7/30/2014

1

TOPIC 1: MASS, MOMENTUM AND

ENERGY

OBJECTIVES

1. Extend continuity and momentum principles to

non-uniform velocity

2. Apply continuity and Bernoulli’s equation to flow

measurement and tank-emptying

3. Learn methods for dealing with non-ideal flow

NOTATION

Geometry

x (x, y, z) position

t time

Field Variables

u (u, v, w) velocity (V for average speed in a duct)

p pressure

p – patm gauge pressure

p* = p + ρgz piezometric pressure

T temperature

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FLUID PROPERTIES

ρ density

γ ρg specific weight

s.g. ρ/ρref specific gravity (or relative density);

μ dynamic viscosity

ν μ/ρ kinematic viscosity

σ surface tension

K bulk modulus

k conductivity of heat

c speed of sound

NON-DIMENSIONAL PARAMETERS

Reynolds number

Froude number

νμ

ρRe

ULUL

gL

UFr

DEFINITIONS

Fluids / solids

Liquids / gases

Hydrostatics / hydrodynamics

Hydraulics / aerodynamics

Incompressible / compressible

Ideal / real

Newtonian / non-Newtonian

Laminar / turbulent

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STATICS

Hydrostatics zgp ΔρΔ

gz

pρ

d

d

0)ρ(Δ gzp

Ideal Gas Law RTp ρ

DYNAMICS

Continuity (mass conservation) Mass is conserved

For steady flow:

(mass flux)in = (mass flux)out

Momentum principle Force = rate of change of momentum

For steady flow:

Force = (momentum flux)out – (momentum flux)in

Energy principle Change in energy = heat supplied + work done

For incompressible flow:

Change of kinetic energy = work done

BERNOULLI’S EQUATION

streamline a alongconstant 2

21 ρρ Ugzp

Assumes:

no losses

incompressible

steady

streamlineaalongconstantg

Uz

g

p

2ρ

2

energy (per unit mass)

energy (per unit weight)

total head

Losses

Often quantified in terms of a change in head:

headinchangeg

Uz

g

p )

2ρ(Δ

2

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PROPERTIES OF AIR AND WATER

Air:

Density: ρ ≈ 1.2 kg m–3

(Dynamic) viscosity: μ ≈ 1.810–5 kg m–1 s–1 (or Pa s)

Kinematic viscosity: ν ≈ 1.510–5 m2 s–1

Water:

Density: ρ ≈ 1000 kg m–3

(Dynamic) viscosity: μ ≈ 1.010–3 kg m–1 s–1

Kinematic viscosity: ν ≈ 1.010–6 m2 s–1

Properties vary with temperature

QUESTIONS ON HYDROSTATICS

Question

What is an atmospheric pressure of 1 bar equivalent to in:

(a) metres of water;

(b) millimetres of mercury?

Question

The Renold building is 8 storeys high, with each storey

having height 4 m. Find the pressure difference between

ground and roof.

Question

If sea-level pressure is 1 bar and the lower atmosphere is

isothermal, with temperature T = 298 K, calculate the

pressure at a height of 10 km.

EXAMPLE SHEET, Q1

A long bridge with piers 1.5 m wide, spaced 8 m between

centres, crosses a river. The depth of water upstream is

1.6 m and between the piers is 1.45 m.

Calculate the volume flow rate under one arch, assuming

that the river bed is horizontal, the banks are parallel and

frictional effects are negligible.

Find the maximum height to which water rises at the front

of the piers.

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FLOW RATE

Volume flow rate: uAQ

Mass flow rate: uAQm ρρ

uA

EXAMPLE, PAGE 8

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.

(a) Determine the velocity of flow of the mixture at the exit.

(b) Determine the density of the mixture at the exit.

0.5 m

2 m/s

4 m/s

0.5 m

0.5 m

p =15 kN/m1

2

CONTINUITY

(MASS OR VOLUME CONSERVATION)

In steady flow: total flow in = total flow out

Volume flow rate, Q

uAQ (uniform flow)

AuQ d (non-uniform flow)

An integral is just a sum!

uA

outin

uAuA

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SPECIAL CASES

2-dimensional:

Axisymmetric:

ywA dd

yuwQ d

rrA dπ2d

rruQ dπ2

AuQ d

w

dyu(y)

drr

u(r)

EXAMPLE, PAGE 9

The distribution of velocity in a rectangular channel of

width w = 800 mm and depth h = 200 mm is given by

where u0 = 8 m s–1. What is:

(a) the quantity of flow;

(b) the average velocity?

