FLUID MECHANICS{for energy conversion}

Keith Vaugh BEng (AERO) MEng SME ASME

}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.

BULK PROPERTIES OF FLUIDS

Pressure (P) - Force per unit area in a fluid

Viscosity - Force per unit area due to internal friction

ρ = mV

⎛⎝⎜

⎞⎠⎟Density (ρ) - Mass per unit volume}

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.

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.

MASS CONTINUITY

Fluid stream velocity (u)

Stream tube

streamlines

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.

EXAMPLE 1An 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

u2 = u1A1A2

= 2 ms( ) 1.20.25

⎛⎝⎜

⎞⎠⎟ = 9.6

ms

}The potential energy (PE) of liquid per unit mass

ENERGY OF A MOVING LIQUID

and kinetic energy (KE) of liquid per unit mass

total energy of liquid per unit mass

PE = g z + pρg

⎛⎝⎜

⎞⎠⎟

KE = u2

2

Etotal = g z + pρg

+ u2

2g⎛⎝⎜

⎞⎠⎟

z

pρg

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.

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;

pρ+ gz + 1

2u2 = const

For a stationary fluid, u = 0

pρ+ gz = const

ENERGY CONSERVATION IN AN IDEAL FLUID}

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.

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

Work done in pushing elemental mass δm a small distance δs

δs = uδt

δW1 = p1A1δs1 = p1A1u1δt

Similarly at P2

δW2 = p2A2u2δt

The net work done is

δW2 +δW2 = p1A1u1δt + p2A2u2δt

By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore;

p1A1u1δt + p2A2u2δt =

δmg z2 − z1( ) + 12δm u2

2 − u12( )

Putting δm = ρu1A1δt = ρu2A2δt and tidying

p1ρ+ gz1 +

12u12 = p2

ρ+ gz2 +

12u22

Finally, since P1 and P2 are arbitrary

pρ+ gz + 1

2u2 = const

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

No losses between stations ① and ②

Losses between stations ① and ②

Energy additions or extractions

z1 +p1ρg

+ u12

2= z2 +

p2ρg

+ u22

2 (1)

z1 +p1ρg

+ u12

2= z2 +

p2ρg

+ u22

2+ h (2)

z1 +p1ρg

+ u12

2+ win

g= z2 +

p2ρg

+ u22

2+ wout

g+ h (3)

①

②

z1

z2p1

p2

u1

u2

Consider stations ① and ② 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 ① and ② 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

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)

}

EXAMPLE 2A 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.

}

p0ρ

+ 12U 2 = ps

ρ

U =2 ps − p0( )

ρ⎡

⎣⎢

⎤

⎦⎥

12

Applying bernoulli’s equation

pρ+ gz + 1

2u2 = const

Bernoulli’s equation

Rearranging

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.

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.

VELOCITY DISTRIBUTION & FLOW RATE}

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

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 =

VA=

au∑A

The velocity profile for a circular pipe is the same across any diameter

Δ V = ur × Δa where ur is the velocity at radius r

V = ur∑ × Δa and V =

ur∑ × Δaπr2

The velocity at the boundary of any duct or pipe is zero.

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.

}

p1ρ+ 12u12 = p2

ρ+ 12u22

From Bernoulli’s equation

Also by mass continuity

u2 = u1A1A2

Eliminating u2 we obtain the volume flow rate as

V = A1u1 = A1A2 A1

2 − A22( )−12⎛

⎝⎞⎠2 p1 − p2( )

ρ⎡

⎣⎢

⎤

⎦⎥

12

Applying Bernoulli’s equation to stations ① and ②

z1 +p1ρg

+ u12

2= z2 +

p2ρg

+ u22

2

Given that meter is horizontal, this equation reduces to:

p1ρg

+ u12

2= p2ρg

+ u22

2

Mass continuity A1u1 = A2u2 ⇒ u2 = u1A1A2

∴ p1ρg

− p2ρg

= h = u12

2gA1A2

⎛⎝⎜

⎞⎠⎟

2

− u12

2g= u1

2

2gA12 − A2

2

A22

⎛⎝⎜

⎞⎠⎟

① ②

u1 u2

h

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

∴u1 =A2

A12 − A2

2( )2gh( )

V = A1u1 =A1A2

A12 − A2

2( )2gh( ) = k h since A1 and A2 are constants

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 = cdk h

The pressure differential between the main and throat sections is generally measured with a mercury-and-water U-tube, therefore

If a Venturi meter is inclined to the horizontal, i.e station ① has a height z1

and station ② a height z2 then;

h = x S −1( )

V = k h − z2 − z1( )( )

Show that the manometer reading for an inclined Venturi is the same as for a horizontal Venturi for a given flow rate.

ORIFICE PLATE

}p1ρg

+ u12

2= p2ρg

+ u22

2

Applying Bernoulli’s equation to stations ① and ②

Mass continuity

A1u1 = A2u2 ⇒ u2 = u1A1A2

∴ V = A1u1 = A1A2 A1

2 − A22( )−12⎛

⎝⎞⎠2 p1 − p2( )

ρ⎡

⎣⎢

⎤

⎦⎥

12

= k h

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

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.

}

Mass flow rate -

The actual discharge is

m = ρ V⇒∴ V=0.033m3 s

V = k h

V =

π × 0.052

4⎛⎝⎜

⎞⎠⎟

π × 0.042

4⎛⎝⎜

⎞⎠⎟

π × 0.042

4⎛⎝⎜

⎞⎠⎟

2

− π × 0.042

4⎛⎝⎜

⎞⎠⎟

2⎛

⎝⎜

⎞

⎠⎟

122 × 9.81× h

V = 0.007244 h

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 = 103g × x∴ x = 64 ×103m

The motion of a viscous fluid is more complicated than that of an inviscid fluid.

DYNAMICS OF A VISCOUS FLUID}

u y( ) =U yd

(0 ≤ y ≤ d)

The viscous shear force per unit area in the fluid is proportional to the velocity gradient

FA= −µ du

dy= µ u

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

Laminar flow in a pipe Turbulent flow in a pipe

Typical flow regimes

Reynolds Number, Re = ρULµ

= ULv

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

Flow around a cylinder for an inviscid fluid

Flow around a cylinder for a 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.

Lift and Drag

Lift = 12CLρU2A Drag = 12CDρU

2A

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.

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.

Bulk properties of fluidsStreamlines & stream-tubesMass continuityEnergy conservation in an ideal fluidBernoulli’s equation for steady flow

Applied example’s (pitot tube, Venturi meter & questionsDynamics of a viscous fluid

Flow regimesLaminar and turbulent flowVortices Lift & Drag

Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and impacts, Oxford University PressBacon, D., Stephens, R. (1990) Mechanical Technology, second edition, Butterworth HeinemannBoyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University PressÇengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw HillTurns, S. (2006) Thermal fluid sciences: An integrated approach, Cambridge University PressTwidell, 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