INSTITUTE OF SPACE TECHNOLOGY, ISLAMABAD

DEPARTMENT OF AERONAUTICS AND ASTRONAUTICS

SPACE MISSION ANALYSIS AND DESIGN

CLASS ASSIGNMENT: 02

Prepared By: - Shakir Hussain Chaudhry

Reg No. 150512011 MS (AE-08)

Submitted To: - Dr Salimuddin Zahir

PoP Aerospace Engineering

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Q. No 1: A Two stage rocket used LOX and RP-1 in both stages. The details on this vehicle are as follows: - Stage 1: Stage gross mass 22,993 kg; empty mass 1505 kg; exhaust velocity 2940 m/sec. Stage 2: Stage gross mass 4177 kg; empty mass 350 kg; exhaust velocity 3107 m/sec, Payload: 500 kg (due East LEO) (a). Determine what is the ∆V for each stage and for the total vehicle? Solution: We can solve by using exhaust velocity as well as the Isp

First by using the given values of exhaust velocity, the forumula is: ∆V = Ve × lnm0

mf

for stage 1: m0 = 1st stage gross mass + 2nd stage gross mass + + Payload mass = 22,993 + 4177 + 500 = 27670 kg and mf = 1st stage empty mass = 1505 kg and Ve = 2940 m/sec Now putting values in above equation, we get: -

∆V = 2940 × ln27670

1505 = 8559.97 𝑚/ 𝑠𝑒𝑐 𝐴𝑛𝑠𝑤𝑒𝑟

for stage 2: m0 = 2nd stage gross mass + Payload mass = 4177 + 500 = 4677 kg and mf = 2nd stage empty mass = 350 kg and Ve = 3107 m/sec Now putting values in above equation, we get: -

∆V = 3107 × ln4677

350 = 8054.83 𝑚/ sec 𝐴𝑛𝑠𝑤𝑒𝑟

So the total ∆V becomes: -

∆𝑉𝑡 = 8559.97 + 8054.83 = 16614.8 m/ sec 𝐴𝑛𝑠𝑤𝑒𝑟

---------------------------------------------------------------------------------------------------------------------------------------

Now using the given values of Isp the formula is: - ∆V = g × Isp × lnmstart

mend

Since Isp = 310 sec for “LOX & RP-1”, and massses are known. Putting the values, we get:- for Stage 1:

∆V = 9.81 × 310 × ln27670

1505= 8854.33 𝑚/𝑠𝑒𝑐 𝐴𝑛𝑠𝑤𝑒𝑟

similarly for Stage 2:

∆V = 9.81 × 310 × ln4677

350= 7883.99 𝑚/𝑠𝑒𝑐 𝐴𝑛𝑠𝑤𝑒𝑟

So the total ∆V becomes: -

∆𝑉𝑡 = 8854.33 + 7883.99 = 16738.32 m/ sec 𝐴𝑛𝑠𝑤𝑒𝑟

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(b). Calculate what is the inert mass fraction δ for each stage?

Solution: 𝒊𝒏𝒆𝒓𝒕 𝒎𝒂𝒔𝒔 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏, 𝜹𝒊 = 𝑴𝒊𝒏𝒆𝒓𝒕

𝑴𝒕𝒐𝒕𝒂𝒍

for stage 1: mtotal = 1st stage gross mass + 2nd stage gross mass + Payload mass = 22,993 + 4177 + 500 = 27670 kg and minert = 1st stage emtpy mass = 1505 kg Now putting values in above equation, we get: -

δ1 = 1505

27670= 0.05439 Answer

for stage 2: mtotal = 2nd stage gross mass + Payload mass = 4177 + 500 = 4677 kg and minert = 2nd stage empty mass = 350 kg Now putting values in above equation, we get: -

δ2 = 350

4677= 0.07483 Answer

(e). Repeat all calculations for LOX and LH2 in both stages.

Solution.

