mass momentum and energy

Hydraulics 2 T1-1 David Apsley

TOPIC T1: MASS, MOMENTUM AND ENERGY AUTUMN 2006 (REVISED) Objectives (1) Extend the basic principles of mass and momentum to flows with non-uniform velocity

and pressure. (2) Extend the continuity principle to time-varying flows. (3) Apply continuity and Bernoulli’s eqn for flow measurement and tank-emptying problems. (4) Learn methods for quantifying losses. Contents 0. Revision of basic concepts 0.1 Notation 0.2 Dimensionless parameters 0.3 Definitions 0.4 Basic principles of fluid mechanics 0.5 Physical constants 0.6 Properties of common fluids 1. Continuity (conservation of mass) 1.1 Mass and volume fluxes 1.2 Flows with non-uniform velocity 1.3 Time-dependent flow 2. Forces and momentum 2.1 Control-volume formulation of the momentum principle 2.2 Fluid forces 2.3 Boundary layers and flow separation 2.4 Drag and lift coefficients 2.5 Calculation of momentum flux 2.6 Calculation of pressure forces 2.7 The wake-traverse method for measurement of drag 3. Energy 3.1 Bernoulli’s equation 3.2 Fluid head 3.3 Static and stagnation pressure 3.4 Flow measurement 3.5 Tank emptying 3.6 Summary of methods for incorporating losses References Hamill (2001) – Chapters 1, 2, 4, 5, 7 Chadwick and Morfett (2004) – Chapters 1, 2, 3 Massey (1998) – Chapters 1, 2, 3, 4 White (2002) – Chapters 1, 2, 3


0. REVISION OF BASIC CONCEPTS 0.1 Notation Geometry � x ≡ (x, y, z) position; (z is usually vertical) t time Field Variables � u ≡ (u, v, w) velocity (also use V for average velocity in a pipe or conduit) p pressure p – patm is the gauge pressure p* = p + � gz is the piezometric pressure T temperature Fluid Properties � density � dynamic (or absolute) viscosity � ≡ � /� kinematic viscosity � ≡ � g specific weight (weight per unit volume) s.g. ≡ � /�

ref specific gravity (or relative density); “ ref” = water (for liquids) or air (for gases)

� surface tension (force per unit length) K bulk modulus (pressure change divided by volumetric strain) c speed of sound k conductivity (heat flux per unit area divided by temperature gradient) 0.2 Dimensionless Parameters

��

�Re

ULUL ≡≡ Reynolds1 number (viscous flow)

gL

U≡Fr Froude2 number (open-channel hydraulics)

c

U≡Ma Mach3 number (compressible flow)

�

�We

2LU≡ Weber4 number (surface tension)

U

fL=St Strouhal5 number (vortex shedding; f = shedding frequency)

Here, U is a representative velocity scale and L is a representative length scale. There are many other important dimensionless combinations (see Topic T3: “Dimensional Analysis” ).

1 Osborne Reynolds (1842-1912); appointed first Professor of Engineering at Owens College (precursor of the University of Manchester). 2 William Froude (1810-1879), British naval architect; developed scaling laws for the model testing of ships. 3 Ernst Mach (1838-1916), Austrian physicist and philosopher. 4 Moritz Weber (1871-1951), developed modern dimensional analysis; actually named the Re and Fr numbers. 5 Vincenz Strouhal (1850-1922), Czech physicist; investigated the “singing” of wires.


0.3 Definitions A fluid is a substance that continuously deforms under a shear stress, no matter how small. A solid will reach equilibrium under such a stress. Fluids may be liquids (definite volume; free surface) or gases (expand to fill any container). Hydrostatics is the study of fluids at rest. Hydrodynamics is the study of fluids in motion. Hydraulics is the study of the flow of liquids (usually water). Aerodynamics is the study of the flow of gases (usually air). All fluids are compressible to some degree, but their flow can be approximated as incompressible (that is, hydrodynamic pressure changes don’ t give rise to density changes) for velocities much less than the speed of sound (∼ 1480 m s–1 in water, 340 m s–1 in air). An ideal fluid is one with no viscosity. It doesn’ t exist, but it can be a good approximation. A Newtonian fluid is one for which viscous stress is proportional to velocity gradient (rate of strain):

y

u

d

d�� =

where � is the viscosity. Most fluids of interest in hydraulics (including air and water) are Newtonian, but there are some important non-Newtonian fluids (e.g. mud, blood, paint). Real flows may be laminar (adjacent layers slide smoothly over each other) or turbulent (subject to random fluctuations about a mean flow). If the viscosity is too small to maintain a smooth, orderly flow, then a laminar flow undergoes transition and becomes turbulent. Although transition to turbulence is dependent on a number of factors, including surface roughness, the primary determinant is the Reynolds number

��

�Re

ULUL ≡≡ (1)

U and L are typical velocity and length scales of the flow. In general: “high” Re � turbulent “ low” Re � laminar Typical critical Reynolds numbers for transition are: pipe flow: ReD ≈ 2300; circular cylinder: ReD ≈ 3×105; flat plate: Rex ≈ 5×105 – 3×106. Important. The Reynolds number and its critical value depend on which velocity and length scale are used to define it – which should be stated. For example, you could choose to use either radius or diameter for flow in a pipe, and they would obviously give a factor-of-2 difference in Reynolds number even though the flow is the same. (Why are the values quoted for circular cylinder and flat plate above so much larger than that for pipe flow?) The vast majority of civil-engineering and environmental flows have high Reynolds numbers and are fully turbulent.


