978-0131206687 Chapter 9 Part 2

= (0.02)(20)/(1) = 0.4. Hence, L < Lmax so the flow cannot reach Me = 1. To compute the

exit Mach number we have

05288.04.045288.0

D

fL

D

fL

D

fL

2

max

e

max =−=−

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

From which we find Me = 1.2636. Now at the nozzle throat, M = 1 so

p

t

⎞

⎛

()( )

s/kg02250.0

3333.833287.0

=

⎥

⎦

⎢

⎣

(b)

p

2

⎞

⎛

27448.0

p

2

⎠

⎝

⎠

⎝

⎠

⎝

(c) In this case a shock stands at the nozzle exit – station 2. We will call the duct inlet,

(on the downstream side of the shock), station 3. Now at M2 = 2.60147, from the

normal shock relations M3 = 0.50374 and p3/p2 = 7.7289. Therefore,

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189

Also from the Fanno relations at M3, (fLmax/D)3 = 1.03895. Since fL/D = 0.4 it

follows that

63895.04.003895.1

D

3

e

⎟

⎠

⎜

⎝

=

⎟

⎠

⎜

⎝

From which we find Me = 0.5668.

12146.2

p

3

⎠

⎝

⎠

⎝

⎠

⎝

(d) In this case a shock appears just downstream of the nozzle throat. Consequently,

subsonic flow exits the nozzle. For A2/A1 = A2/A* = 2.9, M2 = 0.2046. Therefore,

(fLmax/D)2 = 13.7780. Since, fL/D = 0.4, then (fLmax/D)e = 13.7780 – 0.4 = 13.3780

from which we find Me = 0.2072

1

p

2o

2

e

⎞

⎛

⎞

⎛

⎞

⎛

⎟

⎞

⎜

⎛

⎟

⎞

⎜

⎛

Problem 12. – In which configuration of Figure P9.12, (a) or (b), will the high-pressure

tank empty faster? Explain.

(a)

(b)

Figure P9.12

High

Pressure

Tank

2 3

4

High

Pressure

Tank

2 1

3

4

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190

Problem 13. – Air (γ = 1.4 and R = 0.287 kJ/kg · K) flows through a converging-

diverging nozzle with area ratio of 2.9 (Figure P9.13), which exhausts into a constant-

area insulated duct with a length of 50 cm and diameter of 1 cm. If the system back

pressure is 50 kPa, determine the range of reservoir pressures over which a normal shock

will appear in the duct. Let f = 0.02 in the duct.

Figure P9.13

Shock at Duct Inlet

From which we find Me = 0.8455. Consequently,

7289.7

05.0

p

2

1

1or =

⎠

⎝

⎠

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Shock at Duct Exit

pr

pb = 50 kPa

1 2 e

i

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191

Problem 14. – A converging-diverging nozzle with area ratio of 3.2 (Figure P9.14)

exhausts air (γ = 1.4 and R = 0.287 kJ/kg · K) into a constant-area insulated duct with a

length of 50 cm and diameter of 1 cm. If the reservoir pressure is 500 kPa, determine the

range of back pressures over which a normal shock will appear in the duct (f = 0.02).

Figure P9.14

Shock at Duct Inlet

For A1/At = A1/A* = 3.2, M1 = 2.7056. From the isentropic relations at this Mach

number we obtain p1/po1 = 0.04258. From the normal shock relations at this Mach

number we obtain M2 = 0.4952 and p2/p1 = 8.3737. From Fanno flow relations at M2 we

obtain (fLmax/D)2 = 1.1090. Therefore,

From which we find Me = 0.7645. Consequently,

p

1

⎞

⎛

1 2

pr = 500 kPa

pb

e

i

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() ()

kPa9199.1112761.178

1598.2

1

3559.1p

p

p2

2

e

e=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=∗

∗

Shock at Duct Exit

Problem 15. – A converging-diverging nozzle is connected to a reservoir containing gas

(γ = 1.4). The area ratio of the nozzle is such that the Mach number is 3.5 exiting the

nozzle and entering a constant-area duct of length-to-diameter ratio, L/D, of 100 to 1 and

friction coefficient of 0.01. (a) Determine the normal shock location, if the Mach number

at the exit is 0.75. (b) With the shock at this location, how much longer can the duct be

made before choking occurs at the exit with no change of Mi? Refer to Figure 9.19 for

the nomenclature.

(a) Determining shock location:

5409.05864.01273.00.1

=−+=

The value of M1 can be obtained by numerically solving Eq.(9.33) using a spreadsheet

program that implements the Newton-Raphson method. The following table contains the

history of the iteration process:

Iteration M f(M) f(M+∆M) f(M-∆M) ∆f/∆M Mnew

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193

At M1 = 2.5097, M2 is found from Eq.(9.31) to be 0.5121. At these Mach numbers,

(fLmax/D)1 = 0.4340 and (fLmax/D)2 = 0.9749. The shock location is determined from

Eq.(9.34)

(b) Determining the duct length to accelerate the flow to Mach 1 for the same shock

location determined above.

