978-0131206687 Chapter 9 Part 1

Chapter Nine

F

FL

LO

OW

W

WI

IT

TH

H

F

FR

RI

IC

CT

TI

IO

ON

N

Problem 1. – Draw the T-s diagram for the adiabatic flow of a gas with γ = 1.4 in a

constant diameter pipe with friction. The reference Mach number, M1, for the flow is 3.0.

⎭

⎩

In this expression there are two values of To/T that will cause ∆s/cp to vanish. Clearly,

both will cause the argument of the natural log function to be exactly equal to 1. One

value occurs at To/T1, i.e., when T = T1. Because of the nonlinearity of the function

involving To/T, the other value must be found numerically. This is readily accomplished

using a spreadsheet program to implement the Newton-Raphson method. Setting the

argument of the natural log function to unity gives

1TT

TT

1o

o

⎥

⎦

⎢

⎣

⎟

⎠

⎜

⎝

−

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Rearranging this produces

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170

To/T f f ′ (To/T)new Mmin

So

T=∆

The coordinates for the Fanno-Line at this reference state are shown in the following

table. The figure shown below is a plot of this data.

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0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

2.75

3.00

Problem 2. – Draw the T-s diagram for the adiabatic flow of a gas with γ = 1.3 in a

constant diameter pipe with friction. The reference Mach number, M1, for the flow is 4.0.

Following the same procedure as indicated in Problem 1

To/T f df/dt (To/T)new Mmin

1

T

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The coordinates for the Fanno-Line at this reference state are shown in the following

table. The figure shown below is a plot of this data.

M ∆s/cp T/T1

4.00 0.0000 1.0000

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

1

T

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173

Problem 3. – Air (γ = 1.4) flows into a constant-area insulated duct with a Mach number

of 0.20. For a duct diameter of 1 cm and friction coefficient of 0.02, determine the duct

length required to reach Mach 0.60. Determine the length required to attain Mach 1.

Finally if an additional 75 cm is added to the duct length needed to reach Mach 1, while

the initial stagnation conditions are maintained, determine the reduction in flow rate that

would occur.

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0333.16

1

D

R1

=

⎟

⎠

⎜

⎝

From this value we find that M1R = 0.1917. Note the subscript R has been added to

indicate the reduced value. Using the isentropic flow relations we have

⎭

⎬

⎫

=γ

=

4.1

1917.0M R1

9927.0

T

9747.0

p

0333.16

D

1o

R1

1o

R1

max

=

⎟

⎠

⎜

⎝

The original mass flow rate and the reduced flow rate may be written respectively as

1R

RT

⎠

⎝

Since the stagnation conditions are maintained we may write the following

T

M

p

T

M

T

p

T

p

m

1o

R1

1o

1

1R

1o

1

1o

R1

1o

1

1o

R1

1

1R

1o

R1

1o

1

1o

1

1o

R1

R

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

&

Problem 4. – Air (γ = 1.4 and R = 0.287 kJ/kg · K) enters a constant-area insulated duct

with a Mach number of 0.35, a stagnation pressure of 105 kPa, and stagnation

temperature of 300 K. For a duct length of 50 cm, duct diameter of 1 cm, and friction

coefficient of 0.022, determine the air force on the duct wall.

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12airon wall2211 VVmFApAp

Thus to compute the force we must first determine the entry and exit values of the static

pressure and velocity as well as the mass flow rate.

Using the Fanno flow and isentropic flow relations we have at the upstream location

35.0M1

=

⎪

⎨

⎧

==

ρ

=

⎟

⎠

⎞

⎜

⎝

⎛

∗

6400.2

V

0922.3

p

4525.3

D

fL

1

max

3525.2

1

4525.3

D

1

2

⎟

⎠

⎜

⎝

=

⎟

⎠

⎜

⎝

From this value we find that M2 = 0.3976. Using the Fanno and isentropic flow relations

we have

⎪

⎩

=

9694.0

T

2o

2

Since po1 = 105 kPa,

p

1

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176

Because the flow is adiabatic: To1 = To2 = 300 K.

s/kg01082.0

1

=

⎦

⎣

⎠

⎝

Finally,

N7585.01716.09302.0

kN

4

=−=

⎠

⎝

Problem 5. – Hydrogen (γ = 1.4 and R = 4124 J/kg · K) enters a constant-area insulated

duct with a velocity of 2600 m/s, static temperature of 300 K, and stagnation pressure of

520 kPa. The duct is 2 cm in diameter, and 10 cm long. For a friction coefficient of 0.02,

determine the change of static pressure and temperature in the duct and the exit velocity

of the hydrogen.

