Section 3: Lecture 2 Alternative Forms of the One

MAE 5420 - Compressible Fluid Flow 1

Section 3: Lecture 2 ��Alternative Forms of the One-Dimensional

Energy Equation

Anderson: Chapter 3 pp. 71-86


Review: 1-D, Steady, Flow: Collected Equations

• Continuity

• Momentum

• Energy

→ ρiVi = ρeVe

→ pi + ρiVi2 = pe + ρeVe

2

→ q•+ hi +

Vi2

2= he +

Ve2

2


Review:��Effect of Mach Number on Flow Properties

Mach number is a measure of the ratio of the fluid Kinetic energy to the fluid internal energy (direct motion To random thermal motion of gas molecules)

V 2

2e

=V 2

2cvT

=V 2

2Rgγ −1⎛

⎝⎜⎞

⎠⎟T=

γγ

V 2

2Rgγ −1⎛

⎝⎜⎞

⎠⎟T=γ γ −1( )2

V 2

γ RgT=γ γ −1( )2

M 2


Review:��Effect of Mach Number on Flow Properties

dVdA

⎛⎝⎜

⎞⎠⎟=

1M 2 −1( )

VA

dpdA

=−1

M 2 −1( )ρV 2

A

M < 1→ dVdA

⎛⎝⎜

⎞⎠⎟< 0→ dp

dA⎛⎝⎜

⎞⎠⎟> 0

M > 1→ dVdA

⎛⎝⎜

⎞⎠⎟> 0→ dp

dA⎛⎝⎜

⎞⎠⎟< 0


Review:��Fundamental Properties of Supersonic��

and Supersonic Flow


Stagnation Temperature for the��Adiabatic Flow of a Calorically Perfect Gas • Consider an adiabatic flow field with a local gas Temperature T(x), pressure p(x), and a velocity V(x)

x

T(x)

V(x)

p(x)

• Since the Flow is adiabatic

h(x) + V (x)2

2= cpT (x) +

V (x)2

2= Const

Stephen Whitmore


Stagnation Temperature for the��Adiabatic Flow of a Calorically Perfect Gas

(cont’d) • Now define a condition a some location x’, within this flow field where The gas velocity is reduced to zero with no heat loss

• Since the Flow is adiabatic

y

To(y)

V(y)=0

po(y)

cpT (x) +V (x)2

2= cpT0 → To = T (x) +

V (x)2

2cp

T0(x’)

P0(x’) V (x’)=0

x’

Stephen Whitmore


Stagnation Temperature for the��Adiabatic Flow of a Calorically Perfect Gas

(cont’d) • From Earlier Analysis

V 2

2cvT

=γ γ −1( )2

M 2 →

V 2

2γ cv=

V 2

2cpcvcv=V 2

2cp= T

γ −1( )2

M 2

• Therefore

To = T (x) + T (x)γ −1( )2

M (x)2Holds anywhere Within an adiabatic Flow field

Stephen Whitmore


Stagnation Temperature for the��Adiabatic Flow of a Calorically Perfect Gas (cont’d)

• In general for an adiabatic Flow Field the Stagnation Temperature is defined by the relationship

• Stagnation Temperature is Constant Throughout An adiabatic Flow Field

• T0 is also sometimes referred to at Total Temperature

• T is sometimes referred to as Static Temperature

T0T= 1+

γ −1( )2

M 2


Stagnation Temperature for the��Adiabatic Flow of a Calorically Perfect Gas (cont’d)

• Stagnation temperature is a measure of the Kinetic Energy of the flow Field.

• Largely responsible for the high Level of heating that occurs on high speed aircraft or reentering space Vehicles …

T0 = T ⋅ 1+γ −12

⋅M 2⎛⎝⎜

⎞⎠⎟


Hypersonic Flow of a “Real Gas” ?

Let’s Revisit Hypersonic Flow


Hypersonic Flow of a Real Gas (cont’d)


Hypersonic Flow of a Real Gas (cont’d)

Stephen Whitmore


Hypersonic Flow of a Real Gas (cont’d) ��


Hypersonic Flow of a Real Gas (cont’d) ��•Hypersonic vehicles dissipate so much kinetic energy and Produce such high temperatures due to shock wave that they cause chemical changes to occur in the fluid through which they fly.

