Compressible Flow Fundamentals Questions and Answers

Answer: C

The point of maximum entropy on the Fanno curve is the sonic state, M = 1.

Fanno flow is adiabatic flow with friction in a constant-area duct. Friction is irreversible, so entropy must rise along the duct. The Fanno curve has a subsonic upper branch and a supersonic lower branch, and friction drives both toward the point of maximum entropy, where the Mach number is 1 and the flow is choked.

It is not a maximum or minimum Mach number: subsonic flow accelerates toward 1 and supersonic flow decelerates toward 1. A normal shock joins a point on the supersonic branch to one on the subsonic branch with the same mass flux and stagnation enthalpy; it is not the maximum-entropy point.

Answer: C

The critical Mach number is reached when the Mach number somewhere on the airfoil is unity.

Flow accelerates over the upper surface, so the local Mach number there is higher than the freestream value. The critical Mach number is defined as the freestream Mach number at which the local flow first reaches sonic speed at some point on the surface, usually near the point of minimum pressure. It is always below 1 for a lifting airfoil.

A sonic or supersonic freestream is well past the critical condition, and a flow that is supersonic everywhere on the airfoil describes the supersonic regime, not the onset of local sonic flow.

Answer: A

The Mach angle is 30°.

The Mach angle is the half-angle of the Mach cone, defined by

\[ \begin{aligned} \mu &= \sin^{-1}\frac{1}{M} \\ &= \sin^{-1}\frac{1}{2} = 30^{\circ} \end{aligned} \]

45° corresponds to \(M = \sqrt{2} \approx 1.41\), and 90° to \(M = 1\), where the Mach wave stands normal to the flow. 60° would need \(M = 1/\sin 60^{\circ} \approx 1.15\). As Mach number rises the Mach angle shrinks, so Mach 2 gives the smallest of these.

Answer: A

The critical Mach number is likely to decrease.

The critical Mach number is the free-stream Mach number at which the flow somewhere on the airfoil first reaches the speed of sound. A thicker airfoil accelerates the flow more over its surface, so the peak local velocity is higher for the same free-stream speed and sonic flow appears at a lower free-stream Mach number.

That is why high-speed aircraft use thin airfoils: they raise the critical Mach number. An increase would need a thinner section, and thickness clearly affects the peak velocity, so 'unchanged' and 'undefined' are wrong.

Answer: C

For a completely subsonic flow, pressure at the nozzle exit = back pressure.

When the exit flow is subsonic, pressure disturbances travel upstream from the exit, so the jet adjusts until its static pressure matches the surroundings. A subsonic jet cannot sustain a pressure difference across its boundary.

An exit pressure above the back pressure happens only in a choked nozzle, where the exit is sonic and the flow cannot adjust further. An exit pressure below the back pressure cannot be sustained in a convergent nozzle. And the exit pressure equals the total pressure only when the gas is not moving at all.

Answer: B

The mass flow rate becomes half of \(\dot{m}_{o}\).

For choked flow through a nozzle of throat area \(A^{*}\), the mass flow rate is

\[ \begin{aligned} \dot{m} &= \frac{A^{*}P_{o}}{\sqrt{T_{o}}} \cdot K \\ K &= \sqrt{\frac{\gamma}{R}}\left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma + 1}{2(\gamma - 1)}} \end{aligned} \]

where \(K\) depends only on the gas. With \(P_{o}\) and the geometry unchanged, \(\dot{m} \propto 1/\sqrt{T_{o}}\). Raising the total temperature to \(4T_{o}\) multiplies the flow by \(1/\sqrt{4} = 1/2\).

Hotter gas is less dense at the same pressure, and although the sonic speed at the throat rises by \(\sqrt{4} = 2\), the density falls by 4, so the net effect is a halving. 'Unchanged', 'twice' and 'four times' ignore or invert this.

Answer: C

The relation is \(c(p_{2}^{(\gamma-1)/\gamma} - p_{1}^{(\gamma-1)/\gamma})\) = \(\frac{V_{1}^{2}}{2} - \frac{V_{2}^{2}}{2}\).

Along a streamline in steady inviscid flow, Euler's equation is \(dp/\rho + V\,dV = 0\). For isentropic flow \(p/\rho^{\gamma} = K\), so \(\rho = (p/K)^{1/\gamma}\) and

\[ \begin{aligned} \int \frac{dp}{\rho} &= K^{1/\gamma}\int p^{-1/\gamma}\,dp \\ &= \frac{\gamma}{\gamma - 1}K^{1/\gamma} \\ &\quad \times p^{(\gamma-1)/\gamma} \end{aligned} \]

Integrating Euler's equation from point 1 to point 2 then gives the relation above, with \(c = \frac{\gamma}{\gamma-1}K^{1/\gamma}\). It is the compressible form of Bernoulli's equation.

The exponents \(\gamma\), \(\gamma - 1\) and \(\gamma/(\gamma - 1)\) in the other options do not come out of the integration: the power of p is fixed at \(1 - 1/\gamma = (\gamma - 1)/\gamma\) by the isentropic relation.

Answer: D

The best section is (D), the supercritical airfoil.

At a cruise Mach number of 0.8 the flow over the upper surface of a conventional airfoil goes supersonic and ends in a strong shock, which brings wave drag and can separate the boundary layer. A supercritical airfoil has a flattened upper surface, so the local supersonic region is weak and ends in a weak shock or none, and its camber is moved to the rear, with the cusp-like lower surface near the trailing edge seen in (D) recovering the lift lost by flattening the top. This raises the drag-divergence Mach number, which is why transport aircraft cruising near Mach 0.8 use such sections.

(A) is a thick symmetric section that gives no lift at zero incidence and has a strongly curved upper surface. (B) is a highly cambered low-speed, high-lift section whose curved upper surface would produce strong shocks at Mach 0.8. (C) is a flat-bottomed section of the kind used on light, low-speed aircraft.

Answer: B

The Mach numbers are equal: \(M_{2} = M_{1}\).

The speed of sound in an ideal gas is \(a = \sqrt{\gamma R T}\), and \(\gamma\) and \(R\) are the same on both days. So

\[ \begin{aligned} M_{2} &= \frac{\sqrt{1.2}\,V_{1}}{\sqrt{\gamma R (1.2 T_{1})}} \\ &= \frac{V_{1}}{\sqrt{\gamma R T_{1}}} = M_{1} \end{aligned} \]

The speed and the speed of sound both rise by the factor \(\sqrt{1.2}\), so their ratio is unchanged. \(\sqrt{1.2}M_{1}\) accounts for the higher speed but forgets the higher speed of sound; \(M_{1}/\sqrt{1.2}\) does the reverse.

Answer: C

The flow at the throat can either be sonic or supersonic.

In isentropic flow, \(dA/A = (M^{2} - 1)\,dV/V\). With \(M \gt 1\), a decreasing area slows the flow, so the Mach number falls through the convergent part, but it can only reach 1 where the area is a minimum, at the throat. If the throat area equals the critical (sonic) area for the entering flow, the flow is just sonic there; if the throat is larger, it is still supersonic there and simply speeds up again in the divergent part.

It cannot be subsonic at the throat: in isentropic flow the Mach number cannot pass through 1 inside a converging section, and slowing to subsonic before the throat would need a shock, which is not isentropic. So 'only supersonic', 'only sonic' and 'only subsonic' are all wrong.