0.4 Basic Principles of Fluid Mechanics Hydrostatics

In stationary fluids, pressure forces balance weight. Hydrostatic Principle

zgp��

−= or gz

p �d

d −= (2)

The same equation holds in a moving fluid if there is no vertical component of acceleration. For a constant-density fluid the hydrostatic equation can also be written

constantgzp =+ � p + � gz is called the piezometric pressure, p* . It represents the combined effect of pressure and weight. Thermodynamics For compressible fluids thermodynamics and heat input are important and one requires, in addition, an equation of state; e.g. Ideal Gas Law p = � RT (3) The gas constant R is a constant for any particular gas and is given by R = R0 / m, where R0 is the universal gas constant and m is the mass of one mole. For dry air, R = 287 J kg–1 K–1. T is the absolute temperature in Kelvin: 15.273)C()K( +°= TT (4) Fluid Dynamics Continuity (Mass Conservation) Mass is neither created nor destroyed. For steady flow, flow in = flow out Momentum Principle Force = rate of change of momentum. For steady flow, force = (momentum flux)out – (momentum flux)in Energy Change in energy = heat supplied + work done For incompressible flow, change of kinetic energy = work done


The Energy Equation (i) Incompressible Fluids For incompressible fluids the energy equation is a purely mechanical equation, equivalent to, and directly derivable from, the Momentum Principle. Bernoulli’s Equation Steady incompressible flow without losses:

eamlinelong a strconstant aUgzp =++ 2

21

�

More generally, in steady incompressible flow,

fluidondoneworkUgzp =++ )�(

� 221 (5)

Here,

�( ) means “change in” and the RHS of (5) represents the energy (per unit mass) input

by pumps or removed by turbines or friction. (ii) Compressible Fluids For compressible fluids the energy equation involves thermodynamics. The energy per unit mass is supplemented by the internal energy e and energy can also be transferred to the fluid as heat. (5) becomes

fluidon doneworkfluidtosuppliedheat)�(� 2

21 +=+++ Ugz

pe (6)

The quantity �/pe+ is called enthalpy. 0.5 Physical Constants Gravitational acceleration: g = 9.81 m s–2 (at British latitudes) Universal gas constant: R0 = 8314 J kg–1 K–1 Standard atmospheric pressure: 1 atmosphere = 1.01325×105 Pa = 1.01325 bar Standard temperature and pressure (s.t.p): IUPAC (International Union of Pure and Applied Chemistry): 0° C (273.15 K) and 105 Pa. ISO (International Standards Organisation): 0° C (273.15 K) and 1 atm (1.01325×105 Pa). Since two major international organisations can’ t agree it’s probably better to specify reference conditions explicitly.


0.6 Properties of Common Fluids Properties are given at 1 atmosphere and 20 ºC unless otherwise specified. Air Density: � = 1.20 kg m–3 (� = 1.29 kg m–3 at 0 ºC) Specific weight: � = 11.8 N m–3 Dynamic viscosity: � = 1.80×10–5 kg m–1 s–1 (or Pa s) Kinematic viscosity: � = 1.50×10–5 m2 s–1

Specific heat capacity at constant volume: cv = 718 J kg–1 K–1

Specific heat capacity at constant pressure: cp = 1005 J kg–1 K–1 Gas constant: R = 287 J kg–1 K–1 Speed of sound: c = 343 m s–1

Water Density: � = 998 kg m–3 (� = 1000 kg m–3 at 0 ºC) Specific weight: � = 9790 N m–3 Dynamic viscosity: � = 1.003×10–3 kg m–1 s–1 (or Pa s) Kinematic viscosity: � = 1.005×10–6 m2 s–1

Surface tension: � = 0.0728 N m–1

Speed of sound: c = 1482 m s–1

Mercury Density: � = 13550 kg m–3 Ethanol Density: � = 789 kg m–3 Fluid properties – especially viscosity – change significantly with temperature.

Water Air T (°C) � (kg m–3) � (Pa s) � (m2 s–1) � (kg m–3) � (Pa s) � (m2 s–1)

0 1000 1.788×10–3 1.788×10–6 1.29 1.71×10–5 1.33×10–5

20 998 1.003×10–3 1.005×10–6 1.20 1.80×10–5 1.50×10–5 50 988 0.548×10–3 0.555×10–6 1.09 1.95×10–5 1.79×10–5 100 958 0.283×10–3 0.295×10–6 0.946 2.17×10–5 2.30×10–5

As temperature increases: viscosities of liquids decrease; viscosities of gases increase. (Explain why.) The viscosity of gases may be approximated by Sutherland’s law:

��

��

�

++

��

��

�=

ST

ST

T

T 0

2/3

00�

�

(7)

For air: T0 = 273 K, � 0 = 1.71×10–5 Pa s, S = 110.4 K.


1. CONTINUITY (CONSERVATION OF MASS) The rate at which something crosses a surface is called its flux. “Mass flux” (or “mass flow rate”) is the mass crossing a given surface per unit time. Conservation of mass can be applied to the fluid in or passing through the surface of an arbitrary control volume: Steady flow: (mass flux)in = (mass flux)out

Unsteady flow: outin mass fluxmass fluxmasst

)()()(d

d −=

1.1 Mass and Volume Fluxes If velocity u is uniform over a section and normal to area A, then Volume flux: uAQ = (m3 s–1) Mass flux: uAQm �� ==� (kg s–1)

Q is also called the quantity of flow, (volumetric) flow rate or discharge. A stream tube is a bundle of streamlines across which there is no flow. For an incompressible flow, � is constant along a streamline and mass conservation implies volume conservation; i.e.:

21 QQ = If u is uniform over the cross-section then

2211 AuAu = If u is not uniform or if there is more than one inlet or outlet, then

total flow in = total flow out

outinQQ �� =

and the total volume flux or mass flux must be obtained by summation or, for continuously-varying quantities, by integration.

uA

u1 u2

12

m

mout

min


Example. The figure shows a converging two-dimensional duct in which flow enters in two layers. A fluid of specific gravity 0.8 flows as the top layer at a velocity of 2 m s–1 and water flows along the bottom layer at a velocity of 4 m s–1. The two layers are each of thickness 0.5 m. The two flows mix thoroughly in the duct and the mixture exits to atmosphere with the velocity uniform across the section of depth 0.5 m.

0.5 m

2 m/s

4 m/s

0.5 m

0.5 m

p =15 kN/m12

(a) Determine the velocity of flow of the mixture at the exit. (b) Determine the density of the mixture at the exit. (c) If the pressure p1 at the upstream section is 15 kPa, what is the force per unit width

exerted on the duct? (Do this part after Section 2 on the Momentum Principle).

