978-0131206687 Chapter 12 Part 2

vvv

y

=Φ

Now use the chain rule and write

v

yv

xvv

∂

The above pair may be written in matrix-vector form as

⎥

⎦

⎢

⎣

=

⎥

⎦

⎢

⎣

φ

⎥

⎦

⎢

⎣

ΦΦ

0

xy

vvuv

Solving this for φxx and φxy yields

vuuu

vv

vuuu

vv

vu

xx

0

1

ΦΦ

Φ

=

ΦΦ

Φ

=φ

vvuv

vuuu

vvuv

vuuu

xy

ΦΦ

Following the same procedure

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262

u

yu

xuu

∂

The above pair may be written in matrix-vector form as

⎥

⎦

⎢

⎣

=

⎥

⎦

⎢

⎣

φ

⎥

⎦

⎢

⎣

ΦΦ

1

yy

vvuv

Solving this for φyx and φyy yields

vuuu

uu

yy

ΦΦ

Φ

==φ

vvuv

vuuu

yx

ΦΦ

Because of irrotationality

xyyx

0

y

u

x

vφ−φ==

∂

−

∂

So φxy = φyx. Accordingly, from the above we see that

vuuu

vu

yx

vuuu

uv

xy

ΦΦ

Φ−

=φ=

ΦΦ

Φ−

=φ

vvuv

Finally then we obtain the expressions requested i.e.,

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263

2

22

2

u

J

1

y

vuJ

1

yx

v

J

1

x

∂

Φ∂

=

∂

φ∂

∂∂

Φ∂

∂∂

φ∂

∂

Φ∂

=

∂

φ∂

(c) Equation (12.27)

0

y

v

a

v

1

y

u

x

v

a

uv

x

u

a

u

12

2

22

2

=

∂

⎟

⎠

⎞

⎜

⎝

⎛−+

⎟

⎠

⎞

⎜

⎝

⎛

∂

+

∂

−

∂

⎟

⎠

⎞

⎜

⎝

⎛−

may be written as

0

ya

v

1

yx

a

uv

2

xa

u

1

ya

v

1

xyyx

a

uv

xa

u

1

2

22

2

222

22

2

=

∂

φ∂

⎟

⎠

⎞

⎜

⎝

⎛−+

∂∂

φ∂

−

∂

φ∂

⎟

⎠

⎞

⎜

⎝

⎛−

=

∂

φ∂

⎟

⎠

⎞

⎜

⎝

⎛−+

⎟

⎠

⎞

⎜

⎝

⎛

∂∂

φ∂

+

∂∂

φ∂

−

∂

φ∂

⎟

⎠

⎞

⎜

⎝

⎛−

The derivatives of the potential may be replaced by using the expressions from part (b)

0

u

J

1

a

v

1

vuJ

1

a

uv

2

v

J

1

a

u

12

2

22

2

=

∂

Φ∂

⎟

⎠

⎞

⎜

⎝

⎛−+

∂∂

Φ∂

−

∂

Φ∂

⎟

⎠

⎞

⎜

⎝

⎛−

Assuming that J ≠ 0 this becomes

0

ua

v

1

vu

a

uv

2

va

u

12

2

22

2

2=

∂

Φ∂

⎟

⎠

⎞

⎜

⎝

⎛−+

∂∂

Φ∂

+

∂

Φ∂

⎟

⎠

⎞

⎜

⎝

⎛−

Problem 11. – Prove Eq.(12.49).

Start with Eq.(12.48)

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264

1

2

o

a

2

1−γ

−

⎟

⎠

⎜

⎝

−=

ρ

Differentiate this with respect to V, i.e.,

1

2

o

2

o

1

2

o

a

V

2

1

a

V

a

2

a

2

V

1

d

−

−γ

−

⎟

⎠

⎞

⎜

⎝

⎛−γ

−

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

ρ

=

⎠

⎝

⎠

⎝

⎠

⎝

⎤

⎡−γ−

⎞

⎛−γ

⎞

⎛

ρ

But 222

oV

2

1

aa −

γ

+= or

o

a

2

a

Hence,

1

2

o

2

o

a

V

2

1

a

a−

⎟

⎠

⎞

⎜

⎝

⎛−γ

−=

Finally then

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

ρ

=

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

ρ

=

⎟

⎠

⎞

⎜

⎝

⎛−γ

−

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

ρ

=

⎟

⎠

⎞

⎜

⎝

⎛

ρ

ρ−

2

o

2

o

2

o

1

2

o

2

o

oo

a

V

a

V

a

V

2

1

a

V

dV

d

Problem 12. – In the solution of problems using the hodograph equations, the resulting

expressions often contain the following two parameters: τ = (V/Vmax)2 and β = 1/(γ – 1).

