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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

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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