978-0131206687 Chapter 8

149

Chapter Eight

A

AP

PP

PL

LI

IC

CA

AT

TI

IO

ON

NS

S

I

IN

NV

VO

OL

LV

VI

IN

NG

G

S

SH

HO

OC

CK

KS

S

A

AN

ND

D

E

EX

XP

PA

AN

NS

SI

IO

ON

N

F

FA

AN

NS

S

Problem 1. – A supersonic inlet (Figure P8.1) is to be designed to handle air (γ = 1.4,

R = 287 J/kg·K) at Mach 1.75 with static pressure and temperature of 50 kPa and 250 K.

Determine the diffuser inlet area Ai if the device is to handle 10 kg/s of air.

Figure P8.1

Using the oblique shock solution method we obtain

Ai

Ae

M1 = 1.75

14°

123

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

p

2

⎞

⎛

⎞

⎛

()

3

2m 0222.0

s/m4015.503)kg/m 8964.0(

Problem 2. – The diffuser in Problem 1 is to further decelerate flow after the normal

shock so that the velocity entering the compressor is not to exceed 25 m/s. Assuming

isentropic flow after the shock, determine the area Ae required. For this condition, find

the static pressure pe. Take γ = 1.4 and cp = 1.004 kJ/kg·K.

6202.0

T

1

1ooioe =

⎠

⎝

⎟

⎠

⎜

⎝

For M2 = 1.5090, the Mach number downstream of the normal shock is found to be

6979.0M3=. Hence, the area ratio for this Mach number can be obtained from the

isentropic flow tables, 0959.1

A

*

3

3=. And since the flow downstream of the normal

shock is assumed to be isentropic *

3

*

eAA =. Now

2

eTcTc

V=+

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

151

1

A

3

*

3

*

e

e=

⎞

⎛

i

ie m 1892.05231.80222.0

A

⎟

⎠

⎜

⎝

Using the various Mach numbers that have been determined we can find the following

corresponding pressure ratios from isentropic flow and normal shock relations

p

7223.0

p

,6979.0M

9973.0

p

,0621.0M

3

3o

3

oe

e

==

Problem 3. – Compare the loss in total pressure incurred by a one-shock spike diffuser

with that incurred by a two-shock diffuser operating at Mach 2.0. Repeat at Mach 4.0 (see

Figure 8.5). Assume that each oblique shock turns the flow through an angle of 10°. Take

γ = 1.3.

Figure 8.5 Flow Regions within the Spike Diffusers of Example 8.1

M1

3

2

1

M1

3 1 2 4

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

152

From the oblique shock solver at γ = 1.3, M1 = 2.0 and δ = 10°, the weak solution yields

θ = 38.8127°. Moreover the Mach number downstream of the shock is, M2 = 1.6765. For

the one-shock diffuser,

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

o1

o2

o3

shock–one

o1

o3

p

From the oblique shock relations at M1 = 2.0, po2/pol = 0.9861 and from the normal shock

relations at M2 = 1.6765, po3/po2 = 0.8570. Hence,

p

shockone

o1

⎟

⎠

⎜

⎝

−

For the two-shock inlet, M2 = 1.6765. At the latter Mach number and δ = 10°, the wave

angle for the weak shock solution is θ = 47.3152º, po3/po2 = 0.9889 and M3 = 1.3533. At

M3 from the normal shock relations po4/po3 = 0.9677. Thus,

()( )( )

9437.09861.09889.0967.0

p

o1

o2

o3

o4

shocks–two

o1

o4 ==

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

=

⎟

⎠

⎜

⎝

Now at M1 = 4.0 and δ = 10°, the weak solution yields θ = 21.8411°, po2/pol = 0.9301

and M2 = 3.4050. From the normal shock relations at M2 = 3.4050, po3/po2 = 0.1853.

Therefore, for the one oblique shock diffuser,

p

o1

o2

shockone

o1

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

−

For the two-shock inlet, M2 = 3.4050. At M2 = 3.4050 and δ = 10°, θ = 24.4808°,

po3/po2 = 0.9533 and M3 = 2.9186. Using M3 in the normal shock relations gives

po4/po3 = 0.3065. For this case,

p

o1

o2

o3

o4

shockstwo

o1

o4 ==

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

−

Problem 4. – A converging nozzle is supplied from a large air (γ = 1.4, R = 287 J/kg·K)

reservoir maintained at 600K and 2 MPa. If the nozzle back pressure is 101 kPa,

determine the pressure and Mach number that exist at the nozzle exit plane. Since the

nozzle is operating in the underexpanded regime, expansion waves occur at the nozzle

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

exit. Determine the flow direction after the initial expansion fans and the flow Mach

number.