7

1

0

h

yuu

AVERAGE VELOCITY

For non-uniform velocity you will need to find

the volume flow rate in order to calculate the

average velocity

AuQ avarea

rateflow

A

Quav

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EXAMPLE, PAGE 10

Fully-developed laminar flow in a pipe of radius R

has velocity profile:

Find the average velocity in terms of u0.

)/1( 22

0 Rruu

MOMENTUM

1. Newton’s 2nd Law (“Momentum Principle”)

force = rate of change of momentum

applied to all the fluid passing through a control volume.

2. Newton’s 3rd Law (“Action / Reaction”)

The force of a fluid on its containment … is equal and

opposite … to the reaction of the containment on the fluid.

3. Force and momentum are vector quantities.

MOMENTUM PRINCIPLE IN CONTINUOUS

FLOW

For steady flow and fixed control volume:

momentumofchangeofrateforce

)( inoutρQ uuF

Momentum flux = (mass flux) x (velocity) = ρQu

= (ρuA)u if uniform

= (ρuA)u if non-uniform

F

uout

uin

time

velocity in change enteringmass

velocity in changefluxmass

inout ρQρQ )()( uu

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MOMENTUM PRINCIPLE FOR STEADY FLOW

Force = (momentum flux)out – (momentum flux)in

Force = (rate that momentum leaves CV) – ( rate that momentum enters CV)

Momentum flux = (ρuA)u For non-uniform flow ... sum (integrate)

In 1-d: ρu2A

EXAMPLE, PAGE 8 (REPRISE)

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.

(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?

0.5 m

2 m/s

4 m/s

0.5 m

0.5 m

p =15 kN/m1

2

FORCES ON FLUIDS

Body forces (proportional to volume)

– weight

– centrifugal and Coriolis forces

Surface forces (proportional to area)

– pressure forces

– viscous forces

Reaction forces (from solid boundaries)

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SURFACE FORCES

Surface forces are usually expressed in terms of stress:

area

forcestress areastressforce

Pressure, p

force = pressure area

net force (x direction) = (pL − pR)A

Shear stress, τ

force = shear stress area

net force (x direction) = (τT − τB)A

Ap A

Rp A

L

A T

B

A

A

VISCOUS STRESS

y

u

d

dμτ

Note: what we usually call stress, τ, is actually the result of the

net effect of turbulent fluctuations, not molecular viscosity.

yu(y)

BOUNDARY LAYERS

What is a boundary layer?

A layer of slow-moving fluid close to a solid boundary

Occurs because viscosity imposes a no-slip condition

For high Re/small viscosity it is usually extremely thin

Free stream

Boundary layer

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FLOW SEPARATION

Requires:

(1) a layer of slow-moving fluid (boundary layer)

(2) free-stream deceleration (adverse pressure gradient)

There is near-surface flow reversal

To satisfy continuity, flow breaks away from the surface

adverse pressuregradient

backflow

flowseparation

speeds up ... ... slows down

FLOW SEPARATION

Occurs on sufficiently convex surfaces

Almost inevitable at sharp corners

A viscous effect … but causes large pressure drag

H L

Highpressure

Lowpressure

LH H

LH H

LH H

H LH L

LAMINAR vs TURBULENT BOUNDARY

LAYERS Laminar Turbulent

Adjacent layers don’t mix

Momentum transfer by

viscous stress:

Blasius velocity profile

Adjacent layers mix (a lot)

Mean momentum transfer by

net effect of mixing:

Logarithmic mean-velocity profile

Important:

Turbulent boundary layers are less likely to separate

Controlling separation can greatly reduce drag

y

u

d

dμτ

y

u

d

dμτ

u(y)

y

u(y)

y

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FORCES ON OBJECTS

Drag = component of force parallel to approach flow.

Lift = component of force perpendicular to approach flow.

