fluid mechanics kundu cohen 6th edition solutions sm ch (15)

Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.1. Use (15.4), (15.5), and (15.6) to derive (15.7) when the body force is spatially uniform and the effects of viscosity are negligible. Solution 15.1. The goal is develop an equation with an inhomogeneous-medium convected-wave operator acting on p on the left side with source terms involving q, fi, and ∂ui/∂xj on the right.

First, use (15.6), DpDt

= c2 DρDt

, to eliminate Dρ/Dt from (15.4):

1ρDρDt

+∂ui∂xi

= q = 1ρc2

DpDt

+∂ui∂xi

, or 1ρc2

DpDt

= q− ∂ui∂xi

,

where the final equation is just a rearrangement of the first. Apply D/Dt to the second equation to find:

DDt

1ρc2

DpDt

!

"#

$

%&=

DqDt

−DDt

∂uj∂x j

. (a)

Here, the summed-over index in the final term has been switched from 'i' to 'j'. Next, apply –∂/∂xj to (15.5) to produce:

−∂∂x j

DujDt

−∂∂x j

1ρ∂p∂x j

#

$%%

&

'((= −

∂gj∂x j

−∂∂x j

1ρ

∂τ ij∂xi

#

$%

&

'(−

∂fi∂x j

.

Note that ∂gj/∂xj = 0 because the body force is spatially uniform, drop the viscous stress term, and move the remaining term that explicitly involves uj to the right side:

−∂∂x j

1ρ∂p∂x j

#

$%%

&

'((= −

∂fi∂x j

+∂∂x j

DujDt

. (b)

Add equations (a) and (b): DDt

1ρc2

DpDt

!

"#

$

%&−

∂∂x j

1ρ∂p∂x j

!

"##

$

%&&=

DqDt

−∂fi∂x j

+∂∂x j

DDt

−DDt

∂∂x j

!

"##

$

%&&uj , (c)

The remaining steps involve simplifying the final term. The first operator inside the final large parentheses may be rewritten:

∂∂x j

DDt

=∂∂x j

∂∂t+ui

∂∂xi

"

#$

%

&'=

∂2

∂t∂x j+∂ui∂x j

∂∂xi

+ui∂2

∂xi∂x j=∂ui∂x j

∂∂xi

+DDt

∂∂x j

.

Thus: ∂∂x j

DDt

−DDt

∂∂x j

#

$%%

&

'((uj =

∂ui∂x j

∂uj∂xi

, so (c) reduces to (15.7)

DDt

1ρc2

DpDt

!

"#

$

%&−

∂∂x j

1ρ∂p∂x j

!

"##

$

%&&=

DqDt

−∂f j∂x j

+∂ui∂x j

∂uj∂xi

.


Exercise 15.2. Derive (15.12) through the following substitution and linearization steps. Set q and fi to zero in (15.7) and insert the decompositions (15.9). Treat Ui, p0, ρ0 and T0 as time-invariant and spatially uniform, and drop quadratic and higher order terms involving the fluctuations !ui , p´, ρ´, and T´. Solution 15.2. Start with (15.7) and set q and fi to zero. This leaves:

DDt

1ρc2

DpDt

!

"#

$

%&−

∂∂x j

1ρ∂p∂x j

!

"##

$

%&&=

∂ui∂x j

∂uj∂xi

.

The decomposition equations are: ui =Ui + !ui , p = p0 + p´ , ρ = ρ0 + ρ´ , and T = T0 + T´ ,

and all derivatives of Ui, p0, ρ0, and T0 are zero. Consider the first term: DpDt

=∂∂t+ (Ui + "ui )

∂∂xi

#

$%

&

'( p0 + "p( ) = ∂

"p∂t

+Ui∂ "p∂xi

+ "ui∂ "p∂xi

.

Here the final term is quadratic in the small fluctuations, so DpDt

≅∂∂t+Ui

∂∂xi

#

$%

&

'( )p ,

and this term is linear in the fluctuation quantity p´. Thus, only the leading-order portion of the operator coefficient of Dp/Dt need be retained because any extra first-order contributions from this coefficient will wind up being second order when multiplied with Dp/Dt. Therefore, the first term of (15.7) reduces to:

DDt

1ρc2

DpDt

!

"#

$

%& ≅

∂∂t+Ui

∂∂xi

!

"#

$

%&1

ρoc2∂∂t+Ui

∂∂xi

!

"#

$

%& )p = ∂

∂t+Ui

∂∂xi

!

"#

$

%&

2

)p .

where c is the speed of sound at pressure p0 and temperature T0. The second term simplifies in a similar manner:

∂p∂x j

=∂∂x j

p0 + "p( ) = ∂"p

∂x j.

Again this term is linear in the fluctuation quantity p´, so its coefficient cannot contain any first-order contributions from fluctuation quantities

∂∂x j

1ρ∂p∂x j

"

#$$

%

&'' ≅

∂∂x j

1ρ0

∂ )p∂x j

"

#$$

%

&''=

1ρo

∂2 )p∂x j

2 .

The lone remaining right-side term is quadratic in the fluctuation velocity, ∂ui∂x j

∂uj∂xi

=∂ Ui + "ui( )

∂x j

∂ Uj + "uj( )∂xi

=∂ "ui∂x j

∂ "uj∂xi

,

and therefore should be dropped. So, term-by-term replacements in (15.7) using the results above lead to:

1c2

∂∂t+Ui

∂∂xi

!

"#

$

%&

2

'p − ∂ 2 'p∂xi∂xi

≅ 0 ,

which is (15.2)


Exercise 15.3. The field equation for acoustic pressure fluctuations in an ideal compressible fluid is (15.13). Consider one-dimensional solutions where p = p(x,t) and x = x1. a) Drop the x2 and x3 dependence in (15.13), and change the independent variables x and t to

€

ξ = x − ct and

€

ζ = x + ct to simplify (15.13) to

€

∂ 2 # p ∂ξ∂ζ

= 0 .

b) Use the simplified equation in part a) to find the general solution to the original field equation:

€

" p (x, t) = f (x − ct) + g(x + ct) where f and g are undetermined functions. c) When the initial conditions are: p´ = F(x) and ∂p´/∂t = G(x) at t = 0, show that:

€

f (x) =12

F(x) − 1c

G(x )dx 0

x∫

$

% & '

( ) , and

€

g(x) =12

F(x) +1c

G(x )dx 0

x∫

#

$ % &

' ( ,

where x is just an integration variable. Solution 15.3. a) The starting point is (15.13) simplified for no x2 and x3 dependence. This is the classical one-dimensional acoustic wave equation,

€

1c 2∂ 2 # p ∂t 2

−∂ 2 # p ∂x 2

= 0.

Use

€

ξ = x − ct and

€

ζ = x + ct , and convert the partial derivatives with respect to x and t into partial derivatives with respect to ξ and ζ.

€

∂∂x

=∂ξ∂x

∂∂ξ

+∂ζ∂x

∂∂ζ

=∂(x − ct)

∂x∂∂ξ

+∂(x + ct)

∂x∂∂ζ

=∂∂ξ

+∂∂ζ

, and

€

∂∂t

=∂ξ∂t

∂∂ξ

+∂ζ∂t

∂∂ζ

=∂(x − ct)

∂t∂∂ξ

+∂(x + ct)

∂t∂∂ζ

= −c ∂∂ξ

+ c ∂∂ζ

.

Thus:

€

∂ 2

∂x 2=

∂∂ξ

+∂∂ζ

%

& '

(

) * ∂∂ξ

+∂∂ζ

%

& '

(

) * =

∂ 2

∂ξ 2+ 2 ∂ 2

∂ξ∂ζ+∂ 2

∂ζ 2, and

€

∂ 2

∂t 2= c 2 ∂

∂ξ−∂∂ζ

&

' (

)

* + ∂∂ξ

−∂∂ζ

&

' (

)

* + = c 2

∂ 2

∂ξ 2− 2 ∂ 2

∂ξ∂ζ+∂ 2

∂ζ 2&

' (

)

* +

So,

€

∂ 2 # p ∂x 2

=1c 2∂ 2 # p ∂t 2

€

→ ∂2 $ p ∂x 2 =

∂ 2 $ p ∂ξ 2 + 2 ∂

2 $ p ∂ξ∂ζ

+∂ 2 $ p ∂ζ 2 =

1c 2 ⋅ c

2 ∂2 $ p ∂ξ 2 − 2 ∂ 2 $ p

∂ξ∂ζ+∂ 2 $ p ∂ζ 2

)

* +

,

- . =

1c 2∂ 2 $ p ∂t 2 .

