HYDRAULICS 1 (HYDRODYNAMICS)
SPRING 2005
Part 1. Fluid-Flow Principles
1. Introduction
1.1 Definitions
1.2 Notation and fluid properties
1.3 Hydrostatics
1.4 Fluid dynamics
1.5 Control volumes
1.6 Visualising fluid flow
1.7 Real and ideal fluids
1.8 Laminar and turbulent flow
2. Continuity (mass conservation)
2.1 Flow rate
2.2 The steady continuity equation
2.3 Unsteady continuity equation
3. The Equation of Motion
3.1 Forms of the equation of motion
3.2 Fluid acceleration
3.3 Bernoulli’s equation
3.4 Application to flow measurement
3.5 Other applications (flow through an orifice; tank-emptying)
4. The Momentum Principle
4.1 Steady-flow momentum principle
4.2 Applications (pipe contractions; pipe bends; jets)
5. Energy
5.1 Derivation of Bernoulli’s equation from an energy principle
5.2 Fluid head
5.3 Departures from ideal flow (discharge coefficients; loss coefficients; momentum & energy coefficients)
Part 2. Applications (Separate Notes)
1. Hydraulic Jump
2. Pipe Flow (Dr Lane-Serff)
Recommended Reading
Hamill, 2001, Understanding Hydraulics, 2nd Edition, Palgrave, ISBN 0-333-77906-1
Chadwick, Morfett and Borthwick, 2004, Hydraulics in Civil and Environmental
Engineering, 4th Edition, Spon Press, ISBN 0-415-30609-4
Massey, 1998, Mechanics of Fluids, 7th Edition, (Revised by Ward-Smith, J.), Stanley
Thornes, ISBN 0-748-74043-0
White, 2003, Fluid Mechanics, 5th Edition, McGraw-Hill, ISBN 0-07-240217-2
Hydraulics 1
1
David Apsley
1. Introduction and Basic Principles
1.1 Definitions
A fluid is a body of matter that can flow; i.e. continues to deform under a shearing force.
Fluids may be liquids (definite volume; free surface) or gases (expand to fill any container).
Fluids obey the usual laws of Newtonian mechanics, but as a continuum. Unlike rigid bodies,
fluid particles may move relative to each other, interacting via internal forces. These are
usually expressed in terms of stresses (stress = force / area), the principal ones being:
pressure, p – normal stress: pushing or pressing;
shear stress, – tangential stress: frictional drag; opposing relative motion.
An ideal fluid has no viscosity. This is never exactly true, but is often a useful approximation.
Fluids may be regarded as incompressible (density or volume not changed by the flow) at
speeds much less than that of sound. This is usually the case in hydraulics.
Fluid flow may be described as laminar (adjacent layers “slide” smoothly past each other) or
turbulent (irregular, with constant intermingling of adjacent layers).
1.2 Notation and Fluid Properties
The main flow variables are:
p
pressure (force/area)
u
velocity
These are field variables because they are functions of position and time t; e.g. u = u(x,t)
Vector quantities like position and velocity are often decomposed into components:
x ≡ ( x, y , z )
u ≡ (u , v, w)
U (or occasionally V) will be used for the magnitude of velocity.
Important fluid properties are:
density = mass / volume
dynamic viscosity; defined by Newton’s viscosity law:
du
=
dy
= / is called the kinematic viscosity
surface tension = force / length (or surface energy/area)
F= l
K
bulk modulus = pressure change / volumetric strain
− V
p = K(
)
V
Hydraulics 1
2
David Apsley
Most gases at normal temperatures and pressures satisfy the ideal gas law, which in fluid
mechanics is usually written as
p = RT
The gas constant R depends on the particular gas (R = R*/m = universal gas constant / molar
mass). It has a value of 287 J kg–1 K–1 for dry air.
1.3 Hydrostatics
Hydrostatics concerns the balance of forces in a fluid at rest.
The principal forces are:
pressure;
weight;
surface tension;
p is the normal force per unit area
≡ g is the specific weight (weight per unit volume)
is the force per unit length
In the interior of a stationary fluid, pressure forces balance weight. As
a result the change in pressure with depth is given by the hydrostatic
equation
p=− g z
(1)
Pressure increases as height decreases and vice versa. As a result there
is a net buoyancy force on an immersed body equal to the weight of
fluid displaced (Archimedes’ Principle); i.e. the immersed weight of a
body is less than its actual weight.
p
If density is constant then
(2)
p + gz = constant
The quantity on the LHS is called the piezometric pressure.
Hydrostatic principles are used to measure:
h
Pressure in a flowing liquid, using a piezometer tube.
The pressure (relative to atmospheric pressure) is given by
p = gh
(3)
pA
Pressure differences using a U-tube manometer.
The difference in pressure between points A and B is proportional to
the difference in heights of the manometer fluid in the two arms:
p = ( m − ) gh
(4)
where m is the density of the manometer fluid and the density of the
working fluid.(The latter can be ignored if the working fluid is a gas).
>
ρ
ρm
Note that, in incompressible flow, absolute pressure is unimportant: pressure differences
determine the net force. In particular, it is usual to work with the gauge pressure – i.e.
difference between actual and atmospheric pressures. The pressure at a free surface is then
p = 0, whilst a sub-atmospheric (suction or vacuum) pressure has p < 0.
Hydraulics 1
3
David Apsley
pB
h
Surface tension is vital to survival if you are a water-hopping
insect but it can usually be ignored in large-scale hydraulics.
The capillary rise in a thin-bore tube is given by
h=
cos
gd
where d is the tube diameter and
for water on glass).
θ
(5)
h
is the contact angle (≈ 0
1.4 Fluid Dynamics
Fluid dynamics concerns fluids in motion. Almost all hydraulic problems can be addressed by
the application of one or more of three equations/physical principles:
•
continuity (mass conservation);
•
momentum;
•
energy.