U0

lift F

drag

DRAG AND LIFT COEFFICIENTS

Drag coefficient

Lift coefficient

AUcD 2

021 ρ

drag

AUcL 2

021 ρ

lift

SOURCE OF DRAG FORCE

Drag coefficient AU

cD 2

021 ρ

drag

Bluff bodies

force is predominantly pressure drag

A is projected area (normal to flow)

cD = O(1)

Streamlined bodies

force is predominantly viscous drag

A is plan area (parallel to flow)

cD << 1

AU0

AU0

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MOMENTUM FLUX

uuA)ρ(

velocityfluxmassfluxmomentum

Au dρ 2flux momentum

For a continuously-varying profile:

MOMENTUM FLUX

Au dρ 2flux momentum

Velocity profile Area elements Momentum flux

Uniform

Area A

2-dimensional

dA = w dy

Axisymmetric

dA = 2πr dr

AU 2ρ

yuw dρ 2

drru π2ρ 2

w

dyu(y)

drr

u(r)

EXAMPLE, PAGE 18

A cylinder spans a wind tunnel of rectangular cross section and height h = 0.3 m, as shown in the

figure. The spanwise width w = 0.6 m and the cylinder diameter is 90 mm. The upstream velocity is

uA and is uniform. The velocity profile measured a short distance downstream of the cylinder is

symmetric about the centreline and is given by

where u is the velocity in m s–1 and y is the distance from the centreline in m.

(a) Assuming that the downstream velocity profile has no discontinuities, what is the value of uB?

(b) Calculate the upstream velocity uA.

(c) Assuming that Bernoulli’s theorem is applicable outside the wake of the cylinder calculate the

pressure difference between upstream and downstream sections.

(d) Neglecting drag on the walls of the tunnel, calculate the total drag force on the cylinder.

(e) Define a suitable drag coefficient and calculate its value.

1.0if,

1.0if,1.0

41.0

6110

32

yu

yyy

u

B

y

h

cylinderuA

uB

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EXAMPLE, PAGE 18

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:

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 ρ = 1000 kg m–3 and

ν = 1.110–6 m2 s–1 respectively.

)/1( 220 Rruu

FORCES ON IMMERSED BODIES

Measure the change of momentum of the fluid

... deduce the force on the body

FForce on BODY

F

Force on FLUID

FORCES ON IMMERSED BODIES

Momentum principle:

inoutoutin

AuAuApApF dρdρdd 22

out

in

oncancellatiofamountlarge

outin

Auuu

AuAuF

d)(ρ

dρdρ 22

forcepressurenet

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

inflow wake

streamline

body

inflow wake

body

force on fluid = (momentum flux)out – (momentum flux)in

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FORCES ON IMMERSED BODIES

Force deduced from downstream measurements only:

out

inin AppuuuF d)]()(ρ[

For unconstrained case or streamlined bodies:

out

AuuuF d)(ρ

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

inflow wake

streamline

body

inflow wake

body

EXAMPLE SHEET, Q8

A hydraulic jump occurs in an open channel of width 1.0 m (see figure). Upstream of

the jump the depth is 0.1 m and the velocity is uA (uniform). The velocity profile just

downstream of the jump is of the form

where u is the velocity at a distance y from the bed of the channel, uB is the velocity

near the bed and D (= 0.8 m) is the depth downstream of the jump.

(a) Determine uB, leaving your answer as a function of uA.

(b) Calculate the difference between the hydrostatic pressure forces on the fluid

cross-sections upstream and downstream of the jump.

(c) Neglecting viscous stresses on the channel bed or the free surface, use the

momentum principle to find the upstream velocity uA.

]π

cos1[2

B

D

yuu

0.1 m

uB

0.8 muA

BERNOULLI’S EQUATION: DERIVATION

Mechanical-energy equation (rate form):

rate of change in (KE + PE) = rate of working by non-con. forces

Steady flow; thin stream tube:

(energy flux)2 − (energy flux)1 = rate at which work is done on fluid

WpAUpAUUgzQUgzQ 211

2

21

2

2

21 )()()(ρ)(ρ

Divide by mass flow rate, ρQ = (ρUA)1 = (ρUA)2

Q

WppUgzUgz

ρ)

ρ()

ρ()()( 211

2

21

2

2

21

massunitperdoneworkUgzp

)ρ

(Δ 2

21

Incompressible: density ρ constant along a streamline

volumeunitperdoneworkUgzp )ρρ(Δ 2

21

u1

u2

12

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BERNOULLI’S EQUATION

constantUgzp 2

21 ρρ

Assumptions:

applies along a streamline

steady

incompressible

inviscid (no losses)

COMPRESSIBLE FLOW

Energy equation (rate form):

rate of change in (IE + KE + PE) = rate of working + supplying heat

Steady flow; thin stream tube:

(energy flux)2 − (energy flux)1 = rate of working + supplying heat

HQWpAUpAUUgzeQUgzeQ 211

2

21

2

2

21 )()()(ρ)(ρ

Divide by mass flow rate, ρQ = (ρUA)1 = (ρUA)2

Q

QWppUgzeUgze H

ρ)

ρ()

