978-0131206687 Chapter 3 Part 2

Problem 16. – Steam is to be expanded to Mach 2.0 in a converging-diverging nozzle

from an inlet velocity of 100 m/s. The inlet area is 50 cm2; inlet static temperature is 500

K. Assuming isentropic flow, determine the throat and exit areas required. Assume the

steam to behave as a perfect gas with constant γ = 1.3.

100

exit

t

*cm 1389.27A so7732.1

A

Problem 17. – Write a computer program that will yield values of T/To, p/po, and A/A*

for isentropic flow of a perfect gas with constant γ = 1.27. Use Mach number increments

of 0.05 over the range M = 0 to M = 2.0.

1

b+γ

−

γ

=,

b1

T

2

−

=, b2

1b

T

p

+

⎟

⎞

⎜

⎛

=,

()

[

]

1Mb1

Ab2

1

2−+

=

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0.95 0.89139 0.58230 1.00226

1.00 0.88106 0.55121 1.00000

1.05 0.87045 0.52067 1.00215

1.10 0.85959 0.49081 1.00844

1.15 0.84851 0.46177 1.01864

1.20 0.83724 0.43362 1.03264

Problem 18. – A gas is known to have a molecular mass of 18, with cp = 2.0 kJ/kg · K.

The gas is expanded from negligible initial velocity through a converging-diverging

nozzle with an area ratio of 5.0. Assuming an isentropic expansion in the nozzle with

initial stagnation pressure and temperature 1 MPa and 1000 K, respectively, determine

the exit nozzle velocity.

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Problem 19. – A jet plane is flying at 10 km with a cabin pressure of 101 kPa and a cabin

temperature of 20°C. Suddenly a bullet is fired inside the cabin and pierces the fuselage;

the resultant hole is 2 cm in diameter. Assume that the temperature within the cabin

remains constant and that the flow through the hole behaves as that through a converging

nozzle with an exit diameter of 2.0 cm. Take the cabin volume to be 100 m3. Calculate

the time for the cabin pressure to decrease to one-half the initial value. At 10 km, p =

26.5 kPa and T = 223.3 K.

Because the back pressure to cabin pressure is 26.5/101 = 0.2624, which is less than

0.5283 the critical pressure ratio at γ = 1.4, the flow is choked and Me = 1. Hence, the

mass flow rate is

p104186.7

c

7

−

×=

In the cabin,

()

dt 104186.7

RT

p

dp

m

RT

dt

dmRT

dt

dp

mRTp

7

c

−

×

∀

−=

∀

−=

∀

=

∴

=∀

&

Integration produces,

RT

p

final

c−

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47

Problem 20. – A rocket nozzle is designed to operate isentropically at 20 km with a

chamber pressure of 2.0 MPa and chamber temperature of 3000 K. If the products of

combustion are assumed to behave as a perfect gas with constant specific heats (γ = 1.3

and MM = 20), determine the design thrust for a nozzle throat area of 0.25 m2.

T

e

o

ee ==

⎟

⎠

⎜

⎝

At design

20

⎠

⎝

Now at the throat M t = 1, so (p/po)t = 0.5457 and (T/To)t = 0.8696.

2

⎥

⎤

⎢

⎡

0.8480MN

=

Problem 21. – A converging nozzle has a rectangular cross section of a constant width of

10 cm. For ease of manufacture, the sidewalls of the nozzle are straight, making an angle

of 10° with the horizontal, as shown in Figure P3.21. Determine and plot the variation of

M, T, and p with x, taking M1 = 0.4, Po1 = 200 kPa, and To1 = 350 K. Assume the

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working fluid to be air, which behaves as a perfect gas with constant specific heats

(γ = 1.4), and that the flow is isentropic.

10 cm

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0.0

0.2

0.4

0.6

0.8

1.0

0.0 2.0 4.0 6.0 8.0 10.0 12.0

x

400

x

Problem 22. – A spherical tank contains compressed air at 500 kPa; the volume of the

tank is 20 m3. A 5-cm burst diaphragm in the side of the tank ruptures, causing air to

escape from the tank. Find the time required for the tank pressure to drop to 200 kPa.

