essential fluids

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FLUID MECHANICS {for energy conversion}

Keith Vaugh BEng (AERO) MEng

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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

OBJECTIVES

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BULK PROPERTIES OF

FLUIDS

ρ = mV

⎛⎝⎜

⎞⎠⎟

Density (ρ) - Mass per unit volume

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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

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STREAMLINES & STREAM-TUBES

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Fluid stream

Stream tube

MASS CONTINUITY

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Fluid stream

Stream tube

streamlines

MASS CONTINUITY

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Fluid stream velocity (u)

Stream tube

streamlines

MASS CONTINUITY

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Fluid stream velocity (u)

Stream tube

streamlines

!m = ρuA

MASS CONTINUITY

kgs

= kg

m3

⎛

⎝⎜

⎞

⎠⎟ m s⎛⎝⎜

⎞⎠⎟m2( )

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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

u2= u

1

A1

A2

= 2m s( ) 1.20.25

⎛⎝⎜

⎞⎠⎟= 9.6m s

EXAMPLE 1

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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

ENERGY OF A MOVING

LIQUID

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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;

For a stationary fluid, u = 0

ENERGY CONSERVATION

IDEAL FLUID

pρ+ gz + 1

2u2 = const

pρ+ gz = const

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BERNOULLI’s EQUATION FOR

STEADY FLOW

(Andrews & Jelley 2007)

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Work done in pushing elemental mass δm a small distance δs

δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t


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δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t

Similarly at P2

δW2= p

2A2u2δ t


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δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t

Similarly at P2

δW2= p

2A2u2δ t

The net work done is

δW2+δW

2= p

1A1u1δ t + p

2A2u2δ t


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δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t

Similarly at P2

δW2= p

2A2u2δ t


δW2+δW

2= p

1A1u1δ t + p

2A2u2δ t

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

p1A

1u

1δ t + p

2A

2u

2δ t =

δmg z2− z

1( )+ 12δm u

22 − u

12( )


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δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t

Similarly at P2

δW2= p

2A2u2δ t


δW2+δW

2= p

1A1u1δ t + p

2A2u2δ t


p1A

1u

1δ t + p

2A

2u

2δ t =

δmg z2− z

1( )+ 12δm u

22 − u

12( )

Putting δm = ρu1A

1δ t = ρu

2A

2δ t and tidying

p1

ρ+ gz

1+ 12u12 =p2

ρ+ gz

2+ 12u22


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δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t

Similarly at P2

δW2= p

2A2u2δ t


δW2+δW

2= p

1A1u1δ t + p

2A2u2δ t


p1A

1u

1δ t + p

2A

2u

2δ t =

δmg z2− z

1( )+ 12δm u

22 − u

12( )

Putting δm = ρu1A

1δ t = ρu

2A

2δ t and tidying

p1

ρ+ gz

1+ 12u12 =p2

ρ+ gz

2+ 12u22

Finally, since P1 and P

2 are arbitrary


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δ s = uδ t

δW1= p

1A1δ s1= p

1A1u1δ t

Similarly at P2

δW2= p

2A2u2δ t


δW2+δW

2= p

1A1u1δ t + p

2A2u2δ t


p1A

1u

1δ t + p

2A

2u

2δ t =

δmg z2− z

1( )+ 12δm u

22 − u

12( )

Putting δm = ρu1A

1δ t = ρu

2A

2δ t and tidying

p1

ρ+ gz

1+ 12u12 =p2

ρ+ gz

2+ 12u22

Finally, since P1 and P

2 are arbitrary

pρ+ gz + 1

2u2 = const


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No losses between stations ① and ②

z1+p1

ρg+u12

2= z

2+p2

ρg+u22

2(1)

①

②

z1

z2p1

p2

u1

u2

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Losses between stations ① and ②

z1+p1

ρg+u12

2= z

2+p2

ρg+u22

2(1)

z1+p1

ρg+u12

2= z

2+p2

ρg+u22

2+ h (2)

①

②

z1

z2p1

p2

u1

u2

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Losses between stations ① and ②

Energy additions or extractions

z1+p1

ρg+u12

2= z

2+p2

ρg+u22

2(1)

z1+p1

ρg+u12

2= z

2+p2

ρg+u22

2+ h (2)

z1+p1

ρg+u12

2+win

g= z

2+p2

ρg+u22

2+wout

g+ h (3)

①

②

z1

z2p1

p2

u1

u2

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(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 2

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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+½pU2 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.