Cancel common terms across the middle equality to find:

€

2 ∂2 # p

∂ξ∂ζ= −2 ∂

2 # p ∂ξ∂ζ

which implies

€

∂ 2 # p ∂ξ∂ζ

= 0 .

b) Use the result of part a) and integrate with respect to ξ to find:

€

∂ # p ∂ζ

= C ζ( ) where C(ζ) is a

function of integration that cannot depend on ξ. Now integrate this result with respect to ζ to find:

€

" p = C ζ( )∫ dζ + f (ξ) where f(ξ) is a second function of integration that cannot depend on ζ. Since C is undetermined at this point, a new function g can be defined to make the last equation look a little nicer:

€

g(ζ ) = C ζ( )∫ dζ . Thus, the general solution of the field equation for

€

" p (x, t) is:

€

" p (x, t) = f (ξ) + g(ζ ) = f (x − ct) + g(x + ct) . c) Use the result of part b) and the condition p´ = F(x) at t = 0 to find:

€

" p (x,0) = f (x) + g(x) = F(x) . (1)


Differentiate the result of part b) with respect to time and use the ∂p´/∂t = G(x) at t = 0 to find:

€

∂ # p ∂t$

% & '

( ) t= 0=

dfd(x − ct)

∂(x − ct)∂t

+df

d(x + ct)∂(x + ct)

∂t$

% & '

( ) t= 0= c −

df (x)dx

+dg(x)

dx+

, -

.

/ 0 = G(x) .

Use the last equality, divide by c, and integrate in x to find:

€

− f (x) + g(x) =1c

G(x )dx 0

x∫ + D, (2)

where D is a constant. Adding equations (1) and (2) together produces:

€

g(x) =12

F(x) +1c

G(x )dx 0

x

∫ + D#

$ %

&

' ( ,

while subtracting (2) from (1) produces:

€

f (x) =12

F(x) − 1c

G(x )dx 0

x

∫ −D$

% &

'

( ) .

Given that D drops out of the general solution,

€

" p (x, t) = f (x − ct) + g(x + ct) =12

F(x − ct) + F(x + ct) − 1c

G(x )dx 0

x−ct

∫ +1c

G(x )dx 0

x +ct

∫ −D + D%

& '

(

) *

=12

F(x − ct) + F(x + ct) +1c

G(x )dx x−ct

x +ct

∫%

& '

(

) * ,

it may be ignored without loss of generality, a situation that is also true for the lower limit of the x-integral. Therefore:

€

f (x) =12

F(x) − 1c

G(x )dx 0

x

∫$

% &

'

( ) and

€

g(x) =12

F(x) +1c

G(x )dx 0

x

∫#

$ %

&

' ( ,

where the integration lower limit, 0, is chosen for convenience.


Exercise 15.4. Starting from (15.15) use (15.14) to prove (15.16) Solution 15.4. Equation (15.15) is

€

" p (x, t) = f (x − ct) + g(x + ct) , and (15.14) is the linearized integrated Euler equation without the body force:

€

" u 1(x, t) = −1ρo

∂ " p ∂x

dt∫ .

Substitute (15.15) into (15.14), define the variables

€

ξ = x − ct and

€

ζ = x + ct , and note that

€

∂ξ ∂x = ∂ζ ∂x =1:

€

" u 1(x, t) = −1ρo

∂∂x

f (x − ct) + g(x + ct)( )dt∫ = −1ρo

dfdξ

∂ξ∂x

+dgdζ

∂ζ∂x

)

* +

,

- . dt∫ = −

1ρo

dfdξ

+dgdζ

)

* +

,

- . dt∫ .

Separate the two terms of the integration, change integration variables to ξ and ζ, and integrate to reach (15.16).

€

" u 1(x, t) = −1ρo

dfdξ∫ dt − 1

ρo

dgdζ

dt∫ = −1ρo

dfdξ∫ dξ

−c(

) *

+

, - −

1ρo

dgdζ

dζc

(

) *

+

, - ∫ =

1ρoc

f (ξ) − 1ρoc

g(ζ )

=1ρoc

f (x − ct) − g(x + ct)[ ]

From a strictly mathematical point of view, the integrations involved in these final steps should produce constants of integration. However, the average of an acoustic fluctuation – such as u´– is zero, so the constants of integration are zero.


Exercise 15.5. Consider two approaches to determining the upper Mach number limit for incompressible flow. a) First consider pressure errors in the simplest-possible steady flow Bernoulli equation. Expand (15.29) for small Mach number to determine the next term in the expansion:

€

p0 = p + 12 ρu

2 + ... and determine the Mach number at which this next term is 5% of p when γ = 1.4. b) Second consider changes to the density. Expand (15.30) for small Mach number and determine the Mach number at which the density ratio ρ0/ρ differs from unity by 5% when γ = 1.4. c) Which criterion is correct? Explain why the criteria for incompressibility determined in a) and b) differ, and reconcile them if you can. Solution 15.5. a) Start with (15.29),

€

p0 = p 1+γ −12

M 2$

% & '

( )

γ (γ −1)

,

and expand using the Taylor series

€

(1+ ε)β =1+ βε + 12 β(β −1)ε

2 + ... for ε << 1:

€

p0 = p 1+γ

γ −1$

% &

'

( ) γ −12

M 2$

% &

'

( ) +12

γγ −1$

% &

'

( )

γγ −1

−1$

% &

'

( ) γ −12

M 2$

% &

'

( ) 2

+ ...*

+ ,

-

. / .

Keep the first three terms inside the brackets and simplify using M2 = u2/c2 and c2 = γRT:

€

p0 ≅ p 1+γ2M 2 +

γ8M 4$

% & '

( ) = p 1+

γ2u2

γRT+γ8M 4

$

% &

'

( ) = p +

12

pRT*

+ ,

-

. / u2 +

γ8pM 4 .

Use the perfect gas law p = ρRT to further simplify the approximate equation:

€

p0 ≅ p +12ρu2 +

γ8M 4 p . ($)

The final term will be 5% of p when

€

γM 4 8 = 0.05 , or

€

M = 8(0.05) 1.4( )1 4 = 0.73. b) Start with (15.30) and expand as in part a),

€

ρ0ρ

= 1+γ −12

M 2%

& ' (

) *

1 (γ −1)

= 1+1

γ −1γ −12

M 2 +12

1γ −1+

, -

.

/ 0 1

γ −1−1

+

, -

.

/ 0 γ −12

M 2+

, -

.

/ 0 2

+ ...%

& '

(

) * .

Keep the first three terms inside the brackets and simplify.

€

ρ0ρ≅1+

M 2

2+2 − γ8

M 4

This time the third term is not needed. The ratio ρ0/ρ will differ from unity by 5% when

€

M 2

2= 0.05 , or

€

M = 2(0.05) = 0.32.

c) The two Mach number criteria differ by more than a factor of two, and the second one is correct. The main reason that the approximate Bernoulli equation ($) retains its accuracy to higher Mach number is that the second term on the right in ($) contains the gas density, ρ, which is not constant, and this term retains some Mach number dependence that is not explicitly shown. When ($) is expanded using ρ0, it becomes:

€

p0 − p ≅12 ρou

2 1−M 2 2( ) , which implies a 5% error in the pressure difference p0 – p when M = 0.32, the same as the part b) result.


Exercise 15.6. The critical area A∗ of a duct flow was defined in §4. Show that the relation between A∗ and the actual area A at a section, where the Mach number equals M, is that given by (15.31). This relation was not proved in the text. [Hint: Write

€

AA*

=ρ*c*

ρu=ρ*

ρ0

ρ0ρc*

ccu

=ρ*

ρ0

ρ0ρ

T*

T0T0T1M

.