Show that

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265

(a)

()

β

τ−=

ρ

1

o

(b)

τ

−

β

τ

=1

2

M2.

From 222

oV

2

1

aa −

γ

+= when a = 0, V = Vmax; therefore omax a

1–

2

V ⎟

⎟

⎠

⎞

⎜

⎝

⎛

γ

=.

(a) So

ρ

o

(b) From static to total density relation and the result of part (a), we have

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Problem 13. – Using Eq. (12.79), show that Eq.(12.77) can be written as

⎟

⎠

⎞

⎜

⎝

⎛+

θ

θ+

θ

⎥

⎦

⎤

⎢

⎣

⎡⎟

⎠

⎞

⎜

⎝

⎛

θ

−−

−γ

=

⎟

⎠

⎞

⎜

⎝

⎛

θ

+

⎟

⎠

⎞

⎜

⎝

⎛

θr

r

2

r

2

r

2

r

2

max

2

r

2

r

2

rv2

d

dv

cot

d

vd

d

dv

vV

2

1

d

vd

v

d

dv

Verify that this is identically satisfied by a uniform stream given by vr = Vcosθ.

Equation (12.77), the Taylor-Maccoll equation is

or

⎟

⎠

⎞

⎜

⎝

⎛+

θ

θ+

θ

=

⎟

⎠

⎞

⎜

⎝

⎛+

θ

⎟

⎠

⎞

⎜

⎝

⎛

θr

r

2

r

2

r

2

r

2

rv2

d

dv

cot

d

vd

av

d

vd

d

dv

Next combine Eq.(12.78)

θ

=

θ

d

dv

vr

and Eq.(12.79)

−

γ

−

γ

22

2

22

2

1

Using this to replace the a2 in the above gives the desired expression

⎟

⎞

⎜

⎛+

θ+

⎥

⎤

⎢

⎡⎟

⎞

⎜

⎛

−−

−γ

=

⎟

⎞

⎜

⎛

+

⎟

⎞

⎜

⎛

r

2

r

2

r

2

rv2

dv

cot

vd

dv

vV

1

vd

v

dv

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267

Now suppose vr = Vcosθ. So dvr/dθ = −Vsinθ and d2vr/dθ2 = −Vcosθ . Substitution into

the above brings

2

or

00

1

22

=

−

γ

Problem 14. – A steady, two-dimensional, supersonic flow, uniformly streams along a

horizontal wall that is aligned with the x-axis. The stream encounters a sharp corner

located at x = 0. Show that the maximum angle through which the flow may turn is given

by π/2b – π/2, where b2 = (γ – 1)/(γ + 1). Also show that the maximum angle can only

occur if the original flow is sonic.

From Eq.(12.66)

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

+γ

−γ

⎞

⎛−−

⎞

⎛−=

⎞

⎛µ−

π

−−

1

2121

The maximum value of ν will occur at M → ∞. As Mach number goes to infinity the

argument of the inverse tangent becomes infinite, i.e., tan-1(∞) = π/2. Therefore, the

above becomes

1

max

π

−

π

=ν

The angle through a flow is turned may be expressed as

12

ν

−

ν

=

∆

The maximum turning angle will occur when ν2 =νmax and ν1 = 0. As shown above νmax

occurs when M2 → ∞; The upstream Prandtl-Meyer function, ν1, i.e.,

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268

(

)

(

)

1Mtan1Mbtan

b

12

1

12

1

1−−−=ν −−

is seen to vanish when M1 = 1.

Problem 15. – Uniform supersonic flow at Mach 3.0 and p = 20 kPa passes over a cone

of semi-vertex angle of 20° aligned parallel to the flow direction. Determine the shock

wave angle, the Mach number of the flow along the cone surface, and the surface

pressure. Take γ = 1.3.

Except for the ratio of specific heats this is identical to Example 12.8.

