For a normal shock at \( M_1 = 1.5 \), the static pressure ratio is nearest to
Answer: C
Using \( 1 + 1.1667 (M_1^2 - 1) \) with \( M_1^2 - 1 = 1.25 \) gives \( 1 + 1.458 = 2.458 \). Comparing with the value of 4.5 at Mach 2 shows how sharply shock strength climbs with Mach number, and with it the stagnation pressure loss.
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For a normal shock at \( M_1 = 2 \), the stagnation pressure ratio \( p_{02}/p_{01} \) is nearest to
Answer: D
Combining the two factors of the normal shock stagnation relation gives \( 2.667^{3.5} \times (2.4/10.8)^{2.5} = 30.96 \times 0.02328 = 0.721 \). Nearly twenty-eight per cent of the stagnation pressure has been destroyed by a single shock, which is the loss a supersonic intake exists to avoid.
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For the same normal shock at \( M_1 = 2 \), the stagnation temperature ratio is
Answer: A
A shock is adiabatic, so the stagnation temperature is identical on both sides and the ratio is exactly one, whatever happens to pressure. Confusing this with the stagnation pressure ratio of 0.721 is among the commonest errors in the subject, so the two are worth quoting together.
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An oblique shock in a stream at \( M_1 = 3 \) makes an angle of \( 30 \) degrees with the flow. The Mach number of the component normal to the shock is
Answer: B
The normal component is \( M_{n1} = M_1 \sin\beta = 3 \times \sin 30^\circ = 3 \times 0.5 = 1.5 \). Every normal shock relation may now be applied with 1.5 in place of the Mach number, which is what makes oblique shocks so little extra work once normal shocks are understood.
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Continuing that case, the Mach number normal to the shock downstream is nearest to
Answer: C
The normal component behaves exactly as it would through a normal shock at \( M = 1.5 \), so it emerges at \( 0.701 \). Note that this is not the downstream Mach number: the tangential component is unchanged, and the resultant downstream Mach number is obtained by dividing 0.701 by the sine of the shock angle less the deflection.
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In a stream at \( M_1 = 3 \), the smallest angle an oblique shock can make with the flow is nearest to
Answer: D
The limiting case is a vanishingly weak wave, whose angle is the Mach angle \( \mu = \arcsin(1/3) = 19.47 \) degrees. Any real oblique shock at this Mach number lies between that value and ninety degrees, the latter being the normal shock.
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A wedge deflects a stream at \( M_1 = 2 \) through \( 20 \) degrees, producing an attached weak oblique shock at \( \beta = 53.4 \) degrees. The normal component Mach number is nearest to
Answer: A
The normal component is \( M_1 \sin\beta = 2 \times \sin 53.4^\circ = 2 \times 0.803 = 1.606 \). Because this exceeds one, the shock is a genuine compression; had the geometry given a normal component below one, no shock could have existed there at all.
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The maximum flow deflection achievable through an attached oblique shock at \( M_1 = 2 \) is nearest to
Answer: B
For air at Mach 2 the maximum deflection is close to 23 degrees. A wedge of larger half-angle cannot hold an attached shock, and a detached curved bow wave forms ahead of it instead. The maximum grows with Mach number, approaching about 45 degrees as the Mach number becomes very large.
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As \( M_1 \) grows without bound, the limiting density ratio across a normal shock in air is
Answer: C
The limit is \( (\gamma+1)/(\gamma-1) = 2.4/0.4 = 6 \). Pressure and temperature ratios keep growing indefinitely, but the density cannot: the gas simply cannot be packed more tightly by a shock. At genuinely hypersonic speeds dissociation lowers the effective specific heat ratio and pushes the limit higher.
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A normal shock at \( M_1 = 2 \) stands in air at \( 250 \) K. The static temperature downstream is nearest to
Answer: D
The temperature ratio at Mach 2 is 1.688, so \( 250 \times 1.688 = 422 \) K. Note that the stagnation temperature, which is \( 250 \times 1.8 = 450 \) K on both sides, is unchanged by the shock; only the split between thermal and kinetic energy has moved.
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