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 116EP
To determine

Mach number, pressure and temperature of air after the compression.

Expert Solution & Answer
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Answer to Problem 116EP

Downstream weak shock for pressure and temperature after the compression.

  P2=17.6 psia

  T2=610 R

Downstream strong shock for pressure and temperature after the compression.

  P2=34.9 psia

  T2=798 R

Mach numbers is given by,

Weak shock Ma2=1.45

Strong shock Ma2=0.644

Explanation of Solution

Given:

Temperature T1 = 480R

Velocity V1=100 m/s

Pressure P1=8 psia

Mach number Ma1=2.0

Force undergo compression turn of 15.

Calculation:

The properties of air,

  cp = 1.039 kJ/kg.K

  k=1.4

k is the specific heat ratio.

FLUID MECHANICS FUND. (LL)-W/ACCESS, Chapter 12, Problem 116EP , additional homework tip  1FLUID MECHANICS FUND. (LL)-W/ACCESS, Chapter 12, Problem 116EP , additional homework tip  2

Will take the angle of deflection as equal to the wedge half-angle, i.e., θδ= 15°.

So, the two values of oblique shock angle β is calculated from

  tanθ=2( M a 1 2 si n 2 β-1)/tanβMa12( k+cos2β)+2tan15o=2( 2 2 si n 2 β-1)/tanβ22( 1.4+cos2β)+2

So, simplify in β. Therefore, by solving in an iterative procedure or using β-θ curves.

  βweak=45.34βstrong=79.83

Then the upstream "normal" Mach number Ma1,n will be given by

  Weak shock   Ma1,n=Ma1sinβ=2sin45.34o=1.423Strong shock   Ma1,n=Ma1sinβ=2sin79.83o=1.969

The downstream "normal" Mach number Ma2,n will be given by

  Weak shock   Ma2,n= ( k-1 )M a 1,n 2 +2 2kM a 1,n 2 -k+1Ma2,n= ( 1.4-1 )1.42 3 2 +2 2×1.4×1.42 3 2 -1.4+1=0.7304Strong shock   Ma2,n= ( k-1 )M a 1,n 2 +2 2kM a 1,n 2 -k+1Ma2,n= ( 1.4-1 )1.96 9 2 +2 2×1.4×1.96 9 2 -1.4+1=0.5828

Downstream pressure and temperature for each case is given by

  Weak shock   P2=P12kMa 1,n2-k+1k+1=6psia2×1.4×1.4232-1.4+11.4+1=17.6 psiaT2=T1P2P1ρ1ρ2=T1P2P12+( k-1)Ma 1,n2( k+1)Ma 1,n2=T1P2P1ρ1ρ2=480 R17.682+( 1.4-1)1.4232( 1.4+1)1.4232=609.5R=610RStrong shock   P2=P12kMa 1,n2-k+1k+1=8psia2×1.4×1.9692-1.4+11.4+1=34.85 psia=34.9 psiaT2=T1P2P1ρ1ρ2=T1P2P12+( k-1)Ma 1,n2( k+1)Ma 1,n2=T1P2P1ρ1ρ2=(480 R)34.8582+( 1.4-1)1.9692( 1.4+1)1.9692=797.9R=798 R

Downstream Mach number is given by

  Weak shock   Ma2=Ma 2,nsin( β-θ)=0.7304sin( 45.3 4 o -1 5 o )=1.45Strong shock   Ma2=Ma 2,nsin( β-θ)=0.5828sin( 79.8 3 o -1 5 o )=0.644

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Chapter 12 Solutions

FLUID MECHANICS FUND. (LL)-W/ACCESS

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