Because Me = 1.0, (fLmax/D)e = 0.0. Also, because the shock location is fixed

F(M1) = 0.54085 and for the same inlet Mach number, i.e, Mi = 3.5, (fLmax/D)i = 0.5864;

hence,

()

5864.00.0

D

L01.0

−+=

Therefore, L/D = 112.7280 or an additional 12.7280D must be added to the original

length to produce sonic conditions for the same Mi and shock location as in part (a).

Problem 16. – Air (γ = 1.4) enters a pipe of diameter 2 cm at a Mach number of Mi = 3.0.

A normal shock wave stands in the pipe at a location where the Mach number on the

upstream side of the shock is M1 = 2.0. The Mach number exiting the pipe is Me = 1.0.

For steady, adiabatic, one-dimensional flow in the pipe, i.e., Fanno flow, determine the

location of the shock and the total length of the pipe. Assume f = 0.02.

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194

2

fL

D

max

⎤

⎡⎟

⎞

⎛

⎞

⎛

⎞

⎛

cm5023.80

02.0

=

Problem 17. – A rocket nozzle is operating with a stretched out throat, (L = 50 cm and

D = 10 cm) as shown in Figure P9.17. If the inlet stagnation conditions are po1 = 1 MPa

and To1 = 1500 K, determine the nozzle exit velocity and mass flow for a back pressure of

30 kPa. The diameter of the nozzle at the exit station is the same as at the inlet station:

30 cm. Treat the exhaust gases as perfect, with γ = 1.4 and R = 0.50 kJ/kg · K. Assume

isentropic flow in variable-area sections and Fanno flow in constant-area sections with

f = 0.22.

Figure P9.17

i

1 2

e

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Assume the system is choked so that M2 = 1 and (fLmax/D)2 = 0. Also, (p/po)2 = 0.5253

and p2 = p*. Now fL/D = (0.022)(50)/(10) = 0.110 = (fLmax/D)1. The corresponding

Mach number to this value is M1 = 0.7637. At this Mach number from the isentropic

relations: (p/po)1 = 0.67966 and from the Fanno flow relations: (p/p*)1 = 1.35745.

Now for an area ratio of 9, from the isentropic relations we find that Me = 3.8061. At this

Mach number from the isentropic relations: (p/po)e = 0.0.008558. Therefore,

kPa1108.8

35745.1

5283.0

=

⎠

⎝

⎠

⎝

If there is a normal shock at the exit, the pressure ratio across the shock is: (pe2/pe1) =

16.7337. Therefore, pe2 = pb = 135.7230 kPa. Because this is well above the stated back

pressure of 30 kPa, pe = 8.1108 kPa and the flow is further compressed outside the nozzle

by oblique shocks.

At Me = 3.8061 from the isentropic relations, (T/To)e = 0.2566. Hence,

Problem 18. – Air (γ = 1.4 and R = 0.287 kJ/kg · K) flows adiabatically in a tube of

circular cross section with an initial Mach number of 0.5, initial T1 = 500 K, and

pl = 600 kPa. The tube is to be changed in cross-sectional area so that, taking friction into

account, there is no change in the temperature of the stream. Assume the distance

between inlet and exit, L, is equal to 100 Dl, with Dl = initial duct diameter; f = 0.02. Find

the following:

(a) Mach number M2

(b) D2/Dl

(c) Static pressure p2

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196

(b) From Eq.(9.42)

Integration yields

1

12 D

4

⎠

⎜

⎝

Hence,

4

M

D

D2

2

1

2=+=

γ

(c) Since both the static temperature and the Mach number are constant, then so are the

speed of sound and the velocity. Accordingly, p1A1 = p2A2. So that

175.1

D

A

2

⎠

⎝

⎠

⎝

Problem 19. – In a rocket nozzle of area ratio 8 to 1, combustion gases (γ = 1.2 and

R = 0.50 kJ/kg · K) are expanded from a chamber pressure and temperature of 5 MPa and

2000 K. For a nozzle coefficient Ch equal to 0.96, determine the rocket exhaust velocity

in space.

From the isentropic relations with γ = 1.2 and an area ratio, A/A* = 8, the supersonic

solution is M = 3.1219. The corresponding temperature ratio T/To = 0.5064.

