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Basics of Fluid Mechanics, 2014a

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12.2. OBLIQUE SHOCK 511<br />

M x M ys M yw θ s θ w δ<br />

P 0y<br />

P 0x<br />

3.0000 0.48013 2.7008 87.8807 23.9356 6.0000 0.99105<br />

The transition for shock AB is<br />

M x M ys M yw θ s θ w δ<br />

P 0y<br />

P 0x<br />

3.0000 0.47641 2.8482 88.9476 21.5990 3.0000 0.99879<br />

For the shock BC the results are<br />

M x M ys M yw θ s θ w δ<br />

P 0y<br />

P 0x<br />

2.8482 0.48610 2.7049 88.8912 22.7080 3.0000 0.99894<br />

And the isentropic relationships for M =2.7049, 2.7008 are<br />

M<br />

T<br />

T 0<br />

ρ<br />

ρ 0<br />

A<br />

A ⋆<br />

P<br />

P 0<br />

A×P<br />

A ∗ ×P 0<br />

2.7049 0.40596 0.10500 3.1978 0.04263 0.13632<br />

2.7008 0.40669 0.10548 3.1854 0.04290 0.13665<br />

The combined shocks AB and BC provide the base <strong>of</strong> calculating the total pressure<br />

ratio at zone 3. The total pressure ratio at zone 2 is<br />

P 02<br />

= P 02 P 01<br />

=0.99894 × 0.99879 = 0.997731283<br />

P 00 P 01 P 00<br />

On the other hand, the pressure at 4 has to be<br />

P 4<br />

= P 4 P 04<br />

=0.04290 × 0.99105 = 0.042516045<br />

P 01 P 04 P 01<br />

The static pressure at zone 4 and zone 3 have to match according to the government<br />

suggestion hence, the angle for BE shock which cause this pressure ratio needs to be<br />

found. To do that, check whether the pressure at 2 is above or below or above the<br />

pressure (ratio) in zone 4.<br />

P 2<br />

= P 02 P 2<br />

=0.997731283 × 0.04263 = 0.042436789<br />

P 02 P 00 P 02<br />

Since P 2<br />

P 02<br />

< P 4<br />

P 01<br />

a weak shock must occur to increase the static pressure (see Figure<br />

11.13). The increase has to be<br />

P 3 /P 2 =0.042516045/0.042436789 = 1.001867743<br />

To achieve this kind <strong>of</strong> pressure ratio the perpendicular component has to be

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