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AblationHeatShieldsduringSatelliteandMissilere-entrytotheEarth’sAtmosphere

Research·August2015

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KailashKumarJainMunoth

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Ablation Heat Shields during Satellite and

Missile re-entry to the Earth’s Atmosphere

By

Kailash Kumar Jain Munoth

Contents:

1. Introduction

2. Problem statement

3. Discretization for the first transient state

4. Results from the first transient state

5. Discretization for second transient state

6. Results for the second transient state

7. Cumulative result table

8. References

9. Appendix

Introduction

Ablation Heat shield

In spacecraft design, ablation is used to both cool and protect mechanical parts and/or payloads that

would otherwise be damaged by extremely high temperatures. Two principal applications are heat

shields for spacecraft entering a planetary atmosphere from space and cooling of rocket engine nozzles.

In a basic sense, ablative material is designed to slowly burn away in a controlled manner, so that heat

can be carried away from the spacecraft by the gases generated by the ablative process while the

remaining solid material insulates the craft from superheated gases.

Figure 1: Typical Ablation shield for space craft entering planetary atmosphere from space.

Problem statement

Aerodynamic heating is a major problem in satellite and missile re-entry to the earth’s atmosphere.

Ablation heat shields have proved successful in protecting surfaces. In this application the high rate of

heat transfer experienced at the surface first causes an initial transient temperature rise until the surface

reaches the melting point temperature, Tm. Ablation (melting of the surface) starts after this heat-up

period, and there follows a second short transient period after which a steady-state ablation velocity is

reached. The melted material is assumed to run-off immediately. We idealize the problem at the

stagnation point of a missile nose as a 1-D conduction problem with the origin x=0 at the liquid solid

interface. Then the wall appears moving to the left at the ablation velocity u. The initial transient before

the surface reaches Tm is governed by

1

𝛼

𝜕𝑇

𝜕𝑡=

𝜕2𝑇

𝜕𝑥2

And during the second transient

ρc∂T

∂t+ uρc

∂T

∂x=

∂

∂x(k

∂T

∂x)

Where,

(q

A)

0= 106

Btu

ft2. hr

c = 0.3 Btu

lbm. F , k = 0.5

Btu

hr. ft. F , ρ = 100

lbm

ft3 , F = 4,000

Btu

lbm,

Tm = 3,0000F

Where F is the heat of ablation (Btu/lbm of material). An energy balance requires the following relation

to hold:

(𝑞

𝐴)

0+ 𝜌𝑢𝐹 = −𝑘 (

𝜕𝑇

𝜕𝑥)

0

Let L=3 in, T(x,0)= 600F & T(L,t)= 600F.

Results for part one of the project

I have tried both the Implicit method of discretization and Crank-Nicolson method for the first transient

stage. They both converge to the same result for time taken to reach melting point i.e 0.3669 sec,

accurate to 4-decimal places (verified from the analytical solution). This means the ablation shield’s

surface took 0.3669 sec to reach its melting point from the point of re-entry into the atmosphere. The

code for both the methods is attached in the appendix.

Table 1: First transient stage results using Implicit and Crank-Nicolson methods

Sl.No Name of the scheme used

No. of Equations in each time step

dt Dx Time taken to reach melting point(Seconds)

1. Implicit Scheme 2500 0.0001 0.0001 0.3669

2. Crank Nicolson 2500 0.0001 0.0001 0.3669

Table 2: Implicit method with different time step and ‘dx’.

Sl.no Number of equations dt dx Time taken to reach melting point (seconds)

1. 2500 0.001 0.0001 0.3680

2. 250 0.001 0.001 0.3960

3. 25 0.001 0.01 1.7134

4. 25 0.00001 0.01 1.7134

Table 3: Crank-Nicolson Method with different time steps and 'dx'.

Sl.no Number of equations dt dx Time taken to reach melting point (seconds)

1. 2500 0.001 0.0001 0.367

2. 250 0.001 0.001 0.3950

3. 25 0.001 0.01 1.7134

4. 25 0.0001 0.001 0.3949

Results for part 2 of the project:

Assumption: As the melting point for the material is reached, the oncoming gases washes away the

metal completely without forming a film on the shield.

The second transient part has a flow term which has to be dealt with carefully. I have taken the positive

direction of velocity in +ve x-direction and then have applied Crank-Nicolson method for discretization.

Even Implicit method can be applied in similar fashion as part one.

The steady state velocity is 2.05 ft/hour.

Amount of Material ablated during second transient = 0.027 inches.

Amount of material ablated over next 20 seconds=0.13666668 inches

Cumulative Result table:

Table 4: Results obtained from the entire project analysis

Sl.no Questions needed to be answered Result

1. How long it takes to reach Tm ? 0.3669 seconds

2. What is the steady state velocity uss ? 2.05 ft/hr ~0.0006ft/s

3. How much material was ablated during second transient before steady state is reached?

0.00225 feet ~0.027 inches

4. How much material is ablated over next 20 seconds i.e after steady state is reached ?

0.01138889 feet~0.13666668 inches

5. Time taken to reach steady state approx.? 2 seconds

References:

1. Course notes for Computational Heat Transfer by Dr. George Gogos.

2. Atmospheric Entry - http://en.wikipedia.org/wiki/Atmospheric_entry

3. Analysis of Galileo Probe Heatshield Ablation and Temperature Data F. S. Milos; Y.-K. Chen; T. H.

Squire; R. A. Brewer.