Continuity is almost invariably required and is usually applied first to reduce the number of
unknowns.
Continuity (Mass Conservation)
•
Mass is neither created nor destroyed.
•
For a steady flow,
mass flow rate in = mass flow rate out
or, if the flow is also incompressible,
volume flow rate in = volume flow rate out
•
Any mathematical expression of this is called the continuity equation.
Q2
Q1
Q3
Qin = Qout
Hydraulics 1
Q1 = Q2 + Q3
4
David Apsley
Momentum
Based on Newton’s Laws of Motion; specifically:
•
Second Law:
force = rate of change of momentum.
•
Third Law (Action and reaction):
force exerted by a fluid on its containment is equal and opposite to the force
exerted by the containment on the fluid.
force on
FLUID
force on
FLUID
force on
BODY
wake
force on
PIPE
Energy
•
Energy principle:
change of energy = work done + heat input
•
For incompressible fluids we need only consider the mechanical energy principle:
change of kinetic energy = work done
•
For ideal fluids (no losses) this is expressed as Bernoulli’s equation
p + gz + 12 U 2 = constant (along a streamline)
In particular:
velocity increases
↔
pressure decreases
small loss
of head
LIFT
•
Bernoulli’s equation is strictly for ideal fluids (no friction or other fluid energy
changes), but it can be amended to account for:
energy put in (by pumps) or taken out (by turbines);
frictional losses (e.g. pipe flow).
•
Because all forms of fluid energy can be converted to an equivalent gravitational
potential energy it is common in hydraulics to measure energy in height units or fluid
head.
Hydraulics 1
5
David Apsley
1.5 Control Volumes
Although it is possible to write the equations of fluid mechanics in terms of differential
equations (describing what is happening at every point), engineers are more often concerned
with the behaviour of fluid “in bulk” and the usual method of analysis is to apply mass and
momentum principles, etc., to all the fluid in a control volume.
For example, consider suitable control volumes for the two pipe fittings shown.
F
Mass conservation:
Net mass flux out of control volume
= (mass flow rate) B − (mass flow rate) A
=0
Momentum principle:
Net force on fluid in control volume
= rate of change of momentum
= (mass flow rate) × change in velocity
1.6 Visualising Fluid Flow
Velocity vectors are arrows of
length proportional to magnitude.
Streamlines are everywhere tangential to the instantaneous flow direction. The flow is fastest
where the streamlines are closest together
(why?). Streamlines can never cross (else
the velocity would be 2-valued) and cannot
terminate in the interior of the flow. In
steady flow one streamline always coincides
with a solid boundary or free surface.
Hydraulics 1
6
David Apsley
Path lines or particle paths are the lines followed by individual elements of fluid. (They are
what you would see on a long-exposure photograph).
Streaklines are swept out by all the fluid elements which have passed through a particular
point; e.g. dye injection.
In steady (i.e. not time-varying) flow streamlines, streaklines and particle paths are identical.
Stream tubes are bundles of streamlines (“virtual pipes”). By
definition, there is no flow across the sides and the same flow
must enter at one end as leaves at the other.
1.7 Real and Ideal Fluids
Ideal fluids have no viscosity – there is no internal friction or loss of mechanical energy. No
such fluid exists, but many flows can be approximated as ideal if viscous forces are small and
do not cause major flow phenomena such as boundary-layer separation.
Real fluids have non-zero viscosity. This has two important consequences:
(1)
They satisfy the no-slip condition at solid boundaries.
i.e. the (relative) velocity at the boundary is zero.
ideal
(2)
real
There are frictional forces between adjacent layers of fluid moving at different speeds
and between the fluid and a boundary.
shear stress of
UPPER fluid
on LOWER
shear stress of
LOWER fluid
on UPPER
For Newtonian fluids (which includes most fluids of interest), the stress (force per unit
area) of the upper fluid on the lower fluid is given by Newton’s viscosity law:
du
=
(6)
dy
An equal and opposite force is exerted by the lower fluid on the upper fluid.
Hydraulics 1
7
David Apsley
1.8 Laminar and Turbulent Flow
At low flow speeds viscosity ensures that adjacent layers of fluid slide smoothly over one
another in a steady fashion without intermingling. This flow regime is called laminar.
At higher speeds viscosity is insufficient to smooth out minor
perturbations to the flow, which grow rapidly to create
unsteadiness and eddying motions. This flow regime is called
turbulent.
turbulent
laminar
The transition between laminar and turbulent flow can be
observed in the rising plume from a cigarette. There is an initial
laminar flow which wavers and then breaks up into turbulent
eddies as the flow accelerates.
The pioneering experiments determining the occurrence of laminar or turbulent flow in pipes
were first performed by Osborne Reynolds (Professor of Engineering at Owens College –
soon to become the University of Manchester). He demonstrated that which regime occurred
depended on a dimensionless parameter later to become known as the Reynolds number:
UL
UL
Re =
or
(7)
where U and L are typical velocity and length scales of the flow, is the density, is the
dynamic viscosity and (= / ) the kinematic viscosity. The Reynolds number is probably
the single most important parameter in fluid mechanics!
laminar
turbulent
The Reynolds number can be regarded as a measure of the ratio of the total force
(= mass × acceleration) to the viscous force. For a block of fluid, volume L3, using order-ofmagnitude estimates:
U
mass × acceleration ∼ ( L3 ) ×
= U 2 L2
L /U
whilst
U
viscous force ∼ A ∼
× L2 = UL
L
Hence,
mass × acceleration
U 2 L2
UL
∼
=
= Re
viscous force
UL
When the Reynolds number is large the effects of viscosity are small and conversely. In
general,
Hydraulics 1
8
David Apsley
low Re
high Re
laminar flow
turbulent flow
Note that “low” and “high” is all relative and depends on the particular flow and the choice of
velocity and length scales (which should be stated). For pipes it is conventional to take U as
average velocity and L as the diameter; then:
Re < 2000 (laminar flow)
Re > 4000 (turbulent flow)
(The commonly-accepted critical Reynolds number for transition in a round pipe is 2300).