Then use the other relations given in Section 15.4.] Solution 15.6. Start with the hint and continue the equality, noting that M* = 1 and using (15.28) and (15.30)

€

AA* =

ρ*

ρ0

ρ0

ρT*

T0

T0

T1M

= 1+γ −1

2M*2%

& ' (

) *

−1γ −1

1+γ −1

2M 2%

& ' (

) *

1γ −1

1+γ −1

2M*2%

& ' (

) *

−12

1+γ −1

2M 2%

& ' (

) *

12 1M

= 1+γ −1

2%

& ' (

) *

−1γ −1

1+γ −1

2M 2%

& ' (

) *

1γ −1

1+γ −1

2%

& ' (

) *

−12

1+γ −1

2M 2%

& ' (

) *

12 1M

=γ +1

2%

& ' (

) *

−1γ −1

1+γ −1

2M 2%

& ' (

) *

1γ −1 γ +1

2%

& ' (

) *

−12

1+γ −1

2M 2%

& ' (

) *

12 1M

=γ +1

2%

& ' (

) *

−1γ −1

+12

+

, -

.

/ 0

1+γ −1

2M 2%

& ' (

) *

+1γ −1

+12

+

, -

.

/ 0 1M

=2

γ +11+

γ −12

M 2+

, -

.

/ 0

%

& '

(

) *

+12γ +1γ −1+

, -

.

/ 0

.


Exercise 15.7. A perfect gas is stored in a large tank at the conditions specified by p0, T0. Calculate the maximum mass flow rate that can exhaust through a duct of cross-sectional area A. Assume that A is small enough that during the time of interest p0 and T0 do not change significantly and that the flow is isentropic. Solution 15.7. The mass flux through the tube will be:

!m = ρUA = ρρ0ρ0Uccc0c0A = 1+

γ −12

M 2"

#$%

&'

−1 (γ−1)

M 1+ γ −12

M 2"

#$%

&'

−1 2

ρ0 γRT0A ,

where the subscript '0' indicates reservoir (or stagnation conditions) and (15.28) and (15.30) have been used to introduce the factors involving the Mach number M. This equation can be simplified by collecting like factors, and by using p0 = ρ0RT0 to eliminate ρ0.

!m =M

1+ γ −12

M 2"

#$%

&'

12(γ+1)(γ−1)

p0 γ

RT0A .

To determine the maximum flow rate differentiate with respect to M, set the derivative equal to zero, and solve for M, the mach number where the mass flow rate will be maximum.

d !mdM

=1

1+ γ −12

M 2"

#$%

&'

12(γ+1)(γ−1)

p0 γ

RT0A− 1

2(γ +1)(γ −1)

M

1+ γ −12

M 2"

#$%

&'

12(γ+1)(γ−1)

+1

γ −122M

(

)*

+

,-p0 γ

RT0A = 0 .

Divide out common factors and simplify:

1+ γ −12

M 2 −12(γ +1)(γ −1)

M γ −1( )M =1+ γ −12

M 2 +γ +12

M 2 =1−M 2 = 0 .

The two roots are M = ± 1, corresponding to the maximum mass flux out-of or into the tank. Here we presume outflow is of interest M = +1, so:

!mmax =1

1+ γ −12

"

#$%

&'

12(γ+1)(γ−1)

p0 γ

RT0A = 2

γ +1"

#$

%

&'

12(γ+1)(γ−1) γ p0

RT0A = 0.685 p0

RT0A ,

where the final numerical value applies when γ = 1.40.


Exercise 15.8. The entropy change across a normal shock is given by (15.43). Show that this reduces to expressions (15.44) for weak shocks. [Hint: Let M1

2 −1 << 1. Write the terms within the two sets of brackets in equation (15.43) in the form [1 + ε1] [1 + ε2]γ, where ε1 and ε2 are small quantities. Then use the series expansion ln(1 + ε) = ε – ε2/2 + ε3/3 + … . This gives equation (15.44) times a function of M1 which can be evaluated at M1 = 1.] Solution 15.8. From (15.43)

S2 − S1cv

= ln 1+ 2γγ +1

M12 −1( )

"

#$

%

&'(γ −1)M1

2 + 2(γ +1)M1

2

"

#$

%

&'

γ()*

+*

,-*

.*.

The term in the first set of [,]-brackets is already in an easily managed form:

€

1+2γγ +1

M12 −1( ) ≡1+ ε1.

The term in the second set of [,]-brackets can be rearranged as follows:

€

(γ −1)M12 + 2

(γ +1)M12 =

(γ +1)M12 − 2M1

2 + 2(γ +1)M1

2 =1− 2(M12 −1)

(γ +1)M12 ≡1−ε2 .

Together these two results produce: S2 − S1cv

= ln 1+ε1[ ] 1+ε2[ ]γ{ }= ln 1+ε1( )+γ ln 1−ε2( ) .

Now expand the natural logarithm functions: S2 − S1

cv

= ε1 −ε1

2

2+ε1

3

3+...+γ −ε2 −

ε22

2−ε2

3

3+...

"

#$

%

&'

≅ (ε1 −γε2 )− 12ε1

2 +γε22( )+ 1

3ε1

3 −γε23( )

.

Evaluate the terms.

€

ε1 − γε2 =2γγ +1

M12 −1( ) − γ 2(M1

2 −1)(γ +1)M1

2 =2γ(M1

2 −1)γ +1

1− 1M1

2

%

& '

(

) * =2γ(M1

2 −1)2

(γ +1)M12

€

12ε12 + γε2

2( ) =12

4γ 2

(γ +1)2M1

2 −1( )2

+ γ4(M1

2 −1)2

(γ +1)2M14

%

& '

(

) * =

2γ(γ +1)2

M12 −1( )

2γ +

1M1

4

+

, -

.

/ 0

€

13ε13 − γε2

3( ) =13

8γ 3

(γ +1)3M1

2 −1( )3− γ8(M1

2 −1)3

(γ +1)3M16

%

& '

(

) * =

8γ3(γ +1)3

M12 −1( )

3γ 2 −

1M1

6

+

, -

.

/ 0

Reconstruct the entropy difference and simplify. S2 − S1

cv

≅2γ (M1

2 −1)2

(γ +1)M12 −

2γ (M12 −1)2

(γ +1)2 γ +1M1

4

#

$%

&

'(+

8γ (M12 −1)3

3(γ +1)3 γ 2 −1M1

6

#

$%

&

'(

= 2γ (M12 −1)2

(γ +1)1M1

2 −γγ +1

−1

(γ +1)M14 +

4(M12 −1)

3(γ +1)2 γ 2 −4(M1

2 −1)3(γ +1)2M1

6

#

$%

&

'(

= 2γ (M12 −1)2

(γ +1)3(γ +1)2M1

2 −3γ (γ +1)M14 −3(γ +1)+ 4(M1

2 −1)γ 2M14 − 4(M1

2 −1)M1−2

3(γ +1)2M14

#

$%

&

'(


S2 − S1

cv

≅2γ (M1

2 −1)2

(γ +1)3(γ +1) (−γM1

2 +1)(M12 −1)#$ %&+ 4(M1

2 −1)γ 2M14 − 4(M1

2 −1)M1−2

3(γ +1)2M14

#

$''

%

&((

= 2γ (M12 −1)3

(γ +1)3(γ +1)(−γM1

2 +1)+ 4γ 2M14 − 4M1

−2

3(γ +1)2M14

#

$'

%

&(

Given that the first term is cubic in the small quantity, the contents of the final [,]-brackets can be evaluated at M1 = 1. This produces:

S2 − S1

cv

≅2γ (M1

2 −1)3

(γ +1)3(γ +1)(−γ +1)+ 4γ 2 − 4

3(γ +1)2

#

$%

&

'(=

2γ (M12 −1)3

(γ +1)3(−γ +1)+ 4γ − 4

3(γ +1)#

$%

&

'(

= 2γ (M12 −1)3

(γ +1)γ −1

3(γ +1)#

$%

&

'(=

2γ (γ −1)3(γ +1)2 (M1

2 −1)3,

which is (15.44a).


Exercise 15.9. Show that the maximum velocity generated from a reservoir in which the stagnation temperature equals T0 is umax = 2cpTo . What are the corresponding values of T and M? Solution 15.9. If the flow starts from zero velocity within a reservoir is T0, then the energy equation is:

€

h0 = h +12u2 .

Thus, the maximum flow speed will occur when h = cpT = 0, and this implies: u = 2cpT0 ,

€

T→ 0, and

€

M→∞. Such a flow (nearly) occurs in practice when a compressed gas expands into the low-temperature vacuum of outer space.


