1. FLUID MECHANICS
{for energy conversion}
Keith Vaugh BEng (AERO) MEng SME ASME

2. OBJECTIVES
Identify the unique vocabulary associated with
fluid mechanics with a focus on energy
conservation
}
Explain the physical properties of fluids and
derive the conservation laws of mass and energy
for an ideal fluid (i.e. ignoring the viscous effects)
Develop a comprehensive understanding of the
effect of viscosity on the motion of a fluid flowing
around an immersed body
Determine the forces acting on an immersed body
arising from the flow of fluid around it
To appreciate energy conversion such as hydro, wave, tidal and wind power a detailed
knowledge of fluid mechanics is essential.
During the course of this lecture, a brief summary of the basic physical properties of fluids
is provided and the conservation laws of mass and energy for an ideal (or inviscid) fluid are
derived.
The application of the conservation laws to situations of practical interest are also explored
to illustrate how useful information about the flow can be derived.
Finally, the effect of viscosity on the motion of a fluid around an immersed body (such as a
turbine blade) and how the flow determines the forces acting on the body of interest.

3. BULK PROPERTIES OF
FLUIDS
Density (ρ) - Mass per unit volume
⎛
⎜
⎝
m⎞
ρ= ⎟
V⎠
}
Pressure (P) - Force per unit area in a fluid
Viscosity - Force per unit area due to internal friction
Density - Mass per unit volume of a fluid. For the purposes of this module,
Density is assumed to be constant i.e. that the flow of the fluid is
incompressible (small variations in pressure arising from fluid motion in
comparison to atmospheric pressure).
Pressure - Force per unit area, and acts in the normal direction to the surface of
a body in a fluid. Its units are the pascal or N-m^2
Viscosity - Force per unit area due to internal friction in a fluid arising from the
relative motion between neighbouring elements in a fluid. Viscous forces act in
the tangential direction to the surface of a body immersed in a flow. (consider a
deck of cards.

4. STREAMLINES &
STREAM-TUBES
}
A useful concept to visualize a velocity field is to imagine a set of streamlines
para;;e; to the direction of motion at all points in the fluid. Any element of mass
in the fluid flows along a notational stream-tube bounded by neighbouring
streamlines. In practice, streamlines can be visualised by injecting small
particles into the fluid e.g. smoke can be used in wind tunnels to visualise fluid
flow around objects.

5. MASS CONTINUITY
Stream tube
streamlines
}
velocity (u)
Fluid stream
ρuA = const
Also known as conservation of mass.
Consider flow along a stream tube in a steady velocity field. The direction of
flow is parallel to the boundries and at any point within the stream tube, the
speed of the fluid (u) and the cross-sectional area (A). Therefore the fluid is
confined to the stream tube and the mass flow per second is constant. Therefore:
Density * Velocity * Cross-sectional Area = Constant
Therefore the speed of fluid is inversely proportional to the cross-sectional area
of the stream tube.

6. EXAMPLE 1
An incompressible ideal fluid flows at a speed of 2 ms-1
through a 1.2 m2 sectional area in which a constriction of
}
0.25 m2 sectional area has been inserted. What is the speed
of the fluid inside the constriction?
Putting ρ1 = ρ2, we have u1A1 = u2A2
A1 ⎛ 1.2 ⎞
u2 = u1 = (2 m s )⎜ ⎟ = 9.6 m s
A2 ⎝ 0.25 ⎠

7. ENERGY OF A MOVING
LIQUID
The potential energy (PE) of liquid
per unit mass
⎛ p⎞
}
PE = g ⎜ z +
ρg ⎟
p
⎝ ⎠
ρg
and kinetic energy (KE) of liquid per
unit mass
u2
z KE =
2
total energy of liquid per unit mass
⎛ p u2 ⎞
Etotal = g⎜ z + +
⎝ ρ g 2g ⎟
⎠
Consider a liquid at a pressure p, moving with a velocity (u) at a height Z above
datum level
If a tube were inserted in the top of the pipe, the liquid would rise up the tube a
distance of p/ρg and this is equivalent to an additional height of liquid relative
to datum level.
Etotal equation represents the specific energy of the liquid. Each of the quantities
in the brackets have the units of length and are termed heads, i.e Z is referred to
as the potential head of the liquid, p/ρg as the pressure head and V2/2g as the
velocity head.

8. ENERGY CONSERVATION IN
AN IDEAL FLUID
In many practical situations, viscous forces are much smaller that
those due to gravity and pressure gradients over large regions of the
flow field. We can ignore viscosity to a good approximation and
}
derive an equation for energy conservation in a fluid known as
Bernoulli’s equation. For steady flow, Bernoulli’s equation is of the
form;
p 1
+ gz + u 2 = const
ρ 2
For a stationary fluid, u = 0
p
+ gz = const
ρ
For a stationary fluid, u = 0 everywhere in the fluid therefore this equation
reduces to
which is the equation for hydrostatic pressure. It shows that the fluid at a given
depth is all the same at the same pressure.