These particular numerical values are for pipe flow only, with a particular choice of velocity
and length scales. For example, choosing radius rather than diameter as a length scale would
immediately halve the Reynolds number, but wouldn’t affect the flow regime.
The numerical size of viscosity for the common fluids, air and water ( air ≈ 1.5×10–5 m2 s–1;
–6
2 –1
water ≈ 1.0×10 m s ), means that most civil-engineering flows are fully turbulent.
Even when the flow is turbulent it usually consists of relatively small fluctuations about a
mean flow, which is what we are actually interested in. In turbulent flow, however, most of
the net transfer of momentum and energy between layers of fluid occurs by the eddying
motions mingling fluid elements and so, as far as the mean flow is concerned, the mean stress
is no longer given by Newton’s viscosity law ( = du/dy) but is substantially increased by
the net effect of turbulent eddies. This mixing of fluid streams leads, for example, to a
turbulent velocity profile in a pipe being much more uniform than a corresponding laminar
velocity profile.
Hydraulics 1
9
David Apsley
2. Continuity (Mass Conservation)
In fluid mechanics any mathematical expression of the conservation of mass is called the
continuity equation.
2.1 Flow Rate
A
Consider fluid passing through an area at uniform velocity u.
In time t the fluid moves a distance u t, and hence the volume of
fluid that has passed through that area is
A×u t
ρ
u
A
u δt
The volumetric flow rate Q (aka “volume flux”, “quantity of flow” or “discharge”) is given by
volume A × u t
Q=
=
time
t
i.e.
(8)
Q = uA
3 –1
–1
–3
3 –1
It may be measured in m s or L s (1 L = 10 m s ).
The mass flow rate (or mass flux) through a section, m , is given by
mass density × volume
m=
=
time
time
i.e.
m = Q = uA
It may be measured in kg s–1.
(9)
Example. A pipe of internal diameter 200 mm carries water (density 1000 kg m–3) at average
velocity 2.5 m s–1. Find the volumetric flow rate and mass flow rate.
Note. In general the flow may be at any angle to the area A. In general the U which appears in
the formula for Q or m should be the component normal to the area.
Hydraulics 1
10
David Apsley
2.2 The Steady Continuity Equation
The continuity equation is invariably applied when either:
•
there is a change of cross section;
•
there is a junction.
Change of Cross Section
Consider a stream tube with a change of cross section from A1 to A2.
ρ1
ρ2
u1
A2
u2
A1
If the flow is steady (i.e. doesn’t change with time) then no mass can accumulate between the
two cross sections and hence
(mass flux )1 = (mass flux ) 2
(10)
=
1u1 A1
2 u 2 A2
In hydraulics the fluid can usually be treated as incompressible (density constant along a
streamline) and hence the volume flow rate is constant:
Q = u1 A1 = u 2 A2
(11)
Junctions
At a junction of more than two pipes, we require that
total flow into junction = total flow out of junction
e.g.
Q2
Q1
Q1 = Q2 + Q3
Q3
Hydraulics 1
11
David Apsley
Example. A circular pipe carrying water undergoes a smooth contraction from internal
diameter 200 mm to one of diameter 80 mm. If the velocity in the larger cross section is
0.6 m s–1 find:
(a)
the velocity in the smaller-diameter pipe;
(b)
the volume and mass flow rates.
Example.
A manifold splits the air supply (unequally) between two components of an engine. The inlet
duct has square section with side 150 mm. The outlet ducts have circular cross-section with
diameters 80 mm and 120 mm. The air velocity in the intake duct is 2 m s–1 and that in the
smaller outlet duct is 4 m s–1. Find the volumetric flow rate and velocity in the larger outlet
duct.
120 mm
150 mm
2 m/s
4 m/s
80 mm
Example. Liquid of controlled density can be supplied by injecting saturated brine (s.g. 1.20)
into a freshwater stream. If brine is injected at 2 L s–1 into a pipe of internal diameter 100 mm
carrying fresh water at 3 L s–1 find, in the well-mixed region downstream:
(a)
the total quantity of flow;
(b)
the mass flow rate;
(c)
the average velocity;
(d)
the density of fluid.
Hydraulics 1
12
David Apsley
2.3 Unsteady Continuity Equation
Mass conservation applied to the fluid in a control volume which is either:
(a) moving; or
(b) changing density;
gives
d
(mass ) = (mass flux) in − (mass flux) out
dt
(12)
In many cases the fluid is incompressible and continuity can equally well be applied to
volume rather than mass:
dV
= Qin − Qout
(13)
dt
i.e.
rate of change of volume = flow rate in – flow rate out
Example. A cylinder of diameter 0.2 m contains water and has intake and exit pipes both with
cross sections of 4×10–3 m2 and fitted with non-return valves. A piston oscillates sinusoidally
up and down in the cylinder with amplitude 80 mm and frequency 2 Hz. Find the maximum
velocity of water in the exit pipe and the volume discharged over a cycle.
Hydraulics 1
13
David Apsley
3. The Equation of Motion
3.1 Forms of the Equation of Motion
The equation of motion is any mathematical expression of Newton’s Second Law.
As you have seen in the Mechanics module, this can be written down in various ways; e.g.