Since the nozzle is operating in the underexpanded flow regime, the nozzle is choked.

Accordingly, the Mach number at the exit is Me = 1.0 and the exit pressure to reservoir

pressure ratio is pe/po = 0.5283 for γ = 1.4. Thus the exit pressure is

The expansion fans turn the supersonic flow and reduce the pressure to that of the back

pressure. Now

0505.05283.0

6.1056

p

o

e

o

From this pressure ratio we can find the corresponding Mach number

Problem 5. – An oblique shock wave occurs in a supersonic flow in which M1 = 3. The

shock turns the supersonic stream through 10°. The shock impinges on a free surface

along which the pressure is constant and equal to p1, i.e., the pressure upstream of the

shock. The shock is reflected from the free surface as an expansion fan. Determine the

Mach number and the angle of the flow just downstream of the fan. Assume γ = 1.4.

free surface

p1 = p3

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

⎬

⎫

°=δ

=

10

0.3M1

5050.2M

3827.27

2

=

°=θ

In region 2 from the Prandtl-Meyer and isentropic relations

p

2o

Because the flow across the expansion fan is isentropic po2 = po3 and because of the

constant pressure free surface p1 = p3, thus we may form the following string of pressure

ratios

0545.2

p

3o

2o

2

1

3o

⎠

⎝

With this pressure ratio, using the static to total pressure-Mach number relation, we

obtain M3 = 2.9750 and therefore from the Prandtl-Meyer relation ν3 = 49.2727˚. Finally

then, for this flow geometry

2323

Accordingly,

Problem 6. – A converging-diverging nozzle is designed to provide exit flow at Mach

2.2. With the nozzle exhausting to a back pressure of 101 kPa, however, and a reservoir

pressure of 350 kPa, the nozzle is overexpanded, with oblique shocks appearing at the

exit. Determine the flow direction, static pressure, and Mach number in regions 1,2, and

3 of Figure P8.6.

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Consider the following geometry

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Because the nozzle is designed for an exit Mach number, Me, of 2.2, it follows that the

static to total pressure ratio at the nozzle exit is

0935.0

p

o

e=

The back pressure is 101 kPa and the reservoir pressure is 2,000 kPa, therefore,

. 0505.0

2000

p

o

Since pb/po < pe/po, the nozzle is underexpanded for this back pressure-reservoir pressure

combination. The following provides nomenclature and a sketch of the flow field.

From the Prandtl-Meyer relation at Me = 2.2, we find that νe = 31.7325˚. Since pb = p1 in

Since αe = 0˚, then it follows that α1 = 9.5719˚. The flow in region 2 must be horizontal,

i.e., α2 = 0˚, and since we must pass through another expansion fan, we may write that

Hence,

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

ratio across the oblique shock, i.e., p3/p2 = 101/49.8 = 2.0281. With this pressure ratio

and M2 = 3.0587 we can first determine the shock angle from Eq.(6.10)

()( )

0587.38.2

Problem 8. – A plug nozzle is designed to produce Mach 2.5 flow in the axial direction at

the plug apex. Flow at the throat cowling must therefore be directed toward the axis.

Determine the flow direction at the throat cowling to produce axial flow at the apex.

Assume γ = 1.4.

α

−

α

=

°

=

−

=

ν−==

Problem 9. – A rocket nozzle is designed to operate with a ratio of chamber pressure to

ambient pressure (pc/pa) of 50. Compare the performance of a plug nozzle with that of a

converging-diverging nozzle for two cases where the nozzle is operating overexpanded;

pc/pa = 40 and pc/pa = 20. Compare on the basis of thrust coefficient; CT = T/(pcAth),

where T is the thrust and Ath is the area of throat. Assume γ = 1.3 and in both cases

neglect the effect of nonaxial exit velocity components.

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8696.0

1

2

T

5457.0

2

1

p

c

th

1

c

th

=

+γ

=

⎟

⎠

⎞

⎜

⎝

⎛+γ

=γ−

γ

Using these values and the values at the exit, we get

5145.1

=

Note, R, pc and Tc drop out of the above expression.

