Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 12, Problem 102P

Air enters a 5-cm-diameter, 4-m-long adiabatic duct with inlet conditions of Ma 1 =  2 . 8, T 1 =  38 0  K , and P 1 = 8 0  kPa . It is observed that a normal shock occurs at a location 3 m from the inlet. Taking the average friction factor to be 0.007, determine the velocity, temperature, and pressure at the duct exit

Chapter 12, Problem 102P, Air enters a 5-cm-diameter, 4-m-long adiabatic duct with inlet conditions of Ma1=2.8,T1=380K , and
FIGURE P12-102

Expert Solution & Answer
Check Mark
To determine

Velocity, temperature and pressure at the duct exit.

Answer to Problem 102P

  T4=813K

  P4=328 kPa

  V4=572 m/s

Explanation of Solution

Given:

The properties of air to be,

   k = 1.4 cp = 1.005 kJ/kg.K R = 0.287 kJ/kg.K.

  Ma1=2.8

Temperature T1=380K

Pressure at Inlet P1=80kPa

The average friction factor is given to be f = 0.007

Calculation:

The Fanno flow functions corresponding to the inlet Mach number of 2.8 are, from Table A-16,

  Ma1=2.8( fL* Dh)1=0.4898T1T*=0.4673

First we check that the flow in everywhere shock at upstream is supersonic. The required length of duct from the inlet L1* for the flow to reach sonic conditions is given by,

  L1*=0.4898DfL1*=0.48980.05m0.007L1*=3.50m

This is higher than the actual length of 3 m.

Therefore, the flow is indeed supersonic when the normal shock appears at the indicated location. Then, using the actual duct length L1,

  fL1Dh=0.007×30.05fL1Dh=0.4200

Taking the value L1=L1*L2*, the function fL*Dh at the exit state and the corresponding Mach number are given by,

  ( fL* D h )2=( fL* D h )1-fL1Dh( fL* D h )2=0.4898-0.4200( fL* D h )2=0.0698Ma2=1.315

From Table A-16, at Ma2 =1.315,

  T2/T &*#x00A0;= 0.8918 P2/P &*#x00A0;= 0.7183

Calculating the temperature, pressure, and velocity before the shock,

  T2T1=T2/T *T1/T *=0.89180.4673T2T1=1.9084T2=1.9084T1T2=1.9084×380KT2=725.2K

  P2P1=P2/P *P1/P *=0.71830.2441P2P1=2.9426P2=2.9426P1P2=2.9426×80kPaP2=235.4kPa

The normal shock functions corresponding to Mach number 1.315 are, from Table A-14,

  Ma2=1.315Ma3=0.7786T3T2=1.2001P3P2=1.8495

Then the temperature after the shock will be,

  T3=1.2001T2T3=1.2001×725.2KT3=870.3K

Pressure after the shock will be,

  P3=1.8495P2P3=1.8495×235.4kPaP3=435.4kPa

Sonic conditions exist at the exit of duct, and the flow downstream the shock will be Fanno flow. From Table A-16,

  Ma3=0.7786T3T*=1.0702P3P*=1.3286Ma4=1;T4T*=1P4P*=1

Calculating the temperature at the duct exit is given by,

  T4T3=T4/T*T3/T*=11.0702T4=T31.0702T4=870.3K1.0702T4=813K

Therefore, the temperature at the duct exit is T4=813K.

Calculating the pressure at the duct exit is given by,

  P4P3=P4/P*P3/P*=11.3286P4=P31.3286P4=435.4kPa1.3286P4=328 kPa

Therefore, the pressure at the duct exit is P4=328 kPa.

Calculating the velocity at the duct exit is given by,

  V4=Ma4c4V4=1kRT4V4=11.4×0.287 kJ/kg.K×813 K( 1000  m 2 / s 2 1 kJ/kg )V4=572 m/s

Therefore, the velocity at the duct exit is V4=572 m/s.

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Fluid Mechanics: Fundamentals and Applications

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