F = ma
or, equivalently, as a momentum principle:
force = rate of change of momentum
or as a mechanical energy principle:
work done = change of kinetic energy
In fluid mechanics, also, the equation of motion can be expressed in different ways. In
Section 3 we examine the first and third of the forms above, leading to an important result
known as Bernoulli’s equation. In Section 4 we apply the momentum principle using control
volumes in order to deduce forces on structures.
3.2 Fluid Acceleration
In order to apply Newton’s Second Law (F = ma) to fluid motion one must know the fluid
acceleration.
Suppose the position of a particle at time t is given by x(t). Its velocity u is given by
dx
u=
dt
du
Its acceleration is defined as a =
, but you will know from Mechanics that this can also be
dt
written in terms of displacement x as
du du dx
a=
=
×
(by the chain rule)
dt dx dt
i.e.
du
a=u
dx
Consider now the fluid flow through a converging section as shown.
u1
u2
u3
a
If the flow is steady then the velocity at any particular position does not change with time; i.e.
∂u / ∂t = 0 . However the fluid is accelerating because as you move with any particular fluid
element along the centreline you are moving with faster velocity. The fluid acceleration – i.e.
Hydraulics 1
14
David Apsley
the acceleration of a particular element of fluid – is given in this instance by u∂u / ∂x .
In general, the acceleration of a fluid element in a flow u(x,t) is the sum of two parts:
∂u
(temporal derivative) because of changes with time at a point;
∂t
∂u
u
(advective derivative) because of changes as you move with the flow.
∂x
The latter usually dominates.
When the velocity is also a function of the other space components y and z the fluid
acceleration can be written as the total (or “material”, or “substantive”) derivative
Du ∂u
∂u
∂u
∂u
a≡
≡
+u
+v
+w
(14)
Dt
∂t
∂x
∂y
∂z
Fortunately, many flows that you will deal with are steady and quasi-1-dimensional, and this
can often be simplified to
∂u
a=u
(15)
∂x
Example. An air supply system consists of ducts with rectangular cross section. The upstream
part of the ducting has a width of 0.4 m and height 0.5 m, whilst the downstream part has the
same width (0.4 m) but a height of 0.2 m. These two sections are connected by a simple
transition duct where the height varies linearly over a distance of 2 m.
0.4 m
0.5 m
0.2 m
x
(a)
If the volumetric flow rate along the duct is 0.6 m3 s–1, find an expression for the
speed of the flow as a function of the distance from the start of the contraction, x.
(b)
Find an expression for the acceleration of the air as a function of distance along the
contraction.
Assume that the flow is quasi-1d.
(Answer: with x in m, u in m s–1, a in m s–2, (a) u =
3
2 .7
; (b) a =
)
1 − 0 .3 x
(1 − 0.3 x) 3
What force provides this acceleration?
Hydraulics 1
15
David Apsley
3.3 Bernoulli’s Equation
A (better) derivation of Bernoulli’s equation from
general energy considerations will be given later.
For now we consider the forces on a short fluid
element, length s, uniform cross-section A and
aligned locally with the flow.
δs
(p+ δp)A
θ
pA
We assume that the flow is inviscid (no friction),
steady (no time derivative) and incompressible (
mg
constant along a streamline). Resolving forces in the
direction of flow:
pA − ( p + p ) A − mg sin = ma
Writing:
dU
a =U
for the acceleration (U is the magnitude of velocity, s is the distance)
ds
m = ( A s ) for the mass
z
sin =
s
gives
dU
− pA − gA z = A sU
ds
dU
d
and, on dividing by the volume (A s) and noting that U
= ( 12U 2 ) :
ds ds
p
z
d 1 2
−
− g
=
( 2U )
s
s
ds
Since the flow is incompressible, is constant along the streamline and can be taken inside
the differential. In the limit as s → 0:
d
( p + gz + 12 U 2 ) = 0
ds
i.e.
p + gz + 12 U 2 = constant
(16)
Remember the qualifiers:
inviscid (no friction)
steady
incompressible
along a streamline
Some useful things to remember when applying Bernoulli’s theorem are:
•
use continuity first to reduce the number of unknowns;
•
at a free surface the pressure is atmospheric: p = 0;
•
for a free discharge to atmosphere the pressure is atmospheric: p = 0;
•
for a large tank/reservoir velocity at the surface is unaffected by the discharge: U = 0.
Hydraulics 1
16
David Apsley
Example (Exam, May 2003 *** modified ***)
Air (density 1.2 kg m–3) is flowing through a smooth contraction from a circular pipe of
diameter 120 mm to a tube of diameter 60 mm as shown in the figure. There are tappings at
four locations along the contraction, where the diameter is 120 mm, 100 mm, 80 mm and
60 mm (labelled A, B, C and D respectively).
(a)
If the air flow along the duct is 0.2 m3 s–1, find the speed of the flow at each location
A, B, C and D.
(b)
Assuming no energy losses, calculate the pressure difference between locations A and
B, and the pressure difference between locations C and D.
(c)
A water manometer is connected across tappings A and B, and another manometer is
connected across C and D (see the Figure). What are the water-level differences ( z1
and z2) in the two manometers.
A
B
120 mm
C
100 mm
∆ z1
Hydraulics 1
D
80 mm
60 mm
∆ z2
17
David Apsley
Example. Water is being siphoned as shown. The internal diameter of the pipe is 20 mm and
that of the nozzle is 15 mm. Assuming no losses find:
(a)
the volumetric flow rate;
(b)
the pressure at the top.
2.5 m
1.5 m
Example (Exam, June 2004)
A pipe carrying water with a flow rate of 0.04 m3 s–1 contracts from a diameter of 100 mm to
60 mm over a distance of 0.5 m, with the diameter changing linearly with distance.
(a)
What is the flow speed as a function of distance along the pipe?