Exercise 15.10. In an adiabatic flow of air through a duct, the conditions at two points are u1 = 250 m/s, T1 = 300 K, p1 = 200 kPa, u2 = 300 m/s, p2 = 150 kPa. Show that the loss of stagnation pressure is nearly 34.2 kPa. What is the entropy increase? Solution 15.10. First, compute the speed of sound at the upstream location:

€

c1 = γRT1 = (1.4)(287)(300) = 347.2ms−1. Therefore, the mach number is: M1 = u1/c1 = 250/347.2 = 0.72. From table 15.1,

€

p1 p01 = 0.7080 , and

€

T1 T01 = 0.9061, so

€

p01 = p1 0.7080 = 200 0.7080 = 282.5kPa, and T01 = T1 0.9061= 300 0.9061= 331.1K . The stagnation enthalpy at location '1' is:

h01 = cpT1 +12 u

2 = (1004)(300)+ 0.5(250)2 = 332.5kJkg−1 . In adiabatic flow, h01 = h02, therefore:

€

h01 = h2 + 12 u2

2 , or

€

h2 = h01 −12 u2

2 = 332.5kJkg−1 − 0.5(300)2 = 287.5kJkg−1. The temperature at location '2' is:

T2 = h2 cp = 287.5 1.004 = 286.4K so the speed of sound at location '2' is:

€

c2 = γRT2 = (1.4)(287)(287.4) = 339.2ms−1. Thus, the mach number at location '2' is: M2 = u2/c2 = 300/339.2 = 0.884. From table 15.1, using the entry for M = 0.88.

€

p2 p02 = 0.6041, so

€

p02 = p2 0.6041=150 0.6041= 248.3kPa . The loss of stagnation pressure and the entropy gain are:

€

p01 − p02 = 282.5 − 248.3 = 34.2kPa, and s2 − s1 = cp ln T02 T01( )− R ln p02 p01( ) = 0− R ln p02 p01( ) = −(287)ln 248.3 282.5( ) = 37.03m2s−2K −1


Exercise 15.11. A shock wave generated by an explosion propagates through a still atmosphere. If the pressure downstream of the shock wave is 700 kPa, estimate the shock speed and the flow velocity downstream of the shock. Solution 15.11. The analysis can be done in a shock-fixed frame of reference, and p1 = 101.3 kPa and T1 = 295K, are taken as the nominal atmospheric conditions. The shock's pressure ratio is:

€

p2p1

=700kPa101.3kPa

= 6.91.

Locating the closest value in Table 15.2 produces: M1 = 2.46, M2 = 0.517, and

€

T2 T1 = 2.098. Therefore:

u1 = shock speed =

€

M1c1 = 2.46 1.4(287)295 = 847ms−1, and T2 = 2.098T1 = 2.098(295K) = 619K.

Thus, u2 = downstream flow speed =

€

M2c2 = (0.517) 1.4(287)619 = 258ms−1. So, in a frame of reference where the upstream air is still, the velocity downstream of the shock will be:

u1 – u2 = 847 – 258 = 589 m/s in the direction of shock propagation.


Exercise 15.12. Prove the following formulae for the jump the conditions across a stationary normal shock wave:

p2 − p1p1

=2γγ +1

M12 −1( ) , u2 −u1

c1= −

2γ +1

M1 −1M1

"

#$

%

&' , and υ2 −υ1

υ1= −

2γ +1

1− 1M1

2

"

#$

%

&' ,

where υ = 1/ρ, and the subscripts '1' and '2' imply upstream and downstream conditions, respectively. Solution 15.12. The pressure ratio condition (15.39) is:

p2p1=1+ 2γ

γ +1M1

2 −1"# $% .

Subtract '1' from both sides, and rearrange on the left to find: p2p1−1= p2 − p1

p1=2γγ +1

M12 −1"# $% .

The velocity ratio condition (15.41) is: u1u2=

(γ +1)M12

(γ −1)M12 + 2

.

Flip this over, subtract '1' from both sides, multiply by u1/c1, and simplify to find: u2

u1

−1"

#$

%

&'u1

c1

=u2 −u1

c1

=(γ −1)M1

2 + 2(γ +1)M1

2 −1"

#$

%

&'u1

c1

=(γ −1)M1

2 + 2− (γ +1)M12

(γ +1)M12

"

#$

%

&'M1

= 2− 2M12

(γ +1)M12

"

#$

%

&'M1 = −

2γ +1

M1 −1M1

"

#$

%

&'.

The specific volume condition can be obtained from the density ratio condition (15.41): ρ2ρ1=υ1υ2

=(γ +1)M1

2

(γ −1)M12 + 2

,

Flip this over, subtract '1' from both sides, and simplify to find: υ2

υ1

−1"

#$

%

&'=

υ2 −υ1

υ1

=(γ −1)M1

2 + 2(γ +1)M1

2 −1= (γ −1)M12 + 2− (γ +1)M1

2

(γ +1)M12

= 2− 2M12

(γ +1)M12 = −

2γ +1

1− 1M1

2

"

#$

%

&'.


Exercise 15.13. Using (15.1i), and (15.43), determine formulae for p02/p01 and ρ02/ρ01 for a normal shock wave in terms of M1 and γ. Is there anything notable about the results? Solution 15.13. Equation (15.1i) applied to stagnation conditions implies:

s2 − s1cp

= ln T02T01

"

#$

%

&'−

Rcpln p02

p01

"

#$

%

&'

Flow through the shock wave is adiabatic so T02 = T01, and

s2 − s1cp

= −Rcpln p02

p01

"

#$

%

&'=

cvcpln 1+ 2γ

γ +1M1

2 −1( )(

)*

+

,-(γ −1)M1

2 + 2(γ +1)M1

2

"

#$

%

&'

γ./0

10

230

40.

A little algebraic rearrangement using (15.1d) and (15.1g) leads to:

p02p01

= 1+ 2γγ +1

M12 −1( )

"

#$

%

&'

−1γ−1 (γ +1)M1

2

(γ −1)M12 + 2

(

)*

+

,-

γγ−1

.

The formula for the stagnation density can be found similarly. Equation (15.1i) applied to stagnation conditions implies:

s2 − s1cv

= ln T02T01

"

#$

%

&'−

Rcvln ρ02

ρ01

"

#$

%

&'

Flow through the shock wave is adiabatic so T02 = T01, and

s2 − s1cv

= −Rcvln ρ02

ρ01

"

#$

%

&'= ln 1+ 2γ

γ +1M1

2 −1( )(

)*

+

,-(γ −1)M1

2 + 2(γ +1)M1

2

"

#$

%

&'

γ./0

10

230

40.

A little algebraic rearrangement using (15.1d) and (15.1g) leads to:

ρ02ρ01

= 1+ 2γγ +1

M12 −1( )

"

#$

%

&'

−1γ−1 (γ +1)M1

2

(γ −1)M12 + 2

(

)*

+

,-

γγ−1

.

Yes, the results are 'notable' because they are exactly the same.


Exercise 15.14. Using dimensional analysis, G. I. Taylor deduced that the radius r(t) of the blast wave from a large explosion would be proportional to (E/ρ1)1/5t2/5 where E is the explosive energy, ρ1 is the quiescent air density ahead of the blast wave, and t is the time since the blast (see Example 1.10). The goal of this problem is to (approximately) determine the constant of proportionality assuming perfect-gas thermodynamics. a) For the strong shock limit where M1

2 >>1, show: <Begin Equation>

ρ2ρ1≅γ +1γ −1

, T2T1≅γ −1γ +1

p2p1

, and u1 =M1c1 ≅γ +12

p2ρ1

.

</End Equation> b) For a perfect gas with internal energy per unit mass e, the internal energy per unit volume is ρe. For a hemispherical blast wave, the volume inside the blast wave will be 2

3 πr3 . Thus, set

ρ2e2 = E23 πr

3 , determine p2, set u1 = dr/dt, and integrate the resulting first-order differential equation to show that r(t) = K(E/ρ1)1/5t2/5 when r(0) = 0 and K is a constant that depends on γ. c) Evaluate K for γ = 1.4. A full similarity solution of the non-linear gas-dynamic equations in spherical coordinates produces K = 1.033 for γ = 1.4 (see Thompson 1972, p. 501). What is the percentage error in this exercise's approximate analysis? Solution 15.14. Start from the normal shock jump conditions (15.39) - (15.42):

€

p2p1

=1+2γγ +1

M12 −1( ) ,

€

M22 = 1+

γ −12

M12$

% &

'

( ) γM1

2 −γ −12

$

% &

'

( ) ,

€

ρ2ρ1

=(γ +1)M1

2

(γ −1)M12 + 2

, and

€

T2T1

=1+2(γ −1)(γ +1)2

γM12 +1M1

2

$

% &

'

( ) M1

2 −1( ) .

