Compressible Flow Free Textbook

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Isothermal Nozzle Attached

In this case the process in nozzle is assumed to isothermal but the process in the chamber is isentropic. The temperature in the nozzle is changing because the temperature in the chamber is changing. Yet, the differential temperature change in the chamber is slower than the temperature change in nozzle. For rigid volume, $\bar{V} =1$ and for isothermal nozzle $\bar{T} = 1$ Thus, equation (11.13) is reduced into

$\displaystyle {d \bar{P} \over d\bar{t}} = \pm f[M] \bar{P} = 0$

(11.34)

Separating the variables and rearranging equation (11.29) converted into

$\displaystyle \int_{1} ^{\bar{P}} {d\bar{P} \over \bar{P}} \pm f[M] \int_{0}^{\bar{t}} d\bar{t} = 0$

(11.35)

Here,

is expressed by equation (11.22). After the integration, equation (11.30) transformed into

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

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

(11.36)

Next: Rapid evacuating of a Up: Rigid Tank with Nozzle Previous: Filling/Evacuating The Chamber Under Index

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