⎟

⎠

⎜

⎝

c

th

cth

design

TT p

p

A

pA

where at Me = 3.1267, Ae/Ath = Ae/A* = 5.9590. So for pc/pa = 40,

1

⎞

⎛−+=

For the plug nozzle,

Flow in the plug nozzle does not continue to expand below ambient pressure, so there is

no pressure term in the expression for thrust.

T

p

e

c===

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

(

)()

4871.1

T4269.0R3.19918.2

T8696.0R3.1A

p5457.0

Vm

Cc

cth

c

eth

T

=

⎥

⎤

⎢

⎡

⎥

⎤

⎢

⎡

== &

Whereas for pc/pa = 20, 5009.0

T

and5773.2M

c

e

e==

Problem 10. – Compute the lift and drag coefficients for a flat plate airfoil of chord

length c = 1m in supersonic flow through air (γ = 1.4) at M∞ = 3 and α = 8°.

Results

Region ∞: freestream

γ M∞ α∞ ν∞ ρv2/(2p∞) p∞/po∞

Region 1: lower region behind oblique shock

γ M∞ δ θ α M1 po1/po∞ p1/p∞

Region 2: upper region behind expansion fan

γ M∞ ν∞ α ν2 M2 p2/po2 p2/p∞

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Region 3: lower region behind expansion fan

γ M1 δ13 ν3 M3 po3/po1 p3/p1

Region 4: upper region behind oblique shock

γ M2 δ24 θ24 M4 po4/po2 p4/p2

L/(cp∞) D/(cp∞)

Mach numbers and pressure ratios 1.2671 0.1781

M∞ M1 M2 M3 M4

Problem 11. – Compute the drag coefficient for a symmetric, diamond-shaped airfoil

(Figure P8.11) with a thickness to chord ratio, t/c, equal to 0.10 flying at Mach 3.5 in air

(γ = 1.4) at 10 km at zero angle of attack.

Figure P8.11

For an oblique shock at the nose of the airfoil,

t

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

161

Using this value of the Prandtl-Meyer function we find the corresponding Mach number

is M3 = 3.8760 and in turn the corresponding static to total pressure ratio is

p3/po3 = 0.007781. Accordingly, we may form the following ratio

Therefore,

p

2

Because of the symmetry and the 0º angle of attack, the lift coefficient is zero. The drag

coefficient may be determined in the following way

()() ()

5.34.1

2

1

cMpγ

2

1

2

D=

∞∞

Problem 12. Compute the lift and drag coefficients for the airfoil described in Problem

11 for an angle of attack of 5°.

Upper Surface

p

003937.0

p

2

2o

4o

2

⎠

⎝

∞

Lower Surface

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7106.107106.55δ

,5.3M

1⎭

⎬

⎫

°=+=∆+α=

=

∞

°=ν=

°

=

0292.47 ,8623.2M

,0309.25θ

11

4950.3M , 4504.584212.110292.472ν323

=

°

=

+

=∆+ν=

()()

3918.2

p

,3939.0

03351.0

1

101320.0

p

p1

1

1o

3o

3

1

3==

⎟

⎠

⎞

⎜

⎝

⎛

==

∞

()

⎟

⎠

⎞

⎜

⎝

⎛

⎥

⎦

⎤

⎢

⎣

⎡

+−

⎥

⎦

⎤

⎢

⎣

⎡

+

=

∞∞ cMpγ

2

1

7106.5cos

7106.10cos

2

c

p

7106.5cos

7106.0cos

2

c

p

7106.5cos

7106.0cos

2

c

p

7106.5cos

7106.10cos

2

c

p

C

2

4321

L

()

(

)

(

)

(

)

()()()

1202.0

5.34.19950.0

9999.09424.09382.09826.03003.03918.2

C2

L=

−

+

−

=

()

(

)

(

)()

()()()

2

4321

D5.34.17106.5cosp

7106.10sinp7106.0sinp7106.0sinp7106.10sinp

C

∞

−

+

=

()()

(

)

(

)

()()()

0228.0

5.34.19950.0

0124.09424.09382.01858.03003.03918.2

C2

D=

−

+

−

=

Problem 13. – Compare the lift to drag ratio of the diamond airfoil in problem 12 with

that of a flat-plate airfoil for the same freestream Mach number of 3.5 and angle of attack

of 5°. Assume γ = 1.4.

