Aeronautical Engineering Chapter 9 Homework Solution Use Frictionless Heat addition Theory

Problem 9.106

Air, from a 3 cubic meter tank initially at 300 kPa and 200C, blows down adiabatically through

a smooth pipe 1 cm in diameter and 2.5 m long. Estimate the time required to reduce the tank

pressure to 200 kPa. For simplicity, assume constant tank temperature and f  0.020.

Solution 9.106*

We know that p2 at the exit is 100 kPa, and we are given f = 0.020, thanks! For the given L and

Problem 9.107

A fuel-air mixture, assumed equivalent to air, enters a duct combustion chamber at V1 = 104 m/s

and T1 = 300 K. What amount of heat addition in kJ/kg will cause the exit flow to be choked?

What will be the exit Mach number and temperature if 504 kJ/kg is added during combustion?

Solution 9.107

Evaluate stagnation temperature and initial Mach number:

Problem 9.108

What happens to the inlet flow of Prob. 9.107 if the combustion yields 1500 kJ/kg heat addition

and po1 and To1 remain the same? How much is the mass flow reduced?

Problem 9.107

A fuel-air mixture, assumed equivalent to air, enters a duct combustion chamber at V1 = 104 m/s

and T1 = 300 K. What amount of heat addition in kJ/kg will cause the exit flow to be choked?

What will be the exit Mach number and temperature if 504 kJ/kg is added during combustion?

Solution 9.108

The flow will choke down to a lower mass flow such that

o2 o

*

T = T :

Problem 9.109

A jet engine at 7000-m altitude takes in 45 kg/s of air and adds 550 kJ/kg in the combustion

chamber. The chamber cross section is 0.5 m2, and the air enters the chamber at 80 kPa and 5°C.

After combustion the air expands through an isentropic converging nozzle to exit at atmospheric

pressure. Estimate (a) the nozzle throat diameter, (b) the nozzle exit velocity, and (c) the thrust

produced by the engine.

Solution 9.109

At 700-m altitude, pa = 41043 Pa, Ta = 242.66 K to use as exit conditions.

Problem 9.110

Compressible pipe flow with heat addition, Sec. 9.8, assumes constant momentum (p +



V2) and

constant mass flow but variable stagnation enthalpy. Such a flow is often called Rayleigh flow,

and a line representing all possible property changes on an temperature-entropy chart is called a

Rayleigh line. Assuming air passing through the flow state p1

=

548 kPa, T1 = 588 K,

V1 = 266 m/s, and A = 1 m2, draw a Rayleigh curve of the flow for a range of velocities from very

low

(Ma 1)

to very high

(Ma 1).

Comment on the meaning of the maximum-entropy point

on this curve.

Solution 9.110

First evaluate the Mach number and density at the reference state:

Problem 9.111

Add to your Rayleigh line of Prob. 9.110 a Fanno line (see Prob. 9.94) for stagnation enthalpy

equal to the value associated with state 1 in Prob. 9.110. The two curves will intersect at state 1,

which is subsonic, and also at a certain state 2, which is supersonic. Interpret these two cases vis-

a-vis Table B.2.

Problem 9.110

Compressible pipe flow with heat addition, Sec. 9.8, assumes constant momentum (p +



V2) and

constant mass flow but variable stagnation enthalpy. Such a flow is often called Rayleigh flow,

and a line representing all possible property changes on an temperature-entropy chart is called a

Rayleigh line. Assuming air passing through the flow state p1

=

548 kPa, T1 = 588 K,

V1 = 266 m/s, and A = 1 m2, draw a Rayleigh curve of the flow for a range of velocities from very

low

(Ma 1)

to very high

(Ma 1).

Comment on the meaning of the maximum-entropy point

on this curve.

