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

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

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

Assumptions
One-dimensional flow. Velocity in the reservoir is negligible and the flow is isentropic, so that p_{r}=p_{t} \text { and } T_{r}=T_{t} . 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 c_{p}=1.005 kJ / kg \cdot K for air. In a one-dimensional isentropic flow of a perfect gas, the total temperature and total pressure remain constant along the flow. Therefore:

T_{t e}=T_{r}=500 K ; \quad \text { and } \quad p_{t e}=p_{r}=500 kPa

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

\frac{p_{t}}{p}=\left(\frac{\rho_{t}}{\rho}\right)^{k}=\left(\frac{T_{t}}{T}\right)^{k /(k-1)}                             (15.9)

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

From Eq. (15.8):

\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}

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.