Answer: (a) 6 m s–1; (b) 933 kg m–3; (c) 7.8 kN 1.2 Flows With Non-Uniform Velocity The continuity principle may be extended to cases where u varies over a cross-section (e.g. flow in pipes or flow in a boundary layer) by considering the flow to be broken down into infinitesimal areas dA, across each of which the velocity is approximately constant:

AuQ dd = The total quantity of flow is found by summation or, in the limit of small areas, integration:

Volume flow rate: �

= AuQ d (8)

The average velocity (sometimes called the bulk velocity) is that constant velocity which would give the same total flow rate; i.e. AuQ av= or

Average velocity: A

Quav = (9)

In other words, to find the average velocity from a non-uniform velocity profile you will first have to find Q.


1.2.1 Two-Dimensional Flow Velocity is a function of height: u ≡ u(y) This is often the case in a wide rectangular channel. A small element of area over which the velocity is uniform has the form of a rectangle, width b and height dy: ybA dd =

Quantity of flow: �

= yubQ d (10)

or:

Flow per unit width: �

= yuq d (11)

Example. The distribution of velocity in a rectangular channel of width b = 800 mm and depth h = 200 mm is given by

7

1

0 ��

��

�=h

yuu

where u0 = 8 m s–1. What is (a) the quantity of flow; (b) the average velocity? Solution. (a)

�=�

=h

yh

yubAuQ

0

7/10 d)(d (b = 0.8 m, h = 0.2 m, u0 = 8 m s–1)

Simplify the integral with a change of variables hyYhyY /dd,/ == :

13

0

1

0

7/80

1

0

7/10

sm12.1

8

7

8

7d

−=

=��

��

�×=�

= bhuYbhuYYbhuQ

(b)

1

0

s m7

8

7area

rateflow

−=

=

==

u

bh

Quav

b

dyu(y)


1.2.2 Axisymmetric Flow Velocity is a function of radius: )(ruu ≡ Examples include pipes and jets. A small element of area over which the velocity is uniform has the form of an infinitesimal hoop of radius r, thickness dr: rrA d�2d = (12)

Quantity of flow: �

= rruQ d�2 (13)

Example. Fully-developed laminar flow in a pipe of radius R has velocity profile:

)/1( 220 Rruu −=

Find the average velocity in terms of u0. Solution. The average velocity can be found by dividing the flow rate by the area. For the flow rate,

�

−=�

=RR

rrRrurruQ0

220

0

d)/1(�2d�2

Substitute RrsRrs /dd,/ == for convenience:

02

1

0

42

02

1

0

30

21

0

20

2

�

2

1

42�2d)(�2d)1(�2

uR

ssuRsssuRsssuRQ

=

��

��

�−=�

−=�

−=

Hence,

0

2

2

1

�

u

R

Quav

=

=

Note. For velocity profiles measured in an experiment (where the integral must usually be evaluated graphically), it is unnatural and inaccurate to have the integrand vanishing at the centre (since this is where velocity is highest) and (13) can be rewritten as

�

= 2d� ruQ (14)

i.e. Quantity of flow = � × (area under a u – r2 graph) See, for example, the pipe-flow laboratory experiment and the Example Sheet.

drr

u(r)


1.3 Time-Dependent Flow Examples of time-dependent flows are: • fluid oscillations; e.g. pressure transients in pipes, vortex shedding, waves; • moving-boundary problems; e.g. pistons, rotating machinery. The mass of fluid (m) inside a control volume can vary if either the density of fluid or the size of the control volume changes. To balance this there must be a net mass flux through the boundaries of the control volume. Thus, in any time interval change of mass = mass that has flowed in – mass that has flowed out or, in rate form,

outin mass fluxmass fluxt

m)()(

d

d −= (15)

For incompressible flows, conservation of mass implies conservation of volume:

outin QQt

V −=d

d (16)

Example. (White, 2002) An incompressible fluid is being squeezed outwards between two large circular discs by the uniform downward motion V0 of the upper disc. Assuming 1-dimensional radial outflow, derive an expression for V(r).

V0

rV(r)h(t)

Solution: For an incompressible fluid, conserving mass is equivalent to conserving volume. Hence, using a control volume consisting of a cylinder of height h and radius r:

outin QQt

V −=d

d

� Vrhhrt

)�2(0)�(d

d 2 −=

� rhVt

hr �2

d

d� 2 −=

� t

h

h

rV

d

d

2−=

But 0d

dV

t

h −= . Hence,

h

rVV

20=


2. FORCES AND MOMENTUM Momentum Principle Force = rate of change of momentum (17) In principle this is a vector equation, but, in practice, often only one direction is relevant. An ideal fluid is one without viscosity. Ideal fluids don’ t exist, but can be a useful approximation. The momentum equation for an ideal fluid is often called the Euler equation6. The momentum equation for a real fluid is often called the Navier7-Stokes8 equation. 2.1 Control-Volume Formulation of the Momentum Principle The equation of motion (17) can be expressed mathematically in many ways, including partial differential equations, velocity potential (a bit like gravitational potential), vorticity (related to local angular momentum), or even in terms of complex variables (see White, 2002). Fortunately, in hydraulics it is usually adequate to work from first principles by considering the momentum balance for a control volume (CV). For a steady flow and fixed control volume:

)� Q(time

velocitychange in massmomentumofchangeofrateforce

inout uu −=

×=

=

This is OK if the velocity at inflow and outflow are uniform, but not very useful if they vary. Instead we define Momentum flux = mass flux × velocity = � Qu (18) Momentum Principle For Steady Flow Force = (momentum flux)out – (momentum flux)in (19) = (rate at which momentum leaves CV) – (rate at which momentum enters CV)

6 Leonhard Euler (1707-1783), Swiss mathematician, later Professor of Physics at the St Petersburg Academy; tackled many problems in fluid mechanics and mathematical physics. 7 Claude Navier (1785-1836), French civil engineer; also known for his strong political views, including opposition to Napoleon’s military aggression. 8 George Gabriel Stokes (1819-1903), Irish mathematician and Lucasian Professor of Mathematics at Cambridge; many important works in hydrodynamics.

uout

uin

F


Example. A jet of fluid flows smoothly onto a stationary curved vane which turns it through 45°. The initial jet has diameter 40 mm and uniform velocity 25 m s–1. The exit jet may be assumed to have uniform velocity 20 m s–1. Calculate the net force on the vane.