Simplify these four equations for

€

M12 >>1,

€

p2p1≅2γγ +1

M12 ,

€

M22 ≅

γ −12

M12 γM1

2 =γ −12γ

,

€

ρ2ρ1≅(γ +1)M1

2

(γ −1)M12 =

γ +1γ −1

, and

€

T2T1≅2(γ −1)(γ +1)2

γM12

M12

%

& '

(

) * M1

2 =2γ(γ −1)(γ +1)2

M12 .

The third equation is in the correct form so it does not need further manipulation. Use the first and fourth equation to eliminate

€

M12 from the expression for the temperature ratio:

€

T2T1≅2γ(γ −1)(γ +1)2

M12 =2γ(γ −1)(γ +1)2

γ +12γ

p2p1

=γ −1γ +1

p2p1

,

which is another result in the correct form. To reach the final part a) result, invert the strong-shock pressure relationship to find and expression for M1:

€

M1 ≅γ +12γ

p2p1

$

% &

'

( )

1 2

, so that u1 =M1c1 ≅γ +12γ

p2p1

"

#$

%

&'

1 2

γRT1 =γ +12γ

p2p1

"

#$

%

&'

1 2

γp1ρ1

=γ +12

p2ρ1

"

#$

%

&'

1 2

,

where R is the gas constant, and the final equality completes the effort for part a). b) As stated in the question, set

ρ2e2 =E

23 πr

3 = ρ2cvT2 = cvp2R=

p2γ −1

, and solve for p2 = γ −1( ) E23 πr

3 .


Put this into the final result of part a) and set u1 = dr/dt to find:

u1 =drdt=γ +12

p2ρ1

!

"#

$

%&

1 2

=γ +12

γ −1( ) E23 ρ1πr

3

!

"##

$

%&&

1 2

, which implies r3 2 drdt= γ 2 −1( ) 3E4ρ1π"

#$

%

&'

1 2

.

This non-linear first-order differential equation integrates to:

r5 2

5 2= γ 2 −1( ) 3E4ρ1π"

#$

%

&'

1 2

t + const , or r(t) = 52

γ 2 −1( ) 34π"

#$

%

&'1 2(

)**

+

,--

2 5Eρ1

"

#$

%

&'

1 5

t2 5 ,

where the initial condition, r(0) = 0, has been used to determine const = 0.

c) For γ = 1.40, K =

€

52

γ 2 −1( ) 34π%

& '

(

) * 1 2+

, -

.

/ 0

2 5

= 1.07, a difference of less than 4% from a similarity

solution of the non-linear gas-dynamic equations. Thus, this simpler analysis has certainly been worthwhile.


Exercise 15.15. Starting from the set (15.45) with q = 0, derive (15.47) by letting station (2) be a differential distance downstream of station (1). Solution 15.15. The equation set (15.45) with q = 0 and the second location a differential distance dx downstream of the first location is:

€

d ρu( ) = 0 ,

€

d p + ρu2( ) = −p1df , and

€

d h + 12 u

2( ) = 0. Use thermodynamic relationships to determine h in terms of p and ρ.

h = cpT = cppρR

=γγ −1

pρ

.

Expand all three equations:

€

ρdu + udρ = 0 ,

€

dp + ρdu2 + u2dρ = −p1df , and

€

γγ −1

dpρ−

γγ −1

pρ2dρ + 1

2 du2 = 0.

Solve the first equation for

€

dρ = −ρdu u , and eliminate dρ from the other two equations:

€

dp + ρdu2 − ρudu = −p1df , and

€

γγ −1

dpρ

+γ

γ −1pρduu

+ 12 du

2 = 0.

Simplify these equations and combine terms:

€

dp + ρudu = −p1df , and

€

γγ −1

dpρ

+pρduu

%

& '

(

) * + udu = 0.

Solve the first of these for

€

du = − 1 ρu( ) p1df + dp( ) and substitute this into the second one:

€

γγ −1

dpρ−pρ1ρu2

p1df + dp( )%

& '

(

) * −1ρp1df + dp( ) = 0 .

Multiply through by ρ, and collect terms:

€

γγ −1

−γ

γ −1pρu2

−1%

& '

(

) * dp −

γγ −1

pρu2

+1%

& '

(

) * p1df = 0.

Separate the terms across the equals sign, use γp/ρ = c2, where c = sound speed, and introduce the Mach number M = u/c:

€

γγ −1

−1

γ −1c 2

u2−1

$

% &

'

( ) dp =

1γ −1

c 2

u2+1

$

% &

'

( ) p1df , or

€

γ −1M 2 − γ +1

$

% &

'

( ) dp =

1M 2 + γ −1$

% &

'

( ) p1df , so

€

M 2 −1( )dp = (γ −1)M 2 +1( )p1df , which implies:

€

−dpp1

=1+ (γ −1)M 2

1−M 2 df .

To reach the second equation of (15.47), multiply by p1 subtract p1df from both sides and divide by ρu to reconstruct

€

du = − 1 ρu( ) p1df + dp( ) on the left:

€

du =1ρu

−dp − p1df( ) =1ρu

1+ (γ −1)M 2

1−M 2 −1%

& '

(

) * p1df =

1ρu

γM 2

1−M 2

%

& '

(

) * p1df .

Divide by u and rearrange the right side:

€

duu

=pρu2

γM 2

1−M 2

%

& '

(

) * p1pdf =

1M 2

M 2

1−M 2

%

& '

(

) * p1pdf =

11−M 2

p1pdf .

Thus, when M < 1, positive friction (df > 0) causes the flow to accelerate.


Exercise 15.16. Starting from the set (15.45) with f = 0, derive (15.48) by letting station (2) be a differential distance downstream of station (1). Solution 15.16. The equation set (15.45) with f = 0 and the second location a differential distance dx downstream of the first location is:

€

d ρu( ) = 0 ,

€

d p + ρu2( ) = 0, and

€

d h + 12 u

2( ) = h1dq . Use h = cpT, expand all three equations, and use the perfect gas law (p = ρRT) to find:

€

ρdu + udρ = 0 ,

€

dp + ρdu2 + u2dρ = 0 , cpdT +12 du

2 = h1dq , and

€

dp = ρRdT + RTdρ . Solve the first equation for

€

dρ = −ρdu u , and eliminate dρ from the second and last equations to reach:

€

dp + ρdu2 − ρudu = dp + ρudu = 0 , and

€

dpp

=dTT−duu

,

where the final form of the last equation is obtained by dividing by p or ρRT. Solve the first of these for dp = –ρudu, and use this to eliminate dp from the second:

€

−ρudup

= −γc 2udu =

dTT−duu

.

where γp/ρ = c2, where c = sound speed, has been used for the first equality. Solve for udu and use M = u/c:

€

−γc 2udu +

duu

= − γM 2 −1( ) duu =dTT

, or

€

udu = 12 du

2 = −u2dT

T γM 2 −1( ).

Substitute this into the differential energy equation (the one that involves cp) and use h1 = cpT1,

cpdT −u2dT

T γM 2 −1( )= h1dq = cpT1dq .

Divide this equation by γR, recognize the factor of M2, and use cp γR =1 γ −1( ) :

cpγR

dT − u2dTγRT γM 2 −1( )

=cpT1γR

dq , or

€

1γ −1

−M 2

γM 2 −1( )$

% & &

'

( ) ) dT =

1γ −1

T1dq .

Solve for dT/T1.

€

1− (γ −1)M2

γM 2 −1( )$

% & &

'

( ) )

dTT1

=−1+ M 2

γM 2 −1$

% &

'

( ) dTT1

= dq, or

€

dTT1

=1− γM 2

1−M 2 dq .

The final equation is the first part of (15.48) which shows that heat addition leads to cooling of the gas when

€

1 γ < M <1. To reach the second part of (15.48), multiply the result for dT/T1 by T1 and substitute this into the differential energy equation:

cpdT +udu = cpT11−γM 2

1−M 2 dq"

#$

%

&'+udu = h1dq .

