Discussion
Consider an isentropic flow of air (ratio of specific heats = 1.4) through a duct as shown in the figure.
The variations in the flow across the cross-section are negligible. The flow conditions at Location 1 are given as follows:
\(P_{1}\) = 100 kPa, \(\rho_{1}\) = 1.2 kg/m3, \(u_{1}\) = 400 m/s
The duct cross-sectional area at Location 2 is given by A2 = 2A1, where A1 denotes the duct cross-sectional area at Location 1. Which one of the given statements about the velocity \(u_{2}\) and pressure \(P_{2}\) at Location 2 is TRUE?
Answer: C
At Location 2, \(u_{2} \gt u_{1}\) and \(P_{2} \lt P_{1}\).
Find the Mach number at Location 1:
\[ \begin{aligned} a_{1} &= \sqrt{\gamma P_{1}/\rho_{1}} \\ &= \sqrt{1.4 \times 10^{5}/1.2} \\ &\approx 341.6\ \text{m/s} \\ M_{1} &= 400/341.6 \approx 1.17 \end{aligned} \]
The flow is supersonic. For isentropic flow in a duct, area and velocity are linked by
\[ \frac{dA}{A} = (M^{2} - 1)\frac{du}{u} \]
With \(M \gt 1\), an increasing area (\(dA \gt 0\)) gives \(du \gt 0\): a supersonic stream speeds up in a diverging duct. In isentropic flow pressure falls as velocity rises, so \(P_{2} \lt P_{1}\).
\(u_{2} \lt u_{1}\) with \(P_{2} \gt P_{1}\) is the subsonic diffuser behaviour, the trap if the Mach number is not checked. The two mixed options break the isentropic link between velocity and pressure.
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