gas dynamics esa 341 chapter 3

Gas DynamicsESA 341Chapter 3

Dr Kamarul Arifin B. Ahmad

PPK Aeroangkasa

Normal shock waves

Definition of shock wave Formation of normal shock wave Governing equations Shock in the nozzle

Definition of shock waveShock wave is a very thin region in a flow where a supersonic flow is decelerated to subsonic flow. The process is adiabatic but non-isentropic.

Shock wave

V

P

T

Formation of Shock WaveA piston in a tube is given a small constant velocity increment to the right magnitude dV, a sound wave travel ahead of the piston.

A second increment of velocity dV causing a second wave to move into the compressed gas behind the first wave.

As the second wave move into a gas that is already moving (into a compressed gas having a slightly elevated temperature), the second waves travels with a greater velocity.

The wave next to the piston tend to overtake those father down the tube. As time passes, the compression wave steepens.

Types of Shock Waves:

Normal shock wave - easiest to analyze

Oblique shock wave - will be analyzed based on normal shock relations

Curved shock wave - difficult & will not be analyzed in this class

- The flow across a shock wave is adiabatic but not isentropic (because it is irreversible). So:

0201

0201

PP

TT

Governing Equations

1

1

1

1

T

P

V

2

2

2

2

T

P

VConservation of mass:

Conservation of momentum:

Rearranging:

Combining:

AVAV 2211

122221

121121

1221

VVVPP

VVVPP

VVmAPP

2

2121

22

1

212121

PP

VVV

PPVVV

21

22

2121

11VVPP

Conservation of energy:

Change of variable:

0

22

2

21

1 22Tc

VTc

VTc ppp

2

2

1

121

22 1

2

PP

VV

combine

22

2

221

1

1

1

2

1

2V

PV

P

Continued:

Multiplied by 2/p1:

Rearranging:

2

2

1

1

2121 1

211

PP

PP

1

2

1

2

1

2

1

2

1

211

P

P

P

P

1

2

1

2

1

2

11

111

P

P

1

2

1

2

1

2

11

111

PP

PP

or

Governing Equations cont.

1

2

1

2

2

1

1

2

11

111

PP

PP

V

V

2

1

1

2

1

2

P

P

T

T

2

1

1

2

1

2

11

11

PP

PP

T

T


From conservation of mass:

From equation of state:


2211 VV

222

211

2

2222

2111

1221

11 MPMP

Pa

VPVP

VVmAPP

22

21

1

2

22

2

21

1

21

1

21

1

22

M

M

T

T

Vh

Vh

C

O

M

BINE

Conservation of mass

Conservation of momentum

Conservation of energy

02

21

1

)2

11(

1

)2

11(

2

11

12

11

1

21

22

21

22

21

22

41

42

222

22

22

221

21

21

222

2

2212

1

1

222

211

1

1

2211

MM

MMMMMM

M

MM

M

MM

MM

MM

M

M

RTMRT

PRTM

RT

P

VV

Expanding the equations


12

212

1

21

2

M

MM

Solution:

Mach number cannot be negative. So, only the positive value is realistic.


1

1

1

2

1

1

121

11

22

11

21

1

21

1

21

1

2

22

21

1

2

21

2

21

21

1

2

22

21

1

2

M

P

P

M

M

P

P

M

MM

T

T

M

M

T

T

2)1(

)1(

121

11

22

11

1221

21

21

1

2

21

2

21

21

21

21

1

1

2

2

1

2

1

2

1

1

2

M

M

M

MM

MM

M

T

T

M

M

V

V

Temp. ratio

Pres. ratio

Dens. ratio

Simplifying:

1

2

3

Stagnation pressures:

Other relations:

1

12

21

1

21

1 21

1

21

22

01

02

1

2

01

1

2

02

01

02

M

M

M

P

P

P

P

P

P

P

P

P

P

2

02

02

01

2

01

1

01

01

02

1

02

P

P

P

P

P

P

P

P

P

P

P

P


Entropy change:

But, S02=S2 and S01=S1 because the flow is all isentropic before and after shockwave.

So, when applied to stagnation points:

But, flow across the shock wave is adiabatic & non-isentropic:

And the stagnation entropy is equal to the static entropy:

So:

Shock wave

1 2

1

2

1

212 lnln

P

PR

T

Tcss p

01

02

01

020102 lnln

P

PR

T

Tcss p

0201 TT

1ln 1201

020102

ss

P

PRss

1exp 12

01

02

R

ss

P

P Total pressure decreases across shock wave !


Group Exercises 3

1. Consider a normal shock wave in air where the upstream flow properties are u1=680m/s, T1=288K, and p1=1 atm. Calculate the velocity, temperature, and pressure downstream of the shock.

2. A stream of air travelling at 500 m/s with a static pressure of 75 kPa and a static temperature of 150C undergoes a normal shock wave. Determine the static temperature, pressure and the stagnation pressure, temperature and the air velocity after the shock wave.

3. Air has a temperature and pressure of 3000K and 2 bars absolute respectively. It is flowing with a velocity of 868m/s and enters a normal shock. Determine the density before and after the shock.

0sM

11 M 12 M

01

01

1

1

1

T

P

T

P

0102

0102

12

12

12

TT

PP

TT

PP

1M 2M1

2

P

P

1

2

T

T

1

2

1

2

a

a

01

02

P

P

1

02

P

P

Stationary Normal Shock Wave Table – Appendix C:

gas dynamics esa 341 chapter 3

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