4540 mm

25 m/s

20 m/s

Solution. Gauge pressures are zero at inlet, outlet and free surface so that there is no net pressure force. Let the force on the vane be ),( yx FF=F . Then the reaction of the vane on the fluid is

),( yx FF −−=− F . A suitable control volume cuts incident and deflected jets where the flow is

uniform. Mass flux:

1-2

2

1

s kg42.314

04.0�

251000

4

�

��

=×××=

= DuQ

Momentum in x direction:

N2.341)252

120(42.31

)45cos(�12

−=−××=

−°=− uuQFx

Momentum in y direction:

N3.444)2

120(42.31

)045sin(�2

=××=

−°=− uQFy

Net force on the vane, (Fx,Fy) = (341, – 444) N, or 560 N at angle 52.5° to the horizontal. If there is more than one inflow or outflow then the net momentum flux must be obtained by summation. In particular, for non-uniform flows it is necessary to work out momentum fluxes (Section 2.5) and fluid forces (Section 2.6) by summation or integration.


2.2 Fluid Forces The total force on the fluid in a control volume is a combination of: • body forces (proportional to amount of fluid); e.g. – weight; • surface forces (proportional to area); e.g. – pressure forces; – viscous forces; • reactions from solid boundaries. (Note that weight acts irrespective of whether the fluid is moving or not and would be balanced by a reaction, or, internally, by a hydrostatic pressure distribution. It can, therefore, be excluded from the analysis if we consider only departures from this hydrostatic state.) These are the only forces we shall consider here. However, other fluid forces exist; e.g.

– Coriolis forces in a rotating frame (e.g in meteorology); – surface tension.

As in structural mechanics, surface forces are usually expressed in terms of stresses:

area

forcestress = (20)

or areastressforce ×=

Pressure (p) is a normal stress directed inward to a surface. For the control volume shown the net pressure force in the x direction from pressures on the left (l) and right (r) faces is

ApAp rl −

Since the net effect of a uniform pressure on all boundaries is zero it does not matter whether absolute, gauge or other relative pressure is used, provided that one is consistent. Shear stresses (� ) act tangentially to surfaces. For the control volume shown the net force in the x direction from shear stresses on the top (t) and bottom (b) faces is

AA bt�� −

What we have called � above is strictly �xy: for complex flows other

components (�xz,

�xx,

�yy, …) may be important. By convention, �

xy is the force per unit area in the x direction that the fluid on the upper (greater y) side of the interface exerts on the fluid on the lower (smaller y) side. Laminar and Turbulent Flow Shear stresses arise from two sources: viscous forces and, in turbulent flow, the net transfer of momentum across an interface by turbulent fluctuations (which, as far as the mean flow is concerned, has the same effect as a force).

Ap Al p Ar

x∆

A

y∆

τt

τb

A

A


For Newtonian fluids, viscous stress is proportional to velocity gradient. If velocity u is a function of y only then, in laminar flow:

y

u

d

d�� = (21)

This defines the dynamic viscosity, � . (In more complex flow fields a more general stress-strain relationship is required.) In turbulent flow one is usually only interested in the mean velocity u . Since momentum transfer between fast- and slow-moving fluid is dominated by the net effect of turbulent fluctuations rather than viscous stresses the mean shear stress is not equal to yu/dd� . 2.3 Boundary Layers and Flow Separation The ideal-fluid (zero-viscosity) approximation is inapplicable if viscous effects have a major effect on the flow. The most important example is boundary-layer separation. In real fluids velocity vanishes at solid boundaries; (the no-slip condition). This gives rise to a boundary layer close to walls where velocity changes rapidly from its value in the free stream to zero at the boundary. At high Reynolds numbers boundary layer are usually extremely thin. In an adverse pressure gradient (where pressure increases and velocity decreases in the direction of flow; for example, in an expanding channel) the net force in the opposite direction to flow actually causes the more-slowly-moving fluid near the boundary to reverse direction. This backflow leads to flow separation.

adverse pressuregradient

backflow

flow separation

speeds up ... ... slows down

y

u

τ


Turbulence in the boundary layer helps to prevent or delay flow separation because it readily transports fast-moving fluid from the free stream into the near-wall region, maintaining forward motion. For bluff bodies with sharp corners flow separation occurs at all but the smallest Reynolds numbers and causes a large increase in pressure (or form) drag. For more streamlined bodies with convex boundaries separation may or may not occur. 2.4 Drag and Lift Coefficients The force on a body in a flow can be resolved into streamwise and cross-stream components. Drag = component of force parallel to the approach flow. Lift = component of force perpendicular to the approach flow. The relative importance of drag or lift forces is quantified by non-dimensionalising them by dynamic pressure ( 2

021 � U ) × area:

Drag Coefficient

AU

dragcD 2

021 �

= (22)

A is a “ representative” area which depends on the body geometry and the nature of the flow (see below). Like the Reynolds number it should always be specified when defining cD. A lift coefficient may be defined in the same manner. Bluff and Streamlined Bodies Bluff bodies (flow separation) • force is predominantly pressure drag • A is the projected area (normal to the flow) • cD = O(1) Streamlined bodies (no flow separation) • force is predominantly viscous drag • A is the plan area (parallel to the flow) • cD << 1

AU0

AU0

lift F

dragU0


2.5 Calculation of Momentum Flux The momentum principle for steady flow may be written for a general control volume:

inout flux) momentum(flux) (momentum Force −=

where

u

u

)�(

)�(

velocityfluxmassfluxmomentum

uA

Q

==

×= (23)

If velocity is not constant then the momentum flux can be calculated (as for mass flux) by breaking the surface into small areas over each of which the velocity is uniform. In particular, if the velocity varies continuously over a cross-section then the sum of contributions from infinitesimal areas can be obtained by integration; e.g. for the x-component:

�

= Au d�flux momentum 2 (24)

Special Cases Velocity profile Momentum flux

(i) Uniform Area A

AU 2�

(ii) 2-dimensional dA = b dy

b

dyu(y)

�

yub d� 2

(iii) Axisymmetric dA = 2 � r dr

drr

u(r)

�

drru �2� 2

Note. As for the mass flux, for experimental measurements and graphical integration rather than theoretical work it is usually more appropriate (and accurate) to write the last of these as

) (��

d��

22

22

graphru aarea under

ruxmentum fluic flow moAxisymmetr

−××=

�

=


Example. (Examination, January 2003) A two-dimensional beam of height h = 100 mm completely spans a square air-conditioning duct of height D = 400 mm (see Figure). The approach flow is uniform (u1 = 0.6 m s–1), whilst the downstream velocity profile is 2-dimensional and may be represented by:

��

��

≥

<−=)2(

)2()2

�

cos4

1

4

3(

2

2

hyu

hyh

yuu

The pressure is uniform over the height of the duct at both sections. Neglecting drag on the walls of the duct find: (a) the value of u2; (b) the difference between pressures at inlet and downstream sections, assuming that

Bernoulli’s equation holds outside the wake region; (c) the force on the beam. Also, (d) define a suitable drag coefficient for the beam and calculate its value. Take the density of air as 1.2 kg m–3.