Recognize h1 = cpT1, and collect terms:

€

uduh1

= 1− 1− γM2

1−M 2

$

% &

'

( ) dq =

(γ −1)M 2

1−M 2 dq .

This is the second equation of (15.48).


Exercise 15.17. For flow of a perfect gas entering a constant area duct at Mach number M1, calculate the maximum admissible values of f and q for the same mass flow rate. Case (a) f = 0; case (b) q = 0. Solution 15.17. From Section 15.6,

€

M2

M1

=1+ γM2

2

1+ γM12 − f

1+ (γ −1) 2( )M12 + q

1+ (γ −1) 2( )M22

$

% &

'

( )

1 2

.

For maximum values of f or q, the flow is choked at the duct exit so M2 = 1. a) f = 0; set M2 = 1 to find:

€

1M1

=1+ γ1+ γM1

2

1+ (γ −1) 2( )M12 + q

(γ +1) 2

$

% &

'

( )

1 2

.

Solve for q by first clearing the square root.

€

1+ γM12

(1+ γ)M1

#

$ %

&

' (

2

=1+ (γ −1) 2( )M1

2 + q(γ +1) 2

, so

€

q =γ +12

1+ γM12

(1+ γ)M1

#

$ %

&

' (

2

− 1+γ −12

M12*

+ ,

-

. / .

(b) q = 0; set M2 = 1 to find:

€

1M1

=1+ γ

1+ γM12 − f

1+ (γ −1) 2( )M12

(γ +1) 2

$

% &

'

( )

1 2

.

Solve for f.

€

1+ γM12 − f = (1+ γ)M1

1+ (γ −1) 2( )M12

(γ +1) 2

$

% &

'

( )

1 2

, or

€

f =1+ γM12 − 2(1+ γ)M1 1+

γ −12

M12$

% & '

( )

1 2

.

Note, f = 0 when M1 = 1.


Exercise 15.18. Show that the accelerating portion of the piston trajectory (0 ≤ xp(t) ≤ cot1) shown in Figure 15.18 is:

xp(t) =γ +1γ −1"

#$

%

&'cot1

tt1

"

#$

%

&'

2γ+1−2cotγ −1

for 1≤ tt1≤

2γ +1"

#$

%

&'

1+γ1−γ

.

Solution 15.18. As described in Example 15.7, the C+ characteristics emanate from the origin during the time that the piston is accelerating. And, as described in the solution to this same example, the C– characteristics originate on the x-axis where u = 0 and c = co, so the I– invariant implies:

I− = u(x, t)−2c(x, t)γ −1

= u(x, 0)− 2c(x, 0)γ −1

= −2coγ −1

or c = co +γ −12

u .

Thus, on the C+ characteristics, dxdt

!

"#

$

%&C+

= (u+ c)C+ = u+ co +γ −12

u!

"#

$

%&C+

= co +γ +12

u!

"#

$

%&C+

=xt

!

"#$

%&C+

where the final equality ensures that the C+ characteristics pass through the origin. To determine the piston trajectory xp(t), require the final equality to hold where any C+ characteristic crosses the piston path. If the time of this crossing is τ, then the last equality implies:

co +γ +12!xp(τ ) =

xp(τ )τ

.

Replace τ with t, and rearrange to find a first-order ordinary differential equation:

!xp(t)−2

γ +1xp(t)t

= −2coγ +1

.

The solution is the sum of homogeneous and a particular solutions,

xp = At2γ+1 −

2cotγ −1

,

where A is an undetermined constant. The initial condition, xp = cot1 at t = t1, allows A to be determined from:

cot1 = At12γ+1 −

2cot1γ −1

, or A = cot1γ +1γ −1"

#$

%

&'t1

−2γ+1 .

Thus,

xp(t) =γ +1γ −1"

#$

%

&'cot1

tt1

"

#$

%

&'

2γ+1−2cotγ −1

for 1≤ tt1≤

2γ +1"

#$

%

&'

1+γ1−γ

,

which is the desired result. In dimensionless form this is:

xp(t)cot1

=γ +1γ −1"

#$

%

&'tt1

"

#$

%

&'

2γ+1−2

γ −1tt1

"

#$

%

&' .


Exercise 15.19. For the flow conditions of Figure 15.18, plot u/co and p/po as functions of x/cot1 for xp(t) < x < cot at t/t1 = 2, 3, and 4 for γ = 1.4, where co and po are the sound speed and pressure of the quiescent gas upstream of any disturbance from the moving piston. Does the progression of these waveforms indicate expansion wave steepening or spreading as t increases? Solution 15.19. Start from the results of Example 15.7,

u(x, t) = 2γ +1

xt− co

"

#$

%

&' and c(x, t) = 2

γ +1co +

γ −1γ +1

xt

,

and rearrange these equations to introduce dimensionless variables U = u/co, C = c/co, X = x/cot1, and T = t/t1:

u(x, t)co

=2

γ +1xcot1

1t t1( )

−1"

#$$

%

&'' or U =

2γ +1

XT−1

"

#$

%

&' , (1)

and c(x, t)co

=2

γ +1+γ −1γ +1

xcot1

1t t1( )

or C = 2γ +1

+γ −1γ +1

XT

.

The final equation can be switched to the dimensionless pressure P = p/po via (15.52):

P = 2γ +1

+γ −1γ +1

XT

"

#$

%

&'

2γγ−1

. (2)

For the given parameters, the plots of (1) and (2) look like:

Examination of both plots suggests that the expansion wave spreads out as time increases.

!1.5%

!0.5%

0.5%

!1% 0% 1% 2% 3% 4%0"

0.5"

1"

1.5"

&1" 0" 1" 2" 3" 4"

U

X

–––– T = 2 – – – T = 3 - - - - T = 4

P

X

–––– T = 2 – – – T = 3 - - - - T = 4


Exercise 15.20. Consider the field properties in Figure 15.19 before the formation of the shock wave. a) Using the piston trajectory from Exercise 15.18, show that the time at which the piston reaches speed co is −t1 (γ +1) 2( )(1+γ ) (1−γ ) = –0.3349t1 for γ = 1.4. b) Plot u/co and p/po as functions of x/cot1 for xp(t) < x < cot1 at: t/t1 = –1/3, –1/6, and –1/25 for γ = 1.4, where co and po are the sound speed and pressure of the quiescent gas upstream of any disturbance from the moving piston. Does the progression of these waveforms indicate compression wave steepening or spreading as t→ 0 ? Solution 15.20. a) For the situation shown in Figure 15.19, the initial piston location is –cot1, so the sign of xp must be changed in the formula given in Exercise 15.18, and the piston starts moving at t = –t1, where t1 is presumed to be positive. Thus, the starting point for this Exercise is:

xp(t) = −γ +1γ −1"

#$

%

&'cot1

t−t1

"

#$

%

&'

2γ+1+2cot1γ −1

t−t1

"

#$

%

&'

where the extra divisor factors of –t1 have been introduced to facilitate the evaluation of xp when t < 0. Time differentiate this formula,

dxp(t)dt

= −γ +1γ −1"

#$

%

&'cot1

2γ +1"

#$

%

&'

t−t1

"

#$

%

&'

2γ+1

−11−t1

"

#$

%

&'−2coγ −1

.

Simplify and set dxp/dt = co:

dxp(t)dt

= co =2

γ −1"

#$

%

&'co

t−t1

"

#$

%

&'

2γ+1

−1

−2coγ −1

=2coγ −1

t−t1

"

#$

%

&'

1−γ1+γ−1

(

)*

+*

,

-*

.*.

Divide out the common factor of co and solve for t:

t = −t1γ +12

"

#$

%

&'

1+γ1−γ

= –0.3349t1,

where the numerical value applies when γ = 1.4. b) The first or lowest C+ characteristic shown on Figure 15.19 has a slope unity. To the right of this characteristic, u = 0 and p = po. Using the part a) result, and equation for the piston's trajectory, the location where the piston speed reaches co can be found:

location =xpcot1, tt1

!

"#

$

%&= −0.7368,−0.3349( ) .

The line from this location to the origin is the last or uppermost C+ characteristic shown on Figure 15.19. To the left of this characteristic, the flow state will be uniform with u = co and p = constant. The region in between the two characteristics must provide the appropriate transition between the left- and right-side flow states. To specify this transition, recognize that x and t are both negative, and start from the results of Example 15.7:


u(x, t) = 2γ +1

xt− co

"

#$

%

&' and c(x, t) = 2

γ +1co +

γ −1γ +1

xt

.