Accordingly, the exit temperature is Te = (0.5064)(2000) = 1012.8 K. The exit velictiy is

readily determined as follows:

This is the exit velocity for isentropic flow. The actual exit velocity is related to this

speed by

ropicexit isent

hV

⎠

⎜

⎝

Hence,

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(

)

s/m4712.23846407.243396.0VCV isentropic,ehactual,e ===

Problem 20. – Air (γ = 1.4) enters a constant-area, insulated duct (Figure 9.9) with a

Mach number of 0.50. The duct length is 45 cm; the duct diameter is 3 cm; and the

friction coefficient is 0.02. Use Euler’s explicit method on a coarse grid containing 11

grid points to determine the Mach number at the duct outlet. Compare the result obtained

to the value obtained using Fanno relations.

Exact Solution

From the Fanno relations at M1 = 0.5, (fLmax/D)1 = 1.0691. Also, from the given

information fL/D = 0.3. Hence,

7691.03.00691.1

D

1

2

⎟

⎠

⎜

⎝

=

⎟

⎠

⎜

⎝

Numerical Solution

Proceeding as in Example problem 9.9, we have

D

f

)M1(4

]M12[M

dx

dM

2

23

−

−γ+γ

The same grid in the example is used here in which the duct length is divided into 10

evenly spaced increments, i.e., ∆x = 4.5 cm. The computations are straightforward and

the results from a spreadsheet program are

pt x Mi F(xi,Mi) Mi+1

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Hence, the error between the results is (0.5421/0.5429 – 1)100 = -0.1429%.

Problem 21. –If the problem described in the above is solved on finer grids using the first

order Euler explicit method, the following results for the exit Mach number are obtained

n ∆x Me

Determine the error when compared to the exact value of the exit Mach number is

0.542923. Use Richardson’s extrapolation method to obtain improved values. Also,

compute the error of these values.

Richardson’s extrapolation for Euler’s explicit method is given by

()

12

2

1

2Euler RR2

1

21

−

Using this relation, the following table is easily prepared

n ∆x Me % error Me (Extr) % error

Problem 22. – Heun’s predictor-corrector method is 2nd order and the Runge-Kutta

method used in this Chapter is 4th order. Obtain an expression for each of these methods

that could be used to perform Richardson’s extrapolation of results, R2 and R1 that were

determined on two grids that differ by a factor of two, i.e., ∆x2 = ∆x1/2.

The extrapolated value, E, in Richardson’s extrapolation method, is given by

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199

where R1 and R2 are the values that have been computed using the same method on two

grids of known width, say ∆x1 and ∆x2. Also, the accuracy of the method is of order n.

For this problem ∆x1 = 2∆x2. Thus,

Problem 23. – An airstream (γ = 1.4, R = 0.287 kJ/kg·K) at Mach 2.0 with a pressure of

100 kPa and a temperature of 270 K, enters a diverging, linear, conical channel with a

ratio of exit area to inlet area of 3.0 (see Figure P9.23). The inlet area is 0.008 m2 and the

length is 10.0 cm. The average friction factor is 0.03. Use Heun’s predictor-corrector

method on a coarse grid of 11 grid points to determine the back pressure, pb, necessary to

produce a normal shock in the channel at 5 cm from the inlet. Assume one-dimensional,

steady flow with the air behaving as a perfect gas with constant specific heats. Compare

results to the pressure value obtained by assuming isentropic flow except across the

normal shock (see Example 4.3). Does friction significantly change the isentropic flow

results?

Figure P9.23

Solution Using Fanno Flow, Isentropic Flow and Normal Shock Relations

Ai

Ae

2 1

ie

is

A3A

A2A

=

e

i

s

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200

2

L

is =

Therefore the area at the shock is As = A1 = πDs2/4 = 0.0149 m2. At Mi = 2.0, from the

isentropic relations with γ = 1.4,

008.0

A

*

1

i

*

1

⎠

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Using the Newton-Raphson iterative procedure and taking the supersonic root because

the flow on the upstream side of the shock must be supersonic, we obtain M1 = 2.6882.

Note the Mach number on the downstream side of the shock is found to be 0.4966. With

the upstream shock Mach number determined, ratios of properties across the shock can be

found from normal shock relations, which are then combined with Eq.(4.21) to give

*

1

2o

A

4278.0

p==

Again, using the Newton-Raphson procedure, this area ratio produces the following

subsonic value at the exit: Me = 0.2800. We can now solve for the exit pressure, pe:

1278.0

p

i

oi

1o

2o

i

⎠

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

With subsonic flow at the channel exit, the channel back pressure is equal to the exit

Solution Obtained by Solving ODE Using Heun’s Method and Normal Shock Relations

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201

The equation which governs the Mach number distribution within the channel when both

area variation and frictional effects are considers is

() ()