(b)
What is the acceleration of the flow as a function of distance along the pipe?
(c)
If the pressure where the diameter is 60 mm is denoted by p0, what is the pressure as a
function of distance along the pipe, assuming no energy loss?
(Answer: With x in m, u in m s–1, Q in m3 s–1, p in Pa,
16
409.6
16
1
1
(a) u =
; (b) Q = 2
; (c) p = p 0 + 12 ( ) 2 ( 4 −
))
2
5
(1 − 0.8 x)
(1 − 0.8 x)
0 .6
(1 − 0.8 x) 4
Example. A river may be approximated by a rectangular channel of width 10 m. The depth of
water is 1.4 m. The abutments of a bridge reduce the channel width to 7 m and the depth of
water to 1.2 m. Find the upstream velocity and quantity of flow.
Explain, physically, why the depth of water decreases between the abutments.
Hydraulics 1
18
David Apsley
3.4 Application to Flow Measurement
Some of the most important applications of Bernoulli’s equation are to flow measurement.
The main reason is that velocity changes can be related to pressure changes and pressure is
usually cheap and easy to measure with a manometer or piezometer tube.
Ideal-flow theory is used to compute a theoretical discharge Q. This can then be corrected for
friction, turbulence, etc. by applying a discharge coefficient cD (see Section 5.3).
3.4.1 Pitot and Pitot-Static Tubes
A Pitot tube is used for measuring velocity in the
flow. It works by converting the kinetic energy into
pressure energy.
Flow is brought to rest at the stagnation point at the
front. By Bernoulli’s theorem this raises the pressure
to the stagnation or Pitot pressure
P0 = P + 12 U 2
(17)
where P and U are the pressure and velocity in the
undisturbed flow. If one also measures the static
pressure P then the difference P0 − P is equal to
piezometer
Pitot tube
p0
ρg
p
ρg
1
2
u
U 2 , from which the velocity can be deduced.
In a pitot-static tube both pitot and static pressures are measured with the same instrument.
The pressure difference can be measured with a manometer: P0 − P = ( m − ) gh .
to manometer
p0
p
h
u
p0
p
P0 − P =
1
2
U2 =(
− ) gh
m
Example. The velocity of water in a conduit is to be measured using a piezometer and Pitot
tube. If the water level in the piezometer rises to 0.3 m and that in the Pitot tube to 0.5 m find
the excess pressure and the flow velocity at the measuring point.
How do you expect the height in the pitot tube to change as the measuring point is
traversed across the conduit?
Hydraulics 1
19
David Apsley
Example. The speed of air is measured with a pitot-static tube connected to a U-tube
manometer containing alcohol (s.g. 0.79).
(a)
Find the air velocity if the difference in fluid levels in the two arms of the manometer
is 150 mm.
(b)
What difference in levels would be expected for a flow speed of 5 m s–1?
How could the accuracy of the reading be enhanced for low flow speeds?
3.4.2 Venturi Meter
A venturi is a smooth contraction in a pipe. The idea is that the contraction causes an increase
in velocity and hence a decrease in pressure, which can be measured.
Continuity and Bernoulli give two equations for two unknowns (U1 and U2) and hence either
can be found and multiplied by the corresponding area to give the volumetric flow rate Q.
p1
p2
u1
u2
A2
A1
Apply Bernoulli between sections 1 and 2:
p1 + 12 U 12 = p 2 + 12 U 22
U 22 − U 12 = 2( p1 − p 2 ) /
(18)
Apply continuity between sections 1 and 2:
U 1 A1 = U 2 A2
A
U 2 = 1 U1
(19)
A2
Substitution in (18) yields an expression for U1, which, when multiplied by A1, gives the
volumetric flow rate (exercise: complete the analysis):
2 p/
Q = A1
= constant × p 1 / 2
( A1 / A2 ) 2 − 1
Notes
1.
This is the ideal flow rate. To account for non-ideal behaviour it can be multiplied by
a discharge coefficient cD, so that
Qactual = c D Qideal
However, for a well-defined venturi meter, cD is 0.98 or more, so is often neglected.
Hydraulics 1
20
David Apsley
2.
The downstream expansion plays no role in the analysis, but should be at a small
angle so as not to provoke flow separation and cause energy loss.
3.
The same principle is used in an orifice meter. This is a sharp contraction in a pipe.
The downstream pressure at the walls is assumed to be the same as at the orifice, due
to the recirculating flow region downstream.
A1
A2
p1
p2
An orifice meter is much cheaper to manufacture and install but causes substantial
energy loss. If the Bernoulli analysis is used to predict an ideal flow rate then a
discharge coefficient with a value of the order 0.6 is needed, mainly due to the
continued convergence of streamlines after passing through the orifice.
3.4.3 Sharp-Crested Weir
Sharp-crested weirs are slotted plates used to provide accurate measurements of quantity of
flow in small open channels – for example, in hydraulics laboratories. (Broad-crested weirs
are used for larger channels and work by a different principle – see open-channel flow in
Hydraulics 3).
crest, or sill
nappe
H
Q
air
The quantity of flow Q is related to the height over the weir, H. For a given rectangular weir,
Bernoulli’s theorem predicts
Q ∝ H 3/ 2
Key to operation is that air at atmospheric pressure reaches the underside of the weir (so that
the pressure across the nappe is approximately atmospheric, p = 0.
•
the weir has a sharp crest (so that the flow breaks away on the downstream side);
•
the backwater level does not approach the crest of the weir;
•
the notch does not span the full width of the channel (contracted weir) or, if it is full
width, then a vent pipe is used to ensure atmospheric pressure underneath the nappe.
