Nitrogen flows from a 2" pipe in which the temperature and pressure are 40°F and 40 psi gage, respectively, into a 1" pipe in which the pressure is 21.3 psi gage. Calculate the velocity in each pipe, assuming isothermal conditions apply and no losses.

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Accepted Answer

Referring to Problem 7.21, equation (C) for isothermal conditions can be solved for [latex]V_{2}[/latex], noting that [latex]z_{1}=z_{2}[/latex],

[latex]\frac{V_{2}^{2}}{2 g}\left[1-\left(\frac{A_{2} p_{2}}{A_{1} p_{1}} ight)^{2} ight]=\frac{p_{1}}{\gamma_{1}} \ln \left(\frac{p_{1}}{p_{2}} ight)=R T \ln \left(\frac{p_{1}}{p_{2}} ight) \quad ext { or } \quad V_{2}=\sqrt{2 g imes \frac{R T \ln \left(p_{1} / p_{2} ight)}{1-\left(A_{2} p_{2} / A_{1} p_{1} ight)^{2}}}[/latex].

Substituting herein, using R = 55.1 ft/°R for nitrogen from Table 1 in the Appendix,

[latex]V_{2}=\sqrt{2 g imes \frac{55.1 imes 500 \ln [(54.7 imes 144) /(36.0 imes 144)]}{1-\left(\frac{1}{2} ight)^{4}[(36.0 imes 144) /(54.7 imes 144)]^{2}}}=873 ft / sec[/latex].

[latex] ext { Also } V_{1}=\left(A_{2} / A_{1} ight)\left(p_{2} / p_{1} ight) V_{2}=\left(\frac{1}{2} ight)^{2}(36.0 / 54.7)(873)=144 ft / sec[/latex].

Question 7.41: Nitrogen flows from a 2" pipe in which the temperature and p...

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