0.6 m/s400 mm

100 mm

Solution. (a) Continuity (per unit width):

)2

1(

)22

3(2

2

�

sin�

2

4

1

4

3

dd)2

�

cos4

1

4

3(d

2

2

2

02

22

2

021

hDu

hDhuhDh

yhyu

yuyh

yuyuDu

h

D

h

h

wake

−=

−+=��

��

��

��

−+��

��

� −=

�

+� −=�

=

Hence

1

2112 s m 6857.0

350

4006.0 −=×=

−=

hD

Duu

Answer: u2 = 0.69 m s–1


(b) By Bernoulli’s equation:

222

12

212

11

�� upup +=+

� Pa 06611.0)6.06857.0(2.1)(� 22212

1222

121 =−××=−=− uupp

Answer: Pressure difference = 0.066 Pa (c) Momentum principle (for steady flow): force = rate of change of momentum = (momentum flux)out – (momentum flux)in

221

222

21

�d� DuyuDFDpDp

wake

−�

=−−

��

��

�

��

��

�

�

−+−= )d1

(� 22121

2

wake

yuD

uppDF

Now

)16

13(

)216

19(

2�

sin�2

1

2

�

sin�

26

2

19

16

1

2d)�

cos2

1

2

1

2

�

cos69(16

1

2d)2

�

cos2

�

cos69(16

1

dd)2

�

cos4

1

4

3(d�

22

22

2

0

22

2

0

22

2

0

222

2

22

2

0

222

2

hDu

hDhu

hDh

yh

h

yhyu

hDyh

y

h

yu

hDyh

y

h

yu

yuyh

yuyu

h

h

h

D

h

h

wake

−=

−+=

��

��

��

��

−+��

��

� +−=

��

��

−+� ++−=

��

��

−+� +−=

�

+� −=�

Hence

N 007759.0)4

1

16

131(6857.06.02.106611.04.0

)]16

131([�

222

22

2121

2

=��

��

��

��

� ×−−+=

��

�� −−+−=

D

huuppDF

Answer: Force on the beam = 0.0078 N (d) Drag coefficient:

90.01.04.06.02.1

007759.0� 2

212

121

=××××

==Dhu

FcD

Answer: Drag coefficient = 0.90


Example. (Examination, January 2004) A long T-shaped element, of depth h / 4 and oriented symmetrically as shown below, completely spans a wind-tunnel duct of depth 2h, where h = 0.2 m. The velocity upstream is uniform: U0 = 40 m s–1. The velocity distribution is measured at a position downstream and is found to be

��

�

��

�

�

��

��

�−−=

2

2

2

max 12

11)(

h

yUyU

where y is the distance from the centreline.

U0 U(y)

yh

(a) Sketch the expected pattern of flow around the T-shaped element, indicating, in

particular, separation and reattachment points, recirculating flow regions and the direction of flow.

(b) Calculate Umax. (c) If the pressure drop from the upstream to the downstream section is 200 Pa, find the

force per unit span on the T-shaped element. (d) Define a suitable drag coefficient for the element and calculate its value. Take the density of air as 1.2 kg m–3.

Answer: (b) 54.6 m s–1; (c) 36.5 N; (d) 0.76 Example. (Examination, January 2002) Water enters a horizontal pipe of diameter 20 mm with uniform velocity 0.1 m s–1 at point A. At point B some distance downstream the velocity profile becomes fully-developed and varies with radius r according to:

)/1( 220 Rruu −=

where R is the radius of the pipe. The pressure drop between A and B is 32 Pa. (a) Find the value of u0. (b) Calculate the total drag on the wall of the pipe between A and B. (c) Beyond point B the pipe undergoes a smooth contraction to a new diameter DC.

Estimate the diameter DC at which the flow would cease to be laminar. [The critical Reynolds number for transition in a circular pipe, based on average velocity and diameter is 2300. Take the density and kinematic viscosity of water as � = 1000 kg m–3 and

� = 1.1×10–6 m2 s–1 respectively.]

Answer: (a) 0.2 m s–1; (b) 0.0090 N; (c) 15.8 mm


2.6 Calculation of Pressure Forces When pressure changes with position the total pressure force can also be found by summing over small contributions

p dA Be very careful about direction in specific situations. Example. Find the hydrostatic force per unit width and the centre of pressure on a plane wall inclined at

� to the vertical.

Solution. The water depth is h and the total inclined length is L. Local depth y and inclined distance s are defined in the diagram: �cossy = From hydrostatics the gauge pressure is �cos�� gsgyp == The total pressure force can be found by summing over contributions from small inclined areas 1dd ×= sA . The force per unit width f is then

�cos�

d�

cosgs�

d

221

0

gL

s

Apf

L

s

=

�=

�

=

=

The centre of pressure, s , is found by equating moments (force × distance, or p dA × s):

�

×=× Apssf d

�

cos�

d�

cosgs��cos�

331

0

2221

gL

ssgLL

s

=

�=×

=

� Ls 32=

Notes. (i) All pressures are relative to a convenient reference – here, atmospheric pressure. (ii) The total force per unit width can be written:

)(

�)

�cos�(

21

21

widthunitperareapressureaverage

Lgh

LgLf

×=×=

×=

θ

y s

Lh

p

dA


(iii) The x component of total force (per unit width) is depthpressureaveragehghf ×=×= ��

cos 21

The y component of total force is abovewaterofweightghLLghf ==×= )

�sin(��

sin��sin 2

121

These are general results, even for curved surfaces, as can be seen by balancing forces on the volume of fluid above the plate.

