Gas Dynamics 1

Isentropic Flow

Gas Dynamics

Agenda

• Introduction• Derivation• Stagnation properties• IF in a converging and converging-diverging

nozzle • Application

Gas Dynamics

Introduction

q=0

P ρ T

Consider a gas in horizontal sealed cylinder with a piston 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.

Gas Dynamics

Isentropiccore flow

Introduction Examples of isentropic flows: Jet or rocket nozzles,

diffusers Airfoils

But in reality there is no real flow is entirely isentropic!!

Gas Dynamics

Derivation

For isentropic flow:

And:

So:

Applying energy eqn to get relation between T & M:

1

1

2

1

1

2

1

2−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγγ

ρρ

PP

TT

RTa γ=

1

1

2

1

1

221

22

1

2−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛==

γγγ

ρρ

PP

aa

TT

22

22

2

21

1VTcVTc pp +=+

( )( )2

22

12

1

1

2

2121

TcVTcV

TT

p

p

++

=

RTV

aVM

γ==

Gas Dynamics

Derivation cont.

But:

And:

So:

1

1

2

1

1

221

22

1

2−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛==

γγγ

ρρ

PP

aa

TT

222

21

22M

cR

RTV

TcV

pp

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡= γγ

γ

22

21

1

2

211

211

M

M

TT

−+

−+= γ

γ

1

21

22

1

2

211

211

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

−+=

γγ

γ

γ

M

M

PP

11

21

22

1

2

211

211

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

−+=

γ

γ

γ

ρρ

M

M

Gas Dynamics

Derivation cont.

To find relation between A & M:

Using relation ρ-M and T-M

Where:

222111 VAVA ρρ = ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

2

1

1

2

VV

AA

ρρ

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

22

11

2

1

1

2

RTMRTM

AA

γγ

ρρ

( )( )12

1

21

22

2

112

1

1

2

2

121

1

21

1

1

2

2

1

1

2

211

211

−+

−+

−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−+

−+=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγ

γγ

γ

γ

γ

M

M

MM

KK

MM

KK

KK

MM

AA

2

211 MK −+= γ

Gas Dynamics

Basic Equations forOne-Dimensional Compressible Flow• Control Volume

Gas Dynamics

Basic Equations forOne-Dimensional Compressible Flow

• Continuity

Momentum

Gas Dynamics

Basic Equations forOne-Dimensional Compressible Flow

Second Law of Thermodynamics

• First Law of Thermodynamics

Gas Dynamics

Basic Equations forOne-Dimensional Compressible Flow

Property Relations

• Equation of State

Gas Dynamics

Isentropic Flow of an Ideal Gas– Area Variation

• Basic Equations for Isentropic Flow

Gas Dynamics

Isentropic Flow of an Ideal Gas– Area Variation

• Isentropic Flow

Gas Dynamics

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

Gas Dynamics

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 = const and 0 = const

(Slow down adiabatically and reversibly)

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

Gas Dynamics

Stagnation condition is a condition that wouldexist if the flow at any point was isentropicallybrought to or come from rest (V =M= 0).

Stagnation values:

0

0

0

0

ρρ ===

==

TTPPMV⎥⎦

⎤⎢⎣⎡ −+= 20

211 M

TT γ

120

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγ

γ MPP

11

20

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγρρ M

Stagnation Properties

Gas Dynamics

Stagnation Properties cont.Examples

10

2

Stagnation point is point 0.

0

0

0

0

ρρ ===

==

TTPPMV

Stagnation point is inside the chamber.

Gas Dynamics

Example:

M T0/T P0/P ρ0/ρ a0/a A/A* θ

0.50 --

2.40

Isentropic Relations in Tabular Form

⎥⎦⎤

⎢⎣⎡ −+= 20

211 M

TT γ

120

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγ

γ MPP

11

20

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγρρ M

( )121

2

2* 1

11

21 −+

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+

+=

γγ

γγ

γM

MAA

21

20

211 ⎥⎦

⎤⎢⎣⎡ −+= M

aa γ

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

Gas Dynamics

Pitot Probe Measurementfor Compressible Flow:

V = 0

P0

PIncompressible flow (Bernoulli eqn):

Compressible flow:

20 2

1 VPP ρ=− ( )ρ

PPV −= 02

120

211

−

⎥⎦⎤

⎢⎣⎡ −+=

γγ

γ MPP

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−⎟

⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

11

21

0γ

γ

γ PPM

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−⎟

⎠⎞⎜

⎝⎛ +−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

111

21

0γγ

γ PPP

aV

Gas Dynamics

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 psiaAssume: Ideal gas behavior for air

Gas Dynamics

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; onethus has a balance between surface and body forces. Consider the linearmomentum equation:

21

Gas Dynamics 22

Gas Dynamics 23

Gas Dynamics

Critical condition is a conditionthat would exist if the flow wasisentropically acceleratedor decelerated until M = 1.

M*=1 T*, P*, ρ*, A*, a* =

Critical Conditions:

⎥⎦

⎤⎢⎣

⎡+−+

+= 2

11

12* M

TT

γγ

γ

12

11

12* −

⎥⎦

⎤⎢⎣

⎡+−+

+=

γγ

γγ

γM

PP

11

2

11

12* −

⎥⎦

⎤⎢⎣

⎡+−+

+=

γ

γγ

γρρ M

21

2

11

12*

⎥⎦

⎤⎢⎣

⎡+−+

+= M

aa

γγ

γ

1

2

3

4

( )( )12

1

2

11

12* −

+−

⎥⎦

⎤⎢⎣

⎡+−+

+=

γγ

γγ

γMM

AA

A=A*M=1

M<1 M>1

0

1

2

3

4

5

6

7

0 1 2 3

Mach number

A/A*

5

Gas Dynamics

Homeworks1. Calculate the Mach number of two aircraft both travelling with anairspeed of 300m/s. One is traveling at sea level (T=250C); theother at an altitude of 11km (T=-160C.)

2. A perfect gas with is traveling at Mach 3 with a statictemperature of 250K, a static pressure of 101kPa, and a staticdensity 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 temperatureis 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 isentropiccompressible flow calculate the stagnation pressure. Assumingincompressible flow, use Bernoulli’s equation to calculate thestagnation pressure. What is the error in assumingincompressible flow at this Mach number?

4.1=γ