Helium gas from a storage tank at 1000 kPa and 310 K is flowing out through a convergent nozzle of exit area 3 cm^2 to another tank. When the mass flow rate is 0.15 kg s^−1, determine the pressure in the second tank.

Question

Helium gas from a storage tank at 1000 kPa and 310 K is flowing out through a convergent nozzle of exit area [latex]3 cm ^{2}[/latex] to another tank. When the mass flow rate is [latex]0.15 kg s ^{-1}[/latex], determine the pressure in the second tank.

Accepted Answer

Given, [latex]p_{0}=1000 kPa ext { and } T_{0}=310 K.[/latex] The maximum mass flow through the nozzle is at choked condition with [latex]M_{e}=1[/latex] and is given by

[latex]\dot{m}_{\max }=\sqrt{\gamma\left(\frac{2}{\gamma+1} ight)^{(\gamma+1) /(\gamma-1)} \frac{p_{0}^{2}}{R T_{0}}} A^{*}[/latex]

For helium gas, [latex]\gamma=1.67[/latex] and the gas constant is given by

[latex]R=\frac{R_{u}}{ ext { molecular weight }}=\frac{8314}{4.003}=2077 J ( kg K )^{-1}[/latex]

Therefore,

[latex]\dot{m}_{\max }=\frac{0.7266 imes 1000 imes 10^{3}}{\sqrt{2077 imes 310}} 3 imes 10^{-4}=0.27 kg s ^{-1}[/latex]

The given mass flowrate [latex]0.15 kg s ^{-1}[/latex] is less than the critical value and hence [latex]M_{e}[/latex] is subsonic.

[latex]\dot{m}=0.15= ho_{e} A_{e} V_{e}[/latex]

[latex]=\left(\frac{ ho_{e}}{ ho_{0}} ight) ho_{0} A_{e} M_{e}\left(\frac{a_{e}}{a_{0}} ight) a_{0}[/latex]

[latex]= ho_{0} a_{0} A_{e} M_{e}\left(\frac{ ho_{e}}{ ho_{0}} ight)\left(\frac{T_{e}}{T_{0}} ight)^{1 / 2}[/latex]

[latex]= ho_{0} a_{0} A_{e} M_{e}\left(1+\frac{\gamma-1}{2} M_{e}^{2} ight)^{\frac{-1}{\gamma-1}}\left(1+\frac{\gamma-1}{2} M_{e}^{2} ight)^{-0.5}[/latex]

For helium, [latex]\gamma=1.67,[/latex] so

[latex]0.15= ho_{0} a_{0} A_{e} M_{e}\left(1+0.33 M_{e}^{2} ight)^{-2}[/latex]

Also,

[latex] ho_{0}=\frac{p_{0}}{R T_{0}}[/latex]

[latex]=\frac{1000 imes 10^{3}}{2077 imes 310}[/latex]

[latex]=1.553 kg m ^{-3}[/latex]

[latex]a_{0}=\sqrt{\gamma R T_{0}}[/latex]

[latex]=\sqrt{1.67 imes 2077 imes 310}[/latex]

[latex]=1036.95 m s ^{-1}[/latex]

[latex]A_{e}=3 imes 10^{-4} m ^{2}[/latex]

Thus,

[latex]0.15=\frac{0.4831 M_{e}}{\left(1+0.335 M_{e}^{2} ight)^{2}}[/latex]

Now let us solve for [latex]M_{e}[/latex] by trial and error.

Trial 1 Let [latex]M_{e}=0.3.[/latex]

RHS [latex]\approx 0.145[/latex]

Trial 2 Let [latex]M_{e}=0.32.[/latex]

RHS [latex]\approx 0.15[/latex]

Hence, [latex]M_{e}=0.32.[/latex]

By isentropic relation, we have

[latex]\frac{p_{0}}{p_{e}}=\left(1+\frac{\gamma-1}{2} M_{e}^{2} ight)^{\frac{\gamma}{\gamma-1}}[/latex]

[latex]=\left(1+0.335 M_{e}^{2} ight)^{2.493}[/latex]

[latex]= 1.087[/latex]

Thus, the pressure in the second tank is

[latex]p_{e}=\frac{p_{0}}{1.087}[/latex]

[latex]=\frac{1000 imes 10^{3}}{1.087}[/latex]

[latex]= 919.96 kPa[/latex]

Question 2.1: Helium gas from a storage tank at 1000 kPa and 310 K is flow...

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