Chapter 12, Problem 116EP

To determine

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

Answer to Problem 116EP

Downstream weak shock for pressure and temperature after the compression.

  ${\text{P}}_{\text{2}}\text{=17}\text{.6 psia}$

  ${\text{T}}_{\text{2}}\text{=610 R}$

Downstream strong shock for pressure and temperature after the compression.

  ${\text{P}}_{\text{2}}\text{=34}\text{.9 psia}$

  ${\text{T}}_{\text{2}}\text{=798 R}$

Mach numbers is given by,

Weak shock $\text{M}{\text{a}}_{\text{2}}\text{=1}\text{.45}$

Strong shock $\text{M}{\text{a}}_{\text{2}}\text{=0}\text{.644}$

Explanation of Solution

Given:

Temperature ${\text{T}}_1\text{ = 480R}$

Velocity ${\text{V}}_1\text{=100 m/s}$

Pressure ${\text{P}}_1\text{=8 psia}$

Mach number $\text{M}{\text{a}}_{\text{1}}\text{=2}\text{.0}$

Force undergo compression turn of ${15}^{\circ }.$

Calculation:

The properties of air,

  ${\text{c}}_{\text{p}}\text{ = 1}\text{.039 kJ/kg}\text{.K}$

  $\text{k=1}\text{.4}$

k is the specific heat ratio.

Will take the angle of deflection as equal to the wedge half-angle, i.e., $\text{θ}\simeq \text{δ= 15°}\text{.}$

So, the two values of oblique shock angle $\text{β}$ is calculated from

  $\begin{array}{l}\text{tanθ=}\frac{\text{2}\left(\text{M}{\text{a}}_{\text{1}}{}^{\text{2}}\text{si}{\text{n}}^{\text{2}}\text{β-1}\right)\text{/tanβ}}{\text{M}{\text{a}}_{\text{1}}{}^{\text{2}}\left(\text{k+cos2β}\right)\text{+2}}\\ \text{tan1}{\text{5}}^{\text{o}}\text{=}\frac{\text{2}\left({\text{2}}^{\text{2}}\text{si}{\text{n}}^{\text{2}}\text{β-1}\right)\text{/tanβ}}{{\text{2}}^{\text{2}}\left(\text{1}\text{.4+cos2β}\right)\text{+2}}\end{array}$

So, simplify in $\text{β}$. Therefore, by solving in an iterative procedure or using $\text{β-θ}$ curves.

  $\begin{array}{l}{\text{β}}_{weak}={45.34}^{\circ }\\ {\text{β}}_{strong}={79.83}^{\circ }\end{array}$

Then the upstream "normal" Mach number $M{a}_{1,n}$ will be given by

  $\begin{array}{l}\text{Weak shock M}{\text{a}}_{\text{1,n}}\text{=M}{\text{a}}_{\text{1}}\text{sinβ=2sin45}\text{.3}{\text{4}}^{\text{o}}\text{=1}\text{.423}\\ \text{Strong shock M}{\text{a}}_{\text{1,n}}\text{=M}{\text{a}}_{\text{1}}\text{sinβ=2sin79}\text{.8}{\text{3}}^{\text{o}}\text{=1}\text{.969}\end{array}$

The downstream "normal" Mach number $\text{M}{\text{a}}_{\text{2,n}}$ will be given by

  $\begin{array}{l}\text{Weak shock M}{\text{a}}_{\text{2,n}}\text{=}\sqrt{\frac{\left(\text{k-1}\right)\text{M}{\text{a}}_{\text{1,n}}{}^{\text{2}}\text{+2}}{\text{2kM}{\text{a}}_{\text{1,n}}{}^{\text{2}}\text{-k+1}}}\\ \text{M}{\text{a}}_{\text{2,n}}\text{=}\sqrt{\frac{\left(\text{1}\text{.4-1}\right)\text{1}\text{.42}{\text{3}}^{\text{2}}\text{+2}}{\text{2×1}\text{.4×1}\text{.42}{\text{3}}^{\text{2}}\text{-1}\text{.4+1}}}\text{=0}\text{.7304}\\ \text{Strong shock M}{\text{a}}_{\text{2,n}}\text{=}\sqrt{\frac{\left(\text{k-1}\right)\text{M}{\text{a}}_{\text{1,n}}{}^{\text{2}}\text{+2}}{\text{2kM}{\text{a}}_{\text{1,n}}{}^{\text{2}}\text{-k+1}}}\\ \text{M}{\text{a}}_{\text{2,n}}\text{=}\sqrt{\frac{\left(\text{1}\text{.4-1}\right)\text{1}\text{.96}{\text{9}}^{\text{2}}\text{+2}}{\text{2×1}\text{.4×1}\text{.96}{\text{9}}^{\text{2}}\text{-1}\text{.4+1}}}\text{=0}\text{.5828}\end{array}$

