposite arm rises to a level H above that
in the arm connected to the pump, what must be the pressure generated by the pump? We
consider that the downward force acting on two surfaces at equal heights within the liquid
in each of the two arms must be the same if the liquid is at rest. If we choose to compare
the air-water interface in the pump arm with a surface within the water and at the same
height in the opposite arm, it is clear that the downward force on each surface must be the
same, and thus the pressure acting on each surface must be the same. The pressure acting
on the air-water interface in the pump arm is Patm + Ppump. The pressure acting on an
internal
Patm +
Ppump
Patm
Patm
Patm
Patm +
Ppump
H
Patm
H
Ppump < 0
Ppump > 0
surface in the opposite is Patm plus the pressure exerted by the weight of the column of
water. The two must be equal for the water to be at rest:
Patm  Ppump  Patm  gH
(9)
or solving for the height of the column of water:
H  Ppump / g
(10)
If the pump is run so as to decrease the pressure in the pump arm beneath atmospheric
pressure, by removing air molecules, the water will move up in the pump arm, and down
in the arm exposed to atmosphere, so that H will be negative. Note that in the extreme
limit, as all of the gas is removed from the pump arm, the total pressure in the pump arm
goes to zero. This is because in a gas, the ideal gas law relates the pressure to the number
of molecules N in a fixed volume V at a given temperature T:
PV  NkT
P
NkT
V
(11)
In this case the depth of the air-water interface on the atmosphere side beneath the
vacuum-water interface on the pump side is:
H 0   Patm / g
(12)
Thus if we think of the pump as drawing the water up into the pump arm, the maximum
height to which this “suction” pump can draw the water, above a reservoir exposed to
atmosphere is:
H max_ suction   H 0  Patm / g
(13)
This in fact is the basis for a common type of barometer, which measures the height of a
column of fluid in an arm in which all of the air has been removed, immersed in a
reservoir exposed to atmosphere. To keep the height of the liquid practical, mercury is
generally used instead of water, as its density is considerably greater (14.0 g/cm3)
Fluids in motion
The motion of a fluid can in principle be described by dividing the fluid up into a very
large number of small volume elements, and following the motion of each under the
influence of applied external, and internal forces. This is a computationally extremely
difficult task, and fortunately is generally not necessary. A much simpler approach is to
treat the fluid as a continuum and describe its motion by specifying its density and
velocity as a function of spatial coordinate and time. It is also common to describe the
flow according to four characteristics:
Fluid flow can be steady, meaning that the velocity at each point is constant with time, or
nonsteady. In starting a pump, there is always a transient regime in which the flow is
nonsteady. For some types of pump the flow becomes steady, for others not.
Fluid flow is also classified as to whether it is rotational or irrotational. If the velocity
varies along the direction perpendicular to the flow, or if vortices, whirlpools or eddy’s
are set up then the flow is rotational. If not, it is irrotational. A small paddlewheel
immersed in an irrotational flow will not rotate.
Fluid flow can further be classified as either compressible or incompressible. Water flow
is essentially incompressible, while air flow is compressible.
Lastly, fluid flow can be viscous or nonviscous. Viscosity is analogous to the frictional
forces which result when solids slide past one another. If we think of the fluid as flowing
in neighboring layers, generally with different velocities, the layers are coupled due to the
motion of molecules from one layer to a neighboring layer, imparting a change in
momentum along the direction of the flow. This corresponds to forces applied between
layers in relative motion, and results in dissipative energy losses and heating of the fluid
and vessel.
Bernoulli’s Equation
A particularly simple equation can be derived for nonviscous, steady incompressible flow
of a fluid through a vessel, which comes from the work-energy relation we discussed
earlier. Consider the motion of a fluid through a tube of varying diameter and height, and
in response to different external pressures at the two ends of the tube, p1 and p2, with
p1>p2. The associated forces are F1=p1A1 and F2=p2A2. If in a time t, a portion of fluid
at side 1 moves through a distance l1, and at the second side an equal volume of fluid
moves through l2, the total work done by the two pressure forces is
W = p1A1l1 – p2A2l2,
(14)
With the second term negative since p2 opposes the flow. If the flow is incompressible,
then the volumes represented by the areas times the distance must be equal:
A1l1 = A2l2 = V
(15)
so that the work done by the pressure forces is
W = (p1-p2)V = (p1-p2)m/,
(16)
Where m is the mass of one of the volume elements, and  is the mass density.
Other than the small elements A1l1 and A2l2, the velocity and height of the fluid is the
same at the end of the time interval At as before. The work done in moving the fluid can
thus be taken as equal to the difference in the kinetic energy of the two volume elements
plus the difference in potential energy of the two elements:
W  ( p1  p2 )m /  
1
1
2
2
mv2  mv1  mgy2  mgy1
2
2
(17)
so that
p1  p2 
1
1
v2 2  v12  gy 2  gy1 ,
2
2
(18)
and finally:
p1 
1 2
1
v1  gy1  p2  v2 2  gy2 = constant
2
2
(19)
The last expression is known as Bernoulli’s equation, which relates the pressure, velocity
and height at any point within the tube.
Laminar Flow
In fact fluid flow through a tube is generally not uniform along the direction
perpendicular to the flow. Immediately adjacent to the inner surface of the tube there is a
so-called boundary layer, which does not move relative to the tube. Due to the exchange
of fluid molecules between adjacent layers, momentum is transferred between them,
resulting in a force. (By analogy, think of two freight cars on adjacent rail road tracks,
one of which is moving and the other initially is not. If workers on each car shovel coal
across to the adjacent car, each at right angles to the track in his own reference frame, the
component of the momentum along the track of the coal from the moving car is added to
that of the originally stationary car, and it begins to move forward. The overall
momentum of the moving car is decreased as moving coal is replaced by initially
stationary coal.) In the simplest case, called laminar flow, the velocity varies
continuously from zero at the edge of the tube to a maximum value in the center with an
approximate quadratic dependence:
v  v0 [1  (r / R) 2 ] ,
(20)
where r is the radial distance from the center of the tube, Ri is the inner radius of the tube,
and v0 is the velocity measured at the center. We can think of the flow as consisting of
concentric layers of fluid, each with a different velocity. The viscosity of the fluid, which
microscopically is related to the exchange of molecules between layers, couples the
velocities of adjacent layers.
The volumetric flow rate can be defined as the volume of fluid moving past a plane
oriented at right angles to the flow direction. If we consider this to be the sum of the
contributions from each concentric layer, the overall flow will be:
qv   vi Ai
(21)
i
where vi is the velocity of the ith layer and Ai is its cross-sectional area. If we allow the
thickness of each layer to be infinitesimally small, the sum can be replaced by an integral,
and solved analytically, with the result:
qv  Ri vav ,
2
(21)
with vav  v0 / 2 .
Measuring the flow rate in response to a difference in pressures allows a determination of
the power supplied by the force creating that pressure difference. Imagine that a pump
creates a pressure Ppump + Patm at the inlet of a horizontal tube of cross sectional area A,
and that the pressure is Patm at the outlet. The magnitude of the net force on the water
associated with this pressure difference is
F  PA  Ppump A
(22)
The power, or rate at which this force does work is
 