Assume the temperature of the air in the tank remains constant at 280 K, the ambient

pressure is 101 kPa and that the airflow through the opening can be treated as isentropic

flow through a converging nozzle with a 5-cm exit diameter.

For

()

so choked5283.0 505.0

200

101

p

,Pak 200p

o

b

tank <===

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In the tank,

po∀= mRT

01838.0

−

Problem 23. – A converging-diverging nozzle has an area ratio of 3.3 to 1. The nozzle is

supplied from a tank containing a gas at 100 kPa and 270 K (see Figure P3.23).

Determine the maximum mass flow possible through the nozzle and the range of back

pressures over which the mass flow can be attained assuming the gas is (a) helium (γ =

1.67, R = 2.077 kJ/kg·K) and (b) hydrogen (γ = 1.4, R = 4.124 kJ/kg·K).

Figure P3.23

(a) Helium: 3.3

*A

A

,67.1 e==γ

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51

throat

p

mγ=

&

(b) Hydrogen: 3.3

*A

A

,40.1 e==γ

Problem 24. – Superheated steam is stored in a large tank at 6 MPa and 800°C. The

steam is exhausted isentropically through a converging-diverging nozzle. Determine the

velocity of the steam flow when the steam starts to condense, assuming the steam to

behave as a perfect gas with γ = 1.3.

Solution Using Steam Table Data

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52

Interpolation gives

kJ

8.2638h ,C77T Pa,k 42p 222 =°==

Solution Assuming Steam is a Perfect Gas

42

p

heats, the first answer is more accurate.

Problem 25. – Air is stored in a tank 0.037661m3 in volume at an initial pressure of

5,760.6 kPa and a temperature of 321.4K. The gas is discharged through a converging

nozzle with an exit area of 3.167×10-5 m2. For a back-pressure of 101 kPa, assuming a

spatially lumped polytropic process in the tank, i.e., pvn = constant, and isentropic flow in

the nozzle, i.e., pvγ = constant, compare predicted tank pressures to the measured values

contained the following table. Try various values of the polytropic exponent, n, from 1.0

(isothermal) to 1.4 (isentropic). Perform only a Stage I analysis, i.e., the nozzle is

choked.

Now from the continuity equation

eeee VAm

dt

For polytropic expansion within the tank

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53

1o

o

opp

=

So

1

1o

1oo p

⎠

⎜

⎝

And for isentropic expansion in the nozzle

⎟

⎠

⎜

⎝

ρ=ρ

1o

o

oe p

For a choked flow: Me = 1, Ve = ae = √γpe/ρe and

oe 2

⎠

⎝

Therefore,

1

() ()

n2

1

1o

o

12

1

21

1o

12

1o

21

oe

e

p

2

1

2

1

pA

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛+γ

ρ

⎟

⎠

⎞

⎜

⎝

⎛+γ

ργ=

ρ

γ−γ−

γ

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54

Using ao1 = √γpo1/ρ01 the mass flow rate at the exit can be written as

1n

1

+

+γ

Now the time rate of change of the mass within the tank is given by

(

)

dt

dp

p

pn

dt

d

dt

dm o

n

n1

o

n1

1o

1oo

−

⎥

⎦

⎤

⎢

⎣

⎡∀ρ

=

ρ

∀=

Equating this to the exiting flow rate gives

1o

2

⎠

⎝

∀

Integration yields, (note: n2

n1

1

n2

n31

−

=+

−)

n1

1

n1

−

+γ

−−

Rearrangement brings

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n1

n2

1

1o

2

p

+γ

⎥

⎦

⎤

⎢

⎣

⎡

⎠

⎝

⎟

⎠

⎜

⎝

∀

⎠

⎝

Note this is not valid for n =1, the isothermal case which must be treated separately. For

n = 1

o

1o

1oee

1o

2

p

dt

pdt

dt

⎠

⎝

Canceling, separating variables, integrating and rearranging yields,

A spreadsheet program was written and run for various n. A table of the results is as

follows

n = 1.0 1.1 1.2 1.3 1.4

t po/po1 po/po1 po/po1 po/po1 po/po1 po/po1 (exp)

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