Problem 9.94

Compressible pipe flow with friction, Sec. 9.7, assumes constant stagnation enthalpy and mass

flow but variable momentum. Such a flow is often called Fanno flow, and a line representing all

possible property changes on a temperature-entropy chart is called a Fanno line. Assuming a

perfect gas with k = 1.4 and the data of Prob. 9.86, draw a Fanno curve of the flow for a range of

velocities from very low

(Ma <<1)

to very high

(Ma >>1).

Comment on the meaning of the

maximum-entropy point on this curve.

Problem 9.86

Air enters a 3–cm diameter pipe 15 m long at V1 = 73 ms, p1 = 550 kPa, and T1 = 60C.

The friction factor is 0.018. Compute V2, p2, T2, and p02 at the end of the pipe. How much

additional pipe length would cause the exit flow to be sonic?

Solution 9.111*

For T1 = 588 K and V1 = 266 m/s, the stagnation temperature is

Problem 9.112

Air enters a duct at V1 = 144 m/s, p1 = 200 kPa, and T1 = 323 K. Assuming frictionless heat

addition, estimate (a) the heat addition needed to raise the velocity to 372 m/s; and (b) the

pressure at this new section 2.

Solution 9.112

First establish the Mach number and stagnation temperature at section 1:

Problem 9.113

Air enters a constant-area duct at p1 = 90 kPa, V1 = 520 m/s, and T1 = 558°C. It is then cooled

with negligible friction until it exists at p2 = 160 kPa. Estimate (a) V2; (b) T2; and (c) the total

amount of cooling in kJ/kg.

Solution 9.113

We have enough information to estimate the inlet Ma1 and go from there:

Problem 9.114

The scramjet of Fig. C9.8 operates with supersonic flow throughout. Assume that the heat

addition of 500 kJ/kg, between sections 2 and 3, is frictionless and at constant area of 0.2 m2.

Given Ma2 = 4.0, p2 = 260 kPa, and T2 = 420 K. Assume airflow at k = 1.40. At the combustion

section exit, find (a) Ma3, (b) p3, and (c) T3.

Solution 9.114

Use frictionless heat-addition theory to get from section 2 to section 3:

Problem 9.115

Air enters a 5-cm-diameter pipe at 380 kPa, 3.3 kg/m3, and 120 m/s. Assume frictionless flow

with heat addition. Find the amount of heat addition for which the velocity (a) doubles;

(b) triples; and (d) quadruples.

Solution 9.115

First find the conditions at the entrance, which we will call section 1:

Problem 9.116

An observer at sea level does not hear an aircraft flying at 12000 ft standard altitude until it is

5 (statute) miles past her. Estimate the aircraft speed in ft/sec.

Solution 9.116

The average temperature over this range is 498°R, hence

Problem 9.117

A tiny scratch in the side of a supersonic wind tunnel creates a very weak wave of angle 17, as

shown in Fig. P9.117, after which a normal shock occurs. The air temperature in region (1) is

250 K. Estimate the temperature in region (2).

Solution 9.117

The weak wave is a Mach wave, hence the Mach number in region 1 is Ma1 = 1/sin(17) = 3.42.

Problem 9.118

A particle moving at uniform velocity in sea–level standard air creates the two disturbance

spheres shown in Fig. P9.118. Compute the particle velocity and Mach number.

Solution 9.118

If point “a” represents t = 0 units, the particle reaches point “b” in 8 − 3 = 5 units. But the distance

from a to b is only 3 units. Therefore the (subsonic) Mach number is

Problem 9.119

The particle in Fig. P9.119 is moving supersonically in sea-level standard air. From the two given

disturbance spheres, compute the particle Mach number; velocity; and Mach angle.

Solution 9.119

If point “a” represents t = 0 units, the particle reaches point “b” in 8 – 3 = 5 units. But the distance

Problem 9.120

The particle in Fig. P9.120 is moving in sea-level standard air. From the two disturbance spheres

shown, estimate (a) the position of the particle at this instant; and (b) the temperature in °C at the

front stagnation point of the particle.