Downstream pressure and temperature for each case is given by

  $\begin{array}{l}\text{Weak shock }{\text{P}}_{\text{2}}\text{=}{\text{P}}_{\text{1}}\frac{\text{2kM}{\text{a}}_{\text{1,n}}{}^{\text{2}}\text{-k+1}}{\text{k+1}}\text{=6psia}\frac{\text{2×1}\text{.4×1}\text{.42}{\text{3}}^{\text{2}}\text{-1}\text{.4+1}}{\text{1}\text{.4+1}}\text{=17}\text{.6 psia}\\ {\text{T}}_{\text{2}}\text{=}{\text{T}}_{\text{1}}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}{\text{T}}_{\text{1}}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\frac{\text{2+}\left(\text{k-1}\right)\text{M}{\text{a}}_{\text{1,n}}{}^{\text{2}}}{\left(\text{k+1}\right)\text{M}{\text{a}}_{\text{1,n}}{}^{\text{2}}}\text{=}{\text{T}}_{\text{1}}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=480 R}\frac{\text{17}\text{.6}}{\text{8}}\frac{\text{2+}\left(\text{1}\text{.4-1}\right)\text{1}\text{.42}{\text{3}}^{\text{2}}}{\left(\text{1}\text{.4+1}\right)\text{1}\text{.42}{\text{3}}^{\text{2}}}\text{=609}\text{.5R=610R}\\ \text{Strong shock }{\text{P}}_{\text{2}}\text{=}{\text{P}}_{\text{1}}\frac{\text{2kM}{\text{a}}_{\text{1,n}}{}^{\text{2}}\text{-k+1}}{\text{k+1}}\text{=8psia}\frac{\text{2×1}\text{.4×1}\text{.96}{\text{9}}^{\text{2}}\text{-1}\text{.4+1}}{\text{1}\text{.4+1}}\text{=34}\text{.85 psia=34}\text{.9 psia}\\ {\text{T}}_{\text{2}}\text{=}{\text{T}}_{\text{1}}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}{\text{T}}_{\text{1}}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\frac{\text{2+}\left(\text{k-1}\right)\text{M}{\text{a}}_{\text{1,n}}{}^{\text{2}}}{\left(\text{k+1}\right)\text{M}{\text{a}}_{\text{1,n}}{}^{\text{2}}}\text{=}{\text{T}}_{\text{1}}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{=}\left(\text{480 R}\right)\frac{34.85}{\text{8}}\frac{\text{2+}\left(\text{1}\text{.4-1}\right)\text{1}\text{.96}{\text{9}}^{\text{2}}}{\left(\text{1}\text{.4+1}\right)\text{1}\text{.96}{\text{9}}^{\text{2}}}\text{=797}\text{.9R=798 R}\end{array}$

Downstream Mach number is given by

  $\begin{array}{l}\text{Weak shock M}{\text{a}}_{\text{2}}\text{=}\frac{\text{M}{\text{a}}_{\text{2,n}}}{\text{sin}\left(\text{β-θ}\right)}\text{=}\frac{\text{0}\text{.7304}}{\text{sin}\left(\text{45}\text{.3}{\text{4}}^{\text{o}}\text{-1}{\text{5}}^{\text{o}}\right)}\text{=1}\text{.45}\\ \text{Strong shock M}{\text{a}}_{\text{2}}\text{=}\frac{\text{M}{\text{a}}_{\text{2,n}}}{\text{sin}\left(\text{β-θ}\right)}\text{=}\frac{\text{0}\text{.5828}}{\text{sin}\left(\text{79}\text{.8}{\text{3}}^{\text{o}}\text{-1}{\text{5}}^{\text{o}}\right)}\text{=0}\text{.644}\end{array}$