2
P  F  vav  Fvav  p pump Avav  p pumpRi vav  p pumpqv ,
(23)
i.e., it is equal to the product of the pressure difference times the volumetric flow rate.
h
h0
Ppump+
Patm
Patm
The flow rate can be measured in a number of different ways. A common way
relies on the placement of small tubes in openings at the top of the main water tube, to
allow a determination of the local pressure. Consider first a so-called piezotube, which
extends through the wall of the water flow tube just to its inner radius. If a pressure is
applied at the inlet of the water flow tube then water will rise in response in the piezo
tube. When the weight of the water in the piezotube just balances the force associated
with this pressure distributed over the area of the piezotube, it will be in equilibrium.
Papp Apiezo  mg  pghApiezo
(24)
Papp  gh
(25)
h  Papp / g
(26)
Usually, were going to be interested in the pressure, and flow at the center of the waterflow tube, and so we’ll be interested in the pressure at the center of the tube. This means
adding the weight of the water whose height is the inner radius of the water flow tube:
P0 Apiezo  ( gh  gRi ) Apiezo  gh0 Apiezo
p0  gh0
(28)
h0  p0 / g
(29)
(27)
h
Ppump+
Patm
vo
Patm
If a so-called Pitot tube, which is L-shaped is now introduced downstream of the
piezotube, with its horizontal arm centered and oriented to intercept the flow, the added,
so-called dynamic pressure associated with the flow will cause water to rise even higher
within this tube. Equilibrium will be set up when the weight associated with the
additional water in this tube just balances the added pressure, bringing the flow locally to
a halt. The difference in the height of the water in the Pitot and piezo tubes is clearly
related to the dynamic pressure by:
pdyn  gh .
(30)
We can now use a form of Bernoulli’s equation to derive the velocity of the water at the
center of the tube.
p piezo  v piezo / 2  p pitot  v pitot / 2
2
2
where v piezo = v0, v pitot = 0 and Pdyn  p pitot  p piezo , so that
pdyn  v0 / 2
2
(32)
and substituting from equation (30) gives:
gh  v0 2 / 2
so that the velocity at the center of the water flow tube is given by:
(31)
v0  2 gh
(33)
Fluid flow with dissipation
As fluid moves through a tube it interacts with the walls due to viscosity as discussed
above. The viscous forces are velocity dependent, and can often be approximated by a
power law dependence on velocity. Your book discusses a so called head loss which is
proportional to the cube of the flow rate. This corresponds to a dissipative force which is
proportional to the cube of the average velocity, as we show below.
Consider the flow of a water through a tube, which for simplicity is taken to have a
constant cross sectional area A. Consider a difference in pressure such that the value at
the entrance to the tube is p1 > p2, that at the outlet of the tube. The work done by the
overall force on the water in moving through a path length l is given by:
W  ( p1  p2 ) Al  Fdis l  KE  PE

