Rocket Propulsion

Q1 (14)

(a) Rocket equation $\Delta v = g_0 I_{sp}\ln(MR)\Rightarrow MR=\exp\!\big(\Delta v/(g_0 I_{sp})\big)$.

• Stage 1: $g_0 I_{sp}=9.80665\times311=3049.9\ \mathrm{m/s}$; $\Delta v_1=4000$. $MR_1=e^{4000/3049.9}=e^{1.3115}=3.712$. (2)
• Stage 2: $\Delta v_2=9400-4000=5400\ \mathrm{m/s}$; $g_0 I_{sp}=9.80665\times450=4412.99$. $MR_2=e^{5400/4412.99}=e^{1.2237}=3.400$. (2)

(b) For each stage with structural fraction $\sigma=0.08$ of stage total, and initial mass $m_0$, final $m_f$: propellant $m_p$, structure $m_s=\sigma(m_p+m_s)$... work top-down.

Stage 2: payload above it $m_{pl}=2000$. Let stage-2 total (struct+prop) $=S_2$, final mass $m_f=m_{pl}+m_{s2}$, initial $m_0=m_{pl}+S_2$. $m_{s2}=0.08 S_2$, $m_{p2}=0.92 S_2$. $MR_2=\dfrac{m_{pl}+S_2}{m_{pl}+0.08S_2}=3.400$. $2000+S_2=3.400(2000+0.08S_2)=6800+0.272S_2$ $\Rightarrow 0.728S_2=4800\Rightarrow S_2=6593.4\ \mathrm{kg}$. Propellant $m_{p2}=0.92\times6593.4=6066.0\ \mathrm{kg}$; structure $527.5\ \mathrm{kg}$. (3)

Mass above stage 1 = $m_{pl}+S_2 = 2000+6593.4=8593.4\ \mathrm{kg}$ (this is the "payload" for stage 1).

Stage 1: $MR_1=\dfrac{m_{above}+S_1}{m_{above}+0.08S_1}=3.712$. $8593.4+S_1=3.712(8593.4+0.08S_1)=31898.7+0.29696S_1$ $0.70304S_1=23305.3\Rightarrow S_1=33149\ \mathrm{kg}$. Propellant $m_{p1}=0.92\times33149=30497\ \mathrm{kg}$; structure $2652\ \mathrm{kg}$. (3)

Lift-off mass $=8593.4+33149=41742\ \mathrm{kg}$ (≈41.7 t). Total propellant $=6066+30497=36563\ \mathrm{kg}$. (1)

(c) Payload fraction $=2000/41742=0.0479\approx4.8\%$. (1) The high-$I_{sp}$ LH2 stage on top provides the large final $\Delta v$ most efficiently and does not have to be lifted along the whole trajectory; putting the dense high-thrust lower-$I_{sp}$ stage first handles the heavy early boost. (2)

Q2 (14)

(a) Isentropic $P_c/P_e=(1+\tfrac{\gamma-1}{2}M_e^2)^{\gamma/(\gamma-1)}$. $P_c/P_e=7.0\times10^6/1.01\times10^5=69.31$. $(P_c/P_e)^{(\gamma-1)/\gamma}=69.31^{0.2/1.2}=69.31^{0.16667}=2.0207$. $1+0.1 M_e^2=2.0207\Rightarrow M_e^2=10.207\Rightarrow M_e=3.195$. (4)

(b) Area ratio: $\varepsilon=\frac{1}{M_e}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M_e^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}.$ $\frac{2}{\gamma+1}=1.8182\times...$ compute: $\frac{2}{2.2}=0.90909$. Bracket $=0.90909\times2.0207=1.8370$. Exponent $=\frac{2.2}{0.4}=5.5$. $1.8370^{5.5}=\exp(5.5\ln1.8370)=\exp(5.5\times0.60826)=\exp(3.3454)=28.37$. $\varepsilon=28.37/3.195=8.88$. (4)

(c) $\frac{R_u T_c}{M}=\frac{8314\times3400}{22}=1.2849\times10^6$. $\frac{2\gamma}{\gamma-1}=\frac{2.4}{0.2}=12.0$. $(P_e/P_c)^{(\gamma-1)/\gamma}=(1/69.31)^{0.16667}=1/2.0207=0.49488$; bracket $=1-0.49488=0.50512$. $v_e=\sqrt{12.0\times1.2849\times10^6\times0.50512}=\sqrt{7.789\times10^6}=2791\ \mathrm{m/s}$. (4)

(d) At $P_a=20$ kPa the fixed nozzle now has $P_e=101\,\text{kPa}>P_a$: it is under-expanded. Prandtl–Meyer expansion fans form at the lip and the plume continues to expand outside the nozzle → efficiency loss vs an ideal nozzle sized for that altitude. (2)

