tutorial 1 in compressible flow

8/8/2019 Tutorial 1 In Compressible Flow

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Incompressible flow Example sheet 1

1

1.

From resolving the forces we have the following:

Dont forget to convert the angle into radians when calculating the gradient

The real curve may not be linear, due to three dimensional effects.

L

D

R

N

A

U

RTP = RTa = KsmR22/287=

2.1=NC

= 12

03.0=AC

d

dCa L=

cossin AND +=

12cos03.012sin2.1 +=DC

cossin AND CCC +=

279.0978.003.0208.02.1 =+=DC

sincos ANL =

12sin03.012cos2.1 =LC

168.1208.003.0978.02.1 ==LC

sincos ANL CCC =

CL

12

1.168

For symmetrical aerofoil

258.5 =

d

dCL


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2

2.

Plot CM(C/4) Vs CL

44' cLE ML

c

M=+

0' ==+ cpcpLE MLcxM

Subtracting second from the one;

4

)4

( ccp Mxcc

L = dividing by Scq

4)25.0(cMcpL cxc = rearranging

L

cM

cp c

cx 425.0 =

CL CM(0.25)c xcp

-2 0.05 -0.042 1.09

0 0.25 -0.040 0.41

2.0 0.44 -0.038 0.336

4.0 0.64 -0.036 0.306

6.0 0.85 -0.036 0.292

8.0 1.08 -0.036 0.283

10.0 1.26 -0.034 0.277

12.0 1.43 -0.030 0.271

14.0 1.56 -0.025 0.266

Mc/4

ac

c/4

acxc

L


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3

For most conventional airfoils, the aerodynamic centre is close to, but not necessarily

exactly at, the quarter-chord point.

4/)4/( cacac McxcLM +=

Dividing by Scq

, we have

Scq

Mx

Sq

L

Scq

M cac

ac

+

=

4/)25.0(

Or

4/,, )25.0( cmaclacm cxcc +=

Differentiating with respect to angle of attack , we have

d

dcx

d

dc

d

dc cmac

lacm 4/,, )25.0( +=

Howeverd

dc acm ,is zero by definition of the aerodynamic centre (Pitching moment is

constant with incidence, hence the derivative is zero).

d

dcx

d

dc cmac

l 4/,)25.0(0 +=

For airfoils below the stalling angle of attack, the slope of the lift coefficient is

constant and also the variation in the second term is negligible compared to the slopeof lift curve.

Designating these slopes by

ad

dcl

; 0

4/,

d

dc cm

We arrive at the following;

0)25.0(0+=

acxa

25.0=acx


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4

-0.045

-0.04

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

cl

Cmc/4

0

0.2

0.4

0.6

0.8

1

1.2

-4 -2 0 2 4 6 8 10 12 14 16

aoa(degrees)

Xcp


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5

3.

For dynamic similarity we must satisfy three criterions:

1. Same Mach number

2. Same Reynolds number3. Geometric similarity

For aerofoil (1) we have the following data:

For aerofoil (2) we have the given values:

Assume that 21

T

Since 21

T we have that:

1

2

1

2

T

T=

By definition1

11

a

VM =

2

2

2a

VM =

Where a1 and a2 are the speed of sound for the two different aerofoils.

1

1

1RT

VM

=

2

22

RT

VM

=

i.e. the Mach numbers are the same so we have so far satisfied the first

condition for dynamic similarity.

Now moving on to explore the Reynolds numbers

KT o200=

smV /100=

3/23.1 mkg=

12

2

200200

800100

12

21

2

2

1

1

2

1=====

TV

TV

RT

V

RT

V

M

M

11 RTa =

KT o800=

smV /200=

3/615.0 mkg=

22 RTa =

21 MM =


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6

By definition we have:

1

1111Re

cV=

2

222

2Re

cV=

Reynolds numbers are the same

Since the bodies are geometrically similar and the M and Re are the same, we have

satisfied all the criterions, hence the two flows are dynamically similar.

4.

Need the Re number and M number doubled. Also 2/1T

55 12

2

1 ccc

c==

12

1

2

22

2

1

200

100

615.0

23.1

200

800

2 1

1

2

1

2

1

1

2=

==

c

c

V

V

T

T

=

1

2

1

2

2

1

2

1 2c

c

V

V

T

T

2

222

1

111 2T

cV

T

cV =

2221

1112

2

222

1

111

2

1

Re

Re

cV

cV

cV

cV

==

KT o2231 =

smV /2501 =

3

1 /414.0 mkm=

kmh 10=

(1)

25

2 /1001.1 mNP =

?2 =T

?2 =V

?2 =

(2)

1/5th

scale model

21 2MM =

21 Re2Re =


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7

21 2MM = 2

2

1

1 2a

V

a

V=

2

2

1

1 2RT

V

RT

V

=

1

2

1

2

2 T

T

V

V=

=

5

1

22

1

2

2

1

2

1

T

T

T

T

5

1

2

1=

3

12 /07.2414.055 mkg===

222 RTP = RPT 222 /=

KT 56.1702=

(= -102.6 C, Cryogenic tunnel required)

437.02232

56.170

2 1

2

1

2===

T

T

V

V smV /32.1092 =

5.

The zeppelin is a symmetric wing.

315000mV = md 14max =

smV /30=

3/9.0 mkg=

05.0=LC

SV

LC

L 2

21

=

Where2

max

2

4drS

==

NL 3117=

Because we have a free stream velocity we have lift but if it was not moving then it

would float due to the effect of buoyancy.

Buoyancy force = weight of displaced air

VgFb =

NFb

1324359.01500081.9 ==


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rceBuoyancyFoLiftWeight +=

N1355521324353117 =+=

tonnesgW 8.13/ =

Top View of the

Zeppelin

-7 7

tutorial 1 in compressible flow

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