1
2
2
m(v2  v1 )  mg (h2  h1 )
2
,
(34)
where as before, Fdis is the dissipative force acting upon the surface, m is the mass of an
element of water bounded by the walls of the tube and of length l and h1 and h2 are the
heights of the midpoint of the tube at the entrance and exit and v1 and v2 are the velocities
at the entrance and exit.
The Bernoulli equation would now appear differently:
p1  Fdis / A 
1
1
v12  gh1  p2  v2 2  gh2
2
2
(35)
And since the area to be constant then since water is in compressible, qv1=qv2 and v1=v2
we can rewrite equation (34):
(36)
( p1  p2 ) Al  Fdisl  mg(h2  h1 )

The overall output power supplied by pressure difference force and the dissipative force
during the time intervalt is
Pout  ( p1  p2 ) Av  Fdisv 
or equivalently
m
g (h2  h1 )  Avg(h2  h1 )
t
(37)
Fdis  ( p1  p2 ) A  Ag (h2  h1 )
(38)
If we take the dissipative force as cubic in the velocity we can write this as
Fdis   disvss  ( p1  p2 ) A  Ag (h2  h1 )
3
(39)
so that the steady state velocity is given by
1/ 3
 ( p  p2 ) A  Ag (h2  h1 ) 

vss   1
 dis


(40)
Head vs. flow rate
In pump engineering texts, when treating the lifting of a fluid it is common to divide the
power by the product of the fluid weight flow rate to obtain a quantity with units of
height.
Head = Power/qw
(41)
Where the weight flow rate is given by
qw 
(mg )
V
 g
 gAv
t
t
(42)
Rewriting equation (37) for the output head, for steady state flow, gives:
H out 
Pout
 h2  h1
Avg
(43)
while the input head, required to be supplied by the pump is given by
H req 
H req
Pin
F
( p  p2 )
 1
 (h2  h1 )  dis
Avg
g
Ag
 v 3
 (h2  h1 )  dis ss
Ag
(44)
which is usually instead written in terms of the volumetric flow rate, qv=Av,
H req  (h2  h1 )  disqv
dis 
3
 dis
A4 g
(45),
which does indeed have a term proportional to the cube of the volumetric flow rate. The
power which can be supplied by the pump also has a dissipation term, but of the opposite
sign, since it corresponds to dissipation within the pump. The available head is related to
the maximum difference in pressure that can be generated by the pump and to the internal
pump dissipation by
 qv ,
H available  pmax  dis
3
(46)
where in general    , as the available cross sectional area, and the smoothness of the
housing walls are different within the pump housing than in the outlet tube.
The operation of the pump, in steady state will be defined by equating the required and
available head.
pmax/g
Head
h2-h1
Operation
Point
Volume Flow Rate
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