(iv) The centre of pressure, on the line of action of the resultant, is at 2/3 of total depth. The pressure force on a small element is directed normal to that element. The pressure force on the 2-d element of length ds shown has individual x and y components

directionxthetonormalareaprojectedpressure

yp

spf x

×===

d

�cos)d(d

directionythetonormalareaprojectedpressure

xp

spf y

×==

=

d

�sin)d(d

Thus, for inclined surfaces, one can find components of force by either: calculating the force, then taking components in x and y directions; or multiplying pressure by x and y components of area. The latter is more convenient for curved surfaces. e.g. on aerofoil sections the pressure force (per unit span) in the upward y direction is

� −=�

−�

xppxxpxxp UL

surfaceupper

U

surfacelower

L d)(d)(d)(��

This can be used to calculate the lift force. 2.7 The Wake-Traverse Method for Measurement of Drag Objects in a flow experience a hydrodynamic force. If the fluid exerts a force F on the body then the body exerts a reaction force –F on the fluid. By measuring the change in momentum and pressure one can use the momentum principle to deduce the force on the body. Suitable control volumes for constrained (e.g. wind tunnel) and unconstrained flow are shown below. Upper and lower boundaries are streamlines, across which there is no mass or momentum flux.

p (x)L

p (x)U

x

FForce on BODY

F

Force on FLUID

ds dy

dx

θ


Fluid passing close to a body forms a wake of low-velocity downstream. If the flow is constrained by boundaries then the velocity outside the wake must increase slightly, with a compensating fall in pressure. In the unconstrained case, upper and lower boundaries should be sufficiently far away that pressure is equal to that in the free stream.

The steady-state momentum principle gives: force = (momentum flux)out – (momentum flux)in

�

−�

=�

−�

+−in

in

outoutin

in AUAuApAPF d�d�dd 22 (25)

Since the inflow velocity is uniform (Uin) the last term can be converted to an integral over the outflow by using continuity:

�

=�

=�

=�

out

in

out

in

in

inin

in

in AuUAuUAUUAU d�d�d�d� 2

Substituting in (25) and rearranging gives

�

−+−=out

inin ApPuUuF d)]()(�[ (26)

For unconstrained flows, pressures upstream and downstream are equal, as are the free-stream velocities: Uin = Uout. In the constrained case, it can be shown that, provided the wake is narrow (so that the difference in free-stream velocities is small), any pressure difference approximately9 compensates for the change in free-stream velocity U � . In both cases, then,

�

−= ∞

out

AuUuF d)(� (27)

Thus, hydrodynamic forces may be deduced indirectly by measuring momentum deficit in the wake, rather than directly using a force balance. This is called the wake traverse method and you will have an opportunity to use it in the wind-tunnel laboratory experiment.

9 By Bernoulli, )(� 2221

inoutoutin UUPP −=− ; the error in (27) is �

− AUU outin d)(� 221 , which is second

order in the (small) free-stream velocity difference Uin – Uout.

inflow wake

(i) constrained (change in free-stream velocity)

inflow wake

(ii) unconstrained (no change in free-stream velocity)

streamline

body

body


3. ENERGY 3.1 Bernoulli’s Equation Consider a thin stream tube, with varying cross-sectional area A. In the absence of thermal effects the energy of fluid passing through it is changed by the work done by pressure and viscous forces from the adjacent fluid and energy supplied or removed by external agents (e.g. pumps and turbines). Since the sides are locally parallel to the flow, energy only flows in or out of the control volume through ends 1 and 2. Hence, for steady flow, ( ) ( ) work doing of rate1 passingenergy of rate2 passingenergy of rate =− (28) The mechanical energy consists of kinetic energy (½mU2) and potential energy (mgz). The rate at which forces do work (i.e. power) is given by force × velocity (in direction of force). The rate of working of pressure at end 1 is therefore (pAU)1 and at end 2 is –(pAU)2. where U is flow speed. Pressure does no work on the sides because force and velocity are perpendicular there. (28) can then be written

WpAUpAUgzUQgzUQ �+−=+−+ 2112

21

22

21 )()()(�)(�

where W� consists of the rate of working of friction forces (i.e. losses) and pumps/turbines etc. � Q = � AU is the mass flow rate, which must be constant along the stream tube. Dividing by � Q = � AU gives

Q

WppgzUgzU �)�()�()()( 211

221

22

21

�+−=+−+

or, rearranging, Bernoulli’s Equation With Losses

�=++ )�(� 2

21 gzU

p (29)

∆ means “change in” and � is the work done per unit mass of fluid passing through this length of stream tube. Notes. (i) � represents all non-pressure work done on the system. It is composed of frictional

losses (due to viscosity) and any work done on or by the flow via pumps, turbines etc. (ii) Each term in the equation represents some form of energy or work done per unit

mass. (iii) (29) can easily be extended to thermal flows (boilers, condensers, refrigerators, ...) by

adding the internal energy e to the LHS and the rate of heat input QH to the RHS.

u1 u2

12


(iv) If there are no losses and no external sources of energy then (29) reduces to Bernoulli’s equation10:

=++ gzUp 2

21

� constant (along a streamline) (30)

For incompressible flows, � is also constant along a streamline and hence this equation is often applied as

=++ gzUp �� 221 constant (along a streamline) (31)

Note the assumptions: • inviscid (no losses) • incompressible • steady • along a streamline (different streamlines may have a different “constant” )

Example. Water is emptied from a tank through a horizontal pipe with centreline h below the level of water in the tank. The pipe has a severe constriction where the diameter is D1 and the water exits to the atmosphere through a nozzle of diameter D2 (> D1).

hD2D1

(1) (2)

(a) Assuming no losses, formulate an expression for the gauge pressure at the constricted section (1) in terms of diameters D1 and D2 and the tank level h.

(b) Cavitation (i.e. the formation of bubbles of vapour) will occur if the absolute pressure

falls below the vapour pressure for water (2.3 kPa at 20 °C). If the tank level h = 2 m and the constriction diameter D1 = 25 mm, calculate the exit diameter D2 at which this begins to occur. To avoid cavitation should you increase or decrease D2?

