The following data is given mPL = 10 000 kg πPL = 0.05 ε = 0.15 (a) Isp = 350 sg0 = 0.00981 km/s² Calculate the payload velocity vbo at burnout, the empty mass of the launch vehicle and the propellant mass for (a) a single stage and (b) a restricted, two-stage vehicle.

Question

The following data is given
[latex]m_{PL}=10000 kg[/latex]
[latex]\pi_{PL}=0.05 [/latex]
[latex]\varepsilon=0.15[/latex] (a)
[latex]I_{sp}=350 s[/latex]
[latex]g_{0}=0.00981 km/s^{2}[/latex]
Calculate the payload velocity [latex]v_{bo}[/latex] at burnout, the empty mass of the launch vehicle and the propellant mass for (a) a single stage and (b) a restricted, two stage vehicle.

Accepted Answer

(a) From Equation 11.43 we find
[latex]v_{bo}=I_{sp}g_{0}\ln \frac{1}{\pi _{PL}(1-\varepsilon)+\varepsilon}[/latex] (11.43)
[latex]v_{bo} = 350 · 0.00981\ln \frac{1}{0.05(1 + 0.15) + 0.15}=\underline{5.657 km/s}[/latex]
Equation 11.40 yields the gross mass
[latex]\pi _{PL}=\frac{m_{PL}}{m_{0}}[/latex] (11.40)
[latex]m_{0}=\frac{10000}{0.05}=200000 kg[/latex]
from which we obtain the empty mass using Equation 11.34,
[latex]\varepsilon=\frac{m_{E}}{m_{E}+m_{P}}=\frac{m_{E}}{m_{0}-m_{PL}}[/latex] (11.34)
[latex]m_{E} = \varepsilon(m_{0} − m_{PL}) = 0.15(200 000 − 10 000) =\underline{28 500 kg}[/latex]
The mass of propellant is
[latex]m_{p} = m_{0} − m_{E} − m_{PL} = 200 000 − 28 500 − 10 000 =\underline{161 500 kg}[/latex]
(b) For a restricted two-stage vehicle, the burnout speed is given by Equation 11.50,
[latex]v_{bo_{2-stage}}= I_{sp}g_{0} \ln\left[\frac{1}{\pi _{PL}^{\frac{1}{2}}(1-\varepsilon )+\varepsilon }\right]^{2}[/latex] (11.50)
[latex]v_{bo_{2-stage}}= 350 · 0.00981 \ln\left[\frac{1}{0.05^{\frac{1}{2} }(1-0.15)+0.15} \right]^{2}=\underline{7.407 km/s}[/latex]
The empty mass of each stage is found using Equations 11.51,
[latex]\begin{matrix}m_{E_{1}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)^{\varepsilon}}{\pi _{PL}}m_{PL} & m_{E_{2}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)^{\varepsilon}}{\pi_{PL}^{\frac{1}{2}}}m_{PL} \end{matrix}[/latex] (11.51)
[latex]m_{E_{1}}=\frac{(1-0.05^{\frac{1}{2}})· 0.15}{0.15}· 10 000 =\underline{23 292 kg}[/latex]
[latex]m_{E_{2}}=\frac{(1-0.05^{\frac{1}{2}})· 0.15}{0.15^{\frac{1}{2}}}· 10 000 =\underline{5208 kg} [/latex]
For the propellant masses, we turn to Equations 11.53
[latex]\begin{matrix}m_{P_{1}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)(1-\varepsilon)}{\pi _{PL}}m_{PL} & m_{P_{2}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)(1-\varepsilon)}{\pi_{PL}^{\frac{1}{2}}}m_{PL} \end{matrix}[/latex] (11.53)
[latex]m_{p_{1}} =\frac{(1-0.05^{\frac{1}{2}})·(1- 0.15)}{0.15}· 10 000 =\underline{131990 kg}[/latex]
[latex]m_{p_{1}} =\frac{(1-0.05^{\frac{1}{2}})·(1- 0.15)}{0.15^{\frac{1}{2}}}· 10 000 =\underline{29513 kg}[/latex]
The total empty mass, [latex] m_{E} = m_{E_{1}} + m_{E_{2}}[/latex], and the total propellant mass, [latex]m_{p} = m_{p_{1}}+m_{p_{2}}[/latex], are the same as for the single stage rocket. The mass of the second stage, including the payload, is 22.4 percent of the total vehicle mass.

Question 11.2: The following data is given mPL = 10 000 kg πPL = 0.05 ε = 0...

Related Answered Questions

(a) Find the extrema of the function z = −x² − y². (b) Find the extrema of the same function under the constraint y = 2x + 3. ...

Verified Answer:

A sounding rocket of initial mass m0 and mass mf after all propellant is consumed is launched vertically (γ = 90°). The propellant mass flow rate me is constant. Neglecting drag and the variation of gravity with altitude, calculate the maximum height h attained by the rocket. ...

Verified Answer:

Repeat Example 11.2 for the restricted three-stage launch vehicle. ...

Verified Answer:

Verified Answer: