978-0131206687 Chapter 11

Chapter Eleven

E

EQ

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ON

NS

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OF

F

M

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IO

ON

N

F

FO

OR

R

M

MU

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LT

TI

ID

DI

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Problem 1. – Prove that for a perfect gas

(a) e1)(γρp−=

(b) 2

V

1)(γγ

a

e

22

t+

+

=

(a) For a perfect gas

RTρp

=

(1)

Tce v

=

(2)

R

Substitute Eq. (4) into Eq. (1) to the result for part (a)

(b) The definition of the total (internal) energy is

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Problem 2. – According to the generalized continuity equation given by Eq. (11.1), for

steady, incompressible, one-dimensional flow, 0xu

=

∂

, or, in other words, u is equal

to a constant. Previously, for incompressible flow, however, it has been customary to

assume that, for steady, one-dimensional flow, the product of velocity and cross-sectional

area (AV) is a constant. Explain this seeming contradiction.

Problem 3. – The continuity equation for steady two-dimensional flow is

y

x

∂

A function ψ (the compressible stream function) may be defined so that this equation is

automatically satisfied. Show that the following accomplish this

y

ψ

ρ

∂

ψ

ρ

∂

where ∞

ρ is a constant that is inserted so that the stream function has the same units as

the incompressible flow stream function. What are the units of the stream function? What

are the units of the velocity potential,

φ

?

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Therefore

⎟

⎠

⎞

⎜

⎝

⎛

∂∂

∂

−

∂∂

∂

=

∂

+

∂

∞xy

ψ

yx

ψ

ρ

y

v)(ρ

x

u)(ρ22

But for a continuous stream function with continuous derivatives

ψ

ψ22

∂

=

∂

and, hence, the continuity equation is automatically satisfied.

The dimensions of both the stream function and velocity potential are: L2T-1.

Problem 4. – Expand Eq.(11.10) into the three component equations, and show that

Eqs.( 11.7), (11.8), and (11.9) result.

Equation (11.10) can be written as

kjikji

t

w

t

v

t

u

z

p

y

p

x

p

ρ

1

⎤

⎡⎟

⎞

⎛

∂

+

∂

+

∂

+

∂

=

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

+

∂

−

kji

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

+

∂

+

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

+

∂

+

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

+

∂

z

w

y

w

v

x

w

u

z

v

w

y

v

x

v

u

z

u

w

y

u

v

x

u

This implies that

⎪

⎨

⎧

∂

+

∂

+

∂

+

∂

=

∂

−

∂

+

∂

+

∂

+

∂

=

∂

−

w

p

1

z

v

w

y

v

x

v

u

t

v

y

p

ρ

1

z

u

w

y

u

v

x

u

t

u

x

p

ρ

1

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Problem 5. – Show that

V

dp 2

⎞

⎛

0

2

ρ

⎟

⎠

⎜

⎝

+∇

are equivalent.

Take the dot product of Eq. (2) with the differential displacement vector,

kjir dzdydxd ++= to get

V

p2=•

⎤

⎡

⎟

⎞

⎜

⎛

∇

V

12=•

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240

Substitution of these into Eq.(3) produces Eq. (1).

Problem 6. – Derive Eq.(11.8), i.e., prove that

Dt

Dp

qρ

Dt

Dh

ρ+= &

Consider the continuity equation

0ρ

D

t

Dρ=•+ V∇

or

Dρ

1

−=• V∇ (1)

Dt

ρ

Dt

Substitution of Eqs.(1) and (2) in Eq.(11.17), then leads to

Dt

ρ

Dt

ρ

Dt

which simplifies to give Eq.(11.18).

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Problem 7. – Under what conditions can it be assumed that 2

V

ρp

2

+ is equal to a

constant?