(Take atmospheric pressure = 101 kPa.)

Answer: (a) ])(1[ 4

1

21 D

Dghpp atm −ρ=− ; (b) 39 mm; decrease D2.

10 Daniel Bernoulli (1700-1782) Swiss-Dutch mathematician, a member of an illustrious family of well-known mathematicians.


3.2 Fluid Head In Bernoulli’s equation:

volumeunitperchangeenergyUgzp =++ )��(� 2

21

Each term has dimensions of pressure, or of energy per unit volume. If one divides by the specific weight � g these become energies per unit weight:

weightunitperchangeenergy

fluid headchange in g

Uz

g

p

=

=++ )2�(

�2

(32)

Energy per unit weight has dimensions of length and is called fluid head.

g

p� = pressure head

g

U

2

2

= dynamic head

g

Uz

g

p

2�

2

++ = total (or available) head

In hydraulics, energy and pressure are both often expressed in length units; e.g. “metres of water” or “millimetres of mercury” . Losses due to friction and the capabilities of pumps are typically specified in terms of head; that is, work done per unit weight. For the latter the rate of working (i.e. power) is given by gQHpower �= (33) where Q is the quantity of flow and H is the change in head. These will be examined further in Topic 2 (Pipe flow) and Topic 4 (Pumps). 3.3 Static and Stagnation Pressure A stagnation point is a point on a streamline where the velocity is reduced to zero. In general, any non-rotating solid obstacle in a stream produces a stagnation point next to its upstream surface, where the flow streamlines must split to pass around the obstacle. The stagnation pressure (or Pitot pressure) p0 is that pressure which would arise if the flow were brought instantaneously to rest. By Bernoulli’s

equation it is given (for incompressible fluids) by 221 � Up + . We define:

stagnation pressure 221 � Up +

static pressure p

dynamic pressure 221 � U

The dynamic pressure (and hence the flow velocity) is found by the difference between stagnation and static pressures (see the wind-tunnel laboratory experiment).

stagnation point(highest pressure)

U = 0P = P0

P = P + U012 ρ 2


3.4 Flow Measurement 3.4.1 Measurement of Pressure – Manometry Principles Three basic rules apply in a stationary fluid: (1) Same fluid, same height � same pressure (2) Same fluid, different height � zgp

��−=

(3) Different fluids: pressure is continuous at an interface U-Tube Manometer By rule (1) the pressure at level C is the same in both arms of the manometer. By rules (2) and (3) it can be found from pA and pB respectively by summing the changes in pressure over the heights of columns of fluid:

�� armright

mBC

armleft

A ghgyppyhgp ��)(� ++==++

Cancelling � gy and subtracting � gh gives the pressure difference BA ppp −=

�:

Differential Manometer Equation ghp m )��(

�−= (34)

If the working fluid is a gas then � � �

m and (34) can be approximated by ghp m

��= (35)

Inclined Manometer Differences in head may be small and difficult to measure accurately. The movement of the manometer fluid may be amplified by inclining the manometer. It is the vertical difference in height which is proportional to pressure differences: this is given in terms of the much larger length L by

�sinLh = (36)

L (large)

h (small)

θ

A B

h

C

y


3.4.2 Measurement of Velocity Basic idea: bring the fluid to rest at one point and measure the difference between static and stagnation (Pitot) pressures:

pressuredynamic

pressurestatic

pressurePitot

Upp 221

0�=−

Examples. (1) Open channel flow. (2) Pitot tube and piezometer (3) Pitot-static tube – measures both stagnation and static pressures in the same instrument.

static holes

stagnation point

static head tube

total head tube

free surfaceU /2g2

stagnation point

U

piezometer Pitot tube

Ug22


3.4.3 Measurement of Quantity of Flow Venturi Flowmeter A venturi is a localised smooth constriction in a duct. In the throat region: U increases (by continuity) p decreases (by Bernoulli) The difference in pressure between the main flow and the throat can be measured by a differential manometer and converted to quantity of flow. Bernoulli: 2

221

2212

11

�� UpUp +=+

� )(� 21

222

121 UUpp −=−

There are two unknowns on the RHS but this can be reduced to one (the velocity U1 which is required) by using continuity:

2211 AUAU =

� 2

112 A

AUU =

Hence, substituting for U2 in Bernoulli’s equation:

��

��

�−= 1)(�� 2

2

1212

1

A

AUp

Rearranging for U1 and taking Q = U1A1,

2/1

221

21 �

�

1)/(

2

��

��

−= p

AA

AQ (37)

2/1)

�( pconstantQ ×= (38)

Thus, the flow rate can be found by measuring the pressure difference �

p. In accurate experiments a coefficient of discharge, cd, is included to represent departures from non-ideality. cd is the ratio of the actual flow rate to the ideal flow rate which would be computed from Bernoulli’s equation and continuity: idealdQcQ = (39)

Design Features • A large convergence angle is advantageous as it tends to make the flow more uniform. • A small divergence angle is necessary to prevent flow separation. • The throat must be long enough for parallel flow to be established. • For a well-designed flowmeter a typical value of the coefficient of discharge is ~ 0.98.

12


Orifice Flowmeter An orifice is an aperture of negligible streamwise thickness through which fluid passes. The streamwise thickness determines frictional losses. The fluid cannot turn immediately, so that the emerging stream tube continues to contract up to the vena contracta – the section of minimum area.

An orifice meter is a means of measuring the flow rate in a duct by measuing the differential pressure across an orifice. It is basically an extreme variant of the venturi meter with the divergent region omitted. The basic premise is that the pressure throughout the recirculating eddy is essentially equal to that at the vena contracta.

