The wave angle, Mach number and temperature after shock.

Question

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Answer 1

Question

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer

Answer to Problem 72EP

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

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

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Answer 2

Question

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer

Answer to Problem 72EP

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

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

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Answer 3

Question

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer

Answer to Problem 72EP

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

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

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Answer 4

Question

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer

Answer to Problem 72EP

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

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

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Answer 5

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer

Answer to Problem 72EP

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

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

Answer 6

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Expert Solution & Answer

Answer to Problem 72EP

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

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

Answer 7

Chapter 12, Problem 72EP

To determine

The wave angle, Mach number and temperature after shock.

Answer 8

Expert Solution & Answer

Answer to Problem 72EP

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

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

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

Concept used:

Relation of wave angle is expressed as follows:

$tanθ=2(Ma12 sin2β−1)tanβ(Ma12( k+cos2β)+2)$ ........ (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:

$(Ma2)n=( k−1) ( ( M a 1 ) n )2+22k ( ( M a 1 ) n )2−k+1$ ........ (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:

$Ma2=( M a 2 )n(sinβ weak−θ)$ ........ (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:

$p2=p1(2k ( ( M a 1 ) n )2−k+1k+1)$ ........ (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:

$T2=T1(p2p1)(2+( k−1) ( ( M a 1 ) n )2( k+1) ( ( M a 1 ) n )2)$ ........ (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).

$tan8°=2( ( 2 ) 2 sin 2 β−1)tanβ( ( 2 ) 2 ( 1.4+cos2β )+2)0.1405(tanβ( 4( 1.4+cos2β )+2))=(8 sin2β−2)0.787tanβ+0.562tanβcos2β+1.124tanβ=(8 sin2β−2)$

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).

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

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

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

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

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

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

Conclusion:

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

Answer 12

Ch. 12 - What is dynamic temperature?Ch. 12 - Calculate the stagnation temperature and pressure...Ch. 12 - Prob. 6P Ch. 12 - Prob. 7P Ch. 12 - Prob. 8EP Ch. 12 - Prob. 9P Ch. 12 - Products of combustion enter a gas turbine with a...Ch. 12 - Is it possible to accelerate a gas to a supersonic...Ch. 12 - Prob. 72EP Ch. 12 - Prob. 73P

The wave angle, Mach number and temperature after shock.

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The wave angle, Mach number and temperature after shock.

Answer to Problem 72EP

Explanation of Solution

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