FLUID MECHANICS FUND. (LL)-W/ACCESS
FLUID MECHANICS FUND. (LL)-W/ACCESS
4th Edition
ISBN: 9781266016042
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 12, Problem 138P
To determine

The velocity, temperature, and pressure at the duct exit.

Expert Solution & Answer
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Answer to Problem 138P

  T4=727K

  P4=430 kPa

  V4=520 m/s

Explanation of Solution

Given:

The properties of air:

  R =0.280 kJ/kg.K

  cp=1.005 kJ/kg.K

  k=1.33

Inlet Temperature T1=510K

Inlet Pressure P1=180kPa

Mach number Ma1=2.0

Shock is located distance of 2m from inlet.

Diameter D=0.1m

Average friction factor f=0.010

Calculation:

Fanno flow equations related to the Mach number of inlet are from Table A-16.

  Ma1=2( fL* Dh)1=0.3402T1T*=0.7018P1P*=0.4189V1V*=0.2405

  L1=0.3402DfL1=0.3402×0.10.01L1=3.402 m

Actual duct length is given by,

  ( f L 1 Dh)=0.010×20.10( f L 1 Dh)=0.200

We know,

  L1=L1*-L2*

Then,

  ( fL* D h )2=( fL* D h )1-fL1Dh( fL* D h )2=0.3402-0.2000( fL* D h )2=0.1402

From Table A-16

  Ma2=1.476 we have,

  T2T*=0.8568P2P*=0.6720

Temperature before the shock is given by

  T2T1=T2/T*T1/T*T2T1=0.85680.7018T2T1=1.2209T2=1.2209T1T2=1.2209×510T2=622.7K

Pressure before the shock is given by

  P2P1=P2/P*P1/P*P2P1=0.62700.4189P2P1=1.4968P2=1.2209P1P2=1.2209×180P2=269.4 kPa

From normal shocks Table A-16 we have

  Ma2=1.476Ma3=0.7052T3T2=1.2565P3P2=2.3466

Temperature after the shock is given by

  T3T2=1.2565T3=1.2565×622.7T3=782.4K

Pressure after the shock is given by

  P3P2=2.3466P3=2.3466×269.4P3=632.3 kPa

Then,

  Ma3=0.7052T3T*=1.0767P3P*=1.4713Ma4=1T4T*=1P4P*=1

Temperature at duct exit is given by

  T4T3=T4/T*T3/T*T4T3=11.0767T4=782.41.0767T4=727K

Pressure at duct exit is given by

  P4P3=P4/P*P3/P*P4P3=11.4713P4=632.31.4713P4=430 kPa

Velocity at duct exit is given by

  V4=Ma4c4=1kRT4V4=1.33×0.280( 727)( 1000 J/kg.K 1kJ/kg.K )V4=520 m/s

Conclusion:

Therefore, it is given L3 &*#x00A0;= 2.13 m, and for the duct the total length is 4.13 m. If the duct is going to be further enlarged, then normal shock is going to move upstream further and eventually to the inlet of the duct.

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Chapter 12 Solutions

FLUID MECHANICS FUND. (LL)-W/ACCESS

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