9. BERNOULLI’s EQUATION
FOR STEADY FLOW
Consider the steady flow of an ideal fluid in the control volume shown.
The height (z), cross sectional area (A), speed (u) and pressure are denoted.
The increase in gravitational potential energy of a mass δm of fluid between z1
and z2 is δmg(z2-z1)
In a short time interval (δt) the mass of fluid entering the control volume at P1
is δm=ρu1A1δt and the mass exiting at P2 is δm=ρu2A2δt

10. By energy conservation, this is equal to
the increase in Potential Energy plus the
increase in Kinetic Energy, therefore;
p1 A1u1δ t + p2 A2u2δ t =
Work done in pushing elemental mass δm
δ mg ( z2 − z1 ) + δ m ( u2 − u12 )
1 2
a small distance δs
2
δ s = uδ t
Putting δ m = ρu1 A1δ t = ρu2 A2δ t and tidying
δW1 = p1 A1δ s1 = p1 A1u1δ t p1 1 p 1 2
+ gz1 + u12 = 2 + gz2 + u2
Similarly at P2 ρ 2 ρ 2
δW2 = p2 A2u2δ t
Finally, since P1 and P2 are arbitrary
The net work done is
p 1
δW2 + δW2 = p1 A1u1δ t + p2 A2u2δ t + gz + u 2 = const
ρ 2
In order for the fluid to enter the control volume it has to do work to overcome
the pressure p1 exerted by the fluid. The work done in pushing the elemental
mass δm a small distance δs = uδt
at P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity *
Change in time. Remember F = P*A
Similarly, the work done in pushing the elemental mass out of the control
volume at P2 is δW2=-p2A2u2δt.
The net work done is δW1+δW2

11. No losses between stations and
u2 p1 u12 2
p2 u2
p2 z1 + + = z2 + + (1)
ρg 2 ρg 2
u1
p1 Losses between stations and
z2
z1
p1 u12 2
p2 u2
z1 + + = z2 + + +h (2)
ρg 2 ρg 2
Energy additions or extractions
p1 u12 win 2
p2 u2 wout
z1 + + + = z2 + + + +h (3)
ρg 2 g ρg 2 g
Consider stations and of the inclined pipe illustrated. If there are no
losses between these two sections, and no energy changes resulting from heat
transfer of work, then the total energy will remain constant, therefore stations
and can be set equal to one another, equ (1)
If however losses do occur between the two stations, then equation equ (2)
When energy is added to the fluid by a device such as a pump, or when energy
is extracted by a turbine, these need also be accounted for equ (3). w is the
specific work in J/kg
NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation

12. QUESTION’s
(a) The atmospheric pressure on a surface is 105 Nm-2.
Assuming the water is stationary, what is the pressure
at a depth of 10 m ?
}
(assume ρwater = 103 kgm-3 & g = 9.81 ms-2)
(b) What is the significance of Bernoulli’s equation?
(c) Assuming the pressure of stationary air is 105 Nm-2,
calculate the percentage change due to a wind of
20ms-1.
(assume ρair ≈ 1.2 kgm-3)

13. EXAMPLE 2
}
A Pitot tube is a device for measuring the velocity in a fluid.
Essentially, it consists of two tubes, (a) and (b). Each tube
has one end open to the fluid and one end connected to a
pressure gauge. Tube (a) has the open end facing the flow
and tube (b) has the open end normal to the flow. In the case
when the gauge (a) reads p = po+pU2 and gauge (b) reads
p = po, derive an expression for the velocity of the fluid in
terms of the difference in pressure between the gauges.

14. Bernoulli’s equation
p 1
+ gz + u 2 = const
ρ 2
p0 1 2 ps
Applying bernoulli’s equation + U =
ρ 2 ρ
⎡ 2 ( ps − p0 ) ⎤
1
2
Rearranging U=⎢ ⎥
⎣ ρ ⎦
Consider the fluid flowing along the stream-line AB. In case (a), the fluid slows
down as it approaches the stagnation point B. Putting u = U, p = po at A and u
= 0, p = ps at B and applying bernoulli’s equation we get
In example (and the previous question), ps is measured by tube (a) and p0
measured by tube (b).
NOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2,
p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and total
pressure respectively.

