Question 11.7: Find the optimal mass for a three-stage launch vehicle that ...
Find the optimal mass for a three-stage launch vehicle that is required to lift a 5000 kg payload to a speed of 10 km/s. For each stage, we are given that
Stage 1 I_{s p 1}=400 s \left(c_{1}=3.924 km / s \right) \quad \varepsilon_{1}=0.10
Stage 2 I_{s p 2}=350 s \left(c_{2}=3.434 km / s \right) \quad \varepsilon_{2}=0.15
Stage 3 I_{s p 3}=300 s \left(c_{3}=2.943 km / s \right) \quad \varepsilon_{3}=0.20
Substituting this data into Equation 11.86, we get
\sum\limits_{i=1}^{N} c_{i} \ln \left(c_{i} \eta-1\right)-\ln \eta \sum\limits_{i=1}^{N} c_{i}-\sum\limits_{i=1}^{N} c_{i} \ln c_{i} \varepsilon_{i}=v_{b o} (11.86)
3.924 ln (3.924 η – 1) + 3.434 ln (3.434 η – 1) + 2.943 ln (2.943 η – 1) – 10.30 ln η + 7.5089 = 10
As can be checked by substitution, the iterative solution of this equation is
η = 0.4668
Substituting η into Equations 11.87 yields the optimum mass ratios,
n_{i}=\frac{c_{i} \eta-1}{c_{i} \varepsilon_{i} \eta}, i=1,2, \ldots, N (11.87)
n_{1}=4.541 \quad n_{2}=2.507 \quad n_{3}=1.361For the step masses, we appeal to Equations 11.88 to obtain
\begin{aligned}&m_{N}=\frac{n_{N}-1}{1-n_{N} \varepsilon_{N}} m_{P L} \\&m_{N-1}=\frac{n_{N-1}-1}{1-n_{N-1} \varepsilon_{N-1}}\left(m_{N}+m_{P L}\right) \\&m_{N-2}=\frac{n_{N-2}-1}{1-n_{N-2} \varepsilon_{N-2}}\left(m_{N-1}+m_{N}+m_{P L}\right) \\&\vdots \\&m_{1}=\frac{n_{1}-1}{1-n_{1} \varepsilon_{1}}\left(m_{2}+m_{3}+\ldots m_{P L}\right)\end{aligned} (11.88)
m_{1}=165,700 kg \quad m_{2}=18,070 kg \quad m_{3}=2477 kgThe total mass of the vehicle is
m_{o}=m_{1}+m_{2}+m_{3}+m_{P L}=191,200 kgUsing Equations 11.89 and 11.90, the empty masses and propellant masses are found to be
m_{E i}=\varepsilon_{i} m_{i} (11.89)
m_{p i}=m_{i}-m_{E i} (11.90)
m_{E 1}=16,570 kg \quad m_{E 2}=2710 k \quad m_{E 3}=495.4 kg
m_{p 1}=149,100 kg \quad m_{p 2}=15,360 kg \quad m_{p 3}=1982 kg
The payload ratios for each stage are
\lambda_{1}=\frac{m_{2}+m_{3}+m_{P L}}{m_{1}}=0.1542
\lambda_{2}=\frac{m_{3}+m_{P L}}{m_{2}}=0.4139
\lambda_{3}=\frac{m_{P L}}{m_{3}}=2.018
The overall payload fraction is
\pi_{ PL }=\frac{m_{P L}}{m_{o}}=\frac{5000}{191,200}=0.0262Finally, let us check Equation 11.93,
\eta c_{i}\left(\varepsilon_{i} n_{i}-1\right)^{2}+2 \varepsilon_{i} n_{i}-1>0, \quad i=1, \ldots, N (11.93)
\eta c_{1}\left(\varepsilon_{1} n_{1}-1\right)^{2}+2 \varepsilon_{1} n_{1}-1=0.4541
\eta c_{2}\left(\varepsilon_{2} n_{2}-1\right)^{2}+2 \varepsilon_{2} n_{2}-1=0.3761
\eta c_{3}\left(\varepsilon_{3} n_{3}-1\right)^{2}+2 \varepsilon_{3} n_{3}-1=0.2721
A positive number in every instance means we have indeed found a local minimum of the function in Equation 11.85.
h=\sum\limits_{i=1}^{N}\left[\ln \left(1-\varepsilon_{i}\right)+\ln n_{i}-\ln \left(1-\varepsilon_{i} n_{i}\right)\right]-\eta\left\lgroup v_{b o}-\sum\limits_{i=1}^{N} c_{i} \ln n_{i} \right\rgroup (11.85)