Hydraulics 1
21
David Apsley
1
Idealised Analysis
h1
Velocity is not uniform over the weir but is a
function of height. Let h be the distance from the
free surface. Apply Bernoulli between points 1
upstream and 2 over the crest of the weir:
p1 + gz1 + 12 U 12 = p 2 + gz 2 + 12 U 22
gh1 + g ( H 0 − h1 ) +
1
2
2 p=0
h
p=0
H0
p hydrostatic
U 12 = 0 + g ( H 0 − h) +
1
2
U2
U 2 = 2 gh + U 12
As a first approximation it is usually assumed that the approach flow U1 is small; (the notch
of the weir is considerably smaller than the cross-section of the channel). Then
U = (2 gh)1 / 2 ,
0<h<H
Since the velocity of flow varies with height the total quantity of flow must be obtained by
summation over small elements of area A = b h, where b is the width of the weir:
H
Q=
U A =
(2 gh)1 / 2 b h → b(2 g )1 / 2
h1 / 2 dh
0
= b( 2 g ) H 3 / 2
This is usually multiplied by a discharge coefficient cD (∼0.62) to account for non-ideal flow:
Q = c D 23 b(2 g )1 / 2 H 3 / 2
(20)
3/2
Most important, however, is the power-law dependence, Q ∝ H ; the constant of
proportionality can be obtained by calibration.
1/ 2
2
3
Notes
1.
For accurate work the upstream velocity U1 (= Q/A) can be included in the formula
for U and the integration leads to an implicit equation for Q which can be solved
iteratively; (see Hamill for details and an example).
2.
Another common device – you can see one in the second-year centrifugal-pump
experiment – is the V-notch weir. This has the advantage of large sensitivity for small
flow rates.
Q
H
The velocity U is still given by (2gh)1/2, but since the width of flow at any level,
b ∝ h, integration for the quantity of flow produces a different power-law relationship,
Q ∝ H 5 / 2 ; (exercise: prove it!).
Hydraulics 1
22
David Apsley
H
3.5 Other Applications
3.5.1 Discharge Through an Orifice
Consider water discharging from a small hole at depth h
below the surface of a large tank. Apply Bernoulli’s theorem
between free surface (1) and exit point (2):
( p + gz + 12 U 2 )1 = ( p + gz + 12 U 2 ) 2
At the free surface, p1 = 0 (free surface) and U1 = 0 (large
reservoir).
At the exit point, p2 = 0 (free discharge).
1
h
2
For convenience, take all heights relative to the orifice. Then
2
0 + gh + 0 = 0 + 0 + 12 U exit
Hence
U exit = 2 gh
(21)
Notes
1.
This result is called Torricelli’s formula and it is basically a statement of energy
conservation (potential energy kinetic energy).
2.
If the orifice is not small then the depth below the free surface and hence the exit
velocity will vary. The quantity of flow must then be obtained by integration.
3.
The exit velocity can be multiplied by the exit area to compute discharge. However, in
practice, this ideal volumetric flow rate must be multiplied by a discharge coefficient
(Section 5.3) to account for losses and flow convergence after exit.
Example. Water flows from a 3-cm-diameter hole 2 m below the surface of a large tank.
(a)
Assuming no losses calculate the exit velocity and volumetric flow rate.
(b)
Using the results of (a) determine the subsequent path of the jet.
(c)
If a discharge coefficient cD = 0.6 is applied, calculate the discharge.
Hydraulics 1
23
David Apsley
u exit
3.5.2 Tank Emptying
By equating the rate of change of fluid volume in a tank to the volumetric flow rate,
dV
= −Qout
dt
the time to empty the tank can be found.
(22)
Using Torricelli’s formula for the exit velocity, and allowing for a discharge coefficient cD
the RHS of (22) can be expanded to give
dV
= −c D Aout 2 gh
(23)
dt
The volume of fluid V is a function of the depth of water h, so this becomes a differential
equation in h.
Note that one assumption of Bernoulli’s theorem – that of steady-state conditions – is strictly
not met here!
Example. A tank has a rectangular base with sides 2 m by 3 m and initially contains water to
a depth of 0.5 m. When the plug is pulled out water exits from the base through a hole of area
10–3 m2. Assuming no losses, estimate the time to empty the tank.
Hydraulics 1
24
David Apsley
4. The Momentum Principle
Consideration of momentum is necessary whenever one is considering forces. Two
mechanical principles are involved.
MOMENTUM PRINCIPLE (Newton’s 2nd Law):
Force = Rate of change of momentum
ACTION/REACTION (Newton’s 3rd Law):
The force exerted by a fluid on its containment is equal and opposite to that exerted
by its containment on the fluid.
These principles are applied to the whole body of fluid in a control volume. Only external
forces need be considered (since internal forces cancel in action/reaction pairs).
4.1 Steady-Flow Momentum Principle
F
Consider a control volume consisting of a (thin) stream
tube carrying mass flow rate m .
u2
u1
Since every element of fluid has its velocity changed from
u1 to u2 the net rate of change of momentum in this control
volume is m(u 2 − u1 ) . Hence,
F = m(u 2 − u1 )
(24)
Note:
1.
F is the sum of all external forces on the control volume, including
– reaction from solid boundaries
– fluid forces from adjacent fluid (pressure, viscous forces etc.)
2.
Force, velocity and momentum are all vector quantities and each direction must be
considered.
(*) does not generalise easily to non-uniform flow (where velocity varies over a crosssection). Since mass flux is constant along the stream tube (continuity) a better way of
writing it is
F = (mu) 2 − (mu)1
or
force = (momentum flux)out – (momentum flux)in
where
momentum flux = mass flux × velocity
= ( uA)u
Can you see how this would generalise to non-uniform inflow or outflow velocity profiles
and to time-dependent flow? (These will be addressed in Hydraulics 2).