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

9756.1M1

=

⎪

⎨

⎧

==

ρ

=

⎟

⎠

⎞

⎜

⎝

⎛

∗

6166.0

V

4155.0

p

2977.0

D

fL

1

max

1977.0

2

2977.0

D

1

2

⎟

⎠

⎜

⎝

=

⎟

⎠

⎜

⎝

From this value we find that M2 = 1.6712. Using the Fanno and isentropic flow relations

we have

6712.1M2

=

⎪

⎧

ρ

=

∗

V

5251.0

p

2

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T

2

⎞

⎛−=−

Problem 6. – A constant-area duct, 25 cm in length by 1.3 cm in diameter, is connected

to an air reservoir through a converging nozzle, as shown in Figure P9.6. For a constant

reservoir pressure of 1 MPa and constant reservoir temperature of 600 K, determine the

flow rate through the duct for a back pressure of 101 kPa. Assume adiabatic flow in the

tube with f = 0.023.

Figure P9.6

First determine the exit pressure assuming the duct is choked. Therefore,

()

1

2

1

D

3.1

D

⎠

⎝

⎠

⎝

⎠

⎝

⎠

⎝

From this value we can determine that M1 = 0.6129. At this Mach number using the

isentropic and Fanno flow pressure relations we may write that

7239.1

p

1o

1

⎠

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Since the back pressure is well below this value the assumption that the duct is choked is

correct and we may proceed to determine the flow rate. Now at M1 = 0.6129,

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()

()( ) ()()()()

s/kg1867.0

406.558287.0

1.776

RTAM

RT

p

AVρm

2

11

1

11

=

⎠

⎞

⎝

⎛π

⎥

⎦

⎤

⎢

⎣

⎡

γ

⎟

⎠

⎞

⎜

⎝

⎛

==

&

Problem 7. – Find the time required for the pressure in the tank filled with Nitrogen

(γ = 1.4 and R = 296.8 J/kg · K) shown in Figure P9.7 to drop from 1 MPa to 500 kPa.

The tank volume is 8 m3 and the tank temperature is 300K. Assume the tank temperature

remains constant and the flow in the 3 m long, 1 cm diameter connecting tube is adiabatic

with f = 0.018. The back pressure is 101 kPa.

Figure P9.7

First determine the exit pressure assuming the duct is choked. Therefore,

From this value we can determine that M1 = 0.2979. At this Mach number using the

isentropic and Fanno flow pressure relations we may write that

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()

r1o

1o

1

1o

1

p9403.0p

p

K7800.2943009826.0T

T

=

⎟

⎠

⎞

⎜

⎝

⎛

=

==

⎟

⎠

⎞

⎜

⎝

⎛

=

Thus,

()

2

r

11

1

11

p9403.0

RTAM

RT

p

AVρm

⎞

⎛π

⎤

⎡

γ

⎟

⎠

⎞

⎜

⎝

⎛

==

&

Now within the reservoir,

r

Taking the time derivative of this expression gives

dt

8

dt

r

⎟

⎠

⎜

⎝

∀

From a mass balance on the reservoir,

dt

dmr&

Therefore,

()

r

rp10x7950.9p10x8006.81300.11

dt

Integration gives

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Problem 8. – A converging-diverging nozzle has an area ratio of 3.3, i.e., the exit and

therefore the duct area is 3.3 times the throat area, which is 60 cm2. The nozzle is

supplied from a tank containing air (γ = 1.4 and R = 0.287 kJ/kg · K) at 100 kPa at 270 K.

For case A of Figure P9.8, find the maximum mass flow possible through the nozzle and

the range of back pressures over which the mass flow can be attained. Repeat for case B,

in which a constant-area insulated duct of length 1.5 m and f = 0.022 is added to the

nozzle.

Case A Case B

Figure P9.8

Case A

The maximum flow rate will occur when the throat Mach number is 1. At this Mach

number, the throat static to total pressure and temperature ratios are: 0.5283 and 0.8333,

respectively. Accordingly, the flow rate is computed to be

()( )()

s/kg4760.1

2708333.0287.0

T

RAM

T

R

p

RTAM

RT

p

VAρm

o

t

o

t

o

t

tt

t

tttmax

=

⎥

⎦

⎢

⎣

⎟

⎠

⎞

⎜

⎝

⎛

γ

⎟

⎠

⎞

⎜

⎝

⎛

=γ

⎟

⎠

⎞

⎜

⎝

⎛

==

&

For A/A* = 3.3, we can determine that the exit Mach number is 0.1787. At this value the

exit static to total pressure ratio is 0.9780. Thus, the maximum flow rate will occur for

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182

Case B

Here too, the maximum flow rate will occur when the throat Mach number is 1. At this

Mach number, the throat static to total pressure and temperature ratios are: 0.5283 and

0.8333, respectively. Accordingly, the maximum flow rate will be the same as that in

Case A, viz., 1.4760 kg/s.

