2 Isentropic Flow

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GasDynamics 1

Isentropic Flow


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GasDynamics

Agenda

Introduction

Derivation

Stagnation properties IF in a converging and converging-diverging

nozzle

Application


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Introduction

q=0

P T

Consider a gas in horizontal sealed cylinder with apiston at one end. The gas expands outwards

moving the piston and performing work. The walls

of the piston are insulated and no heat transfer takes

place. Is this an isentropic process?

The mathematical relationships between

pressure, density, and temperature are

known as the isentropic flow relations.


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GasDynamics

Isentropiccore flow

Introduction

Examples of isentropic flows: Jet or rocket nozzles,

diffusers Airfoils

But in reality there is no real flow is entirely

isentropic!!


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GasDynamics

Derivation

For isentropic flow:

And:

So:

Applying energy eqn to get relation between T & M:

1

1

2

1

1

2

1

2

=

=

P

P

T

T

RTa =

1

1

2

1

1

2

2

1

2

2

1

2

=

==

P

P

a

a

T

T

22

2

22

2

11

VTc

VTc pp +=+

( )( )2

2

2

1

2

1

1

2

21

21

TcV

TcV

T

T

p

p

+

+=

RT

V

a

VM

==


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GasDynamics

Derivation cont.

But:

And:

So:

1

1

2

1

1

2

2

1

2

2

1

2

=

==

PP

aa

TT

222

2

1

22M

c

R

RT

V

Tc

V

pp

=

=

2

2

2

1

1

2

2

11

2

1

1

M

M

T

T

+

+

=

1

2

1

2

2

1

2

2

11

2

11

+

+

=

M

M

P

P

1

1

2

1

2

2

1

2

2

11

2

11

+

+

=

M

M


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Derivation cont.

To find relation between A & M:

Using relation -M and T-M

Where:

222111 VAVA =

=

2

1

2

1

1

2

V

V

A

A

=

22

11

2

1

1

2

RTM

RTM

A

A

( )

( )121

2

1

2

2

2

112

1

1

2

2

12

1

1

21

1

1

2

2

1

1

2

2

11

2

11

+

+

+

+

=

=

=

M

M

M

M

K

K

M

M

K

K

K

K

M

M

A

A

2

2

11 MK

+=


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Basic Equations for

One-Dimensional Compressible Flow

Control Volume


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GasDynamics

Basic Equations for


Continuity

Momentum


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Basic Equations for


Second Law of Thermodynamics

First Law of Thermodynamics


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GasDynamics

Basic Equations for


Property Relations

Equation of State


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Isentropic Flow of an Ideal Gas

Area Variation

Basic Equations for Isentropic Flow


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GasDynamics

Isentropic Flow of an Ideal Gas

Area Variation

Isentropic Flow


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Stagnation Conditions

Total (Stagnation) conditions :

A point (or points) in the flow where V = 0.

Fluid element adiabatically slow down

A flow impinges on a solid object


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Stagnation Conditions (cont.)

From Energy Equation and the first law of thermodynamics

Total enthalpy = Static enthalpy + Kinetic energy (per unit mass)

Steady and adiabatic flow h0 = const (h01 = h02)

Steady, inviscid, adiabatic flow T0 = const

Isentropic flow P0 = constand 0 = const

(Slow down adiabatically and reversibly)

For a calorically perfect gas , h0 = CPT0 or h = CP T


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GasDynamics

Stagnation condition is a condition that would

exist if the flow at any point was isentropically

brought to or come from rest (V =M= 0).

Stagnation values:

0

0

0

0

=

=

=

==

TT

PP

MV

+=

20

2

11 M

T

T

120

2

11

+=

M

P

P

1

1

20

2

11

+=

M

Stagnation Properties


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GasDynamics

Stagnation Properties cont.

Examples

1

0

2

Stagnation point is point 0.

0

0

0

0

=

=

=

==

TT

PP

MV

Stagnation point is inside the chamber.


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GasDynamics

Example:

M T0/T P0/P 0/ a0/a A/A*

0.50 -

-

2.40

Isentropic Relations in Tabular Form

+=

20

2

11 M

T

T

120

2

11

+=

M

P

P

1

1

20

2

11

+=

M

( )121

2

2

*1

1

1

21 +

+

+

+=

M

MA

A

2

1

20

2

11

+= M

a

a

0.50 1.05000 1.18621 1.12973 1.02470 1.33984

2.40 2.15200 17.08589 7.59373 1.50000 2.63671 36.74650


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GasDynamics

Pitot Probe Measurementfor Compressible Flow:

V = 0

P0

PIncompressible flow (Bernoulli eqn):

Compressible flow:

20

21 VPP = ( ) PPV =

02

120

2

11

+=

M

P

P

=

11

21

0

P

PM

+

=

111

21

0

P

PP

a

V


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GasDynamics

Example (Compressible pitot tube)

20

Given: Air at u = 750 fts , Mercury manometer which reads a

change in height of 8 inches.

Find: Static pressure of air in psia

Assume: Ideal gas behavior for air


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GasDynamics

Analysis: First consider the manometer which is governed by fluid statics. In fluid

statics, there is no motion, thus there are no viscous forces or fluid inertia; one

thus has a balance between surface and body forces. Consider the linear

momentum equation:

21


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GasDynamics

Critical condition is a conditionthat would exist if the flow was

isentropically accelerated

or decelerated until M = 1.

M*=1 T*, P*,*, A*, a* =

Critical Conditions:

+

+

+=

2

1

1

1

2*M

T

T

12

1

1

1

2*

+

+

+=

M

P

P

1

1

2

1

1

1

2*

+

+

+=

M

2

1

2

1

1

1

2*

+

+

+= M

a

a

1

2

3

4

( )( )12

1

2

1

1

1

2* +

+

+

+=

MMA

A

A=A*

M=1

M1

0

1

2

3

4

5

6

7

0 1 2 3

Mach number

A/A*

5


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GasDynamics

Homeworks

1. Calculate the Mach number of two aircraft both travelling with anairspeed of 300m/s. One is traveling at sea level (T=250C); the

other at an altitude of 11km (T=-160C.)

2. A perfect gas with is traveling at Mach 3 with a static

temperature of 250K, a static pressure of 101kPa, and a static

density of 1.4077kg/m3

. Determine the stagnation temperature,pressure, and density values.

3. An aircraft is flying at 80m/s at sea level where the temperature

is 200C, density is 1.225kg/m3 and pressure is 1030.1mbar.

Assuming R=287 J/kgK what Mach number is the aircraft flying?

Air stagnates near the leading edge. Assuming isentropic

compressible flow calculate the stagnation pressure. Assuming

incompressible flow, use Bernoullis equation to calculate the

stagnation pressure. What is the error in assuming

incompressible flow at this Mach number?

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2 Isentropic Flow

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