⎪

⎭

⎪

⎬

⎫

⎪

⎩

⎪

⎨

⎧

−

⎥

⎦

⎤

⎢

⎣

⎡⎟

⎠

⎞

⎜

⎝

⎛

γ+

⎟

⎠

⎞

⎜

⎝

⎛

−

⎪

⎭

⎪

⎬

⎪

⎩

⎪

⎨

==

)M1(

D

f

M

2

1

dx

dA

A

1

2

]M12[M

M,xF

dx

dM

2

2 (9.52)

Since the nozzle is conical, the local cross sectional area is given by

Thus the area term in Eq(9.52) can be written as

() () ()

xD

dx

xD

dx

xA

Consequently, Eq.(9.52) becomes

() () ()

⎪

⎭

⎪

⎬

⎫

⎪

⎩

⎪

⎨

⎧

−

⎥

⎦

⎤

⎢

⎣

⎡⎟

⎠

⎞

⎜

⎝

⎛

γ+

−

⎪

⎭

⎪

⎬

⎪

⎩

⎪

⎨

==

)M1(

D

f

M

2

1

LD

DD2

2

]M12[M

M,xF

dx

dM

2

ie

2

where D = D(x) = Di + (De – Di)x/L. Heun’s method is

2

⎦

⎣

where Mp is the predicted Mach number.

For this problem, we will divide L into 10 pieces of uniform length, i.e., the grid spacing

is therefore ∆x = 1 cm = 0.01 m.

First, we will assume that f = 0. Inserting Mi and the given information into F(x,M) at

x = 0 results in F(0.0,2.0) = 17.5692 and for the grid spacing we can compute the

predicted Mach number, i.e., Mp to be 0.2.1757. This is then used to compute

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202

F(∆x,Mp) = 15.4760. Hence, M2 =2.1602. The static pressures at each x are computed

from

1

2

⎤

⎡

Across the shock located at x = 0.05 m, the normal shock relations are used, viz.,

1

p

5

+γ

The results of the calculations for f = 0 obtained from a spreadsheet program are

presented in the following

The computed exit pressure is 315.6645 kPa, which differs from the value computed

from Fanno relations, i.e., 317.0000kPa by –0.4213%. This is very good for the

particularly coarse mesh used in the computations.

Repeating the calculations for f = 0.03, we obtain,

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203

pt x Di Mi F(xi,Mi) Mp F(xi+1,Mp) Mi+1 pi pi+1

Problem 24. – Helium (γ = 5/3) flows through a symmetrical, C-D nozzle with a circular

cross-section. The shape of the nozzle is given by

t

D

L

x

cos

2

1

12D ⎥

⎦

⎤

⎢

⎣

⎡⎟

⎠

⎞

⎜

⎝

⎛π+=

The nozzle length is two times the throat diameter, i.e., L = 2Dt. Assuming that the

nozzle is choked, determine the Mach number distribution for both subsonic and

supersonic flow in the diverging portion of the nozzle. Assume that the friction

coefficient is 0.4. Use the Method of Beans and the 4th order Runge-Kutta method to

solve this problem on a grid in which ∆x/L = 0.05.

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204

2

t

sp

2

sp

2

L

D

8139.9D

L

x

cos

Ldx

Dd

=

⎟

⎠

⎞

⎜

⎝

⎛π

π

−=

⎟

⎠

⎞

⎜

⎝

⎛

The slopes that are used to begin the solution at the sonic point must be computed. This is

accomplished by solving the following quadratic

0c

dx

b

dx

a

sp

=+

⎟

⎠

⎜

⎝

+

⎟

⎠

⎜

⎝

5176.19

dx

dD

D2

f

dx

dD

D

1

2

dx

Dd

D

2

c

D

sp

2

sp

2

sp

−=

⎟

⎠

⎞

⎜

⎝

⎛γ

−

⎟

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

−=

⎠

⎝

With these coefficients, the two roots for (dM/dx)sp are computed to be

The results of the computations are contained in the following tables

x

i x

i – ∆x/2 xi – ∆x/2 xi – ∆x

M

i M

i Mi – k1∆x/2 Mi – k2∆x/2 Mi – k3∆x

pt x/L Di/Dt Mi F(xi,Mi) k1 k2 k3 k4 Mi-1

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205

16 0.7500 1.2929 0.3391 1.3905 1.3905 1.2387 1.2587 1.1192 0.2765

x

i x

i + ∆x/2 xi + ∆x/2 xi + ∆x

Subsonic decelerating

flow Mi M

i Mi + k1∆x/2 Mi + k2∆x/2 Mi + k3∆x

pt x/L Di/Dt Mi F(xi,Mi) k1 k2 k3 k4 Mi+1

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206

x xi x

i + ∆x/2 xi + ∆x/2 xi + ∆x

Supersonic accelerating flow M Mi M

i + k1∆x/2 Mi + k2∆x/2 Mi + k3∆x

pt x/L Di/Dt Mi F(xi,Mi) k1 k2 k3 k4 Mi+1

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