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 72EP
To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer
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Answer to Problem 72EP

The wave angle, Mach number and temperature after shock are βweak=37.21°andβstrong=85.05°, 1.713 and 567.125°R respectively.

Explanation of Solution

Given:

Air pressure, air temperature and Mach number are 14psia, 40°F and 2 respectively. Flow direction angle is 8°.

Concept used:

Relation of wave angle is expressed as follows:

  tanθ=2(Ma12 sin2β1)tanβ(Ma12( k+cos2β)+2)........ (1)

Here, wave angle is β, angle of deflection is θ and Mach number before ramp is Ma1.

Relation of upstream normal Mach number is expressed as follows:

  (Ma1)n=Ma1sinβweak........ (2)

Here, wave weak angle is βweak, upstream normal Mach number is (Ma1)n and Mach number before ramp is Ma1.

Relation of downstream normal Mach number is expressed as follows:

  (Ma2)n=( k1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2k+1........ (3)

Here, upstream normal Mach number is (Ma1)n and downstream normal Mach number is (Ma2)n.

Relation of the downstream Mach number is expressed as follows:

  Ma2=( M a 2 )n(sinβ weakθ)........ (4)

Here, wave weak angle is βweak, angle of deflection is θ and downstream Mach number is Ma2.

Relation of downstream pressure is expressed as follows:

  p2=p1(2k ( ( M a 1 ) n )2k+1k+1)........ (5)

Here, downstream pressure is p2, upstream normal Mach number is (Ma1)n and pressure before ramp is p1.

Relation of downstream Temperature is expressed as follows:

  T2=T1(p2p1)(2+( k1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)........ (6)

Here, downstream pressure is p2, downstream temperature is T2, Temperature before ramp is T1 upstream normal Mach number is (Ma1)n and pressure before ramp is p1.

Calculation:

Substitute 8° for θ, 2 for Ma1 and 1.4 for k in equation (1).

  tan8°=2( ( 2 ) 2 sin 2 β1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β2)

Solve the above equation by iteration. We get two value of wave angle as βweak=37.21° and βstrong=85.05°.

Substitute 37.21° for βweak and 2 for Ma1 in equation (2).

  (M a 1)n=2sin37.21°(M a 1)n=1.21

Substitute 1.21 for (Ma1)n and 1.4 for k in equation (3).

  (M a 2)n= ( 1.41 ) ( 1.21 ) 2 +2 2×1.4 ( 1.21 ) 2 1.4+1(M a 2)n=0.8360

Substitute 0.8360 for (Ma2)n, 37.21° for βweak and 8° for θ in equation (4).

  Ma2=0.8360( sin( 37.21°8° ))Ma2=1.713

Substitute 14psia for p1, 1.4 for k and 1.21 for (Ma1)n in equation (5).

  p2=14( 2×1.4 ( 1.21 ) 2 1.4+1 1.4+1)p2=21.58psia

Substitute 14psia for p1, 21.58psia for p2, 500R for T1. 1.4 for k and 1.21 for (Ma1)n in equation (5).

  T2=500( 21.58 14)( 2+( 1.41 ) ( 1.21 ) 2 ( 1.4+1 ) ( 1.21 ) 2 )T2=567.125°R

Thus, the wave angle, Mach number and temperature after shock are βweak=37.21°andβstrong=85.05°, 1.713 and 567.125°R respectively.

Conclusion:

The wave angle, Mach number and temperature after shock are βweak=37.21°andβstrong=85.05°, 1.713 and 567.125°R respectively.

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Chapter 12 Solutions

FLUID MECHANICS FUND. (LL)-W/ACCESS

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