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

163

Upper Surface

Using the freestream Mach number

From this and the angle of attack we can find ν2 from which we can find M2

and so M2 = 3.8344. Furthermore using the Mach of the freestream and in region 2 we

can use the isentropic relations to determine the corresponding static to stagnation

pressure ratios. Since the flow from the freestream into region 2 is isentropic

Lower Surface

Because the freestream flow must be turned through 5º as it passes through the oblique

shock

°==

⎟

⎠

⎞

⎜

⎝

⎛

°==

⎟

⎠

⎞

⎜

⎝

⎛5 at 2719.5

0228.0

1202.0

C

,5 at 4301.11

00921.0

1053.0

C

ildiamond fo

D

L

flat plate

D

L

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

164

Problem 14. – Consider a flat-plate supersonic airfoil with a flap, as shown in Figure

P8.10. For a flap angle of 5°, an angle of attack 10°, and a flight Mach number of 2.2,

find the lift and drag coefficients of the airfoil.

Figure P8.14

Except for the trailing edge phenomena, there will be two expansion fans on the top of

the plate and two oblique shocks on the lower portion. The regions for the calculations

are numbered as follows

From this and the angle of attack we can find ν2, which will lead to M2

()

°=−−+=α−α+ν=ν

∴

α+ν=α+ν

∞∞

7325.410000.1007325.31

22

and so M2 = 2.6142. This process is repeated in passing through the expansion fan at the

corner of the flat plate and the flap

()

°=−−−=α−α+ν=ν

∴

α+ν=α+ν

7325.460000.150000.107325.41

4224

4422

10º

5º

4

2

3

1

∞

c

c/3

10º

5º

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

and so M4 = 2.8478. Furthermore using M∞, M2, M4, we can use the isentropic relations

to determine the corresponding static to stagnation pressure ratios. Since the flows from

the freestream into region 2 and from region 2 to 4 are isentropic

1

p

po

2o

22 =

⎞

⎛

∞

09352.0

p

o

2o

4o

⎠

⎝

∞

∞∞

Lower Surface

The freestream flow is turned through 10º as it passes through the first oblique shock.

Therefore,

⎭

⎬

⎫

°=δ

=

∞

10

2.2M

7641.1

p

8228.1M

,7855.35

1

1==

°=θ

∞

The stream in region 1 is turned through 5º as it passes through the second oblique shock

as it flows into region 3. Therefore,

1

And so

()()

2875.27641.12967.1

p

p1

1

33 ===

∞∞

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

−

()()()()

1125.0

0489.00635.02679.01826.01763.03604.0

cMγp

2

1

cMγp

2

1

2

=

+=+=

∞∞∞∞

Problem 15. – Compute the lift and drag coefficients for the supersonic, symmetric

airfoil shown flying in air (γ = 1.4) at Mach 2.5 at an angle of attack of 5° in Figure

P8.15.

Figure P8.l5

Because the angle of attack and the wedge angle have the same value, the flow will

experience only one expansion fan on the upper surface where the slope changes and an

oblique shock on the bottom at the leading edge.

Upper Surface

()

°=−−+=α−α+ν=ν

∴

∞∞

1236.490000.1001236.39

22

and so M2 = 2.9674.

05853.0

1

p

po

o

2o

22 =

⎠

⎞

⎝

⎛

∞

∞∞

Lower Surface

The freestream flow is turned through 5º as it passes through the forward oblique shock.

Therefore,

5º

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

() ()

(

)

()

()()

()

()()

()

2415.05019.03747.1

9962.02

9848.04885.0

.9962.0

5.0

9962.03799.1

cMγp

2

1

5cos 2

10cos cp

5cos 2

cp

5cos cp

C

2

1

L

−−

=

∞∞

∞

Problem 16. – A supersonic jet plane is flying horizontally at 150 m above ground level

at a Mach number of 2.5, as shown in Figure P8.16. The airfoil is symmetric and

diamond shaped, with 2∆ = 10º and a chord length of 4m. As the plane passes over, a

ground observer hears the “sonic boom” caused by the shock waves. Find the time

between the two “booms,” one from the shock at the leading edge and one from the shock

at the trailing edge. Ambient pressure and temperature are 100 kPa and 20°C.

this material may be reproduced, in any form or by any means, without permission in writing from the publisher.