EXAMPLE 3

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pρ+ gz + 1

2u2 = const

Bernoulli’s equation


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Applying bernoulli’s equation

pρ+ gz + 1

2u2 = const



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p0

ρ+ 12U2 =

ps

ρApplying bernoulli’s equation

pρ+ gz + 1

2u2 = const



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p0

ρ+ 12U2 =

ps

ρApplying bernoulli’s equation

pρ+ gz + 1

2u2 = const


Rearranging


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p0

ρ+ 12U2 =

ps

ρ

U =2 p

s− p

0( )ρ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

Applying bernoulli’s equation

pρ+ gz + 1

2u2 = const


Rearranging


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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

(Bacon, D., Stephens, R. 1990)

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The flow rate can be determine;

!V = a1u1+ a

2u2+ a

3u3+ ... = 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∑ × Δaπr 2

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

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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.

EXAMPLE 4

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p1

ρ+ 12u12 =p2

ρ+ 12u22

From Bernoulli’s equation


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p1

ρ+ 12u12 =p2

ρ+ 12u22


Also by mass continuity

u2= u

1

A1

A2 (Andrews & Jelley 2007)

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p1

ρ+ 12u12 =p2

ρ+ 12u22


Also by mass continuity

u2= u

1

A1

A2

Eliminating u2 we obtain the volume flow rate as

!V = A1u1= A

1A2A12 − A

22( )− 12⎛

⎝⎜⎞⎠⎟2 p

1− p

2( )ρ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

12


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p1

ρg+u12

2g=p2

ρg+u22

2g

Applying Bernoulli’s equation to stations ① and ②

ORIFICE PLATE


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p1

ρg+u12

2g=p2

ρg+u22

2g


Mass continuity

A1u1= A

2u2⇒ u

2= u

1

A1

A2

ORIFICE PLATE


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p1

ρg+u12

2g=p2

ρg+u22

2g


Mass continuity

A1u1= A

2u2⇒ u

2= u

1

A1

A2

∴ !V = A1u1= A

1A2A12 − A

22( )− 12⎛

⎝⎜⎞⎠⎟2 p

1− p

2( )ρ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

= k h

ORIFICE PLATE


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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.

EXAMPLE 5

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Mass flow rate - !m = ρ !V⇒∴ !V=0.033m3 s


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Mass flow rate - !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


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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 h0.63× 0.007244 h = 0.0333 ∴h = 53.24m of air


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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 h0.63× 0.007244 h = 0.0333 ∴h = 53.24m of air

Equating pressures at Level XX in the U-tube 1.2g × 53.34 =103 g × x∴ x = 64 ×103m


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

DYNAMICS OF A VISCOUS FLUID


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u y( ) = U yd

(0 ≤ y ≤ d)


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u y( ) = U yd

(0 ≤ y ≤ d)

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


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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 ud

μ - coefficient of dynamic viscosity


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Laminar flow in a pipe Turbulent flow in a pipe

Typical flow regimes


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Laminar flow in a pipe Turbulent flow in a pipe

Typical flow regimes

Reynolds Number, Re = ρULµ

= ULv


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Flow around a cylinder for an inviscid fluid

Flow around a cylinder for a viscous fluid


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Lift and Drag


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Lift and Drag

Lift = 12CLρU2A


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Lift and Drag

Lift = 12CLρU2A Drag = 1

2CDρU2A


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Variation of the lift and drag coefficients CL and CD with angle of attack, (Andrews & Jelley 2007)

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Bulk properties of fluidsStreamlines & stream-tubesMass continuityEnergy conservation in an ideal fluidBernoulli’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

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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 FrancisYoung, D., Munson, B., Okiishi, T., Huebsch, W., 2011 Introduction to Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc.

Illustrations taken from Energy science: principles, technologies and impacts & Mechanical Technology

essential fluids

Engineering

u1 a1 t

u2 a2 t

z1 z2 p1 p2 u1 u2

u1 a1 a2

p1 a1 u1 t andrews jelley

p2 a2 u2 t andrews jelley

p1 a1 s1

u2a2 u2