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HIGH-SPEED AERODYNAMICSMACE 31321

Lecture 3 Basic equations for 1D compressible flows

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OBJECTIVE OF THIS LECTURE

• To derive and discuss the energy equation for steady, inviscid, adiabatic flows

• To derive and discuss the basic relationship for 1D isentropic compressible flows

• To learn how to use the Isentropic Flow Properties Table

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– wherep/ρ = work required to push a unit mass into/out the CVgz : negligible for gases

– Use entropy:

TOTAL ENERGY OF FLOWING FLUID

2

2Uh +=Total energy

ρpeh +=

Flow energyInternal energy Kinetic energy

Potential energy

gzUpe +++=2

2

ρTotal energy

pvEU

Control volume

ρ1

=vNote specific volume

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• Consider 1D steady inviscid flow in a stream-tube

• From mass conservation: δm1 = δm2 =δm• Energy equation

• Along a streamline in a steady, inviscid, adiabatic flow,

mUhmUhQW δδ1

2

2

2

21

21

⎥⎦⎤

⎢⎣⎡ +−⎥⎦

⎤⎢⎣⎡ +=+ &&

0,0 == WQ &&

h1T1U1δm1

h2T2U2δm2

Control volume

1

2

2

2

21

21

⎥⎦⎤

⎢⎣⎡ +=⎥⎦

⎤⎢⎣⎡ + UhUh

.21 2 constUh =+

THE ENERGY EQUATION

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• For 1D steady, inviscid, adiabatic, compressible flow

• Bernoulli Equation

• Estimation of max. temperature changes– At U=10m/s,

– At U=1300m/s,

.const2

2

=+Uh

THE ENERGY EQUATION

.2

2

constUp=+

ρ

.const2

2

=++Upe

ρ

What are the condition for BE to be valid?

CT o05.0=∆

CT o841=∆

.const2

2

=+UTCp

pCUT 2/2

=∆

or

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• For 1D steady, inviscid, adiabatic, compressible flow

• Bernoulli Equation

• Estimation of max. temperature changes– At U=10m/s,

– At U=1300m/s,

.const2

2

=+Uh

THE ENERGY EQUATION

.2

2

constUp=+

ρ

.const2

2

=++Upe

ρ

What are the condition for BE to be valid?

CT o05.0=∆

CT o841=∆

In high-speed flows, the variations in flow velocity will result in a large changes in fluid temperature hence a significant variations in fluid density.

.const2

2

=+UTCp

pCUT 2/2

=∆

or

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STAGNATION CONDITIONS• p, T and ρ are static quantities in the local flow

field.• Stagnation Conditions: the conditions which

exist at a point if the fluid were brought to rest isentropically. It can be either real or imaginary.

• Stagnation values denoted by underscore “o”

opo TChUh ==+ 2

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Stagnation temperature

po C

UTT

2

2

+=

Dynamic temperature

Static temperature

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THE ENERGY EQUATION• For steady, inviscid, adiabatic flows,

• A perfect gas with a constant Cp and Cv is called a calorically perfect gas.

• For a calorically perfect gas,

.constho =

.constTo =

Usually it applies along a streamline. However, if all the streamlines of the flow originate from a common uniform freestream. ho is the same for each streamline. ho then becomes constant for the entire flow field.

x

y

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• Along a streamline in a steady, inviscid, adiabatic flow, we have

BASIC EQUATIONS FOR 1D COMPRESSIBLE FLOW

02

21 TCuTC pp =+

TCu

TT

p21

20 +=

( )

( )( ) ( ) 2

2

2

2

2

211

211

1/21

1/21

Mau

au

RTu

−+=⎟

⎠⎞

⎜⎝⎛−

+=

−+=

−+=

γγ

γ

γγ

RCp 1−=

γγ

RTa γ=

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BASIC EQUATIONS FOR 1D COMPRESSIBLE FLOW

• For isentropic flow

1−⎟⎠⎞

⎜⎝⎛=

γγ

TT

pp oo

11−

⎟⎠⎞

⎜⎝⎛=

γ

ρρ

TToo

11

2

12

211

211

−

−

⎥⎦⎤

⎢⎣⎡ −

+=

⎥⎦⎤

⎢⎣⎡ −

+=

γ

γγ

γρρ

γ

M

Mpp

o

o

2

211 M

TTo −

+=γ

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BASIC EQUATIONS FOR 1D COMPRESSIBLE FLOW

• Ratio to stagnation conditions “o” as function of Mach number

• These ratios are tabulated as functions of M for air (γ=1.4).

11

2

12

2

211

211

211

−

−

⎥⎦⎤

⎢⎣⎡ −

+=

⎥⎦⎤

⎢⎣⎡ −

+=

−+=

γ

γγ

γρρ

γ

γ

M

Mpp

MTT

o

o

o

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IS THE FLOW INCOMPRESSIBLE?• By the rule of thumb

that an air flow can be assumed to be incompressible when M<0.3.

