Chapter 12, Problem 74P

To determine

Pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock and comparison of Helium undergoing the same shock.

Answer to Problem 74P

Downstream pressure ${\text{P}}_2\text{=447}\text{.75kPa}$

Downstream temperature ${\text{T}}_{\text{2}}\text{=604}\text{.34K}$

Downstream velocity ${\text{V}}_{\text{2}}\text{=248}\text{.31m/s}$

Downstream Mach number $\text{M}{\text{a}}_{\text{2}}\text{=0}\text{.5039}$

Downstream stagnation pressure ${\text{P}}_{\text{02}}\text{=532}\text{.69kPa}$

Comparison with helium for the normal shock under same conditions we have

Downstream pressure ${\text{P}}_{\text{2}}\text{=475}\text{.65kPa}$

Downstream temperature ${\text{T}}_{\text{2}}\text{=799}\text{.55k}$

Downstream velocity ${\text{V}}_{\text{2}}\text{=908}\text{.27m/s}$

Downstream Mach number $\text{M}{\text{a}}_{\text{2}}\text{=0}\text{.5455}$

Downstream stagnation pressure ${\text{P}}_{\text{02}}\text{=602}\text{.57kPa}$

Explanation of Solution

Given:

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

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

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

Calculation:

Downstream Mach number is given by $\text{Table A-14,For M}{\text{a}}_1\text{= 2}\text{.6}$ using the below formula.

The properties of air are

  $\text{k=1}\text{.4}$

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

Downstream pressure is given by the formula,

  $\begin{array}{l}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\left(\frac{1+kM{a}_1{}^2}{1+kM{a}_2{}^2}\right)\\ {\text{P}}_2\text{=}{\text{P}}_{\text{1}}\left(\frac{1+kM{a}_1{}^2}{1+kM{a}_2{}^2}\right)\\ {\text{P}}_2\text{=58kPa}\left(\frac{1+1.4\times {2.6}^2}{1+1.4\times {0.5039}^2}\right)\\ {\text{P}}_2\text{=447}\text{.75kPa}\end{array}$

Downstream temperature is given by the formula,

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\left(\frac{\text{2+M}{\text{a}}_{\text{1}}{}^{\text{2}}\left(\text{k-1}\right)}{\text{2+M}{\text{a}}_{\text{2}}{}^{\text{2}}\left(\text{k-1}\right)}\right)\\ {\text{T}}_2\text{=}{\text{T}}_{\text{1}}\left(\frac{\text{2+M}{\text{a}}_{\text{1}}{}^{\text{2}}\left(\text{k-1}\right)}{\text{2+M}{\text{a}}_{\text{2}}{}^{\text{2}}\left(\text{k-1}\right)}\right)\\ {\text{T}}_{\text{2}}\text{=270K}\left(\frac{\text{2+2}\text{.}{\text{6}}^{\text{2}}\left(\text{1}\text{.4-1}\right)}{\text{2+0}\text{.503}{\text{9}}^{\text{2}}\left(\text{1}\text{.4-1}\right)}\right)\\ {\text{T}}_{\text{2}}\text{=604}\text{.34K}\end{array}$

Downstream velocity of speed is given by the formula,

  $\begin{array}{l}{\text{V}}_{\text{2}}\text{=M}{\text{a}}_{\text{2}}{\text{c}}_{\text{2}}\mathrm{......}\text{(1)}\\ \text{where,}\\ {\text{c}}_{\text{2}}\text{=}\sqrt{\text{kR}{\text{T}}_{\text{2}}}\\ \text{equation(1) becomes,}\\ {\text{V}}_{\text{2}}\text{=M}{\text{a}}_{\text{2}}\sqrt{\text{kR}{\text{T}}_{\text{2}}}\\ {\text{V}}_{\text{2}}\text{=0}\text{.5039}\sqrt{\text{1}\text{.4×0}\text{.287×604}\text{.34×}\frac{\text{1000}}{\text{1}}}\\ {\text{V}}_{\text{2}}\text{=248}\text{.31m/s}\end{array}$