Q3 (12)

(a) Mass generation rate $\dot m_{gen}=\rho_p A_b r=\rho_p A_b a P_c^n$. Mass ejected through throat $\dot m_{out}=P_c A^*/c^*$ (from $c^*=P_c A^*/\dot m$). Equilibrium $\dot m_{gen}=\dot m_{out}$: $\rho_p A_b a P_c^n = P_c A^*/c^*$ $\Rightarrow P_c^{1-n}=\dfrac{A_b}{A^*}\rho_p a c^*$ $\Rightarrow P_c=\left(\dfrac{A_b}{A^*}\rho_p a c^*\right)^{1/(1-n)}.$ (5)

(b) $\frac{A_b}{A^*}=0.90/3.0\times10^{-3}=300$. Inside: $300\times1750\times5.0\times10^{-5}\times1550=300\times1750\times0.0775=300\times135.625=40687.5$. Exponent $1/(1-n)=1/0.65=1.53846$. $P_c=40687.5^{1.53846}=\exp(1.53846\times\ln40687.5)=\exp(1.53846\times10.6135)=\exp(16.328)=1.23\times10^{7}\ \mathrm{Pa}\approx12.3\ \mathrm{MPa}$. (4)

(c) Generation scales as $P_c^n$, loss (ejection) scales as $P_c^1$. If $n<1$, a pressure rise increases ejection faster than generation → net restoring effect → stable. If $n\ge1$ a perturbation grows without bound (runaway) → unstable. (3)

Q4 (10)

(a) $I_{sp}=c/g_0=30000/9.80665=3059\ \mathrm{s}$. $F=\dot m c\Rightarrow \dot m=F/c=0.236/30000=7.87\times10^{-6}\ \mathrm{kg/s}$ (≈7.9 mg/s). (4)

(b) $P_{jet}=\tfrac12\dot m c^2=0.5\times7.867\times10^{-6}\times(30000)^2=0.5\times7.867\times10^{-6}\times9\times10^{8}=3540\ \mathrm{W}=3.54\ \mathrm{kW}$. Alternatively $P_{jet}=\tfrac12 F c=0.5\times0.236\times30000=3540\ \mathrm{W}$. Efficiency $\eta=P_{jet}/P_{elec}=3540/10000=0.354=35.4\%$. (4)

(c) For fixed power $P$, $F=2\eta P/c$: raising $c$ (higher $I_{sp}$) reduces thrust, so acceleration drops and trip/burn times grow; there is a power-limited trade between high $I_{sp}$ and usable thrust. (2)

Q5 (10)

(a) $(P_e-P_a)A_e=(70000-101000)\times1.20=-31000\times1.20=-37200\ \mathrm{N}$. $\dot m v_e=F-(P_e-P_a)A_e=845000-(-37200)=882200\ \mathrm{N}$. $v_e=882200/290=3042.1\ \mathrm{m/s}$. (4)

(b) $c=v_e+(P_e-P_a)A_e/\dot m=F/\dot m=845000/290=2913.8\ \mathrm{m/s}$. $I_{sp}=c/g_0=2913.8/9.80665=297.1\ \mathrm{s}$. (4)

(c) $P_e=70$ kPa $<P_a=101$ kPa → over-expanded. Oblique shocks form in the plume (possible flow separation inside the nozzle), reducing effective thrust. (2)

[
 {"claim":"Q1 stage mass ratios and lift-off mass ~41742 kg","code":"g0=9.80665\nMR1=exp(4000/(g0*311)); MR2=exp(5400/(g0*450))\nS2=(2000*MR2-2000)/(1-0.08*MR2)\nabove1=2000+S2\nS1=(above1*MR1-above1)/(1-0.08*MR1)\nliftoff=above1+S1\nresult = abs(float(MR1)-3.712)<0.01 and abs(float(MR2)-3.400)<0.01 and abs(float(liftoff)-41742)<50"},
 {"claim":"Q2 exit Mach 3.195 and area ratio ~8.88","code":"g=1.2; Pr=7.0e6/1.01e5\nMe2=(Pr**((g-1)/g)-1)/((g-1)/2); Me=sqrt(Me2)\nbr=(2/(g+1))*(1+(g-1)/2*Me2)\neps=(1/Me)*br**((g+1)/(2*(g-1)))\nresult = abs(float(Me)-3.195)<0.02 and abs(float(eps)-8.88)<0.15"},
 {"claim":"Q2 exit velocity ~2791 m/s","code":"g=1.2; Ru=8314; Tc=3400