Your Answer : (Not Answered)

Correct Answer : A

The chamber pressure is 3600 kPa and the specific impulse 183.67 s, so the answer is 3600 and 183.67.

The characteristic velocity is defined by \(c^{*} = p_c A_t / \dot{m}\), so

\[ \begin{aligned} p_c &= \frac{c^{*}\dot{m}}{A_t} = \frac{1200 \times 75}{0.025} \\ &= 3.6 \times 10^{6}\ \text{Pa} = 3600\ \text{kPa} \end{aligned} \]

The effective exhaust velocity is \(c = C_F c^{*}\), and the specific impulse in seconds is that divided by g:

\[ \begin{aligned} I_{sp} &= \frac{C_F c^{*}}{g} = \frac{1.5 \times 1200}{9.8} \\ &= 183.67\ \text{s} \end{aligned} \]

The pair '3500 and 183.67' has the right impulse but a wrong pressure, '3600 and 175.23' the right pressure but a wrong impulse, and '4600 and 190.51' gets both wrong.

Your Answer : (Not Answered)

Correct Answer : C

The acceleration at lift-off is \(3g_0\).

The propellant burns at \(\dot{m} = 8000/50 = 160\) kg/s, and the thrust is \(F = I_{sp} g_0 \dot{m}\):

\[ \begin{aligned} F &= 250 \times 160\, g_0 = 40000\, g_0\ \text{N} \\ a &= \frac{F - m g_0}{m} = \frac{40000\, g_0}{10000} - g_0 \\ &= 4g_0 - g_0 = 3g_0 \end{aligned} \]

At lift-off the whole 10-ton mass is still on board, and in vertical flight the weight acts against the thrust.

\(4g_0\) is the thrust-to-mass ratio alone, which forgets gravity. \(g_0\) and \(2g_0\) are too small: the thrust is four times the lift-off weight, so subtracting the weight once leaves \(3g_0\).

Your Answer : (Not Answered)

Correct Answer : C

The second stage, with payload, has a mass of 20 tons.

The payload ratio of a stage is the mass it carries forward divided by its initial mass. For two stages the overall ratio is the product of the stage ratios, and for an optimized rocket with identical stages the two ratios are equal:

\[ \begin{aligned} \lambda^{2} &= \frac{2}{200} = 0.01 \Rightarrow \lambda = 0.1 \\ m_{02} &= \frac{m_{payload}}{\lambda} = \frac{2}{0.1} = 20 \end{aligned} \]

Check: the first stage carries 20 tons out of 200, a ratio of 0.1, matching the second stage's 2 out of 20. With 100, 10 or 50 tons the two stage ratios (for example 100/200 and 2/100) are unequal, so none of them is the optimized split.

Your Answer : (Not Answered)

Correct Answer : B

The thrust is highest when the rocket is operating in vacuum.

Rocket thrust is \(F = \dot{m}V_{e} + (p_{e} - p_{a})A_{e}\). For a given chamber pressure and nozzle, the mass flow, exit velocity and exit pressure are fixed, so the thrust depends only on the ambient pressure \(p_{a}\), and it is largest when \(p_{a} = 0\).

At its design altitude the nozzle is ideally expanded (\(p_{e} = p_{a}\)), which is the best that particular ambient pressure allows, but the thrust is still lower than in vacuum. At sea level the ambient pressure is highest, so thrust is lowest. A normal shock in the nozzle means a badly over-expanded nozzle with large losses.

Your Answer : (Not Answered)

Correct Answer : C

The final velocities are related by \(V_{A} = 0.5V_{B}\).

Without gravity or drag, each missile’s velocity gain follows the rocket equation \(V = I_{sp}g_{0}\ln(m_{0}/m_{f})\), and both have the same \(I_{sp}\) and liftoff mass. Only the propellant burnt differs:

\[ \begin{aligned} m_{p,A} &= 120 \times 30 = 3600\ \text{kg} \\ m_{p,B} &= 70 \times 70 = 4900\ \text{kg} \end{aligned} \]

\[ \begin{aligned} \frac{V_{A}}{V_{B}} &= \frac{\ln(5600/2000)}{\ln(5600/700)} \\ &= \frac{\ln 2.8}{\ln 8} = \frac{1.030}{2.079} \\ &= 0.495 \approx 0.5 \end{aligned} \]

Missile A burns faster, but the burn rate does not matter once gravity is neglected; only the mass ratio does. \(V_{A} = V_{B}\) would need equal propellant loads, and 0.7 is roughly the ratio of the propellant masses, 3600/4900 = 0.73, which forgets the logarithm. 4.1 has no basis.

Your Answer : (Not Answered)

Correct Answer : C

The specific impulse is 200 s.

The rocket burns \(150 - 50 = 100\) kg of propellant in 10 s, a mass flow of 10 kg/s. Specific impulse is thrust per unit weight flow of propellant:

\[ \begin{aligned} I_{sp} &= \frac{F}{\dot{m}g} \\ &= \frac{19620}{10 \times 9.81} = 200\ \text{s} \end{aligned} \]

400 s would need a mass flow of only 5 kg/s, half the actual figure, and 100 s would need 20 kg/s. 300 s corresponds to about 6.7 kg/s, which is what dividing the 100 kg burnt by 15 s instead of 10 s would give. Only 10 kg/s matches the stated masses and burn time.

Your Answer : (Not Answered)

Correct Answer : C

The burnout velocity is about 4245.9 m/s.

