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The temperature and pressure at the stagnation point of a high speed missile are \({934^ \circ }\) R and 7.8 atm, respectively.Calculate the density at this point.

Calculate \({c_p},{c_v},e\,{\rm{and}}\,h\) for

a) The stagnation point conditions given in above problem.

b) Air at standard sea level conditions.

Asked on 26th December 2018 in Aeronautics.
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  • 1 Answer(s)

    \[\begin{array}{l}T = {934^ \circ }R\\p = 7.8\,atm\\Density = \rho  = ?\\p = \rho RT\\\rho  = \frac{P}{{RT}} = \frac{{\left( {7.8 \times 2116} \right)}}{{1716 \times 934}} = 0.0103\,slug/f{t^3}\end{array}\]

    \({c_p},{c_v},e\,{\rm{and}}\,h\) for stagnation point conditions will be

    a)  \[{c_p} = \frac{{\gamma R}}{{\gamma  – 1}} = \frac{{1.4 \times 1716}}{{0.4}} = 6006\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}\]\[{c_v} = \frac{R}{{\gamma  – 1}} = \frac{{1716}}{{0.4}} = 4290\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}\]\[e = {c_v}T = 4290(934) = 4.007 \times {10^6}\frac{{ft\,lb}}{{slug}}\]\[h = {c_p}T = 6006(934) = 5.610 \times {10^6}\frac{{ft\,lb}}{{slug}}\]b) \({c_p}\) and \({c_v}\) For a  calorically perfect gas, \({c_p}\) and \({c_v}\) are constants, \[{c_p} = 6006\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}};\,{c_v} = 4290\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}\]Also at standard sea level \(T = {519^ \circ }R\)\[E = 4290 \times 519 = 2.227 \times {10^6}\frac{{ft\,lb}}{{slug}}\]\[h = 6006\left( {519} \right) = 3.117 \times {10^6}\frac{{ft\,lb}}{{slug}}\]

    Answered on 26th December 2018.
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