∆V = g × Isp × lnminitial

mfinal

As we know that for LOX and LH2 the Isp is given as 455 seconds and g = 9.81 m/s2.

for first stage minitial = 27670 Kg & mfinal = 1505 Kg

Putting the values in above equation, we get,: -

∆V = 9.81 × 455 × ln27670

1505

∆V = 12995.87 m/sec Answer

Now for 2nd stage: minitial = 4677 Kg & mfinal = 350 Kg

Putting the values in above equation, we get: -

∆V = 9.81 × 455 × ln4677

350

∆V = 11571.66 m/sec Answer So the total ∆V become: -

∆𝑉𝑡 = 12995.87 + 11571.66 = 24567.53 m/ sec 𝐴𝑛𝑠𝑤𝑒𝑟

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(c). For a two-stage launch vehicle with the same values of δ and Ve for each stage, find the ∆V split which maximizes payload fraction λo. Compare the total ∆V you found in section – a.

Solution: 𝒑𝒂𝒚𝒍𝒐𝒂𝒅 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏, 𝝀 = 𝑷𝒂𝒚𝒍𝒐𝒂𝒅 𝒎𝒂𝒔𝒔

𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔=

𝟓𝟎𝟎

𝟐𝟕𝟔𝟕𝟎= 𝟎. 𝟎𝟏𝟖𝟏

(d). For a two-stage launch vehicle with the same values of δ and Ve for each stage, find the ∆V split which maximizes the ratio of payload to inert mass λ0/δ0.

Solution: For part c and d MATLAB is written down and their results are shown down by the following methodology:

1=Isp

2=Exhaust Velocity

1 CASE :

Stage 1 Isp

310

Stage 2 Isp

455

Payload Mass

500

Stage 1 Total Mass

22993

Stage 2 Total Mass

4177

Stage 1 Empty Mass

1505

Stage 2 Empty Mass

350

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CASE 2: Exhaust Velocity

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(f). Solve the related equations using MATLAB for both propellants and support your results through plots of lift-off-weight versus stage velocity. Clearly state all assumptions or approximations made for the computations, also give adequate references where necessary.

MATLAB CODE:

clc

choice1=input('Would you like to use "Isp" or "Exhaust Velocity" of System\n 1=Isp\n 2=Exhaust

Velocity \n');

if choice1==1

vexhaust_1=input('Stage 1 Isp\n')*9.8;

vexhaust_2=input('Stage 2 Isp\n')*9.8;

else

vexhaust_1=input('Stage 1 Exhaust Velocity\n');

vexhaust_2=input('Stage 2 Exhaust Velocity\n');

end

mass_payload=input('Payload Mass\n');

mtotal_1=input('Stage 1 Total Mass\n');

mtotal_2=input('Stage 2 Total Mass\n');

mempty_1=input('Stage 1 Empty Mass\n');

mempty_2=input('Stage 2 Empty Mass\n');

inertmassfraction_2=(mempty_2+mass_payload)/(mtotal_2+mass_payload);

inertmassfraction_1=(mtotal_2+mempty_1+mass_payload)/(mtotal_1+mtotal_2+mass_payload);

deltav_1=vexhaust_1*(log(1/inertmassfraction_1));

deltav_2=vexhaust_2*(log(1/inertmassfraction_2));

deltav_total=deltav_1+deltav_2;

ff1=0.1:0.00001:0.9;

for i=1:80001

f1 = ff1(i);

massprop_2(i)=((mass_payload)*(exp(((1-f1)*deltav_total)/vexhaust_2)-1)*(1-

inertmassfraction_2))/(((1-f1)*deltav_total)/vexhaust_2);

%massprop_2(i)=((mass_payload)*(exp(((1-f1)*deltav_total)/vexhaust_2) - 1)*(1-

inertmassfraction_2))/(1 - inertmassfraction_2*(deltav_2/vexhaust_2));

minert_2(i)=(inertmassfraction_2/(1-inertmassfraction_2))*(massprop_2(i)+mass_payload);

stagemass_2(i)=minert_2(i)+massprop_2(i);

massprop_1(i)=((mass_payload+stagemass_2(i))*(exp(((f1)*deltav_total)/vexhaust_2)-1)*(1-

inertmassfraction_2))/(((f1)*deltav_total)/vexhaust_2);