The burn time is \(8400/1200 = 7\) s, and the burnout mass is \(10000 - 8400 = 1600\) kg. With constant gravity and no drag, the rocket equation with a gravity loss term gives

\[ \begin{aligned} v_{b} &= I_{sp}\,g\ln\frac{m_{0}}{m_{b}} - g\,t_{b} \\ &= 240 \times 9.81 \times \ln 6.25 - 9.81 \times 7 \\ &= 2354.4 \times 1.8326 - 68.7 \\ &\approx 4314.6 - 68.7 = 4245.9\ \text{m/s} \end{aligned} \]

4314.6 m/s is the ideal velocity without the gravity loss, a deliberate trap. 4383.3 adds the gravity loss instead of subtracting it, and 3933.7 m/s implies a loss of about 380 m/s, far more than the 68.7 m/s of a 7-second burn.

Your Answer : (Not Answered)

Correct Answer : D

The payload reaches about 385 km.

With all propellant burnt instantaneously there are no gravity losses during the burn, so the full rocket-equation velocity is reached at once. The mass ratio is \(m_{0}/m_{f} = 1/0.4 = 2.5\):

\[ \begin{aligned} \Delta V &= V_{e}\ln\frac{m_{0}}{m_{f}} = 3000\ln 2.5 \\ &= 3000 \times 0.9163 = 2749\ \text{m/s} \end{aligned} \]

The payload then coasts upward against constant gravity with no drag:

\[ \begin{aligned} h &= \frac{\Delta V^{2}}{2g} = \frac{2749^{2}}{2 \times 9.81} \\ &= 3.85 \times 10^{5}\ \text{m} \end{aligned} \]

which is 385 km. The smaller options would need a lower burnout velocity, for example from using a mass ratio below 2.5, which the 60% propellant fraction does not allow.

Your Answer : (Not Answered)

Correct Answer : B

The minimum fuel tank volume is 1.75 m\(^{3}\).

Total propellant flow from the chamber pressure, throat area and characteristic velocity:

\[ \begin{aligned} A_{t} &= \frac{\pi}{4}(0.05)^{2} = 1.963 \times 10^{-3}\ \text{m}^{2} \\ \dot{m} &= \frac{p_{c}A_{t}}{c^{*}} = \frac{4 \times 10^{6} \times 1.963 \times 10^{-3}}{1540} \\ &= 5.10\ \text{kg/s} \end{aligned} \]

With fuel-to-oxidizer ratio 1.8, fuel is \(1.8/2.8\) of the total:

\[ \begin{aligned} \dot{m}_{f} &= 5.10 \times \frac{1.8}{2.8} = 3.28\ \text{kg/s} \\ m_{f} &= 3.28 \times 480 = 1574\ \text{kg} \\ V_{f} &= 1574/900 = 1.75\ \text{m}^{3} \end{aligned} \]

The fuel share is the step to watch: taking the whole flow as fuel gives about 2.72 m\(^{3}\), and reading the ratio the wrong way round (1 part fuel in 2.8) about 0.97 m\(^{3}\). Only 1.8 parts in 2.8 gives 1.75 m\(^{3}\); 1.65, 1.85 and 1.95 m\(^{3}\) do not follow from the data.

Your Answer : (Not Answered)

Correct Answer : B

The throat diameter should be about 38 mm.

At equilibrium the gas generated by the burning grain equals the flow through the nozzle: \(\rho_{p} r A_{b} = p_{c}A_{t}/c^{*}\). First the burning rate at 50 bar (\(5 \times 10^{6}\) Pa):

\[ \begin{aligned} r &= 6.65 \times 10^{-3}(5 \times 10^{6})^{0.45} \\ &= 6.65 \times 10^{-3} \times 1033.6 \\ &= 6.87\ \text{mm/s} \end{aligned} \]

\[ \begin{aligned} \dot{m} &= 1800 \times 0.00687 \times 0.314 = 3.885\ \text{kg/s} \\ A_{t} &= \frac{\dot{m}c^{*}}{p_{c}} = \frac{3.885 \times 1450}{5 \times 10^{6}} \\ &= 1.127 \times 10^{-3}\ \text{m}^{2} \\ d_{t} &= \sqrt{4A_{t}/\pi} = 0.0379\ \text{m} \end{aligned} \]

That is 37.9 mm, so 38 mm. 35 mm would give too small a throat and a higher chamber pressure; 41 mm and 45 mm too large a throat and a lower pressure.

Your Answer : (Not Answered)

Correct Answer : C

The specific impulse is normalized by gravity at earth's standard sea level.

Specific impulse in seconds is defined as \(I_{sp} = F/(\dot{m}g_{0})\), where \(g_{0} = 9.80665\ \text{m/s}^{2}\) is the standard gravity at earth's sea level. It is a fixed conversion constant, chosen so that \(I_{sp}\) in seconds equals the effective exhaust velocity divided by the same number everywhere.

Using local gravity at the launch site, on Mars or anywhere else, would make the same engine report a different specific impulse depending on where it is fired, which defeats the point of the figure. So the crater, Mars 'sea level' and a crater-depth value on earth are all wrong.

Your Answer : (Not Answered)

Correct Answer : B

The answer is combustion chamber and nozzle throat.

The static temperature is highest in the combustion chamber, where the gas is nearly at rest and holds its full stagnation temperature. As the gas accelerates through the nozzle, static temperature falls.

Heat flux to the wall is highest at the throat. The convective heat transfer coefficient scales roughly with mass flux \((\rho V)^{0.8}\), and the mass flux per unit area is greatest at the throat, the smallest cross-section, while the gas is still very hot. That is why throats need the most intensive cooling.

The nozzle exit has the lowest static temperature, and the nozzle throat is not the hottest region in the static sense, which rules out the other three pairs.