Question 11.5: Find the optimal mass for a three-stage launch vehicle which...
Find the optimal mass for a three-stage launch vehicle which 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_{sp_{1}} = 400 s (c_{1} = 3.924 km/s)\varepsilon_{1} = 0.10
Stage 2 I_{sp_{2}} = 350 s (c_{2} = 3.434 km/s) \varepsilon_{2} = 0.15
Stage 3 I_{sp_{3}} = 300 s (c_{3} = 2.943 km/s)\varepsilon_{3} = 0.20
Substituting this data into Equation 11.86, we get
\sum\limits_{i=1}^{N} c_{i}\ln (c_{i}\eta -1)-\ln \eta \sum\limits_{i=1}^{N}c_{i}-\sum\limits_{i=1}^{N}c_{i}\ln c_{i}\varepsilon _{i}=v_{bo} (11.86)
3.924 \ln(3.924η − 1) + 3.434 \ln(3.434η − 1) + 2.943 \ln(2.943η − 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,
\begin{matrix}n_{i}=\frac{c_{i}\eta -1}{c_{i}\varepsilon _{i}\eta }, &i = 1, 2, … , N\end{matrix} (11.87)
\begin{matrix} n_{1} = 4.541&n_{2} = 2.507 & n_{3} = 1.361 \end{matrix}
For the step masses, we appeal to Equations 11.88 to obtain
m_{N-1}=\frac{n_{N-1}-1}{1-n_{N-1}\varepsilon _{N-1}}(m_{N}+m_{PL})
m_{N-2}=\frac{n_{N-2}-1}{1-n_{N-2}\varepsilon _{N-2}}(m_{N-1}+m_{N}+m_{PL}) (11.88)
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m_{1}=\frac{n_{1}-1}{1-n_{1}\varepsilon _{1}}(m_{2}+m_{3}+ \cdots m_{PL})
\begin{matrix}m_{1} = 165 700 kg &m_{2} = 18 070 kg &m_{3} = 2477 kg\end{matrix}
Using Equations 11.89 and 11.90, the empty masses and propellant masses are found to be
m_{E_{i}} = ε_{i}m_{i} (11.89)
m_{p_{i}} = m_{i} − m_{E_{i}} (11.90)
The payload ratios for each stage are
\lambda_{1}=\frac{m_{2}+m_{3}+m_{PL}}{m_{1}}=0.1542
\lambda_{2}=\frac{m_{3}+m_{PL}}{m_{2}}=0.4139
\lambda_{3}=\frac{m_{PL}}{m_{3}}=2.018
The total mass of the vehicle is
m_{0} = m_{1} + m_{2} + m_{3} + m_{PL} = 191 200 kg
and the overall payload fraction is
π_{PL} =\frac{m_{PL}}{m_{0}}=\frac{5000}{191200}=0.0262
Finally, let us check Equation 11.93,
ηc_{i}(\varepsilon_{i}n_{i} − 1)^{2} + 2\varepsilon_{i}n_{i} − 1 > 0, i = 1, … , N (11.93)
η_{c_{1}}(\varepsilon_{1}n_{1} − 1)^{2} + 2\varepsilon_{1}n_{1} − 1 = 0.4541
η_{c_{2}}(\varepsilon_{2}n_{2} − 1)^{2} + 2\varepsilon_{2}n_{2} − 1= 0.3761
η_{c_{3}}(\varepsilon_{3}n_{3} − 1)^{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}[\ln (1-\varepsilon_{i})+\ln n_{i}-\ln (1-\varepsilon _{in_{i}})]-\eta\left(v_{bo}-\sum\limits_{i=1}^{N}c_{i}\ln n_{i}\right) (11.85)