• Gas chemistry effects are important factor in dissipation of heat on hypersonic vehicles



cp =∂h∂T

→ cp ,cv ,γ{ }≠Const



cp =∂h∂T

→ cp ,cv ,γ{ }≠Const Vibration

Dissociation

Ionization



Stephen Whitmore



Vibration Species remain intact

Dissociation

Ionization

Creates different gas species

Creates different, electrically-charges gas species


What was Stagnation Temperature��At Columbia Breakup?

Loss Of Signal at: 61.2 km altitude ~18.0 Mach Number

T∞ ~ 243 ΟK

T0T= 1+

γ −1( )2

M 2

• We’ll revisit this Again later when we talk About Normal Shockwaves And Hypersonic Flow

Shockwaves ar generally “Non-Isentropic”

Hypersonic Shockwaves are “calorically imperfect”

~ 16,000 K ! (15,726 oC)


What was Stagnation Temperature��At Columbia Breakup (cont’d) Wow! Hot!

Ideal gas Variable Real gas

• Real Gas Shockwaves Are “non-adiabatic”

γγ = 1.4 Real Gas Calculations

Vibration mode absorbs significant energy Real gas effects make human spaceflight possible!

Stephen Whitmore


Stagnation Pressure for the��Isentropic Flow of a Calorically Perfect Gas • Now Consider an Isentropic flow field with a local gas Temperature T(x), pressure p(x), and a velocity V(x)

x

T(x)

V(x)

p(x)

• Since the Flow is isentropic, from Section 1

p0p =

T0T⎡

⎣⎢⎤

⎦⎥

γ

γ −1= 1+

γ −1( )2

M 2⎡

⎣⎢

⎤

⎦⎥

γ

γ −1“stagnation” (total) pressure: Constant throughout Isentropic flow field

ρ0ρ

=T0T⎡

⎣⎢⎤

⎦⎥

γ

γ −1= 1+

γ −1( )2

M 2⎡

⎣⎢

⎤

⎦⎥

1γ −1Also, analogous

value for density

Stephen Whitmore


Characteristic or “Sonic” Flow Parameters ��for Isentropic Flow Fields

x

T(x)

V(x)

p(x)

y

To(y)

V(y)=0

po(y)

• Stagnation properties Of a flow field result when Flow is isentropically Slowed from local velocity To zero

y

To(y)

V(y)=0

po(y)

T0(x’)

P0(x’) V (x’)=0

x’


Characteristic or “Sonic” Flow Parameters ��for Isentropic Flow Fields (cont’d)

x

T(x)

V(x)

p(x) • Define a new set of parameters where flow field is Either decelerated (supersonic) or accelerated (subsonic) From local velocity to sonic velocity -- { T*,P*,ρ* …}

yT*P*c*

x’


Characteristic or “Sonic” Flow Parameters ��for Isentropic Flow Fields (cont’d)

T0T= 1+

γ −1( )2

M 2 →

T *

T0=

1

1+γ −1( )2

(1)2=

2γ +1

p*

p0=

1

1+γ −1( )2

(1)2⎡

⎣⎢

⎤

⎦⎥

γ

γ −1

=2

γ +1⎡

⎣⎢

⎤

⎦⎥

γ

γ −1

ρ0ρ

= 1+γ −1( )2

M 2⎡

⎣⎢

⎤

⎦⎥

1γ −1

→

ρ*

ρ0=

2γ +1⎡

⎣⎢

⎤

⎦⎥

1γ −1

Stephen Whitmore


Sonic Velocity, Mach Number Based on T*

• From the definition:

• Useful to define fictitious parameters

• Intermediate analysis tools to be used for normal shock equations

T * =2

γ +1T0

c* = γ RgT* → M * =

Vγ RgT

*


Sonic Velocity, Mach Number Based on T*��(cont’d)

• From Energy equation for adiabatic flow

cpT +V 2

2= cpT

* +c*2

2→

γγ −1

RgT +V 2

2=

γγ −1

RgT* +

c*2

2→

c2

γ −1+V 2

2=

c*2

γ −1+c*2

2=

γ +12 γ −1( )

c*2


Sonic Velocity, Mach Number Based on T*��(concluded)

• Reorganizing terms

2c /V( )2

γ −1+1 = γ +1

γ −1( )c* /V( )2 →

2M 2 γ −1( )

=γ +1

M *2 γ −1( )−1→

1M 2 =

12γ +1M *2 −

γ −1( )2

→

M 2 =1

12γ +1M *2 −

γ −1( )2

=2M *2

γ +1− γ −1( )M *2

• M* = “characteristic Mach number”

MAE 5420 - Compressible Fluid Flow

Characteristic Mach Number (M*)

29

• One undesirable characteristic of the true Mach number is that it is not directly indicative of the local flow velocity.