Now for subsonic flow at the nozzle exit, and the duct inlet, M1 = 0.1787.

8522.18

D

1

⎟

⎠

⎜

⎝

The diameter of the duct is computed as follows

π

Thus

So the exit Mach number is 0.1796. The exit pressure which is equal to the back pressure

is computed as follows

Thus, the maximum flow rate will occur for

Problem 9. – A 3-m3 volume tank, R, is to be filled to a pressure of 200 kPa (initial

pressure 0 kPa). The tank is connected to a reservoir tank, L, containing air at 3 MPa and

300 K, whose volume is also 3 m3. A 30-m length of 2.5 cm-diameter tubing is used to

connect the two vessels, as shown in Figure P9.9. Determine the time required to fill the

tank to 200 kPa. Assume Fanno flow with f = 0.02.

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

Figure P9.9

Because R, the tank on the right, is evacuated, it may be safely assumed that Me = 1.

Therefore,

()

i

e

i

D

025.0

D

⎠

⎝

⎠

⎝

⎠

⎝

⎠

⎝

From this value we find that Mi = 0.1606. Using the isentropic flow relations we have

4.1

=γ

⎪

⎩

⎨

=

⎟

⎠

⎞

⎜

⎝

⎛

9949.0

T

i

o

Thus,

()

2

oR

iii

i

iii

p98215.0

RTMA

RT

p

VAρm

⎞

⎛π

⎤

⎡

γ

⎟

⎠

⎞

⎜

⎝

⎛

==

&

Now in order that there be Fanno flow, To must remain constant. So for tank L

Differentiate this with respect to time to get

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Similarly, for the tank on the right

Therefore,

Clearly,

dt

That is

oL

3

oL

oL p10x9845.8p

03484.0

10x1302.3

dt

dp −

Integration brings

3

dp

10

oL

MPa8.2

3=

⎞

⎛

Problem 10. Find the mass flow rate of air (γ = 1.4 and R = 0.287 kJ/kg · K) through the

system shown in Figure P9.10. Assume Fanno line flow in the duct and isentropic flow in

the converging sections; f = 0.01.

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1 2

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

4761.85.79761.0

D

fL

D

fL

D

fL

2

max

1

max =+=+

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

From this, we find that M1 = 0.2501. Consequently, with M1, M2 and M3 we are able to

compute the following static pressures

p

1

⎞

⎛

8363.0

p

2

2o

3o

3=

⎠

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Too large; therefore M3 needs to be increased. After a few tries M3 = 0.973 and from the

isentropic relations we find

p

A

33 ==

From this we find M2 = 0.5180, from which we obtain

p

fL 2

2

max ===

⎞

⎛

4316.85.79316.0

D

2

1

⎟

⎠

⎜

⎝

=

⎟

⎠

⎜

⎝

From this, we find that M1 = 0.2506. Consequently, with M1, M2 and M3 we are able to

compute the following static pressures

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1

p

2

⎞

⎛

⎟

⎞

⎜

⎛

⎟

⎞

⎜

⎛

8328.0

p

2

2o

3o

3=

⎠

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

⎟

⎠

⎜

⎝

Slightly too high but close enough. Note from the isentropic relations at M1, (T/To)1 =

(0.9876)293 = 289.3668K

() ()( )( )

s/kg0313.0

43668.289287.0

RTAM

RT

p

AVρm

11

1

11

=

⎠

⎝

⎥

⎦

⎢

⎣

γ

⎟

⎠

⎞

⎜

⎝

⎛

==

&

Problem 11. – For the flow of air (γ = 1.4 and R = 0.287 kJ/kg · K) from the reservoir at

650 kPa and 1000 K shown in Figure P9.11, assume isentropic flow in the convergent-

divergent nozzle and Fanno flow in the constant-area duct, which has a length of 20 cm

and a diameter of 1 cm. The area ratio A2/A1 of the C-D nozzle is 2.9. Take the friction

factor to be 0.02.

(a) Find the mass flow rate for a back pressure of 0 kPa.

(b) For part (a), find the pressure at the exit plane of the duct.

(c) Find the back pressure necessary for a normal shock to occur at the exit plane

of the nozzle (2).

(d) Find the back pressure necessary for a normal shock to appear just

downstream of the nozzle throat (1).

1 2 e

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