Hydraulics 1
25
David Apsley
4.2 Applications
(i)
(ii)
(iii)
(iv)
Pipe contraction
Forces on pipe bends
Jets and nozzles
Hydraulic jump (considered as a special application later).
The basic technique is to apply the steady-state momentum principle:
F = Q(u 2 − u1 )
to the fluid in a control volume.
Technique
1.
Choose a suitable control volume; (usually coincides with boundaries, so that there is
only one inflow and outflow; cuts fluid flow where it is simple – uniform, if possible).
2.
Mark in all the forces on the fluid (equal and opposite to that on the boundary; usually
use gauge pressures so that atmospheric pressure is zero).
3.
Mark inflow and outflow velocities (and their directions) and use continuity to relate
them if possible.
4.
Apply the momentum principle in each relevant direction.
4.2.1 Pipe Contraction
Example. For the pipe-reducing section shown, the greater and smaller diameters are 100 mm
and 60 mm, respectively and the water discharges to the atmosphere at section 2. When the
volumetric flow rate is 25 L s–1 the differential height in the mercury manometer is
h = 700 mm. Calculate the total force resisted by the flange bolts. (Mercury has specific
gravity 13.6).
1
2
water
h
mercury
Hydraulics 1
26
David Apsley
4.2.2 Forces on Pipe Bends
Example (Exam, June 2004)
Water is flowing at a rate of 0.001 m3 s–1 through a reducing bend, with the diameter
contracting from 20 mm to 15 mm and the flow direction changing by 45° as shown in the
figure. The bend is in the horizontal plane and the flow exits to the atmosphere at the narrow
end. You may ignore energy losses.
(a)
What is the pressure (relative to atmospheric pressure) in the 20 mm pipe?
(b)
What is the force on the pipe bend (expressed in the x-y coordinate system shown)?
y
x
o
45
Example (Exam, May 2003 *** slightly modified ***)
Water is flowing through a horizontal pipe of diameter 15 mm with a flow rate of
0.003 m3 s-1. At the end of the pipe there is a T-junction (see Figure) with both arms
horizontal and each of diameter 10 mm. The water exits the T-junction directly into the
atmosphere.
(a)
Calculate the velocity of the water in the 15 mm pipe (u1) and in the two branches of
the T-junction (u2).
(b)
If the head-loss coefficient for the T-junction (based on the inflow speed u1) is
K = 1.2, calculate the pressure in the pipe just upstream of the T-junction (ignoring
any other energy losses).
(c)
What is the force exerted by the water on the T-junction?
u2
u1
u2
Hydraulics 1
27
David Apsley
4.2.3 Jets
The most important aspect of these calculations is that the jet is free and the pressure
throughout is atmospheric.
Example. A jet of diameter 80 mm and carrying a flow of 50 L s–1 impacts normally upon a
plate. Calculate the force on the plate.
Example. (from White, 2003) A liquid jet of velocity V and diameter D strikes a fixed hollow
cone and deflects back as a conical sheet at the same speed. Find the cone angle for which
the restraining force F = 32 AV 2 , where A is the cross-sectional area of the jet.
V
V
Hydraulics 1
28
θ
David Apsley
5. Energy
Principle of Conservation of Energy (1st Law of Thermodynamics):
Change of energy = work done + heat input
This is an important equation for mechanical engineers. For incompressible fluids, however,
we do not need the thermal elements of this equation and can work with the much simpler
Mechanical Energy Principle:
Change of kinetic + potential energy = work done by non-conservative forces
You will remember, from Mechanics, that:
work done = force × displacement
rate of doing work (i.e. power) = force × velocity
Remember that “displacement” and “velocity” both refer to components in the direction of
the force.
5.1 Derivation of Bernoulli’s Equation From an Energy Principle
Consider steady, frictionless flow through a
length of a thin stream tube.
u1
1
2
u2
Each mass m of fluid has kinetic energy
1 mU 2 and potential energy mgz. The rate at which energy passes section 1 or 2 is,
2
therefore, given by
m( 12 U 2 + gz )1, 2
and hence the rate at which (kinetic + potential) energy is created in this section is given by
m( 12 U 2 + gz ) 2 − m( 12 U 2 + gz )1
where
m= Q
is the mass flow rate. m is the same at both sections since, by definition, there is no flow
through the long sides of the tube.
In the absence of viscosity, work is done only by pressure forces acting on the two ends.
Rate of working = force × velocity
= p1 A1U 1 − p 2 A2U 2
(No work is done on the long sides of the streamtube because the pressure force is
perpendicular to velocity there).
Hence, applying
rate of working = rate of change of (kinetic + potential) energy
p1 A1U 1 − p 2 A2U 2 = m( 12 U 2 + gz ) 2 − m( 12 U 2 + gz )1
Dividing by m (= 1U 1 A1 = 2U 2 A2 ) ,
p1 p 2
−
= ( 12 U 2 + gz ) 2 − ( 12 U 2 + gz )1
1
Hydraulics 1
2
29
David Apsley
i.e.
(
p
+ 12 U 2 + gz )1 = (
p
+ 12 U 2 + gz ) 2
For incompressible fluids, is also constant along a streamline and hence
p + 12 U 2 + gz = constant (along a streamline)
(25)
Notes
(1) Reminder of the qualifiers:
inviscid (we shall examine how to include losses shortly)
steady (for the unsteady Bernoulli equation see Hydraulics 3)
incompressible (but see note 2 below)
along a streamline
(2) An advantage of the energy-related derivation is that it is easy to incorporate (i) thermal
effects and (ii) frictional losses (or energy input/extraction by pumps/turbines). For thermal
flows the total energy is supplemented by the internal energy (per unit mass) e and equation
generalised to read
h + 12 U 2 + gz = work done + heat input
(26)
where h = e + p/ is the enthalpy and the RHS represents energy transferred to the fluid (per
unit mass) by mechanical means (positive for pumps, negative for frictional forces and
turbines) or by heat transfer. The incompressible assumption can be dropped.