Advantage: cheap. Disadvantage: considerable loss of energy. By the same process as that for the venturi meter one obtains:

2/1

21

21 �

�

1)/(

2

��

��

−= p

AA

AQ

v

where Av is the area of the vena contracta. Av is not obvious from the geometry. If Av is replaced by the area of the orifice then this may be compensated for by a coefficient of discharge, but, in practice, theory is simply used to deduce the form of relationship between flow rate Q and pressure drop

�p:

2/1)�

( pconstantQ ×= (40) with the constant of proportionality determined by calibration.

vena contracta


0.075 m

0.15 m

0.04 m /s3

B

A

C

Example (Massey). A vertical venturi meter carries a liquid of relative density 0.8 and has inlet and throat diameter of 150 mm and 75 mm respectively. The pressure connection at the throat is 150 mm above that at the inlet. If the actual rate of flow is 40 L s–1 and the coefficient of discharge is 0.96 calculate (a) the pressure difference between inlet and throat; (b) the difference in levels in a vertical U-tube manometer connected between these points, the tubes above the mercury being full of the liquid. (Relative density of mercury = 13.56)

Solution. (a) The difference in static pressure comes from Bernoulli’s equation under ideal conditions:

B2

21

A2

21 )��()��( gzVpgzVp llll ++=++

� )(�

)(�

AB2

A2

B21

BA zzgVVpp ll −+−=− (* )

The velocities to be used in (* ) come from the ideal flow rate, which is derived from the actual flow rate via the coefficient of discharge CD = Q/Qideal:

4/�

/

4/� 22ideal

D

CQ

D

QV D==

With Q = 0.04 m3 s–1, CD = 0.96, DA = 0.15 m, DB = 0.075 m, this gives VA = 2.358 m s–1

VB = 9.431 m s–1

The density of the liquid is 3m kg80010008.0� −=×=l

Hence, substituting in (* ) gives:

Pa117715.081.9800)(�

Pa33350)358.2431.9(800)(�

AB

22212

A2

B

=××=−=−××=−

zzg

VV

l

l

And hence

Pa34527

177133350BA

=+=− pp

(b) The height difference h in the U-tube manometer can be established by equating at the common height C the pressures found by applying the hydrostatic law in the two arms of the manometer:

)(��

)(�

CAlACHgCBlB zzgppghhzzgp −+==+−−+

Hence,

Pa33350

)(�

)�()�()��(22

21

=−=

+−+=−

ABl

AlABlBlHg

VV

gzpgzpgh

Then,

m266.0

81.9)80013560(

33350

)��(

33350

=

×−=

−=

gh

lHg


3.5 Tank Emptying For an emptying reservoir, discharging as a free jet, with free surface at a distance h above the discharging fluid, apply Bernoulli’s equation between the free surface and the jet:

2222

121

212

11

��gzUpgzUp ++=++

Here, hzzUppp atm =−=== 21121 ,0, , so that

ghzzgU �)(�� 21222

1 =−=

Hence, we have Torricelli’s Formula11 ghU exit 2= (41)

If the aperture is large (compared with the height to the free surface), then this value will be different for each streamline passing through the orifice (because each will have a different value of h). The total discharge would then have to be found by integration. Ideally the discharge would be

)orifice of area(2 ×= ghQideal (42)

This is not true in practice because of (i) frictional effects (small for a sharp-edged orifice) (ii) contraction (area of vena contract < area of orifice).

If these are significant then a coefficient of discharge cd may be introduced to compensate:

idealdQcQ =

cd must be measured experimentally. For a sharp-edged orifice, cd ≈ 0.6 – 0.65. Contraction effects can be reduced by using a bellmouth exit to minimise rapid changes in direction. However, frictional losses are then greater. Submerged Orifice

If the discharge is not a free jet but into a reservoir of the same fluid it is called a submerged orifice. In Bernoulli’s formula p2 is not then atmospheric, but given by 22

� ghp = . Torricelli’s formula then reads

)(2 21 hhgU exit −= (43)

Bernoulli’s formula is invalid in the reservoir/orifice problem if the free-surface level changes rapidly (since it is then time-dependent). If, however, the tank cross section is much larger than that of the orifice, then a quasi-steady approximation is OK and, by equating the

11 Evangelista Torricelli (1608-1647); Italian mathematician of barometer fame; served as secretary to Galileo.

h2h1

h

1

2


rate at which the volume of fluid in the tank decreases to the rate of discharge from the orifice,

Qt

V −=d

d (44)

we can find how long it takes to drain the tank. Example. A cylindrical tank of base diameter 0.5 m is used to store water. A rupture at the base of the tank allows water to escape through an aperture of area 8 cm2. A discharge coefficient of 0.6 can be assumed for this orifice. If the depth of water in the tank is initially 0.8 m, how long does it take to empty the tank? Solution. The volume of water in the tank is that of a cylinder of base diameter D = 0.5 m and (variable) height h. Its volume is, therefore,

hD

V4

� 2

=

This volume is reduced at a rate equal to the flow rate through the aperture, i.e. exitidealdideald AUcQcQ ==

where cd = 0.6, Aexit = 8×10-4 m2 and ghU ideal 2= by Torricelli’s formula.

Hence,

Qt

V −=d

d

� ghAchD

t exitd 2)4

�

(d

d 2

−=

� ght

h

Ac

D

exitd

2d

d

4

� 2

−=

� thhgAc

D

exitd

dd24

�

2/12

−=−

Integrating between t = 0 (where h = h0 = 0.8 m) and emptying time T (where h = 0):

T

hexitd

thgAc

D

0

0

2/12

0

224

�

��

��

�−=��

��

�

� Tg

h

Ac

D

exitd

−=−22

�

02

Hence,

s165

81.92

8.0

1086.02

5.0�

22

�

4

20

2

=×

××××

×== −g

h

Ac

DT

exitd


3.6 Summary of Methods For Incorporating Losses Many theoretical results are derived for ideal fluids, assuming no frictional losses, simplified geometry and uniform velocity profiles. In practice, compensation is necessary for non-ideal flow. These include the following Discharge coefficients Correct the quantity of flow deduced by Bernoulli’s equation:

idealdQcQ =

Loss coefficients Quantify energy or pressure losses in conduits:

g

VKH

2

� 2

−=

or, equivalently (and with the same K): )�(

� 221 VKP −=

(e.g. pipe friction: D

LK �= , where � is the friction factor).

Momentum and energy coefficients Account for non-uniform velocity profiles when computing total flux.

)�(�d�,

)�(�d�,

33

22

AuAufluxenergy

AuAufluxmomentum

av

av

=�

=�

mass momentum and energy

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