Rearrange and introduce dimensionless variables U = u/co, C = c/co, X = x/cot1, and T = t/t1. u(x, t)co

=2

γ +1xcot1

1t t1( )

−1"

#$$

%

&'' or U =

2γ +1

XT−1

"

#$

%

&' (1)

c(x, t)co

=2

γ +1+γ −1γ +1

xcot1

1t t1( )

or C = 2γ +1

+γ −1γ +1

XT

.

The final equation can be switched to the dimensionless pressure P = p/po via (15.52):

P = 2γ +1

+γ −1γ +1

XT

"

#$

%

&'

2γγ−1

. (2)

The peak pressure then is: P = ppo=

2γ +1

+γ −1γ +1

−0.7368−0.3349"

#$%

&'(

)*

+

,-

2γγ−1

= 3.583, when γ = 1.40.

For the given parameters, the plots of (1) and (2) look like:

Examination of both plots suggests that the compression wave steepens as time increases.

!0.25&

0&

0.25&

0.5&

0.75&

1&

1.25&

!1& !0.8& !0.6& !0.4& !0.2& 0&0"

0.5"

1"

1.5"

2"

2.5"

3"

3.5"

4"

)1" )0.8" )0.6" )0.4" )0.2" 0"

U

X

–––– T = 2 – – – T = 3 - - - - T = 4

P

X

–––– T = 2 – – – T = 3 - - - - T = 4


Exercise 15.21. For the flow conditions of Figure 15.19, assume the flow speed downstream of the shock wave is co and determine the shock Mach number, its x-t location, and the pressure, temperature and density ratios across the shock. Are these results well matched to the isentropic compression that occurred for t < 0? What additional adjustment is needed? Solution 15.21. The shock wave first forms at the origin in Figure 15.19, and the velocity difference across the shock wave is co. This is enough to determine the shock Mach number. Start with the normal shock velocity ratio condition (15.41) for a stationary normal shock wave:

u1u2=

(γ +1)M12

(γ −1)M12 + 2

. (15.41)

When the situation in Figure 15.19 is subject to a Galilean transformation that creates a shock-fixed coordinate system, u1 = –M1co and u2 = (1 – M1)co. Thus, (15.41) becomes:

−M1co(1−M1)co

=(γ +1)M1

2

(γ −1)M12 + 2

or −(γ −1)M12 − 2 = (γ +1)M1(1−M1) ,

which reduces to the quadratic equation:

M12 −

γ +12

M1 −1= 0 , so M1 =12γ +12

±γ +12

!

"#

$

%&2

+ 4!

"

##

$

%

&&.

The physically meaningful root comes from the '+' sign and is M1 = 1.766 for γ = 1.4. Thus, the x-t location of the shock is:

xshock = 1.7662cot, and the pressure, temperature, and density ratios are:

p2p1=1+ 2γ

γ +1M1

2 −1"# $%= 3.473 , (3.583)

ρ2ρ1=

(γ +1)M12

(γ −1)M12 + 2

= 2.305 , and (2.488)

T2T1=1+ 2(γ −1)

(γ +1)2γM1

2 +1M1

2 M12 −1( ) =1.506 . (1.440)

Interestingly, these ratios are all slightly different than the ratios produced by the isentropic compression that occurs before the shock wave forms at t = 0 (provided in parenthesis at the right). [These numbers are obtained from the solution of Exercise 15.20]. Thus, there will be an adjustment region near the origin in Fig. 15.19 that will allow the pressure behind the shock to equilibrate with the adiabatically compressed gas that is between the piston and the shock wave. The net effect of this adjustment will be to shift the shock wave location slightly farther ahead – in the positive x-direction – compared to what is calculated above because the adiabatic compression reaches a higher pressure.


Exercise 15.22. Write momentum conservation for the volume of the small rectangular control volume shown in Figure 4.20 where the interface is a shock with flow from side 1 to side 2. Let the two end faces approach each other as the shock thickness → 0 and assume viscous stresses may be neglected on these end faces (outside the structure). Show that the n component of momentum conservation yields (15.36) and the t component gives u⋅t is conserved or v is continuous across the shock.

Solution 15.22. For a stationary control volume V containing only fluid particles bounded by a surface A, conservation of mass and momentum may be written:

€

ddt

ρV∫ dV = − ρu j

A∫ n jdA , and

€

ddt

ρuiV∫ dV = − ρuiu j + pδij −σ ij[ ]

A∫ n jdA + ρgidV

V∫ .

When applied to a small rectangular control volume like that shown above (a reproduction of Fig. 4.20) in a steady flow, these simplify to:

€

0 = − ρu jA∫ n jdA , and

€

0 = − ρuiu j + pδij −σ ij[ ]A∫ n jdA + ρgidV

V∫ .

For the current situation, let the interface define the location of a shockwave, and let the thickness dimension

€

l→ 0 while requiring the upper rectangular surface to remain in region 2 and the lower rectangular surface to remain in region 1. This means

€

V → 0 so the body force term may be dropped, and that the flux through the sides of the CV may be ignored because the curved-side surface area goes to zero as

€

l→ 0. For this specialized limit, the equations reduce to:

€

ρu j( )1n j = ρu j( )2n j , and

€

ρuiu j + pδij −σ ij[ ]1n j = ρuiu j + pδij −σ ij[ ]2n j , where nj are the components of n, the primarily-upward-pointing unit vector shown in the figure. Here, the two larger CV surfaces are chosen to lie outside the shock structure (which is presumed to be very thin), so that σij = 0. [Inside the shock wave, the viscous stress may be large]. Thus, the momentum equation becomes:

€

(ui)1 ρu j( )1n j + p1ni = (ui)2 ρu j( )2n j + p2ni . (†) For the shock-normal component, take the dot product of (†) with n (which has components ni). The resulting equation is:

ρ(uini )2( )1 + p1 = ρ(uini )

2( )2 + p2 , or p1 − p2 = −ρ1u12 + ρ2u2

2 , where (uini)1 = u1 and (uini)2 = u2, and the second equation is (15.36). For the shock tangent component, take the dot product (†) with either tangent unit vector, generically represented here by t (which has components ti):

(2)!

(1)!

l!

+n!

–n!

u2!

u1!

us!dA!

dA!


€

ti(ui)1 ρu j( )1n j + p1tini = ti(ui)2 ρu j( )2n j + p2tini , or

€

ti(ui)1 ρu j( )1n j = ti(ui)2 ρu j( )2n j , where by definition:

€

t ⋅n = tini = 0. Now use the final form of the mass conservation equation to divide out the terms in larger parenthesis that contain ρ; this leaves:

€

ti(ui)1 = ti(ui)2 , or in vector notation:

€

t ⋅u1 = t ⋅u2. Thus, the tangential components of u are unchanged across a shockwave since t represents either tangent unit vector.


Exercise 15.23. A wedge has a half-angle of 50°. Moving through air, can it ever have an attached shock? What if the half-angle were 40°? [Hint: The argument is based entirely on Figure 15.22.] Solution 15.23. From Figure 15.22, no attached shock can produce a deflection angle of 50°. However, an attached shock with a deflection angle of 40° is possible.


Exercise 15.24. Air at standard atmospheric conditions is flowing over a surface at a Mach number of M1 = 2. At a downstream location, the surface takes a sharp inward turn by an angle of 20°. Find the wave angle σ and the downstream Mach number. Repeat the calculation by using the weak shock assumption and determine its accuracy by comparison with the first method.

Solution 15.24. Using the geometry shown, and assuming M2 > 1, the shock angle σ is approximately 53° based on Figure 15.22. Therefore:

Mn1 = M1sinσ = 2sin(53°) = 1.6, and Mn2 = 0.668 (from Table 15.2).