Downstream stagnation pressure is given by formula,

  $\begin{array}{l}\frac{{\text{P}}_{\text{02}}}{{\text{P}}_{\text{1}}}\text{=}\left(\frac{\text{1+kM}{\text{a}}^{\text{2}}{}_{\text{1}}}{\text{1+kM}{\text{a}}^{\text{2}}{}_{\text{2}}}\right){\left(\text{1+}\frac{\left(\text{k-1}\right)\text{M}{\text{a}}^{\text{2}}{}_{\text{2}}}{2}\right)}^{\text{k/(k-1)}}\\ {\text{P}}_{\text{02}}\text{=58}\times \text{1000}{\left(\frac{\text{1+1}\text{.4×2}\text{.}{\text{6}}^{\text{2}}}{\text{1+1}\text{.4×0}\text{.503}{\text{9}}^{\text{2}}}\left(\text{1+}\frac{\left(\text{1}\text{.4-1}\right)\text{0}\text{.503}{\text{9}}^{\text{2}}}{2}\right)\right)}^{\text{1}\text{.4/(1}\text{.4-1)}}\\ {\text{P}}_{\text{02}}\text{=532}\text{.69kPa}\end{array}$

Analysis for helium is to be found out,so we need to determine the downstream Mach number by the below formula

The properties of helium are

k = 1.667

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

Calculate for the downstream pressure for the helium gas given by the formula,

  $\begin{array}{l}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\frac{\text{1+kM}{\text{a}}^{\text{2}}{}_{\text{1}}}{\text{1+kM}{\text{a}}^{\text{2}}{}_{\text{2}}}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\frac{\text{1+1}\text{.667×2}\text{.}{\text{6}}^{\text{2}}}{\text{1+1}\text{.667×0}\text{.545}{\text{5}}^{\text{2}}}\\ {\text{P}}_{\text{2}}\text{=}{\text{P}}_{\text{1}}\left(\text{8}\text{.2009}\right)\\ {\text{P}}_{\text{2}}\text{=58×1000}\left(\text{8}\text{.2009}\right)\\ {\text{P}}_{\text{2}}\text{=475}\text{.65kPa}\end{array}$

Calculate for the downstream temperature for the helium gas given by the formula,

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{2+M}{\text{a}}^{\text{2}}{}_{\text{1}}\text{(k-1)}}{\text{2+M}{\text{a}}^{\text{2}}{}_{\text{2}}\text{(k-1)}}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{2+2}\text{.}{\text{6}}^{\text{2}}\text{(1}\text{.667-1)}}{\text{2+0}\text{.545}{\text{5}}^{\text{2}}\text{(1}\text{.667-1)}}\\ {\text{T}}_{\text{2}}\text{=}{\text{T}}_{\text{1}}\left(\text{2}\text{.9612}\right)\\ {\text{T}}_{\text{2}}\text{=270}\left(\text{2}\text{.9612}\right)\\ {\text{T}}_{\text{2}}\text{=799}\text{.55k}\end{array}$

Downstream velocity of speed is given by the formula,

  $\begin{array}{l}{\text{V}}_{\text{2}}\text{=M}{\text{a}}_{\text{2}}{\text{c}}_{\text{2}}\\ {\text{V}}_{\text{2}}\text{=M}{\text{a}}_{\text{2}}\sqrt{kR{T}_{\text{2}}}\\ {\text{V}}_{\text{2}}\text{=0}\text{.5455}\sqrt{1.667\times 2.08\times 799.55\times \frac{1000}{1}}\\ {\text{V}}_{\text{2}}\text{=908}\text{.27m/s}\end{array}$

Downstream stagnation pressure is given by formula,

  $\begin{array}{l}\frac{{\text{P}}_{\text{02}}}{{\text{P}}_{\text{1}}}\text{=}\left(\frac{\text{1+kM}{\text{a}}^{\text{2}}{}_{\text{1}}}{\text{1+kM}{\text{a}}^{\text{2}}{}_{\text{2}}}\right){\left(\text{1+}\frac{\left(\text{k-1}\right)\text{M}{\text{a}}^{\text{2}}{}_{\text{2}}}{2}\right)}^{\text{k/(k-1)}}\\ {\text{P}}_{\text{02}}\text{=58}\times \text{1000}{\left(\frac{\text{1+1}\text{.667×2}\text{.}{\text{6}}^{\text{2}}}{\text{1+1}\text{.667×0}\text{.545}{\text{5}}^{\text{2}}}\left(\text{1+}\frac{\left(\text{1}\text{.667-1}\right)\text{0}\text{.545}{\text{5}}^{\text{2}}}{2}\right)\right)}^{\text{1}\text{.667/(1}\text{.667-1)}}\\ {\text{P}}_{\text{02}}\text{=602}\text{.57kPa}\end{array}$

Conclusion:

Therefore, we can concludeMach number and velocity are greater for helium than for air because of the different values of $\text{k }\text{.}$