—> Sonic velocity itself is also function of temperature.

• Consider two fluid particles that are proceeding within the first stage of a compressor and that of a turbine section with the same velocity.

—> Because the compressor particle is exposed to a much lower temperature than the turbine particle, the sonic speeds at these locations will be significantly different, and so will the Mach numbers.

• Sometimes it is advantageous to replace the Mach number with an alternative nondimensional-velocity ratio

—> the magnitude of M* is more directly indicative of the local velocity. ! M* approaches a constant value at high velocities



Characteristic Sonic Velocity (C*)

31

C*

C


Characteristic Sonic Velocity (C*)

32

C*

C


Summary

• Adiabatic Flow Property: T0=Const (stagnation temperature)

• Isentropic Flow Property: P0=Const (stagnation pressure)

T0T= 1+

γ −1( )2

M 2

p0p = 1+

γ −1( )2

M 2⎡

⎣⎢

⎤

⎦⎥

γ

γ −1


34

Summary (concluded) • “Characteristic” Flow Properties

T * =2

γ +1T0 → 0.8333T0

p* = 2γ +1⎡

⎣⎢

⎤

⎦⎥

γ

γ −1p0 → 0.5283T0

a* = γ RgT* = Rg

2γγ +1

T0 → 18.3002 m

sec o K

To

M 2 =2M *2

γ +1− γ −1( )M *2 →2M *2

2.4 − 0.4M *2

“air at normal temperatures”

c*


Revisit De Laval Nozzle

• What Pressure Ratio “Choked” De Laval’s Nozzles?

Assume VI ≈ 0

for Mt =1→ pt = p* pIpt( )

=γ +1

2⎛

⎝⎜

⎞

⎠⎟

γγ+1→ steam ≡

γ ≈ 1.286pIpt=1.824


De Laval Nozzle(cont’d)

• Once Mt=1 throat is “Choked”

Exit pressure no longer has any influence on the massflow through the nozzle, only the upstream pressure value


Homework 3

• Solve for M* in terms of M

• Solve for the Mass Flow per Unit area in a 1-D, steady, isentropic duct flow field as function of T0, P0, M, γ, Rg��(hint start with continuity )

m•

Ac= ρV

AeAI At

then let.... ρ = γ Pγ RgT

! Plot result with M as independent variable

Stephen Whitmore


Homework 3 (cont’d)

• Allowing that for isentropic flow .. Also show that for quasi 1-D flow

m•

A=

γRg

p0T0

M

1+γ −1( )2

M 2⎡

⎣⎢

⎤

⎦⎥

γ +12 γ −1( )

i.e. … Show that …. for Quasi 1-D isentropic flow

P0

T0T= 1+ γ −1

2M 2⎛

⎝⎜

⎞

⎠⎟

P0p= 1+ γ −1

2M 2⎛

⎝⎜

⎞

⎠⎟

γγ−1

→!mA= P0

2γγ −1( ) Rg ⋅T0( )

pP0

⎛

⎝⎜

⎞

⎠⎟

2γ

−pP0

⎛

⎝⎜

⎞

⎠⎟

γ+1γ

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥


Homework 3 (cont’d)

• show that massflow per unit area has a maximum value when M=1

m•

A*=

γRg

2γ +1⎛

⎝⎜⎞

⎠⎟

γ +1γ −1( ) p0

T0

m•

A=

γRg

p0T0

M

1+γ −1( )2

M 2⎡

⎣⎢

⎤

⎦⎥

γ +12 γ −1( )

Show that …. In general for Quasi 1-D flow

Hint:

set..... ∂

∂M!mA

RgT 0

P0

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

∂

∂Mγ ⋅M

1+ γ −12

M 2⎛⎝⎜

⎞⎠⎟

γ +12(γ −1)

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 0....& solve for M


Homework (cont’d)

• Plot as a function of mach number

.. Does this plot agree with result on previous page?

• At what mach number does have the greatest value

• What does this result imply?

Assume γ= 1.4, Rg= 287.056 j/kg-°K

m•

Ac

T0p0

Rg

m•

Ac

T0p0

Rg

Documents

Section 3: Lecture 2 Alternative Forms of the One