MAE 5420 - Compressible Fluid Flow

Homework1, Solution Set• A sample return Probe is being sent on a 1-year mission

from Earth to Mars Via Venus Using “aero-gravity” assist

( both both gravity And aerodynamics at Venus used to turn the

corner to Mars)

• The aero-assist maneuver at Venus is performed at

An altitude of 110 km above the Surface at a peak

Atmospheric velocity of 13.09 km/sec

• At 110 km altitude, the ambient temperature is 147.63°k

• Assume that the Venutian atmospheric

Composition by volume is {97% CO2, 3% N2}

• Calculate the Probe Mach number at the

Venus aero-assist interface

MAE 5420 - Compressible Fluid Flow

Homework1, Solution Set (cont’d)

• First Calculate the Molecular weight of Venutian “air”

-- Composition by volume is {97% CO2, 3% N2}

• Molecular of various gases

Mw =

kg/kg-mole

• Now compute Gas constant

For Venutian “air”

Rg = J/°K-(kg)

MAE 5420 - Compressible Fluid Flow

Homework1, Solution Set (cont’d)

• Calculate cp, cv, of Venutian “air”

-- Composition by volume is {97% CO2, 3% N2}

cv = cp - Rg=

cp At -100 °C

CO2 ~ 0.845kJ/kg °K

N2 ~ 0.995kJ/kg °K

J/°K-(kg)

!

Note Units!

cp

_

=1

Mw

Vfrac jMw j

cp j( )

j

! =

= 847.90

1000

43.520.97 12 2 16!+( ) 0.845 0.03 2 14!( ) 0.995( )+( )

J/°K-(kg)

847.9 191.0481! = 656.85

! =cp

cv

=847.90

656.85= 1.2908

MAE 5420 - Compressible Fluid Flow

Homework1, Solution Set (cont’d)

• Now Compute sonic velocity at 110 km altitude

c = ! RgT =

m/sec

Check units!J

oK ! kg

0K( )

"#$

%&'=

Nt ! moK ! kg

0K( ) =

kg ! msec

2! m

oK ! kg

0K( ) =

m

sec

1.2908 191.0481 147.63! !( )1 2/ = 190.804

MAE 5420 - Compressible Fluid Flow

Homework1, Solution Set (cont’d)

• Finally Compute Mach Number

M =V

! RgT

=

Wow! We’ll find out later that this is “really hot”

13.09 1000!

1.2908 191.0481 147.63! !( )1 2/ = 68.604

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

• Show that for a reversible process

• and that for a reversible, adiabatic process

T2

T1

!

"#

$

%& =

'2

'1

!

"#

$

%&

( )1

s2! s

1= c

vln

T2

T1

"

#$

%

&' + Rg ln

(1

(2

"

#$

%

&'

MAE 5420 - Compressible Fluid Flow

Alternate Derivation

40

Start From Slide 15

Equation of State→p2

p1

=ρ2 ⋅Rg ⋅T2

ρ1 ⋅Rg ⋅T1

=ρ2 ⋅T2

ρ1 ⋅T1

Substitute→p2p1→ s2−s1= cp ⋅ ln

T2T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟−Rg ⋅ ln

ρ2 ⋅T2ρ1 ⋅T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

split lnρ2 ⋅T2

ρ1 ⋅T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= ln

ρ2 ⋅T2

ρ1 ⋅T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= ln

ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟+ ln

T2

T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

MAE 5420 - Compressible Fluid Flow

Alternate Derivation (2)

40

Substitute→ s2−s1= cp−Rg( )⋅ ln T2

T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟−Rg ⋅ ln

ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

From Derivation, Slide 8→ cp−cv = Rg→ cp−Rg = cv

Substitute→ s2−s1= cv ⋅ lnT2

T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟−Rg ⋅ ln

ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

MAE 5420 - Compressible Fluid Flow

Alternate Derivation (3)

41

→ flip sign on final term s2−s1= cv ⋅ lnT2

T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟+ Rg ⋅ ln

ρ1

ρ2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

Isentropic Process→

Δs= 0→ divide by cv→ lnT2

T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟=

Rg

cv

⋅ lnρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟=

cp−cv

cv

⋅ lnρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= γ−1( )⋅ ln ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= ln

ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

γ−1⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

lnT2

T1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟= ln

ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

γ−1⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟→Take exponential of both sides→

T2

T1

=ρ2

ρ1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

γ−1

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

• From Second law, for a reversible process

• From the First law, for a reversible process

• Combining equations

ds =dq

T

de = dq ! pdv

Tds = de + pdv

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

• From definition of cv

• Combining equations

cv=

de

dT

!"#

$%&v

de = cvdT ! > ds =cvdT

T+pdv

T=

cvdT

T+RgT

v

dv

T= cv

dT

T+ Rg

dv

v

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

• Integrating from state 1 to state 2

s2! s

1= cv

dT

T+ Rg

dv

v

"#$

%&'

1

2

( =

cv lnT2

T1

"

#$

%

&' + Rg ln

v2

v1

"

#$

%

&' = cv ln

T2

T1

"

#$

%

&' + Rg ln

)1

)2

"

#$

%

&'

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

• Finally, for an isentropic process, ds = 0

T2

T1

!

"#

$

%& =

'2

'1

!

"#

$

%&

Rg

cv

='2

'1

!

"#

$

%&

cp(cvcv

='2

'1

!

"#

$

%&

) (1

• Solving for T2/T1

cvln

T2

T1

!

"#

$

%& + Rg ln

'1

'2

!

"#

$

%& = 0

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

• Show that for an ideal gas the following

Useful relationships hold

cp=

!

! "1Rg

cv=

1

! "1Rg

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

Rg= c

p! c

v" c

p1!

cv

cp

#

$%%

&

'((= c

p1!

1

)#

$%

&

'( = cp

) !1)

#

$%

&

'(

" cp=

)) !1

Rg

cv= c

p!cv

cp

= cp!1

"=1

"!

"

" #1Rg=

1

" #1Rg

MAE 5420 - Compressible Fluid Flow

Homework, Section 1 (cont’d)

Rg= c

p! c

v" c

p1!

cv

cp

#

$%%

&

'((= c

p1!

1

)#

$%

&

'( = cp

) !1)

#

$%

&

'(

" cp=

)) !1

Rg