ρ()()( 211

2

21

2

2

21

massunitpersuppliedheatdonework )()ρ

(Δ 2

21 Ugz

pe

Density ρ is not constant along a streamline

(or e + pv) is called specific enthalpy ρ

pe

u1

u2

12

ENERGY AND HEAD

2

21 ρρ Ugzp

g

Uz

g

p

2ρ

2

energy per unit volume

energy per unit weight (= total head)

pressure head

dynamic head

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POWER

Flow rate Q, change of head H

time

dtransformeenergyPower

gQHρPower

Qt

mρ (mass flow rate)

t

mgH

QgHρ

STATIC AND STAGNATION PRESSURE

Stagnation pressure

Static pressure

Dynamic pressure

2

21 ρUp

p

2

21 ρU

(= Pitot pressure, p0)

stagnation point (highest pressure)U = 0, P = P0

2

21

0 Uρpp

MANOMETRY PRINCIPLES

In a stationary fluid:

(1) Same fluid, same height same pressure

(2) Same fluid, different height Δp = −ρg Δz

(3) Different fluids: pressure is continuous at an interface

U-Tube Manometer

armright

mBC

armleft

A ghgyppyhgp ρρ)(ρ

ghpp mBA )ρρ(

ghp m )ρρ(Δ

Inclined Manometer

θsinLh

L (large)

h (small)

A B

h

C

y

m

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MEASUREMENT OF VELOCITY

Basic idea: measure the difference between Pitot and static pressures

pressuredynamic

pressurestatic

pressurePitot

Upp 2

21

0 ρ

Free-surface flows

Internal flows

Piezometer

and Pitot tube

Pitot-static

tube

free surfaceU /2g

2

stagnation point

U

piezometer Pitot tube

Ug2

2

static holes

stagnation point

static pressure tube

total pressure tube

MEASUREMENT OF VOLUME FLOW RATE

Basic idea:

Provide constricted section to change speed and measure change in pressure

Use a combination of Bernoulli’s eqn and continuity.

Venturi flowmeter Orifice flowmeter

2

221

2

2

121

1 ρρ UpUp

2211 AUAU

Bernoulli:

Continuity:

2/1

2

21

2

1

ρ

Δ

1)/(

2

p

AA

AQideal

idealdQcQ (cd = discharge coefficient)

12

FREE DISCHARGE UNDER GRAVITY

Bernoulli: 2

2

221

21

2

121

1 ρρρρ gzUpgzUp

ghzzgU ρ)(ρρ 21

2

221

ghU exit 2Torricelli’s Formula:

Ideal quantity of flow:

Actual quantity of flow:

exitexitideal AUQ

idealdQcQ

h

1

2

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TANK FILLING OR EMPTYING

outin QQt

V

d

d

Rate of change of volume = (volume flow rate)in – (volume flow rate)out

hAV sdd

outins QQt

hA

d

d

e.g. Emptying under gravity:

exitexitdout AUcQ ghU exit 2

VolumeV

h(t)

flow in flow out

As

dh

EXAMPLE, PAGE 31

A conical hopper of semi-vertex angle 30º contains water to

a depth of 0.8 m. If a small hole of diameter 20 mm is

suddenly opened at its point, estimate (assuming a

discharge coefficient cd = 0.8):

(a) the initial discharge (quantity of flow);

(b) the time taken to reduce the depth of water to 0.4 m.

0.8 m30

o

EXAMPLE SHEET, Q15

A trough has rectangular cross section (0.6 m wide × 0.8 m deep) and

length 2 m. It is being continuously filled with water via a hose at a rate

of 1.5 L s–1, whilst water leaks from a hole at the bottom with area

8×10–4 m2. The latter can be treated as an orifice with discharge

coefficient cd = 0.6.

(a) Write a numerical expression for the quantity of flow Q (in m3 s–1)

leaking from the bottom of the trough as a function of water depth

h (in m).

(b) Show that the inflow and outflow rates are such that the trough can

not fill to the brim.

(c) Find the maximum depth to which the trough can fill.

(d) Find the time taken to fill the trough to half the depth in part (c).

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METHODS FOR NON-IDEAL FLOW

Discharge coefficients

Loss coefficients

Momentum and energy coefficients

ideald QcQ

g

VKH

2Δ

2

)ρ(Δ 2

21 VKp

e.g pipe friction (λ = friction factor): D

LK λ

Momentum coefficient (β):

)ρ(βdρ 22 AuAu av

Kinetic energy coefficient (α):

)ρ(αdρ 33 AuAu av