Consider the dot product ⎟

⎟

⎠

⎞

⎜

⎝

⎛+• 2

V

ρ

p2

∇V. It can be expanded as follows

2

ρ

ρp

2

V

ρ

p

∇

•−

⎥

⎦

⎤

⎢

⎣

⎡

⎟

⎠

⎞

⎜

⎝

⎛

+•=

VV

But using the vector form of Eq.(11.34)

0

2

ρ

⎟

⎠

⎜

⎝

+∇

for irrotational, steady, frictionless flow with no external forces except pressure.

Therefore the dot product introduced above becomes

2

ρ

ρp

2

V

ρ

p∇

∇•−=

⎟

⎠

⎞

⎜

⎝

⎛+• VV

This means that, if the right hand side term is zero for a flow, then V⊥

⎟

⎠

⎞

⎜

⎝

⎛+2

V

ρ

p2

∇, in

other words, Constant

2

V

ρ

p2

=+ along a streamline. Now, in order for the term 2

ρ

ρp

∇

•V

to vanish, we must have either 0ρ

=

∇ or V

⊥

ρ

∇

. The former is a constant density flow

(for our steady assumption), and we can show that the latter also leads to a constant

density flow:

The continuity equation

=

•

+

•

∇

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Using this, the assumption V⊥ρ∇, which implies 0ρ

=

•

∇

V, leads to 0=• V∇. This

is again the continuity description of a constant density flow.

Problem 8. – Show that Crocco’s equation along a streamline can be written as

o

t

2

∂

Now

()

t

V

2

1

wvu

t2

1

t

w

t

v

t

u

w

v

u

2

222

∂

=++

∂

=

∂

+

∂

+

∂

=

⎞

⎛

∂

V

Take the dot product of Eq.(11.39) with the velocity vector to get

V

∂

t

By definition, the product ωV× is perpendicular to both vectors V and ω. Therefore,

()

0=×• ωVV , and the above equation becomes

Combining the two previous expressions produces

2

hsT

t

V

2

1∇•−∇•=

∂

∂VV

Problem 9. – Using the substantial derivative operator within Crocco’s equation,

Eq.(11.39), develop the following equation for the entropy

⎞

⎛

∂

1

s

Ds

V

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244

f

z

f

y

f

x

f

zyx

f

222222 =

⎟

⎞

⎜

⎛

∂

⎟

⎞

⎜

⎛

∂

⎟

⎞

⎜

⎛

∂

=

kji

∇×∇

Problem 11. – The velocity components for a possible flow field are given by

y2x3u 2+−= and y2x2v += . Is the flow irrotational? If so, determine the velocity

potential.

()

⎥

⎦

⎤

⎢

⎣

⎡⎟

⎟

⎠

⎞

⎜

⎝

⎛

∂

−

∂

+

⎟

⎠

⎞

⎜

⎝

⎛

∂

−

∂

+

⎟

⎠

⎞

⎜

⎝

⎛

∂

−

∂

=

∂

=×∇= kji

kji

Vω

y

u

x

v

x

w

z

u

z

v

y

w

2

1

wvu

zyx2

1

2

1

Since the flow if two-dimensional, the above relation simplifies to

kω⎟

⎟

⎠

⎞

⎜

⎝

⎛

∂

−

∂

=y

u

x

v

2

1

But 2

y

u=

∂

∂ and 2

x

v=

∂

∂. The rotation vector then becomes

2

Therefore, the flow is irrotational.

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Based on definition of the velocity potential, Eq.(11.4), x

u∂

φ

∂

= and y

v∂

φ∂

=. Therefore,

y+==

∂

Substitute this into Eq.(2) to get

Problem 12. – Show that the velocity potential equation, Eq.(11.47) can be written in