Appendix

Code for First Transient state using the Crank-Nicolson Method:

clear all;%Crank Nicolson Method k=0.5/3600;%Thermal Conductivity cp=0.3; density=100; f=4000; tm=3000;%Melting temperature alpha=k/(density*cp);%Thermal Diffusivity l=0.25;%Length of the Ablation Shield in feets dx=0.0001; dt=10^-4;%Time step m=l/dx; A=zeros(m,m); x=zeros(m,1); c=zeros(m,1); theta=zeros(m,367);%Temp's saved at each time step until tm is reached lambda=alpha*dt/dx^2; x(:,1)=60;%Intial temp. a=4*lambda*dx*(10^6)/(k*3600); j=1; for dt=10^-4:10^-4:2 if x(:,1)<=3000 dt A(1,1)=2+2*lambda; A(1,2)=-2*lambda; A(m-1,m-2)=-lambda; A(m-1,m-1)=2+2*lambda; A(m-1,m)=-lambda; A(m,m-1)=-lambda; A(m,m)=2+2*lambda;%Assumption made is tm+1=tm c(1,1)=(2-2*lambda)*x(1,1)+2*lambda*x(2,1)+a; c(m-1,1)=lambda*x(m-2,1)+(2-2*lambda)*x(m-1,1)+lambda*x(m,1); c(m,1)=lambda*x(m-1,1)+(2-2*lambda)*x(m,1)+2*lambda*x(m,1); for i=2:m-2 A(i,i-1)=-lambda; A(i,i)=2+2*lambda; A(i,i+1)=-lambda; end for i=2:m-2 c(i,1)=lambda*x(i-1,1)+(2-2*lambda)*x(i,1)+lambda*x(i+1,1); end x=A\c; x(m,1)=60; theta(:,j)=x(:,1); j=j+1; end plot(0.0001:0.0001:0.25,theta(:,j-1))

hold on xlabel('Position x in feet') ylabel('Temperature T in Fahrenheit') title('Crank Nicholson Method') end

Code for the first Transient state of the project using Implicit Method: clear all; %Implicit Method k=0.5/3600; cp=0.3; density=100; f=4000; tm=3000; alpha=k/(density*cp); l=0.25; dx=0.0001; dt=10^-4; m=l/dx; A=zeros(m,m); x=zeros(m,1); c=zeros(m,1); theta=zeros(m,368); lambda=alpha*dt/dx^2; x(:,1)=60; a=-2*dx*(10^6)/(k*3600); j=1; for dt=10^-4:10^-4:0.4 if x(:,1)<=3000 dt A(1,1)=-2-1/lambda; A(1,2)=2; A(m,m-1)=1; A(m,m)=-2-1/lambda; c(1,1)=a-x(1,1)/lambda; c(m-1,1)=-x(m-1,1)*1/lambda; c(m,1)=lambda*x(m-1,1)+(2-2*lambda)*x(m,1); for i=2:m-1 A(i,i-1)=1; A(i,i)=-2-1/lambda; A(i,i+1)=1; end for i=2:m-2 c(i,1)=(-1/lambda)*x(i,1); end x=A\c; x(m,1)=60; theta(:,j)=x(:,1); j=j+1; end plot(0.0001:0.0001:0.01,x(1:100,1))

hold on xlabel('Position x in feet') ylabel('Temperature T in fahrenheit') title('implicit Method') end

Code for the Second Transient State of the project: clear all;%Crank-Nicolson Method was used for discretization k=0.5/3600;%Thermal Conductivity cp=0.3;%Specific heat hf=10^6/3600;%heat flux density=100; f=4000; tm=3000;%Melting temperature alpha=k/(density*cp);%Thermal Diffusivity dt=10^-4;%Time step l=0.25; dx=0.0001; load ('x_at_melting_point.mat'); x=x(2:2500);%Temperaure values from the first part of project x(2500,1)=60; m=l/dx; A=zeros(m,m); c=zeros(m,1); theta=zeros(m,196331); theta(:,1)=x(:,1); u=zeros(196331,1); n=1; j=1; for dt=0.3669:10^-4:20 lambda=alpha*10^-4/(2*dx^2); r=u(n,1)*10^-4/(4*dx); A(1,1)=1+2*lambda; A(1,2)=-lambda-r; A(m,m-1)=-lambda+r; A(m,m)=1+2*lambda; c(1,1)=(lambda-r)*6000+(1-2*lambda)*x(1,1)+(lambda+r)*x(2,1); c(m,1)=(lambda-r)*x(m-2,1)+(1-2*lambda)*x(m-1,1)+(lambda+r)*2*x(m,1); for i=2:m-1 A(i,i-1)=-lambda+r; A(i,i)=1+2*lambda; A(i,i+1)=-lambda-r; end for i=2:m-1 c(i,1)=(lambda-r)*x(i-1,1)+(1-2*lambda)*x(i,1)+(lambda+r)*x(i+1,1); end x=A\c; x(m,1)=60; n=n+1; u(n,1)=(k/(2*dx*density*f))*((-4*dx*hf/k)-(2*dx*density*f*u(n-1,1)/k)+4*lambda*3000/(lambda-r)+(theta(1,j)+x(1,1))*2*lambda/(r-lambda));

l=l-abs(u(n,1)*10^-4);%Length of the Ablation Shield in feets if dt==0.3669 || 1 || 1.5 || 2 || 2.5 || 3 || 3.5 || 4 y=(0.25-l:dx:0.25); plot(y,x) hold on xlabel('Length of the plate') ylabel('Temperature of the plate') title('tempearture variation in the plate thickness') end j=j+1; dx=l/m; theta(:,j)=x(:,1); end plot(0.3668:0.0001:20,abs(u)) xlabel('time in sec (origin indicates the start of melting)') ylabel('velocity in ft/s') title('Ablation velocity Vs time')

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