Solution 9.120

Given sea-level temperature = 288 K. If point “a” represents t = 0 units, the particle reaches point

Problem 9.121

A thermistor probe, in the shape of a needle parallel to the flow, reads a static temperature of –25°C

when inserted into the supersonic airstream. A conical disturbance of half–angle 17° is formed.

Estimate (a) the Mach number; (b) the velocity; and (c) the stagnation temperature of the stream.

Solution 9.121

If the needle is “very thin,” it reads the stream static temperature, T  –25°C = 248 K. We are

Problem 9.122

Supersonic air takes a 5° compression turn, as in Fig. P9.122. Compute the downstream pressure

and Mach number and wave angle, and compare with small-disturbance theory.

Solution 9.122

From Fig. 9.23,



 25°, and we can iterate Eq. (9.86) to a closer estimate:

1

Ma 3.0, 5 , compute ,



= =  = 23.133

Problem 9.123

The 10 deflection in Example 9.17 caused a final Mach number of 1.641 and a pressure ratio of

1.707. Compare this with the case of the flow passing through two 5 deflections. Comment on

the results and why they might be higher or lower in the second case.

Solution 9.123

The sketch shows both

Problem 9.124

When a sea–level air flow approaches a ramp of angle 20, an oblique shock wave forms as in

Figure P9.124. Calculate (a) Ma1; (b) p2; (c) T2 ; and (d) V2.

Ma= 1.649

Solution 9.124

For sea-level air, take p1 = 101.35 kPa, T1 = 288.16 K, and



1 = 1.2255 kg/m3.

Problem 9.125

We saw in the text that, for k = 1.40, the maximum possible deflection caused by an oblique

shock wave occurs at infinite approach Mach number and is



max = 45.58. Assuming an ideal

gas, what is



max for (a) argon; and (b) carbon dioxide?

Solution 9.125

In the limit as Ma1 → , the normal-velocity ratio across a shock wave, Eq. (9.58), is

Problem 9.126

Airflow at Ma = 2.8, p = 80 kPa, and T = 280 K undergoes a 15º compression turn. Find the

downstream values of (a) Mach number, (b) pressure, and (c) temperature.

Solution 9.126

Fifteen degrees is reasonably small, less than θmax, so we are looking for an attached weak

oblique shock. From Eq. (9.86),

Problem 9.127

Do the Mach waves upstream of an oblique–shock wave intersect with the shock? Assuming

supersonic downstream flow, do the downstream Mach waves intersect the shock? Show that for

small deflections the shock-wave angle



lies halfway between



1 and



2 +



for any Mach

number.

Solution 9.127

Yes, Mach waves both upstream and downstream will intersect the shock:

Problem 9.128

Air flows past a two-dimensional wedge-nosed body as in Fig. P9.128. Determine the wedge

half-angle



for which the horizontal component of the total pressure force on the nose is

35 kN/m of depth into the paper.

Solution 9.128

Regardless of the wedge angle



, the horizontal force equals the pressure inside the shock times

the projected vertical area of the nose:

Problem 9.129

Air flows at supersonic speed toward a compression ramp, as in Fig. P9.129. A scratch on the wall

at a creates a wave of 30 angle, while the oblique shock has a 50 angle. What is (a) the ramp

angle



; and (b) the wave angle



caused by a scratch at b?

Solution 9.129

The two “scratches” cause Mach waves which are directly related to Mach No.:

Problem 9.130

A supersonic airflow, at a temperature of 300 K, strikes a wedge and is deflected 12. If the

resulting shock wave is attached, and the temperature after the shock is 450 K, (a) estimate the

approach Mach number and wave angle. (b) Why are there two solutions?

Solution 9.130

We search through the

weak

Problem 9.131

The following formula has been suggested as an alternate to Eq. (9.86) to relate upstream Mach

number to the oblique shock wave angle



and turning angle



:

2

1

1 ( 1)sin sin

sin 2cos( )

k

Ma





+

=+ −

Can you prove or disprove this relation? If not, try a few numerical values and compare with the

results from Eq. (9.86).

strong