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© D.J.Dunn freestudy.co.uk 1

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA

PRINCIPLES AND APPLICATIONS of FLUID MECHANICS UNIT 13

NQF LEVEL 3

OUTCOME 4 - DYNAMIC FLUID SYSTEMS

TUTORIAL 1 - MODEL TESTING

CONTENT

Be able to determine the parameters of dynamic fluid systems

Model testing: wind-tunnel testing e.g. laminar and turbulent flow, flow around bluff

bodies, dynamic pressure, theoretical and measured drag force, drag coefficient,

application of Reynolds number, operation and use of the Pitot-static tube; test data e.g.

measured drag force, model dimensions, air density, Pitot-static tube reading, density of

manometer fluid

Aerodynamic systems: aerofoil applications e.g. aircraft lift surfaces, helicopter rotor

blades, formula 1 racing cars; system parameters e.g. span, chord, angle of attack, plan

area of lift surfaces, pressure distribution, stalled condition, lift force, drag force, lift and

drag coefficients, airspeed, propeller efficiency, engine power requirements for an aircraft

in level flight

The subject of model testing in wind tunnels and drag on a surface requires very advanced

studies of fluid mechanics so the contents of this tutorial are necessarily over simplified and

only serves as a general introduction. You can find more advanced material on the web site

www.freestudy.co.uk.

http://www.freestudy.co.uk/

D.J.Dunn www.freestudy.co.uk 2

1. DYNAMIC PRESSURE

Dynamic pressure has been covered previously in this unit along with the theory of pitot

static tubes and repeated here for revision.

Consider a stream of air moving at velocity v m/s and with pressure p1. If the stream is

interrupted by any body, it will be diverted around it but at some point on the body the air is

brought entirely to a stop and this point is called the stagnation point. The velocity of the air

is converted into pressure. At this level we will treat the air as having a constant density.

(This becomes less true at higher velocities).

Applying Bernoulli between the air stream and the stagnation point we have:

2

ρvp

2

ρvp

22

2

21

1 but since v2 = 0 this reduces to:

2

21

1 p2

ρvp

121

pp2v

or

2

ρvpp

21

12

p2 - p1 is the indication on the pressure gauge if the air stream is the surrounding atmosphere.

2

ρv21 is the term called the dynamic pressure which results from converting velocity into

pressure.

PITOT STATIC TUBES

Pitot static tubes used for measuring the velocity of air streams.

These days electronic pressure sensors are used to measure the

dynamic pressure as shown in the picture below. Pitot tubes come

in many forms and sizes. The one shown uses a tube inside another

tube that is pointed into an air stream. The pressure difference is

measured at the points shown. Usually a simple hollow needle with

a hole in the side is sufficient and the stream pressure p1 is taken

from the atmosphere. You will see pitot tubes on the leading edges

of aircraft to measure the air speed. They are also used in wind

tunnels for measuring the stream velocity at a given point.


The syllabus mentions manometers because these can be used as simple pressure measuring

devices.

The pressure tapings are connected to the manometer and produces

a difference in the level of the fluid. For measuring air stream

velocities the fluid in the manometer is light oil manufactured with

an accurately known density. This is much better than water. To

calculate the pressure difference we use Δp = ρo g h

ρo is the density of the manometer fluid.

The velocity of the air stream is hence:

ρ

gh2ρ

ρ

pp2v o12

1

The pressure difference is often very small and to obtain accurate measurements one leg of

the manometer is inclined and the other made into a reservoir. The value of h is often given

on the sloping scale in mm of water or the scale might be calibrated to give the velocity.

WORKED EXAMPLE No. 1

A pitot static tube is connected to an inclined manometer and pointed into an air stream.

The reading on the scale is 35 mm of water. Given that the air density is 1.2 kg/m3 and the

density of water is 1000 kg/ m3 calculate the velocity of the air stream.

SOLUTION

23.92m/s1.2

0.035 9.81 1000 2

ρ

gh2ρv o

1 xxx

SELF ASSESSMENT EXERCISE No. 1

1. A pitot static tube is connected to an inclined manometer and pointed into an air stream.

The reading on the scale is 62 mm of water. Given that the air density is 1.05 kg/m3 and

the density of water is 1000 kg/ m3 calculate the velocity of the air stream.

