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5 Isothermal Flow

$\displaystyle \int_{0}^{L} { 4 f dx \over D} = \int_{M^{2}}^{1/k} { 1 - kM{2} \over kM{2}} dM^{2}$ (53)

$\displaystyle {\mbox{$\frac{4fL_{max}}{D}$}} = { 1- k M^{2} \over k M^{2} } + \ln kM^{2}$ (54)

$\displaystyle {P_{0} \over P_{0}^{*}} = {P \over P^{*}} \left[ {1 + { k -1 \ove...
... ^ {2} \over { 1 + {k -1 \over 2k} } } \right] ^ { k \over k -1 } %\label{eq:}
$ (55)

$\displaystyle {P_{0} \over P_{0}^{*}} = {1 \over \sqrt{k}} \left( {2k \over 3k-...
...left( 1 + {k -1 \over 2} M ^{2}\right)^{k \over k-1} { 1 \over M} %\label{eq:}
$ (56)

$\displaystyle {T_{0} \over T_{0}^{*}} = { T \over T^{*}} { 1 + {k -1 \over 2} M...
... {k -1 \over 2k} } = {2k \over 3k -1 } \left( 1 + {k -1 \over 2} \right) M ^{2}$ (57)

$\displaystyle \mbox{$\frac{4fL}{D}$}$$\displaystyle = \left. \mbox{$\frac{4fL_{max}}{D}$}\right\vert _{1} - \left. \m...
... k{M_{2}}^{2} \over k {M_{2}}^{2}} + \ln \left( {M_{1} \over M_{2}} \right)^{2}$ (58)

For the case that $ M_1 > > M_2$ and $ M_1 \rightarrow 1$ equation (58) is reduced into the following approximation

$\displaystyle \mbox{$\frac{4fL}{D}$}$$\displaystyle = 2 \ln M_{1} -1 - \overbrace{ 1 - k{M_{2}}^{2} \over k {M_{2}}^{2}}^{\sim 0}$ (59)

$\displaystyle M_1 \sim \hbox{\huge e}^{{1\over 2}\left(\mbox{$\frac{4fL}{D}$}+1\right)}$ (60)


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Next: 6 Fanno Flow Up: Free textbook pdf Fundamentals Previous: 4 Normal Shock
Genick Bar-Meir ||| www.potto.org
copyright Dec , 2006

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