• Why M=0.3 is used?• Consider air which is

accelerated from rest to Mach number M isentropically,

11

2

211

−

⎥⎦⎤

⎢⎣⎡ −

+=γγ

ρρ Mo

For M<0.3, the variation in ρ/ρ0 < 5%, hence the flow can be treated as incompressible.

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• During re-entry, Space Shuttle Orbiter flies at 1.3 km/s at an altitude 33,000 m. A bow shock forms in front its nose. The corresponding conditions at stagnation point (2) are

• Static pressure at (3)

• The two points are outside the boundary layer• Calculate the local static temperatures and velocities at

points (3) using the Isentropic Flow Properties Table. [Answers: 1742oC, 900m/s]

QUESTION 1

barp 2347.02 =

CT o21452 =

barp ,1240.03 =

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Solution Q1

• Since point 2 is the stagnation point, we have po = p2, T0=T2

• The pressure ratio at point 3 is

• Since the flow is outside the boundary layer, it can be treated as isentropic. From the Isentropic Flow Properties Table,

• From

893.1124.02347.0

3

==ppo

2.13

=TTo KTT o 2015

2.12732145

2.13 =+

==

0233 2

1 TCuTC pp =+

( )( ) sm

TTCpu/9002015273214510052

2 303

=−+××=−=

CT o17423 =

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BASIC EQUATIONS FOR 1D COMPRESSIBLE FLOW

• Along a streamline in a steady, inviscid, adiabatic flow,

• If at a certain point in the flow, the flow velocity reaches the local sonic speed, i.e. u=a, Let u* = a*,

.21 2 constuh =+ 2

22211 2

121 uTCuTC pp +=+

RTa γ=

RCp 1−=

γγ

22

221

1

21

121

1uRTuRT

+−

=+− γ

γγγ

22

222

1

21

21

121

1uaua

+−

=+− γγ

( )( )12

121

121

1

2*2*

2*2

2

−

+=+

−=+

− γγ

γγa

aa

ua

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THE CHARACTERISTIC MACH NUMBER• Along a streamline in a steady, inviscid, adiabatic flow,

– a* is constant along a given streamline.– Note the local sonic velocity a varies with T.

• Substituting the characteristic Mach numberinto (1) it can be approved that

( )( )12

121

1

2*2

2

−

+=+

− γγ

γa

ua

**

auM =

( )( ) 2

22

121*

MMM

−++

=γ

γ

(1)

M*<1 when M<1 M*=1 when M=1M*>1 when M>1

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SONIC CONDITION• Substituting M=1 into

• These relations are useful in determining if the sonic condition is reached at a given point in the flow.

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2

12

2

211

211

211

−

−

⎥⎦⎤

⎢⎣⎡ −

+=

⎥⎦⎤

⎢⎣⎡ −

+=

−+=

γ

γγ

γρρ

γ

γ

M

Mpp

MTT

o

o

o

577.12

1*

894.12

1*

2.12

1*

11

1

=⎥⎦⎤

⎢⎣⎡ +

=

=⎥⎦⎤

⎢⎣⎡ +

=

=+

=

−

−

γ

γγ

γρρ

γ

γ

o

o

o

pp

TTγ=1.4

γ=1.4

γ=1.4

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• Consider a point in an airflow where the local Mach number, static pressure and static temperature are 3.5, 0.3atm and 180K. Calculate the values of a, a* and M*. [answers: 268.9m/s, 456m/s, 2.06]

QUESTION 2

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Solution Q2

• From the local static temperature, the local sonic speed can be found

• From Table A, at M=3.5, T0/T=3.45 and at M=1, T0/T*=1.2, hence

• Thus

875.245.32.1

1**

=×==TT

TT

TT o

o

KTT 5.517875.2* ==

smRTa /9.2681802874.1 =××== γ

smRTa /4565.5172874.1** =××== γ

smMau /9419.2685.3 =×==

06.2456941

** ===

auM

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APPENDIX 1: THERMODYNAMIC PROCESSES

• ADIABATIC = No Heat Transfer• REVERSIBLE = No Energy Dissipation• ISENTROPIC = Adiabatic and Reversible

• ENERGY DISSIPATION phenomena– Viscosity– Mass diffusion– Thermal conductivity

• The flow in a boundary layer is not isentropic.

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APPENDIX 2: SPECIFIC HEAT• Specific heat at constant volume: • Specific heat at constant pressure: • Specific heat ratio: • Additionally, we have

• At moderate temperature (T<1000K for air), Cp and Cv are approximately constant.

• For air at standard conditions, – γ = 1.4– Cp=1005 J/kg.K– R = 287 J/kg.K

pCvC

,RCC vp =−

v

p

CC

=γ

RCRC vp 11,

1 −=

−=

γγγ

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APPENDIX 3: ISENTROPIC RELATIONS

• These are very useful relations

• Condition of validity– Outside the boundary layer

γγγ

ρρ

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

1

21

1

2

1

2

TT

pp

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READING MATERIAL• Lecture notes• “Fundamentals of Aerodynamics” by J D

Anderson, 2nd edition, McGraw-Hill. • §7, p.393-411: review thermodynamics as

necessary • §8.4: Special forms of energy equation • §8.5: when a flow is incompressible