(34 m/s)


2. DRAG

When a fluid flows around the outside of a body, it produces a force that tends to drag the

body in the direction of the flow. The drag acting on a moving object such as a ship or an

aeroplane must be overcome by the propulsion system. Drag takes two forms, skin friction

drag and form drag.

SKIN FRICTION DRAG

Skin friction drag is due to the viscous shearing that takes place between the surface and the

layer of fluid immediately above it. This is the main source of drag on surfaces of objects

that are long in the direction of flow compared to their height. Such bodies are called

STREAMLINED. A thin flat plate is an example of a streamlined object. Consider a stream

passing over a long thin plate as shown.

The stream is undisturbed at a point some distance from the surface but near the surface the

fluid is slowed down in a region called the boundary layer. The slowing down occurs

because of the viscosity of the fluid. The layer next to the surface becomes attached to it (it

wets the surface). This is called the ‘no slip condition’. The layers of fluid close to the

surface are moving so there must be shearing taking place between the layers of the fluid.

The shear stress acting between the wall and the first moving layer next to it is called the

wall shear stress and denoted w.

We often need to use a dimensionless parameter called the Reynolds number. For these cases

it is defined as: μ

ρvLRe

L is the length parallel with the stream. At low Reynolds numbers, the fluid in the boundary

layer may be laminar throughout the entire thickness. At higher Reynolds numbers, it is

turbulent and full of eddies starting at a critical distance from the leading edge.

CALCULATING SKIN DRAG

The skin drag is due to the wall shear stress w and this acts on the surface area (wetted

area).

The drag force is hence: R = w x wetted area = w A.

The dynamic pressure defined earlier is: 2

ρv2

.

In this work it is convenient to use the idea of the drag coefficient. This is defined as:

vρA

2τ

A ρv

2R

area d x wettepressure dynamic

force DragC

2w

2Df

Using this requires a lot of knowledge and much of the work is based on experimental data.

For example, for a smooth surface, it can be shown that CDf = 0.074 (Re)x-1/5



Calculate the drag force on each side of a thin smooth plate 2 m long and 1 m wide with

the length parallel to a flow of fluid moving at 30 m/s. The density of the fluid is 800

kg/m3 and the dynamic viscosity is 8 cP.

SOLUTION

N 2347.2 1 2 1173.6 Area Wetted τ R

Pa 1173.6 10 360 0.00326 pressure dynamic C τ

kPa 360 2

30 x 800

2

ρv pressure Dynamic

0.00326 ) x10(6 x 0.074 C

10 x 6 0.008

2 x 30 x 800

μ

ρvLR

w

3Dfw

22

5

1

6Df

6e

xxx

xxx

This is the force on one side.

Practical objects are not normally thin flat sheets but wider objects like ships hulls or

aeroplane wings. The skin friction is still important but another form of drag starts to take

over. You need to understand that a surface at an angle to the stream will produce different

results.

On a small area the skin drag is dR = w dA. If the body

is not a thin plate and has an area inclined at an angle

to the flow direction, the drag force in the direction of

flow is w dA cos.

The drag force acting on the entire surface area is found by integrating over the entire area.

dA cosR w . Solving this equation is beyond the scope of this unit.


1. A smooth thin plate 5 m long and 1 m wide is placed in an air stream moving at 3 m/s

with its length parallel with the flow. Calculate the drag force on each side of the plate.

The density of the air is 1.2 kg/m3 and the kinematic viscosity is 1.6 x 10

-5 m

2/s.

(0.128 N)


FORM DRAG and WAKES

Form or pressure drag applies to bodies that are tall in comparison to the length in the

direction of flow. Such bodies are called BLUFF BODIES. The extreme bluff body is

shown.

The undisturbed fluid is brought to rest on the front

of the object and flows away around the edges. The

pressure on the front builds up into the dynamic

pressure. The fluid flowing around the edges must

speed up as it is squeezed into a smaller space.

Bernoulli tells us that if it speeds up the pressure

must go down and in fact a vacuum is formed

behind the body and fluid is sucked in and vortices

are formed behind it. The pressure difference acts

on the area of the plate (now normal to the flow)

and produces a drag R, it is all due to the dynamic

pressure. In this extreme case there is no skin drag.