two-dimensional form as

0

ya

v

1

yx

a

vu

2

xa

u

12

2

22

2

2=

∂

φ∂

⎟

⎠

⎞

⎜

⎝

⎛−+

∂∂

φ∂

−

∂

φ∂

⎟

⎠

⎞

⎜

⎝

⎛−

Equation (11.47), in two dimensions, can be written as

2

2=

φ∂

⎤

⎡

φ∂

⎞

⎛

φ∂

⎞

⎛

φ∂

⎟

⎞

⎜

⎛

φ∂

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246

ya

yx

a

xa

2

22

2

∂

⎟

⎠

⎜

⎝

∂∂

∂

⎟

⎠

⎜

⎝

Problem 13. – Consider a steady, uniform flow of air (

γ

= 1.4, R = 0.287 kJ/kg.K) with

velocity components u = 120 m/s and v = w = 0. Determine the velocity potential,

substitute into Eq.(11.48), and find the resultant difference between static and stagnation

temperature.

Using the definition of the velocity potential and the above velocity components, we get

x=

∂

Since the other velocity components are zero, the velocity potential is a function of x

only. Therefore

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(

)

()( )( )

()

K7.1677120

2871.42

11.4

Rγ2

1γ

TT 2

o≅

−

=φ•φ

−

=− ∇∇

Problem 14. – Using the stream function, as defined in Problem 11.3, develop the

following expression for steady, two-dimensional, irrotational flow

0

ya

v

1

yx

a

uv

2

xa

u

12

2

22

2

2=

∂

ψ∂

⎟

⎠

⎞

⎜

⎝

⎛−+

∂∂

ψ∂

−

∂

ψ∂

⎟

⎠

⎞

⎜

⎝

⎛−

From irrotational flow

y

u

x

v=

∂

and from the definition of the stream function

u∂

ψ∂

ρ

=∞

ρ

Hence,

⎞

⎛

ψ∂

ρ

∂

⎞

⎛

ψ∂

ρ

∂∞∞

yyxx

yx 2

∂

ρ

∂

ρ

⎠

⎝

∂

Now for isentropic flow

a

p2

ρ

∂

=

∂

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22 a

y

a

x

yyxx ∂

+

∂

−=

∂

ρ

+

∂

ρ

From Euler’s equations

y

pu

y

v

x

v

uu

x

pv

y

u

v

x

u

uv

∂

ρ

=

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

−

∂

ρ

−=

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

So

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

−

ρ

=

∂

−

∂

+

∂

+

∂

−22

2

22

a

u

y

p

a

v

x

pa

y

v

uv

x

u

uv

y

u

v

x

v

u

Thus,

⎟

⎠

⎞

⎜

⎝

⎛

∂

ρ−

∂

ρ+

∂

ρ+

∂

ρ−=

∂

+

∂

−

y

v

uv

x

u

uv

y

u

v

x

v

u

a

1

a

u

y

p

a

v

x

p

2

22

222

⎟

⎠

⎞

⎜

⎝

⎛

∂

ψ∂

+

∂

ψ∂

ρ=

⎟

⎠

⎜

⎝

∂∂

∂

∞

∞∞∞

2

22

2

yx

yxyx

a

ya

xa

Rearranging produces the result

v

uv2

u

2

22

2

ψ∂

⎞

⎛−+

ψ∂

⎞

⎛−

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249

Problem 15. – Use the technique presented in Example 11.4 to write (a) the velocity

components in terms of the velocity potential in spherical coordinates and (b) the steady

energy equation for three-dimensional, adiabatic flow in cylindrical coordinates.

ϕ∂

b) From Eq.(11.14)

(

)

0pet

=

•

+

•

ρ

VV

∇

r1 VV

=

1h1

=

θ

=

VV2 rh 2

=

z3 VV

=

1h3

=

()

() () ()

0prV

z

pVprV

rr

1

z

e

V

e

r

V

r

e

V

0pe

zr

t

z

tt

r

t

=

⎥

⎦

⎤

⎢

⎣

⎡

∂

+

θ∂

∂

+

∂

+

∂

ρ+

θ∂

∂ρ

+

∂

ρ

=•

+

•

ρ

θ

VV

∇

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=

φ

∂

φ

∂

φ

∂

φ

∂

φ

∂

φ

∂