Compressible Flow Free Textbook

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Adiabatic Isentropic Nozzle Attached

The mass flow out is given by either by Fliegner's equation (4.46) or simply use $c M \rho A^{*}$ and equation (11.17) becomes

$\displaystyle {1 \over k} \bar{P}^{1-k\over k} {d \bar{P} \over d\bar{t}} \pm {\bar{P} ^ {k+1 \over 2k} } (\bar{t}) f[M] = 0$

(11.25)

It was utilized that $\bar{V} =1$ and $\bar{M}$ definition is simplified as $\bar{M} = 1$ . It can be noticed that the characteristic time defined in equation (11.5) reduced into:

$\displaystyle t_c = { V(0) \over A \sqrt{kRT(0)})}$

(11.26)

Also it can be noticed that equation (11.12) simplified into

$\displaystyle f[ M ] = \left[ 1 + {k-1 \over 2} 1^{2}\right] ^ {-(k+1) \over 2(k-1)} = \left( k+1 \over 2 \right) ^ {-(k+1) \over 2(k-1)}$

(11.27)

Equation (11.20) can be simplified as

$\displaystyle {1 \over k} \left( P^{1 - k \over 2k} \right) dP \pm f[m] d\bar{t} = 0$

(11.28)

Equation (11.23) can be integrated as

$\displaystyle \int_{1}^{\bar{P}} { P^{1 - k \over 2k} dP } \pm \int_{0}^{t} dt = 0$

(11.29)

The integration limits are obtained by simply using the definitions of reduced pressure, at $P(\bar{t}=0 )= 1$ and $P(\bar{t}=\bar{t} )= \bar{P}$ . After the integration, equation (11.24) and rearrangement becomes

$\displaystyle \bar{P} = \left[ 1 \pm \left( k -1 \over 2 \right) f[M] \bar{t} \right] ^{ 2k \over 1 -k }$

(11.30)

$\begin{examl} A chamber is connected to a main line with pressure line with a di... ...tion temperature at the main line is the ambient of $27[\celsius]$. \end{examl}$
Solution

The characteristic time is

$\displaystyle t_{max} = {V \over A^{*} c} = {V \over A^{*} c} = {1.0 \over 0.1 \sqrt{1.4\times 287 \times 300 }} = 0.028 [sec]$

(11.31)

And for smaller area

$\displaystyle t_{max} = {1.0 \over 0.001 \sqrt{1.4\times 287 \times 300 }} = 2.8 [sec]$

$\displaystyle \bar{P} = { P (t) \over P(0) } = {4.5 \over 1.5} = 3.0$

The time is

$\displaystyle t = t_{max} \left[ \bar{P} ^{ 1 -k \over k} -1 \right] \left( k+1 \over 2 \right) ^{- \left( \right)}$

(11.32)

Substituting values into equation (11.27) results

$\displaystyle t = 0.028 \left[ 3^{1-1.4 \over 2.8} -1 \right] \left( 2.4 \over 2 \right)^{- 2.4 \over 0.8} = 0.013 [sec]$

(11.33)

Subsections

Filling/Evacuating The Chamber Under Upchucked Condition

Next: Filling/Evacuating The Chamber Under Up: Rigid Tank with Nozzle Previous: Rigid Tank with Nozzle Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21