Again we use the idea of a drag coefficient CDp

area projected x ρv

2R

area projected x pressure dynamic

force DragC

2Dp

If we could find a way of working out the pressure p at any point on the surface at angle θ to

the direction of the stream, the total pressure drag would be given by:

dA pcosθR

Again, solving this is well beyond the

scope of this tutorial. The next diagram

illustrates a symmetrical bluff object. In

this case the fluid sticks to the surface

for a certain length exerting skin drag

and then it breaks away where the

vacuum sucks in fluid to form vortices

and produces form drag. The total drag is

the sum of the two. The degree of each

depends upon the shape of the body.

Solving this would be difficult.

TOTAL DRAG

It has been explained that a body usually experiences both skin friction drag and form drag.

The total drag is the sum of both. This applies to aeroplanes and ships as well as bluff

objects such as cylinders and spheres. The drag force on a body is very hard to predict by

purely theoretical methods. Much of the data about drag force is based on experimental data

and the concept of a drag coefficient is widely used.

The DRAG COEFFICIENT is denoted CD and is defined by the following expression.

Area projected x ρv

2R

Area projected x pressure Dynamic

force ResistanceC

2D

The projected area is the area normal to flow direction.


Data on the drag coefficient for many bodies has been found by researchers and the diagram

below shows the data for cylinders and spheres plotted against Reynolds number.


A cylinder 80 mm diameter and 200 mm long is placed in a stream of fluid moving at

0.5 m/s. The axis of the cylinder is normal to the direction of flow. The density of the

fluid is 800 kg/m3. The drag force is measured and found to be 30 N. Calculate the drag

coefficient.

SOLUTION

Projected area = 0.08 x 0.2 = 0.016 m2

R = 30 N v = 0.5 m/s ρ = 800 kg/m3

Dynamic pressure = 222

1 N/m 1002

0.5 800

2

ρv

x

18.750.016 x 100

30


force ResistanceCD


A sphere 40 mm diameter moves through a fluid of density 750 kg/m3 and dynamic

viscosity 50 cP with a velocity of 0.6 m/s. Calculate the drag on the sphere using the

graphical data given above.

SOLUTION

N 0.136 2

10 x 1.2566 0.6 750 0.8

2

A ρuC R

m10 1.2566 4

0.04

4

d area Projected

Area projected x ρu

2RC

0.8Cgraph thefrom 360 0.05

0.04 0.6 750

μ

ρvdR

3-22D

23-22

2D

De

xxx

x

xx



1. A sphere 50 mm diameter is placed in a wind tunnel and the drag force is measured and

found to be 0.032 N when the velocity is 30 m/s. Calculate the drag coefficient. Take the

density of air to be 1.29 kg/m3.

(0.028 N)

Note these figures do not represent a real situation.

2. Using the graphical data to obtain the drag coefficient of a sphere, determine the drag on

a totally immersed sphere 0.2 m diameter moving at 0.3 m/s in sea water. The density of

the water is 1025 kg/m3 and the dynamic viscosity is 1.05 x 10

-3 Ns/m

2.

(0.639 N)

3. Calculate the drag force for a cylindrical chimney 0.9 m diameter and 50 m tall in a wind

blowing at 30 m/s given that the drag coefficient is 0.8. The density of the air is 1.2

kg/m3.

(19.44 N)

4. Using the graphical data for cylinders to find the drag coefficient, determine the drag

force per metre length acting on an overhead power line 30 mm diameter when the wind

blows at 8 m/s. The density of air may be taken as 1.25 kg/m3 and the kinematic viscosity

as 1.5 x 10-5

m2/s.

(1.8 N)


3. AEROFOILS

Aerofoils are specially shaped objects

designed to produce lift and reduce drag.

The three main dimensions are the

CHORD, THICKNESS and SPAN.

Aerofoils are used for aircraft wings and

many other objects such as turbine

blades and propellers.

LIFT and DRAG FOR AEROFOIL

Lift is a force acting normal to the direction of the stream of fluid and hence normal to the

drag force. This occurs when fluid flows around an object that is either not symmetrical or it

is inclined to the flow. Although this section is about aerofoils, lift can occur on many other

objects. Let's start by considering a symmetrical aerofoil. When there is no inclination the

flow above and below the aerofoil is symmetrical so drag occurs but no lift. When the

aerofoil is inclined to the stream it has an angle of attack defined as the angle between the

chord and the stream direction. The diagram illustrates typically what will happen. The air

passing over the top will separate at some point pulling in a vortex and creating drag. The air

flowing around the bottom will experience a rise in pressure because some of the velocity

will be converted into pressure. The net result is a pressure difference between the top and

bottom that produces a lift force L.

