978-0131206687 Chapter 6 Part 1

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Si

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Problem 1. – Uniform airflow (γ = 1.4, R = 287 J/kg·K) at Mach 3 passes into a concave

corner of angle 15°, as shown in Figure P6.1. The pressure and temperature in the

supersonic flow are, respectively, 72 kPa and 290 K. Determine the tangential and normal

components of velocity and Mach number upstream and downstream of the wave for the

weak shock solution. Also find the static and stagnation pressure ratios across the wave.

How great would the corner angle have to be before the shock would detach from the

corner?

Figure P6.1

1

M = 3

∆

= 15°

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107

cotθ tanθ angle (deg)

weak 1.5855 0.6307 32.24

strong 0.0977 10.2387 84.42

neg root -4.8450 -0.2064 -11.66

So for the weak solution, the shock angle is 32.24°

From the shock tables at this Mach number M n 2 = 0.6683, p2/p1 = 2.8215, T2/T1 = 1.3882

and po2/po1 = 0.8950. From Eq.(6.9b)

() ()

1524.32sin

sin

2=

−

δ−θ

Too be sure, these could also be computed from the oblique shock relations of this

Chapter [Eqs.(6.10), (6.12), (6.13) and (6.17)].

From the isentropic tables at M1 = 3.0, T/To = 0.3571

So the speeds of sound may be computed as

And the normal velocity components are

Also,

The tangential velocity component can be computed from either Eq.(6.6a) or (6.7a)

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108

3.402

a

2

From Table 6.4 for γ = 1.4 and M1 = 3.0, δmax is found to be 34.07°.

Problem 2. – In a helium (γ = 5/3) wind tunnel, flow at Mach 4.0 passes over a wedge of

unknown half-angle aligned symmetrically with the flow. An oblique shock is observed

attached to the wedge, making an angle of 30° with the flow direction. Determine the

half-angle of the wedge and the ratios of stagnation pressure and stagnation temperature

across the wave.

Method 1: Use of normal shock tables.

Using this value we can enter the normal shock table at a γ = 5/3 to find

Entering the isentropic flow table at M1 = 4 we find that

1579.0

T

1o

Thus,

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T

1o

1

2

2o

2==

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

Entering the isentropic flow tables with this temperature ratio provides the downstream

Mach number

Method 2: Oblique shock equations

Problem 3. – A wedge is to be used as an instrument to determine the Mach number of a

supersonic airstream (γ = 1.4); that is, with the wedge axis aligned to the flow, the wave

angle of the attached oblique shock is measured; this permits a determination of the

incident Mach number. If the total included angle of such a wedge is 45°, give the Mach

number range over which such an instrument would be effective.

Refer to Example 6.3. That example concerned the prediction of the minimum upstream

Mach number to produce an attached oblique shock. This is a similar problem; only here,

the half angle ∆ = 45/2 = 22.5°. So for a deflection angle δ of 22.5° and γ = 1.4,

computations following Example 6.3 are as follows:

Iteration θ (deg) θ(rad) f(θ) df/dθ θnew θ (deg) 1/M1

2 M1

1 45.00000 0.78540 0.85858 -1.41421 1.39250 79.78466 0.13234 2.74886

2 79.78466 1.39250 -0.64460 -1.72985 1.01987 58.43444 0.24273 2.02971

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Problem 4. – The leading edge of a supersonic wing is wedge shaped, with a total

included angle of 10° (Figure P6.4). If the wing is flying at zero angle of attack,

determine the lift and drag force on the wing per meter of span. Repeat for an angle of

attack of 3°. Assume the wing is traveling at Mach 2.5.

Figure P6.4

Case I: Zero angle of attack:

First draw a figure (exaggerated) showing the forces acting on the surface:

M∞ = 2.5

2.0 m

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111

∞∞ =°

⎠

⎝

°

5 cos

Case II: Angle of attack = 3

°

:

Using the normal shock relations we obtain

p

∞

For lower surface: δ = 2° and M1 = 2.5, we find θ = 30.0053°. So that

p

∞

Drag °+°=2 sinA p8 sinA p UL

∞

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112

Problem 5. – An oblique shock wave is incident on a solid boundary, as shown in Figure

P6.5. The boundary is to be turned through such an angle that there will be no reflected

wave. Determine the angle β.

