Using \( (1 + 0.2M^2)^{3.5} \), the ram pressure ratio available at Mach 3 is approximately
Answer: A
Substituting gives \( 2.8^{3.5} \approx 36.7 \). A ratio of that size from forward speed alone exceeds what many mechanical compressors deliver, which is precisely why a ramjet needs no rotating machinery at that speed.
Please log in or register to report this question.
At Mach 5 the same relation gives a ram pressure ratio of approximately
Answer: B
Correct answer: 529
For isentropic deceleration of air,
\[ \frac{p_0}{p} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{\frac{\gamma}{\gamma-1}} = \left(1 + 0.2 \times 25\right)^{3.5} = 6^{3.5} \approx 529 \]Ram compression alone therefore delivers a pressure ratio far beyond anything a mechanical compressor achieves, which is why a ramjet needs no rotating machinery at all. The corresponding stagnation temperature, however, is 1320 K from an ambient 220 K, and it is that figure rather than the pressure that sets the ramjet’s limit.
Please log in or register to report this question.
Static temperature is 220 K at Mach 5. Using \( T_0 = T(1 + 0.2M^2) \), the stagnation temperature is
Answer: C
Substituting gives \( 220 \times 6 = 1320 \) kelvin. Decelerating the flow fully to subsonic speed would deliver that temperature to the combustor before any fuel is burned, which is exactly the problem a scramjet avoids.
Please log in or register to report this question.
At Mach 8 the stagnation temperature of air at 220 K is
Answer: D
The factor is \( 1 + 0.2 \times 64 = 13.8 \), giving \( 220 \times 13.8 = 3036 \) kelvin. No material survives that, and the combustion products would dissociate, which places full deceleration entirely out of the question.
Please log in or register to report this question.
A scramjet combustor is \( 1.5 \ \mathrm{m} \) long and the flow passes at \( 1500 \ \mathrm{m/s} \). The residence time is
Answer: A
Correct answer: 1 millisecond
Residence time is length over velocity:
\[ t = \frac{1.5}{1500} = 1 \times 10^{-3} \text{ s} = 1 \text{ millisecond} \]Within that millisecond the fuel must be injected, penetrate the core flow, mix to the molecular level, ignite and release its heat. A conventional gas turbine combustor allows tens of milliseconds for the same tasks at a fraction of the velocity, which is the measure of how severe the scramjet problem is.
Please log in or register to report this question.
A vehicle flies at Mach 7 where the speed of sound is \( 300 \ \mathrm{m/s} \). Its true airspeed is
Answer: B
Correct answer: 2100 m/s
Mach number is the ratio of true airspeed to the local speed of sound, so
\[ V = M a = 7 \times 300 = 2100 \text{ m/s} \]The speed of sound depends only on temperature, and at high altitude it is well below the sea level figure of about 340 m/s, so a given Mach number corresponds to a lower true airspeed there. Dividing rather than multiplying gives 42.9 m/s, which would be a walking pace.
Please log in or register to report this question.
Hydrogen has a heating value near \( 120 \ \mathrm{MJ/kg} \) against kerosene's \( 43 \). The ratio is approximately
Answer: C
Dividing gives \( 120/43 \approx 2.8 \). Hydrogen's advantage per kilogram is large, but its very low density means the tank volume required is several times greater, which is a severe practical penalty on a slender vehicle.
Please log in or register to report this question.
A ramjet is boosted from rest to Mach 2.5 before taking over. If the booster burns for 8 seconds, the average acceleration to reach \( 750 \ \mathrm{m/s} \) is approximately
Answer: D
Correct answer: 94 m/s squared, about 9.6 g
Average acceleration is the velocity change over the time taken:
\[ a = \frac{750}{8} = 93.75 \text{ m/s}^2 \]\[ \frac{93.75}{9.81} \approx 9.6 g \]Around ten g is typical of a rocket booster on a ramjet missile, which is one reason the technique suits missiles rather than manned vehicles. The booster does its work in a few seconds and is then separated, or its casing becomes the ramjet combustor in an integral rocket-ramjet.
Please log in or register to report this question.
A scramjet inlet compresses the flow from Mach 8 to Mach 3 at the combustor entry. The flow remains
Answer: A
Correct answer: Supersonic throughout
Mach 3 at the combustor entry is supersonic, so the flow has been compressed but never brought below sonic speed - that is the definition of scramjet operation. The inlet is a series of oblique shocks and compression surfaces with no normal shock and no throat that would force the flow subsonic. Keeping it supersonic is what keeps the static temperature within reach of the structure and of useful combustion chemistry.
Please log in or register to report this question.
A vehicle of \( 1200 \ \mathrm{kg} \) experiences \( 18 \ \mathrm{kN} \) of thrust and \( 12 \ \mathrm{kN} \) of drag. Its acceleration is
Answer: B
Correct answer: 5 m/s squared
The net force is thrust minus drag, and Newton’s second law gives
\[ a = \frac{F_{net}}{m} = \frac{18\,000 - 12\,000}{1200} = \frac{6000}{1200} = 5 \text{ m/s}^2 \]Using the thrust alone gives 15 m/s squared, which is the usual error: at high speed the drag is a large fraction of the thrust, and the acceleration available is the small difference between two large numbers.
Please log in or register to report this question.