Another simple explanation of lift is that the air flowing over the top has to travel further

from the front to the back than the air underneath so it must speed up. Bernoulli tells us this

will cause a drop in pressure. Conversely the air on the bottom slows down so we get a rise

in pressure. The pressure difference generates the lift. Aerofoils are often made

unsymmetrical as shown in the next diagram.

The angle of attack can be varied to increase or decrease the lift acting on the wing. An

increase in lift often results in an increase in drag.


We define a lift coefficient CL in a similar form to the drag coefficient. The projected area is

based on the chord length and wing span. If the angle of attack is α, the chord length C and

the span S then the projected area is C S cos α.

cosα S C vρ

2L


forceLift C

2L

Usually this is based on S = 1 m.

The lift on an aircraft wing must be sufficient to support the weight of the craft. If the angle

of attack is increased, the lift will increase to a point but after that point it falls again as the

wing becomes more like a bluff object and loses its streamline form. When this happen it

said to reach its stall point and if the lift is insufficient to support the weight, the airplane

falls out of the sky. Knowing the stall angle of attack is extremely important for predicting

the minimum landing and takeoff speeds of an airplane.

The graph shows a typical result for a wing with the lift being based on a span of 1 m. Such

results are obtained in wind tunnel tests with models. The drag coefficient is also shown.

Note this diagram is for illustration purposes only.

The drag coefficient for an aerofoil is also based on the same projected area parallel to the

steam and not the projected area normal to the stream which is used for other objects such as

cylinders. The Drag coefficient is:

cosα S C vρ

2R


force DragC

2D


4. POWER REQUIREMENTS for AEROPLANE ENGINES

When an aeroplane is in level flight, the engine must deliver sufficient power to overcome

the drag. Mechanical power is simply defined as P = Force x Velocity so the minimum

engine power must be the product of drag force and stream velocity. This must be the

powered delivered by the propulsion system. The actual power supplied to the engine in the

form of fuel will be much larger as this must take into account the thermal and mechanical

efficiency of the engine.

If the aeroplane is climbing or descending, the power requirements are affected because

gravity is either being opposed or used to assist motion.

When the engine is propeller driven, the power input to the shaft is greater than the

minimum power because the propeller is not 100% efficient at turning the shaft power into

propulsive power. The efficiency depends on the shape and pitch of the propeller. The shaft

power is:

η

vRS.P. R is the drag force, v the flight speed and η the propeller efficiency.


An aeroplane is propeller driven and has an effective wing length of 6 m with a chord of

1.4 m and has an angle of attack of 10o. Calculate the lift and drag when flying at 150 m/s.

The air density is 1.15 kg/m3. Calculate the minimum power to be delivered by the

propeller shaft if the propeller efficiency is 70%. Use the graph previous to obtain lift and

drag coefficients.

SOLUTION

From the graph CL = 1.45 but remember this is for a span of 1 m.

cosα C ρv

2L1.45C

2Lxx

N/m 864 252

cos10 1.4 150 1.15 1.45

2

cosα C vρ 1.45L

o22

xxxx xx

For a 3 m span this is 77 592 N

From the graph CD = 0.08 but remember this is for a span of 1 m.

cosα S C vρ

2R


force DragC

2D

N/m 427 12

cos10 1.4 150 1.15 0.08

2

cosα C vρ 0.08R

o22

xxxx xx

For a 3 m span this is 4 281 N

Power = R v/η = 4281 x 150/0.7 = 917 357 Watts



1. For the aerofoil data shown on the graph previously, calculate the lift and drag for a wing

2.5 m long and chord 0.8 in an air stream moving at 80 m/s with density 1.21 kg/m3 when

the angle of attack is 5o. Calculate the minimum power required.

(8.1 kN, 1774 N and 142 kW)

2. A wind tunnel test on a scale model of a wing produced the following results.

Chord - 0.1 m

Span - 0.5 m

Angle of attack - 12o

Air stream velocity - 30 m/s

Lift force - 10 N

Drag force - 2 N

Air density 1.2 kg/m3

Calculate the lift and drag coefficients per meter of span.

(0.3786 and 0.075)

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