5.2 Fluid Head
In fluid mechanics, energy is often measured in length units, referred to as fluid head.
If you lift mass m through height H you give it (potential) energy mgH. Thus, the height to
which you lift something is a measure of its energy:
potential energy
H=
= energy per unit weight
mg
The same applies in fluid mechanics. An obvious example is a pumpedstorage power station (e.g. Dinorwig) which pumps water to the top of a
hill during the night, storing energy for release to the turbines during the
day. Another example is the constant-head tank which pressurises the hotwater supply in your house: it is basically the available head which
determines how fast hot water comes out of the taps.
h
Pressure also represents stored energy, and it is common to measure it in length units; e.g
mm Hg (millimeters of mercury) or metres of water. Atmospheric pressure is about
760 mm Hg or 10 m of water.
Consider Bernoulli’s equation:
p + gz + 12 U 2 = constant , p 0 ( the total pressure)
Each term represents a form of energy per unit volume. If we divide by the weight per unit
volume (the specific weight g) we obtain an equivalent form
Hydraulics 1
30
David Apsley
p
U2
+z+
= H , constant
g
2g
(27)
H is called the total head.
p/ g is the pressure head and p/ g + z is the piezometric head.
U2/2g is the dynamic head.
Power
For mass m, energy stored = mgH.
energy mass
Power =
=
× gH
time
time
Hence, in fluid flow if a mass flow rate Q experiences a change of head H then the power
involved is given by
power = QgH
This formula is applicable to both pumps and turbines. It may be multiplied by an efficiency
to account for energy losses.
Example. The Hoover dam in Colorado holds back water to a depth of 180 m. If its Francistype hydroelectric turbines can pass 1400 cubic metres of water per second with a generating
efficiency of 80%, calculate the power available. (Don’t quote me on the flow rate or
efficiency please!)
Key Points
1.
In fluid mechanics pressure and energy are commonly measured in length units,
referred to as fluid head.
2.
Pressure and head are connected (as in hydrostatics) by
p = gh
3.
Bernoulli’s equation describes an interchange between different forms of energy:
pressure energy ↔ gravitational potential energy ↔ kinetic energy
4.
In
p
U2
+z+
=H
g
2g
H is the total head, and is constant unless there are frictional losses (or energy is
input/extracted by pumps/turbines). H can be plotted graphically as the energy line.
5.
U 2 / 2 g is the dynamic head. Many energy losses are quantified as multiples of the
dynamic head (see Section 5.3).
6.
Power = efficiency × gQH.
Hydraulics 1
31
David Apsley
5.3 Departures From Ideal Flow
Many theoretical results are derived for ideal fluids, in particular assuming no frictional
losses and uniform velocity profiles.
In practice, compensation is necessary for non-ideal flow. Important examples include:
•
discharge coefficients to correct the quantity of flow deduced by Bernoulli’s equation;
•
loss coefficients to quantify energy losses in pipelines;
•
momentum and energy coefficients to account for non-uniform velocity profiles.
Such factors are usually determined experimentally and documented in official standards.
5.3.1 Discharge Coefficients
A discharge coefficient cD is the ratio of the actual quantity of flow, Q, to that deduced from
Bernoulli’s equation on the basis of ideal flow. i.e.
Q = c D Qideal
(28)
cD can be nearly 1.0 for a well-designed venturi meter, but as low as 0.6 for a weir or sharpedged orifice.
5.3.2 Loss Coefficients
These are used to quantify energy losses, particularly in pipelines (Dr Lane-Serff will cover
this in detail in the section on “Pipe Flow”).
A loss coefficient K is the ratio of energy loss to the kinetic energy, or head loss h to dynamic
head ( V 2 / 2 g ) :
h=K
V2
2g
(29)
For pipe flow there is a gradual energy loss due to pipe friction over a length L of pipe. This
is quantified in terms of a friction factor :
L V2
L
h=
,
i.e. K =
(30)
D 2g
D
Example. A park’s water feature is supplied with water from a large tank via a uniform pipe
of length 35 m and diameter 150 mm. There is a free discharge at a point 10 m below the
water level in the tank. Find how long it takes to deliver 2000 L of water:
(a)
assuming no losses;
(b)
allowing for an entry loss coefficient K = 1.0 and pipe friction factor = 0.05.
Hydraulics 1
32
David Apsley
5.3.3 Momentum and Energy Coefficients
These may be applied to account for non-uniform velocity profiles in the application of the
Momentum Principle or Bernoulli equation.
In the control-volume application of the Momentum Principle momentum fluxes are
determined as
momentum flux = mass flux × velocity
If the velocity is uniform:
momentum flux = ( uA) × u = u 2 A
If the velocity is not uniform then one has to sum over individual small contributions:
momentum flux =
u 2 dA
(For continuous velocity profiles this sum becomes an integral – see Hydraulics 2).
Since it is inconvenient to write down a sum or integral each time it is common to introduce a
momentum coefficient as the ratio of the actual momentum flux to that calculated by
assuming a uniform velocity equal to the average velocity uav; i.e.
2
momentum flux = ( u av
A)
(31)
For laminar pipe flow (see later) is 4/3. For turbulent pipe flow, however, the velocity
profile is much more uniform and is only slightly more than 1.
Since the kinetic energy flow past a point is given by ( uA) × ( 1 2 u 2 ) a similar energy
coefficient may be defined, which is the ratio of the average value of u 3 to that assuming a
3
uniform velocity, u av
.
Hydraulics 1
33
David Apsley