For air at \( M = 1 \), the ratio of stagnation density to static density is nearest to
Answer: A
The density relation carries the exponent \( 1/(\gamma-1) = 2.5 \), so \( \rho_0/\rho = 1.2^{2.5} = 1.577 \). Comparing the three isentropic ratios at Mach one -- 1.2 for temperature, 1.89 for pressure and 1.58 for density -- shows how much more strongly pressure responds than temperature does.
Please log in or register to report this question.
A section has an incompressible lift coefficient of \( 0.50 \). At \( M = 0.6 \) the Prandtl-Glauert rule gives
Answer: B
The correction divides by \( \sqrt{1 - M^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 \), so \( 0.50 / 0.8 = 0.625 \). Multiplying instead of dividing gives 0.400 and is the commonest error; the coefficient must rise with Mach number, which is a quick way to check the direction.
Please log in or register to report this question.
A flow has a static temperature of \( 250 \) K and a speed of \( 200 \) m/s. With \( c_p = 1005 \) J/kg K, its stagnation temperature is nearest to
Answer: C
The energy equation gives \( T_0 = T + V^2/(2 c_p) \), so the kinetic term is \( 200^2 / (2 \times 1005) = 40\,000 / 2010 = 19.9 \) K. Adding this to 250 gives \( 269.9 \) K. This form is often quicker than the Mach-number relation when the speed is given directly.
Please log in or register to report this question.
At sea level, where the speed of sound is about \( 340 \) m/s, the speed corresponding to the conventional compressibility threshold of \( M = 0.3 \) is nearest to
Answer: D
Multiplying, \( 0.3 \times 340 = 102 \) m/s, or about 200 knots. Below this speed the density varies by less than five per cent through the flow field and an incompressible analysis is adequate, which covers most light aircraft and all low-speed wind-tunnel work.
Please log in or register to report this question.
An aircraft cruises at \( M = 0.85 \) where the speed of sound is \( 295 \) m/s. Its true airspeed is nearest to
Answer: A
True airspeed is the Mach number times the local speed of sound, so \( 0.85 \times 295 = 250.8 \) m/s, about 488 knots. Dividing rather than multiplying gives 347 m/s and would imply the aircraft was supersonic, which the stated Mach number contradicts.
Please log in or register to report this question.
A normal shock reduces the stagnation pressure of a stream by \( 28 \) per cent. The stagnation temperature ratio across it is
Answer: B
A shock is adiabatic, so the stagnation temperature is unchanged whatever happens to the stagnation pressure, and the ratio is exactly one. The 28 per cent loss reflects the entropy rise alone. Confusing the two stagnation quantities is among the most common errors in compressible flow.
Please log in or register to report this question.
Assertion (A): Stagnation temperature is unchanged across a normal shock wave.
Reason (R): A shock wave is an adiabatic process, even though it is strongly irreversible.
Select the correct answer from the options below.
Answer: C
Both statements are true and the second explains the first. The energy equation fixes the stagnation temperature for any adiabatic process without work, whether or not it is reversible. Stagnation pressure is the quantity that suffers, because it responds to the entropy rise that irreversibility produces.
Please log in or register to report this question.
Assertion (A): A supersonic nozzle must diverge if the flow is to accelerate further.
Reason (R): Above Mach one the fractional fall in density outpaces the fractional rise in velocity.
Select the correct answer from the options below.
Answer: D
Both are true and the reason accounts for the assertion. Continuity fixes the product of density, velocity and area, so if density is falling faster than velocity is rising, the area must grow. Below Mach one the balance reverses, which is why the same nozzle converges up to its throat.
Please log in or register to report this question.
Assertion (A): A converging nozzle cannot produce a supersonic exit however low the back pressure is taken.
Reason (R): A transition through Mach one can occur only where the duct area is stationary.
Select the correct answer from the options below.
Answer: A
Both statements are true and the second explains the first. The area-velocity relation permits sonic conditions only at a throat, and a purely converging duct has its minimum area at the exit. The best it can do is choke with Mach one at the exit plane; going further requires a divergent section.
Please log in or register to report this question.
Assertion (A): Lowering the back pressure of a choked nozzle raises its mass flow rate.
Reason (R): Once the throat is sonic, no pressure signal can travel upstream through it.
Select the correct answer from the options below.
Answer: B
The assertion is false and the reason is true, and the reason is precisely why. With the throat sonic, information about the back pressure cannot reach the reservoir, so conditions upstream are frozen and the mass flow is fixed by the stagnation state and the throat area alone. Only raising the stagnation pressure will pass more mass.
Please log in or register to report this question.