%massprop_1(i)=((mass_payload+stagemass_2(i))*((exp((f1*deltav_total)/vexhaust_1)) - 1)*(1-

inertmassfraction_1))/(1 - inertmassfraction_1*(deltav_1/vexhaust_1));

minert_1(i)=(inertmassfraction_1/(1-

inertmassfraction_1))*(massprop_1(i)+mass_payload+stagemass_2(i));

stagemass_1(i)=minert_1(i)+massprop_1(i);

masstotal(i)=stagemass_1(i)+stagemass_2(i)+ mass_payload;

payloadfraction(i)=mass_payload/(masstotal(i));

payloadtoinert(i)=mass_payload/(minert_2(i)+minert_1(i));

inertratio_2(i)=(minert_2(i)+mass_payload)/(stagemass_2(i)+mass_payload);

%inertratio_1(i)=(minert_1(i)+stagemass_2(i)+mass_payload)/(stagemass_2(i)+stagemass_1(i)+mass_payload);

end

[M,I] = max(payloadfraction);

ff1(I);

[N,J] = max(payloadtoinert);

SHAKIR_SMAD _ASSG_2

ff1(J);

[O,P] = min(masstotal);

ff1(P);

subplot(2,3,3)

plot (ff1,masstotal,'m - ','LineWidth',2)

hold on

plot(ff1(P),O,'^','markerfacecolor',[1 0 0]);

indexmin = find(min(masstotal) == masstotal);

xmin = ff1(indexmin);

ymin = masstotal(indexmin);

strmin = [' Minimum at ',num2str(xmin)];

text(xmin,ymin,strmin,'HorizontalAlignment','left');

ylabel('Total SLV Mass (Kg)')

xlabel('Fraction of Delta v Provided by Stage 1')

subplot(2,3,1)

plot (ff1,stagemass_1,'r - ','LineWidth',2)

ylabel('Stage 1 Mass (Kg)')

xlabel('Fraction of Delta v Provided by Stage 1')

subplot(2,3,2)

plot(ff1,stagemass_2,'b - ','LineWidth',2)

ylabel('Stage 2 Mass (Kg)')

xlabel('Fraction of Delta v Provided by Stage 1')

subplot(2,3,4)

plot(ff1,payloadfraction,'g - ','LineWidth',2)

hold on

plot(ff1(I),M,'^','markerfacecolor',[1 0 0]);

indexmax = find(max(payloadfraction) == payloadfraction);

xmax = ff1(indexmax);

ymax = payloadfraction(indexmax);

% strmax = ['Maximum = ',num2str(ymax) num2str(xmax)];

strmax = [' Maximum at ',num2str(xmax)];

text(xmax,ymax,strmax,'HorizontalAlignment','left');

ylabel('Payload Fraction ')

xlabel('Fraction of Delta v Provided by Stage 1')

subplot(2,3,5)

plot(ff1,payloadtoinert,'k - ','LineWidth',2)

hold on

plot(ff1(J),N,'^','markerfacecolor',[1 0 0]);

indexmax1 = find(max(payloadtoinert) == payloadtoinert);

xmax = ff1(indexmax1);

ymax = payloadtoinert(indexmax1);

strmax = [' Maximum at ',num2str(xmax)];

text(xmax,ymax,strmax,'HorizontalAlignment','left');

ylabel('Payload to Inert Mass Fraction ')

xlabel('Fraction of Delta v Provided by Stage 1')

% subplot(2,3,4)

% plot (ff1,(masstotal./max(masstotal)),'m - ','LineWidth',2)

% hold on

% plot (ff1,(stagemass_2./max(stagemass_2)),'m - ','LineWidth',2)

% hold on

% plot (ff1,(stagemass_1./max(stagemass_1)),'m - ','LineWidth',2)

% ylabel('Normalized Mass Scale ')

% xlabel('Fraction of Delta v Provided by Stage 1')