Chapter 12, Problem 75P

To determine

Entropy change of air and helium across the normal shock wave.

Answer to Problem 75P

  ${\text{(}{\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{)}}_{air}\text{ =0}\text{.223kJ/kg}\text{.K}$

  ${\text{(}{\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{)}}_{helium}\text{ =1}\text{.27kJ/kg}\text{.K}$

Explanation of Solution

Given:

The properties of air are:

  $\begin{array}{l}\begin{array}{l}\text{ k = 1}\text{.4}\hfill \\ \text{R = 0}\text{.287 kJ/kg·K,}\hfill \end{array}\\ {\text{c}}_{\text{p}}\text{ = 1}\text{.005 kJ/kg·K}\end{array}$

The properties of helium are,

  $\begin{array}{l}\begin{array}{l}k=1.667\hfill \\ R=2.0769kJ/kg·K.\hfill \end{array}\\ {\text{c}}_{\text{p}}\text{ = 5}\text{.1926kJ/kg·K}\end{array}$

The air properties upstream the shock is, Pressure ${\text{P}}_{\text{1}}\text{=58kPa}$

Temperature ${\text{T}}_{\text{1}}\text{=270K}$

Mach number $\text{M}{\text{a}}_{\text{1}}\text{=2}\text{.6}$

Calculation:

Assumptions

1. Air and helium are ideal gases with same specific heats.
2. Flow through the nozzle is constant, single dimensional and isentropic before the shock occurs.

Figure shows the air flow with the Mach number,

Downstream Mach number is given by,

  $\begin{array}{l}\text{M}{\text{a}}_{\text{2}}\text{=}\sqrt{\frac{\left(k-1\right)M{a}_1{}^2+2}{2kM{a}_1{}^2-k+1}}\\ \text{M}{\text{a}}_{\text{2}}\text{=}\sqrt{\frac{\left(1.4-1\right){2.6}^2+2}{2\times 1.4\times {2.6}^2-1.4+1}}\\ \text{M}{\text{a}}_{\text{2}}\text{=0}\text{.5039}\end{array}$

Downstream pressure ratio is given by,

  $\begin{array}{l}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_1}\text{=}\frac{\text{1+kM}{\text{a}}^2{}_{\text{1}}}{\text{1+kM}{\text{a}}^2{}_2}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_1}\text{=}\frac{\text{1+1}\text{.4}\times \text{2}\text{.}{\text{6}}^2}{\text{1+1}\text{.4}\times \text{0}\text{.503}{\text{9}}^2}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_1}\text{=7}\text{.72}\end{array}$

Downstream temperature ratio is given by,

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{1+M}{\text{a}}^2{}_{\text{1}}(k-1)}{\text{1+M}{\text{a}}^2{}_2(k-1)}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{1+2}\text{.}{\text{6}}^2(1.4-1)}{\text{1+0}\text{.503}{\text{9}}^2(1.4-1)}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}2.2383\end{array}$

Calculate the air, entropy change across the shock ,

  $\begin{array}{l}{\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{ = }{\text{c}}_{\text{p}}\text{ In}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{-RIn}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\\ {\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{ =1}\text{.005In}\left(\text{2}\text{.2383}\right)\text{-0}\text{.287In}\left(\text{7}\text{.7200}\right)\\ {\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{ =0}\text{.223kJ/kg}\text{.K}\end{array}$

Downstream Mach number for helium is given by,

  $\begin{array}{l}\text{M}{\text{a}}_{\text{2}}\text{=}{\left(\frac{M{a}_1{}^2+2/\left(k-1\right)}{2kM{a}_1{}^2/\left(k-1\right)-1}\right)}^{1/2}\\ \text{M}{\text{a}}_{\text{2}}\text{=}{\left(\frac{{2.6}^2+2/\left(1.667-1\right)}{2\times 1.667\times {2.6}^2/\left(1.667-1\right)-1}\right)}^{1/2}\\ \text{M}{\text{a}}_{\text{2}}\text{=0}\text{.5455}\end{array}$

Downstream pressure ratio is given by,

  $\begin{array}{l}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_1}\text{=}\frac{\text{1+kM}{\text{a}}^2{}_{\text{1}}}{\text{1+kM}{\text{a}}^2{}_2}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_1}\text{=}\frac{\text{1+1}\text{.667}\times \text{2}\text{.}{\text{6}}^2}{\text{1+1}\text{.667}\times \text{0}\text{.545}{\text{5}}^2}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_1}\text{=}8.2009\end{array}$

Downstream temperature ratio is given by,

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{1+M}{\text{a}}^2{}_{\text{1}}(k-1)/2}{\text{1+M}{\text{a}}^2{}_2(k-1)/2}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{1+2}\text{.}{\text{6}}^2(1.667-1)/2}{\text{1+0}\text{.545}{\text{5}}^2(1.667-1)/2}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}2.9606\end{array}$

Calculate the helium, entropy change across the shock ,

  $\begin{array}{l}{\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{ = }{\text{c}}_{\text{p}}\text{ In}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{-RIn}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\\ {\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{ =5}\text{.1926In}\left(\text{2}\text{.9606}\right)\text{-2}\text{.0769In}\left(\text{8}\text{.2009}\right)\\ {\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{ =1}\text{.27kJ/kg}\text{.K}\end{array}$

Conclusion:

Hence, the entropy change of air across the normal shock wave is:

  ${\text{(}{\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{)}}_{air}\text{ =0}\text{.223kJ/kg}\text{.K}$

The entropy change of helium across the normal shock wave is:

  ${\text{(}{\text{s}}_{\text{2}}\text{-}{\text{s}}_{\text{1}}\text{)}}_{helium}\text{ =1}\text{.27kJ/kg}\text{.K}$