M2 tan(δs) δs (deg) V2 vr vθ

Spreadsheet calculation results for the first five increments of ∆θ = 0.1˚ for

θs = 29.24443˚, M1 = 3 and γ = 1.3 are as follows

No. θ (deg) (vr)p F[(vr)i,(vθ)i] (vθ)p F[(vr)p,(vθ)p](vr) i+1 (vθ) i+1 V M δ (rad)

1 29.2444 0.6613 -0.1982 0.6613 -0.1982 0.6904 2.4641 -0.2912

6 28.7444 0.6630 -1.2482 -0.1872 -1.2424 0.6630 -0.1872 0.6889 2.4542 -0.2752

Spreadsheet calculation results near the cone surface for ∆θ = 0.1˚, θs = 29.24443˚,

M1 = 3 and γ = 1.3 are contained in the following table

No. θ (deg) (vr)p F[(vr)i,(vθ)i] (vθ)p F[(vr)p,(vθ)p](vr) i+1 (vθ) i+1 V M δ (rad)

92 20.1444 0.6775 -1.3397 -0.0034 -1.3459 0.6775 -0.0034 0.6775 2.3786 -0.0050

Since the velocity at the surface is equal to the radial velocity component, we may readily

compute the Mach number from Eq.(12.84) and the static pressure on the surface.

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269

Problem 16. – Uniform supersonic flow at Mach 4.0 and p = 20 kPa passes over a cone

of semi-vertex angle of 20° aligned parallel to the flow direction. Determine the shock

wave angle, the Mach number of the flow along the cone surface, and the surface

pressure. Take γ = 1.4.

Except for the upstream Mach number this is identical to Example 12.8.

Calculation results near the cone surface for ∆θ = 0.1˚, θs = 26.4850˚, M1 = 4 and γ = 1.4.

No. θ (deg) (vr)p F[(vr)i,(vθ)i] (vθ)p F[(vr)p,(vθ)p](vr) i+1 (vθ) i+1 V M δ (rad)

64 20.1850 0.7908 -1.5610 -0.0051 -1.5680 0.7908 -0.0051 0.7908 2.8891 -0.0064

Since the velocity at the surface is equal to the radial velocity component, we may readily

compute the Mach number on the surface from Eq.(12.84).

θc (deg) Mc Vc vr vθ ps

Problem 17. – Uniform supersonic flow at Mach 3.0 and p = 20 kPa passes over a cone

of semi-vertex angle of 30° aligned parallel to the flow direction. Determine the shock

wave angle, the Mach number of the flow along the cone surface, and the surface

pressure. Take γ = 1.4.

Except for the cone angle this is identical to Example 12.8.

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270

M2 tan(δs) δs (deg) V2 vr vθ

Calculation results near the cone surface for ∆θ = 0.1˚ for θs = 39.7841˚, M1 = 3 and

γ = 1.4 are as follows

No. θ (deg) (vr)p F[(vr)i,(vθ)i] (vθ)p F[(vr)p,(vθ)p](vr) i+1 (vθ) i+1 V M δ (rad)

97 30.1841 0.6335 -1.2566 -0.0041 -1.2602 0.6335 -0.0041 0.6336 1.8310 -0.0064

Since the velocity at the surface is equal to the radial velocity component, we may readily

compute the Mach number on the surface from Eq.(12.84).

θc (deg) Mc Vc vr vθ ps

Problem 18. – A supersonic diffuser contains a conical spike of semi-vertex angle 5°; the

spike is aligned with the flow (Figure P12.18). Determine the Mach number of the flow

along the cone surface and the static pressure at the surface of the cone. Altitude = 5 km

and γ = 1.4.

F

Fi

ig

gu

ur

re

e

P

P1

12

2.

.1

18

8

At an altitude of 5 km the local static pressure is 54.05 kPa.

5º

M∞ = 3.0

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271

Iteration of shock angles continued until a value of θs = 19.75086˚ produced the desired

cone half angle of θc = 5˚. At the free stream Mach number, M1 = 3, and this shock angle,

using Eqs.(12.85), (12.86), (12.87), (12.81) and (12.82), we find M2, V2, δs, vr and vθ,

respectively

M2 tan(δs) δs (deg) V2 vr vθ

Calculation results near the cone surface for ∆θ = 0.1˚ for θs = 19.75086˚, M1 = 3 and

γ = 1.4 are as follows

No. θ (deg) (vr)p F[(vr)i,(vθ)i] (vθ)p F[(vr)p,(vθ)p](vr) i+1 (vθ) i+1 V M δ (rad)

Since the velocity at the surface is equal to the radial velocity component, we may readily

compute the Mach number on the surface from Eq.(12.84).

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