15. VELOCITY DISTRIBUTION
& FLOW RATE
The velocity in a flow stream varies considerably over the cross-
section. A pitot tube only measures at one particular point,
}
therefore, in order to determine the velocity distribution over the
entire cross section a number of readings are required.
The velocity profile is obtained by traversing the pitot tube along the line AA.
The cross-section is divided into convenient areas, a1, a2, a3, etc... and the
velocity at the centre of each area is determined

16. The flow rate can be determine;
V = a1u1 + a2u2 + a3u3 + ... = ∑ au

If the total cross-sectional area if the section is A;
the mean velocity, U =

V
=
∑ au
A A
The velocity profile for a circular pipe is the same across any diameter

ΔV = ur × Δa where ur is the velocity at radius r
V = ∑ ur × Δa
 and V=
∑u r × Δa
πr2
The velocity at the boundary of any duct or pipe is zero.

17. EXAMPLE 3
In a Venturi meter an ideal fluid flows with a volume flow
rate and a pressure p1 through a horizontal pipe of cross
}
sectional area A1. A constriction of cross-sectional area A2 is
inserted in the pipe and the pressure is p2 inside the
constriction. Derive an expression for the volume flow rate
in terms of p1, p2, A1 and A2.

18. From Bernoulli’s equation
p1 1 2 p2 1 2
+ u = + u
ρ 2 1 ρ 2 2
Also by mass continuity
A1
u2 = u1
A2
Eliminating u2 we obtain the volume flow rate as
⎞ ⎡ 2 ( p1 − p2 ) ⎤
1
2
(
V = A1u1 = ⎛ A1 A2 A12 − A )
1
 2 − 2
⎝ 2 ⎠⎢ ρ ⎥
⎣ ⎦

19. Applying Bernoulli’s equation
to stations and
h
p1 u12 2
p2 u2
z1 + + = z2 + +
ρg 2 ρg 2
Given that meter is horizontal,
u1 u2 this equation reduces to:
p1 u12 p2 u2
2
+ = +
ρg 2 ρg 2
A1
Mass continuity A1u1 = A2u2 ⇒ u2 = u1
A2
2
p1 p2 u12 ⎛ A1 ⎞ u12 u12 ⎛ A12 − A2 ⎞
2
∴ − =h= − =
ρg ρg 2g ⎜ A2 ⎟
⎝ ⎠ 2g 2g ⎜ A2 ⎟
⎝ 2
⎠
A venturi meter is a device which is used to measure the rate of flow through a
pipe.
NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation

20. A2
∴u1 = ( 2gh )
( A12 − A2
2
)
A1 A2

V = A1u1 = ( 2gh ) = k h since A1 and A2 are constants
(A 2
1 −A 2
2 )
The pressure at the throat is slightly lower than the the theoretical due to
friction in the tapered pipe. As a consequence h becomes slightly larger
and therefore resulting in volumetric flow rate which is too high. To
compensate for this, a coefficient of discharge, cd is introduced.

V = cd k h

21. The pressure differential between the main and throat sections is generally
measured with a mercury-and-water U-tube, therefore
h = x ( S − 1)
If a Venturi meter is inclined to the horizontal, i.e station has a height z1
and station a height z2 then;

V= k (h − ( z 2 − z1 ) )
Show that the manometer reading for an inclined Venturi is the same as for
a horizontal Venturi for a given flow rate.

22. ORIFICE PLATE
Applying Bernoulli’s equation to
stations and
p1 u12 p2 u2
2
}
+ = +
ρg 2 ρg 2
Mass continuity
A1
A1u1 = A2u2 ⇒ u2 = u1
A2
⎞ ⎡ 2 ( p1 − p2 ) ⎤
1
2
(
∴ V = A1u1 = ⎛ A1 A2 A12 − A )
1
 2 − 2
=k h
⎝ 2 ⎠⎢ ρ ⎥
⎣ ⎦
An Orifice plate is another means of determining the fluid flow in a pipe and
works on the same principle of the Venturi meter.
The position of the Vena contract can be difficult establish and therefore the
area A2. The constant k is determined experimentally when it will incorporate
the coefficient of discharge.
If h is small so that ρ is approximately constant then the this equation can be
used for compressible fluid flow.
NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation

23. EXAMPLE 4
In an engine test 0.04 kg/s of air flows through a 50 mm
diameter pipe, into which is fitted a 40 mm diameter Orifice
}
plate. The density of air is 1.2 kg/m3 and the coefficient of
discharge for the orifice is 0.63. The pressure drop across
the orifice plate is measured by a U-tube manometer.
Calculate the manometer reading.

