Isentropic Discharge from a Tank Air is discharged from a large reservoir (Fig. 15.7) into the atmosphere through a nozzle. The reservoir conditions are pr = 500 kPa and Tr = 500 K, and the atmospheric pressure is 100 kPa. Assuming the flow to be isentropic and the exit pressure to be the same as

Question

Isentropic Discharge from a Tank

Air is discharged from a large reservoir (Fig. 15.7) into the atmosphere through a nozzle. The reservoir conditions are [latex] p_{r}=500 kPa ext { and } T_{r}=500 K [/latex], and the atmospheric pressure is 100 kPa. Assuming the flow to be isentropic and the exit pressure to be the same as the outside pressure, determine (a) the exit velocity of the air. What-if scenario: (b) What would the answer be if the reservoir temperature was 1000 K?

Accepted Answer

Use isentropic relation to obtain the exit temperature and Eq. (15.8) to obtain the exit velocity.

[latex] h_{t}=h+\frac{V^{2}}{2000} ; \quad\left[\frac{ kJ }{ kg } ight] \quad ext { and }, \quad T_{t}=T+\frac{V^{2}}{2000 c_{p}} ; \quad[ K ] [/latex] (15.8)

Assumptions
One-dimensional flow. Velocity in the reservoir is negligible and the flow is isentropic, so that [latex] p_{r}=p_{t} ext { and } T_{r}=T_{t} [/latex]. The pressure at the exit is equal to the outside pressure.

Analysis
From Table C-1, or the gas dynamics TESTcalc, we obtain k = 1.4 and [latex] c_{p}=1.005 kJ / kg \cdot K [/latex] for air. In a one-dimensional isentropic flow of a perfect gas, the total temperature and total pressure remain constant along the flow. Therefore:

[latex]T_{t e}=T_{r}=500 K ; \quad ext { and } \quad p_{t e}=p_{r}=500 kPa [/latex]

The relation between the total and static properties, Eq. (15.9), yields:

[latex] \frac{p_{t}}{p}=\left(\frac{ ho_{t}}{ ho} ight)^{k}=\left(\frac{T_{t}}{T} ight)^{k /(k-1)} [/latex] (15.9)

[latex]\begin{aligned}&\frac{T_{t e}}{T_{e}}=\left(\frac{p_{t e}}{p_{e}} ight)^{(k-1) / k}=\left(\frac{500}{100} ight)^{(1.4-1) / 1.4}=1.584 \&\Rightarrow \quad T_{e}=T_{t e} / 1.584=500 / 1.584=315.7 K\end{aligned}[/latex]

From Eq. (15.8):

[latex]\begin{aligned}&\frac{V_{e}^{2}}{2(1000 J / kJ ) c_{p}}=T_{t e}-T_{e} \&\Rightarrow \quad V_{e}=\sqrt{2(1000)(1.005)(500-315.7)}=608.7 m / s\end{aligned}[/latex]

TEST Analysis
Launch the gas dynamics TESTcalc and select air as the working fluid. Evaluate State-1 as the reservoir state for the given p1, T1, and Vel1 = 0. For the exit, State-2, enter p2, T_t2 = T1 and p_t2 = p1 to obtain Vel1 = 608.5 m/s.

What-if scenario
Change T1 to 1000 K and click Super-Calculate to update the solution. The new exit velocity is calculated as 860.6 m/s. By increasing p1 and updating the solution, the exit velocity can be shown to increase with the chamber pressure as well.

Discussion
Soon we will develop a simpler approach for analyzing isentropic flows and revisit this problem.

Question 15.2: Isentropic Discharge from a Tank Air is discharged from a la...

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