From the geometry: Mn2 = M2sin(σ – δ), so

€

M2 = Mn2 sin(σ −δ) = 0.668 sin(53°− 20°) =1.227 . To develop equivalent results from weak shock theory, start from (15.53) and simplify for δ << 1 and

€

σ → µ = sin−1 1 M1( ) . Thus,

€

M12(γ + cos2σ ) + 2[ ] tanδ = 2cotσ M1

2 sin2σ −1( ) becomes:

€

M12(γ +1− 2sin2σ) + 2[ ]δ ≅ 2 M1

2 −1 M12 sin2σ −1( ) ,

where

€

tanδ ≅ δ and

€

cotσ ≅ cotµ = M12 −1 . Substituting

€

sin2σ ≅ M1−2 on the right side produces

€

M12(γ +1)

2 M12 −1

δ ≅ M12 sin2σ −1, or

€

σ = sin−1 1M1

2 +γ +1

2 M12 −1

δ&

' ( (

)

* + +

1 2

= sin−1 14

+2.4

2 4 −120 π180-

. /

0

1 2

&

' (

)

* +

1 2

= 44.5°.

So, Mn1 = M1sinσ = 2sin(44.5°) = 1.40, and

Mn2 = 0.740 (from Table 15.2). From the geometry:

Mn2 = M2sin(σ – δ), so

€

M2 = Mn2 sin(σ −δ) = 0.740 sin(44.5°− 20°) =1.784 , so the error = (1.784 – 1.227)/1.227 = 0.45 or 45%.


Exercise 15.25. A flat plate is inclined at 10° to an airstream moving at M∞ = 2. If the chord length is b = 3 m, find the lift and wave drag per unit span. Solution 15.25. Considering figure 15.24, and using equations (15.58):

€

L = 2αγM∞2p∞ b M∞

2 −1 = 2 10π 180( )(1.4)22(101.3kPa)(3m) 22 −1 = 342kNm−1 and

€

D =αL = 10π 180( ) = 59.7kNm−1.


Exercise 15.26. Using thin airfoil theory calculate the lift and drag on the airfoil shape given by yu = t sin(πx/c) for the upper surface and y1 = 0 for the lower surface. Assume a supersonic stream parallel to the x-axis. The thickness ratio t/c << 1.

Solution 15.26. From supersonic thin airfoil theory

€

p − p∞p∞

=γM∞

2δ

M∞2 −1

,

where δ is the slope of the streamline on the surface of the foil. On the foil's upper surface:

€

dyudx

=tπccos πx

c#

$ %

&

' ( = tanδ ≅ δ for δ << 1.

On the foil's lower surface dyl/dx = 0. Thus the surface pressures are:

€

pu − p∞p∞

=γM∞

2

M∞2 −1

tπccos πx

c&

' (

)

* + , and pl = p∞.

The lift, L, per unit span will be:

€

L = pl − pu( )ey ⋅ndssurface∫ = pl − pu( )

0

c

∫ dx

where

€

ey ⋅nds = cosδ ⋅ dxcosδ

= dx . Thus,

€

L = −γM∞

2p∞M∞

2 −1tπccos πx

c&

' (

)

* +

0

c

∫ dx = −γM∞

2p∞M∞

2 −1t sin πx

c&

' (

)

* +

-

. /

0

1 2 0

c

= −γM∞

2p∞M∞

2 −1t(0 − 0) = 0 .

The drag, D, per unit span will be:

€

D = pu − p∞( )ex ⋅ndssurface∫ = pu − p∞( )δ

0

c

∫ dx

where

€

ex ⋅nds = sinδ ⋅ dxcosδ

= tanδdx ≅ δdx . Thus,

€

D =γM∞

2p∞M∞

2 −1tπc

&

' (

)

* + 2

cos2 πxc

&

' (

)

* +

0

c

∫ dx =γM∞

2p∞M∞

2 −1tπc

&

' (

)

* + 2

⋅c2

.

The reason for these results is that p > p∞ on the upper foil surface from 0 < x < c/2, and p < p∞ on the upper foil surface from c/2 < x < c with an average value of p∞. Thus, no lift results. However, with a higher pressure on the upstream half and a lower pressure on the downstream half, a drag is produced.


Exercise 15.27. Consider a thin airfoil with chord length l at a small angle of attack in a horizontal supersonic flow at speed M∞. The foil's upper and lower surface contours, yu(x) and yl(x), respectively, are defined by:

yu(x) = t(x)/2 + yc(x) – αx , and yl(x) = –t(x)/2 + yc(x) – αx , where: t(x) = the foil's thickness distribution, α = the foil's angle of attack, and yc(x) = the foil's camber line. Use these definitions to show that the foil's coefficients of lift and drag are:

CL =4αM∞

2 −1, and CD =

4M∞

2 −114

dtdx#

$%

&

'(2

+dycdx

#

$%

&

'(2

+α 2)

*++

,

-...

Solution 15.27. Use x-y coordinates and assume the foil extends x = –l/2 to x = +l/2. For geometrical consistency, the upper and lower foil-surface contours must match at the foil's leading and trailing edges: yu(–l/2) = yl(–l/2), and yu(+l/2) = yl(+l/2), so t(–l/2) = t(+l/2) = 0. And, by definition, yc(–l/2) = yc(+l/2) = 0. To compute the lift coefficient, CL, first determine the pressure coefficient starting from its definition, then eliminate the pressure difference using (15.68):

Cp =p− p∞

(1 / 2)ρ∞U∞2 =

γM∞2δ

M∞2 −1

p∞(1 / 2)ρ∞U∞

2 =γM∞

2δ

M∞2 −1

2γM∞

2 =2δM∞

2 −1.

For small flow deflection angles: δ ≈ sinδ ≈ tanδ = dy/dx and cosδ ≈ 1. Thus,

CL =1l

Cp,l −Cp,u( )cosδ−l/2

+l/2

∫ dx ≅ 1l

−2M∞

2 −1dyldx

%

&'

(

)*−

2M∞

2 −1dyudx

%

&'

(

)*

%

&''

(

)**

−l/2

+l/2

∫ dx .

where the 'l' and 'u' subscripts indicate the lower and upper foil surfaces, and δ is measured from the horizontal and is positive when it leads to flow compression. Here, the integration merely undoes the differentiation of the upper and lower foil contours:

CL ≅ −1l

2M∞

2 −1yl (x)+ yu(x)[ ]−l/2

+l/2

= −2l M∞

2 −10+ 0−αl

2+ 0+ 0−αl

2− 0− 0−αl

2− 0− 0−αl

2$

%&'

()

= 4αM∞

2 −1

.

So, only the angle of attack matters for the production of lift. Neither the thickness nor the chamber is involved.

To determine the coefficient of drag, the local component of the pressure force in the direction of flow must be considered:

CD =1l

Cp,u −Cp,l( )sinδ−l/2

+l/2

∫ dx ≅ 1l

2M∞

2 −1dyldx

%

&'

(

)*

2

−−2M∞

2 −1dyudx

%

&'

(

)*

2%

&''

(

)**

−l/2

+l/2

∫ dx

= 2l M∞

2 −1dyldx

%

&'

(

)*

2

+dyudx

%

&'

(

)*

2%

&''

(

)**

−l/2

+l/2

∫ dx.


Here, dyudx

=12dtdx+dycdx

+α , so dyudx

!

"#

$

%&2

=14

dtdx!

"#

$

%&2

+dycdx

!

"#

$

%&2

+α 2 +dtdx

dycdx

+dtdxα + 2 dyc

dxα , and

dyldx

= −12dtdx+dycdx

+α , so dyldx

!

"#

$

%&2

=14

dtdx!

"#

$

%&2

+dycdx

!

"#

$

%&2

+α 2 −dtdx

dycdx

−dtdxα + 2 dyc

dxα , so

dyldx

!

"#

$

%&2

+dyudx

!

"#

$

%&2

=12

dtdx!

"#

$

%&2

+ 2 dycdx

!

"#

$

%&2

+ 2α 2 + 4 dycdx

α .

The chord-length average of the final term is zero, so

CD =2

l M∞2 −1

12

dtdx#

$%

&

'(

2

+ 2 dycdx

#

$%

&

'(

2

+ 2α 2#

$%%

&

'((

−l/2

+l/2

∫ dx

= 2M∞

2 −112

dtdx#

$%

&

'(

2

+ 2 dycdx

#

$%

&

'(

2

+ 2α 2#

$

%%

&

'

((

= 4M∞

2 −114

dtdx#

$%

&

'(

2

+dycdx

#

$%

&

'(

2

+α 2#

$

%%

&

'

((.

This result suggests that airfoil thickness and camber only lead to drag in supersonic flow. Thus, the wings and control surfaces of supersonic aircraft and missiles are thin and nearly flat.

fluid mechanics kundu cohen 6th edition solutions sm ch (15)

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