24.  
Mass flow rate - m = ρV ⇒∴ V=0.033 m s

3

V= k h
⎛ π × 0.05 2 ⎞ ⎛ π × 0.04 2 ⎞
⎜
⎝ 4 ⎟⎜
⎠⎝ 4 ⎟
⎠

V= 1 2 × 9.81 × h
⎛ ⎛ π × 0.04 2 ⎞ 2 ⎛ π × 0.04 2 ⎞ 2 ⎞
2
⎜⎜
⎝ 4 ⎟ −⎜
⎠ ⎝ 4 ⎟ ⎟
⎠ ⎠
⎝

V = 0.007244 h
The actual discharge is 0.63 × 0.007244 h = 0.0333 ∴ h = 53.24m of air
Equating pressures at Level XX in the U-tube 1.2g × 53.34 = 10 3 g × x
∴ x = 64 × 10 3 m

25. DYNAMICS OF A VISCOUS
FLUID
The motion of a viscous fluid is more complicated than that of
an inviscid fluid. }
The viscous shear force per
unit area in the fluid is
proportional to the velocity
gradient
y
u ( y) = U F du u
d = −µ =µ
A dy d
(0 ≤ y ≤ d)
µ - coefficient of dynamic
viscosity
Let us consider a simple case of laminar flow between two parallel plates
separated by a small distance d.
The upper plate moves at a constant velocity U while the lower plate remains at
rest. At the plate fluid interface in both cases there is no velocity due to the
strong forces of attraction. Therefore the velocity profile in the fluid is given by

26. Typical flow regimes
Laminar flow in a pipe Turbulent flow in a pipe
ρUL UL
Reynolds Number, Re = =
µ v
A viscous fluid can exhibit two different kinds of flow regimes, Laminar and
Turbulent
In laminar flow, the fluid slides along distinct stream-tubes and tends to be quite
stable, Turbulent flow is disorderly and unstable
The flow that exists in an y given situation depends on the ratio of the inertial
force to the viscous force. The magnitude of this ratio is given by Reynolds
number, where U is the velocity, L is the length and v = µ/ρ is the kinematic
viscosity. Reynolds observed that flows at small Re where laminar while flow at
high Re contained regions of turbulence

27. Flow around a cylinder Flow around a cylinder for a
for an inviscid fluid viscous fluid
For the inviscid flow the velocity fields in the upstream and downstream
regions are symmetrical, therefore the corresponding pressure distribution is
symmetrical. It follows that the net force exerted by the fluid on the cylinder is
zero. This is an example of d’Alembert’s paradox
For a body immersed in a viscious fluid, the component of velocity tangential to
the surface of the fluid is zero at all points. At large Re numbers (Re>>1), the
viscous force is negligible in the bulk of the fluid but us very significant in a
viscous boundary layer close to the surface of the body. Rotational components
of flow know as vorticity are generated within the boundary layer. As a certain
point, the seperation point, the boundary layer becomes detached from the
surface and vorticity are discharged into the body of the fluid. The vorticity are
transported downstream side of the cylinder in the wake. Therefore the pressure
distributions on the upstream and downstream sides of the cylinder are not
symmetrical in the case of a viscous fluid. As a result, the cylinder experiences
a net force in the direction of motion known as drag. In the case of a spinning
cylinder, a force called life arises at right angles to the direction of flow.

28. Lift and Drag
Lift = 1 2 C L ρU 2 A Drag = 1 2 C D ρU 2 A
CL and CD are dimensionless constants know as the coefficient of Lift and the
coefficient of Drag respectively.
Lift and drag can be changed by altering the shape of a wing i.e. ailerons. For
small angles of attack, the pressure distribution ont he upper surface of an
aerofoil is significantly lower than that on the lower surface which results is in a
net lift force.

29. Variation of the lift and drag coefficients
CL and CD with angle of attack
CL and CD are dimensionless constants know as the coefficient of Lift and the
coefficient of Drag respectively.
Lift and drag can be changed by altering the shape of a wing i.e. ailerons. For
small angles of attack, the pressure distribution on the upper surface of an
aerofoil is significantly lower than that on the lower surface which results is in a
net lift force.

30. Bulk properties of fluids
Streamlines & stream-tubes
Mass continuity
Energy conservation in an ideal fluid
Bernoulli’s equation for steady flow
 Applied example’s (pitot tube, Venturi meter & questions
Dynamics of a viscous fluid
 Flow regimes
 Laminar and turbulent flow
 Vortices
 Lift & Drag

31. Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and
impacts, Oxford University Press
Bacon, D., Stephens, R. (1990) Mechanical Technology, second edition,
Butterworth Heinemann
Boyle, G. (2004) Renewable Energy: Power for a sustainable future, second
edition, Oxford University Press
Çengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid
sciences, Third edition, McGraw Hill
Turns, S. (2006) Thermal fluid sciences: An integrated approach, Cambridge
University Press
Twidell, J. and Weir, T. (2006) Renewable energy resources, second edition,
Oxon: Taylor and Francis
Illustrations taken from Energy science: principles, technologies and impacts & Fundamentals of thermal fluid science