Figure P6.5

Problem 6. – Explain in physical terms why the angle of incidence and the angle of

reflection of a reflected oblique shock are not equal.

Problem 7. –A converging-diverging nozzle is designed to provide flow at Mach 2.0.

With the nozzle exhausting to a back pressure of 80 kPa, however, and a reservoir

pressure of 280 kPa, the nozzle is overexpanded, with oblique shocks at the exit (Figure

P6.7). Determine the flow direction and flow Mach number in region R with air the

working fluid.

Figure P6.7

θ

= 45°

β

M1 = 3.5

Very Large

Reservoir

po ≈ constant

R

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113

At the exit plane,

p

1

⎞

⎛

Across shock, 2356.2

7840.35

p

e

1

Entering the normal shock tables at this pressure ratio we find, 4350.1M 1n =

⎠

⎝

2

Now at Mn1

T

2o

1o

1

2o

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

At this static to total temperature ratio we can find M2 = 1.4301

Problem 8. –(a) Oblique shock waves appear at the exit of a supersonic nozzle, as shown

in Figure P6.8. Air is the working fluid. If the nozzle back pressure is 101 kPa,

determine the nozzle inlet stagnation pressure. The stagnation temperature of the flow is

500 K. Nozzle throat area is 50 cm2, and nozzle exit area is 120 cm2. (b) Find the

velocity at the nozzle exit plane. (c) Find the mass flow rate through the nozzle.

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114

Figure P6.8

(a) 4.2

50

120

*A

A

Ae

throat

exit ===

(b)

()

K5000.2325004650.0T

T

T hus,T 4650.0

T

o

e

o

e==

⎟

⎠

⎞

⎜

⎝

⎛

==

(c)

()

throat

AVm ρ=

&

Reservoir

30º

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115

()

aMA

TTTR

ppp

m

ttt

oot

1o1ot

⎥

⎦

⎤

⎢

⎣

⎡

=

&

Problem 9. – A supersonic flow leaves a two-dimensional nozzle in parallel, horizontal

flow (region A) with a Mach number of 2.6 and static pressure (in region A) of 50 kPa.

The pressure of the atmosphere into which the jet discharges is 101 kPa. Find the

pressures in regions B and C of Figure P6.9.

Figure P6.9

0200.2

50

101

p

kPa,101p

2

2=== . At this pressure ratio we can find the normal

()

°==

⎟

⎠

⎜

⎝

=θ −− 7719.315265.0sin

M

sin 1

A

°

=

δ

→°=θ= 0346.117719.31 ,6.2M1 So second shock must turn flow back

by 11.0346°. With the flow angles and MA, the temperature ratio across the shock

is found to be

T

o

1

A

o

A

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Using this value with the isentropic flow relations gives

A B C

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116

()

kPa7188.1851018388.1p so8388.1

p

C

B

Problem 10. – For the two-dimensional diffuser shown in Figure P6.10, find Vi and poi,

Figure P6.10

0800.

There is enough information to determine the Mach number downstream of the

shock M2 = 2.1823 as well as several other ratios, viz., 0540.1

T

1

2=and

p

2o = Across the normal shock, at M2 = 2.1823, 6362.0

p

3o = and M3

To

6º

Vi, poi

M∞ = 2.3

p∞ = 50 kPa

T∞ = 0º C

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117

Problem 11. – A two-dimensional supersonic inlet is to be designed to operate at Mach

2.4. Deceleration is to occur through a series of oblique shocks followed by a normal

shock, as shown in Figure 6.12. Determine the loss of stagnation pressure for the cases of

two, three, and four oblique shocks. Assume the wedge turning angles are each 6°.

Case I:

Two oblique shocks:

⎫

=

4.2M1

1589.2M

2

=

°=δ

6

9959.0

p

2o

3o

=

Normal Shock

M

3 = 1.9354, 7510.0

p

3o

4o =

()()()

7440.09948.09959.07510.0

p

1o

2o

3o

1o

Three oblique shocks:

°=δ

6

9966.0

p

3o

4o

=

Normal Shock

M

3 = 1.7240, 8457.0

p

4o

5o =

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