Utilization of a Numerical Method to Solve for One-Dimensional Multi-Layer Heat Transfer

Introduction:

NASA’s space program, although not as illustrious as in its past, still captures the attention of Americans and the rest of the world as well. Something about exploration connects all of us under a universal search of the unknown; however, this curiosity could not be quenched without the space shuttle program. The shuttles break past Earth’s gravity, entering the unthinkable cold of space. Then, like balls of fire, they burst back into Earth’s atmosphere. Collision with gas molecules causes tremendous amounts of heat due to friction and could cause the entire shuttle and its passengers to be destroyed. Fortunately, the aluminum structure of the space shuttles is protected from this catastrophe by a thin sheet of insulation. The goal of my senior project was to determine the thickness of the heat shield on a lifting body, like the space shuttle, in order to create the most efficient and safe missions through Earth’s atmosphere. Creating a python program, I would design a code that solves for heat transfer through the rigid ceramic tile and the minimum thickness necessary at varying locations along the windward side of the lifting body during reentry into Earth’s atmosphere. This insulation is the rigid ceramic tile used on the space shuttles.

The following diagram and equation represent the conservation of energy equation for one dimensional heat transfer.

Figure 1. Control Volume for one-dimensional conduction in rectangular coordinates. (Ref. 1)

1

−KA ∂T∂x

( x )+qG' ' ' A∆ x=−KA ∂T

∂ x( x+∆ x )+ ρcA∆ x ∂T

∂t

K: Thermal Conductivity ρ: Density t: Time

A: Area c: Specific Heat qG'' ': Internal Energy Generation

x: spatial location T: Temperature

When the limit is taken as ∆ x→0, the combination of the first term on the left-hand side and the first term on the right-hand side of the equation becomes the definition of the second derivative of temperature with respect to x. For one-dimensional problems, this equation simplifies down to:

ρ c p∂T∂ t

=K ∂2T∂ x2 +qG

' ' '

Ignoring internal energy generation, when the energy balance is performed on macroscopic scale, this conservation of energy equation is derived:

ρc ∆xT it+∆t−T i

t

∆t=K

T i+1t −T i

t

∆ x+K

T i−1t −T i

t

∆ x

Superscript “t” denotes time increment. Subscript “i” denotes spatial location along the material.

It is possible to solve similar problems analytically with infinite series; however, an analytical solution can only be found in the case of constant properties (K, c) and simplistic boundary conditions. In the case of my problem, I decided to use the explicit time marching technique. This method subdivides a solid and places nodes at the center of each section of material. An energy balance is performed on each node which produces an algebraic equation for the temperature of each node in terms of the adjacent nodal temperatures and thermal properties of the solid. In this manner, the partial differential equation is reduced to simplistic algebraic equations at the nodes. By creating a variable for heat flux, heat is introduced to the material. The variable for thermal conductivity allows heat to be carried through the entire material based on the energy balance equations. To maintain simplicity, it is assumed that a single layer of adhesive is bonding the insulation to the aluminum rather than a layer of strain isolation pad, which compensates for the differences in thermal expansion between the aluminum and insulation, with adhesive. Also, the thermal properties of the adhesive are ignored in this instance.

2

Rate of energy conducted into control volume

Rate of energy stored inside control volume

Rate of energy generated inside control volume

Rate of energy conducted out of control volume

Now I will derive all of the basic equations that will be used to model one-dimensional heat transfer inside of my python code. Beginning with the internal nodal balance, the above equation can be simplified to the form given below:

T 0t+∆t=T0

t + 2K ∆ t(∆ x )2 ρ cp

(T i−1t −2T i

t+T i+1t )

The basic nodal balance for the top boundary condition is:

KAT 1t−T 0

t

∆x +q} ¿ εσA (T ∞

t 4

−T0t 4

) = ρA ∆ xC p

2T0t+∆t−T 0

t

∆ t

ε : Emissivity q} :Heat Flux ( {W} over { {m} ^ {2} } ¿

σ :Stefan Boltzmann Constant (5.57*10−8 Wm2K4 )

I can create an algebraic equation within a for loop inside a python code that will calculate the temperature of node 0 at every time step.

T 0t+∆t=T0

t + 2K ∆ t(∆ x )2 ρ cp

(T1t−T0

t )+ 2∆t∆ xρ cp

¿

For the bottom boundary condition, (the final node (M2)) there is no heat flux or radiation since this node is not exposed to the atmosphere/vacuum. The equations that represents the bottom boundary condition is:

TM 2t+∆t=TM 2

t + 2K ∆t(∆x )2 ρ c p

(TM 2−1t −TM 2

t )

Also, the properties (K, c) vary with temperature, so thermal conductivity and specific heat have to be contained inside arrays. From this equation at node 0, and the assumption that the solid is adiabatic at the 2nd boundary condition, I was able to create all of the algebraic equations that would calculate the nodal temperatures at every time step of the thirty-minute run. One last assumption is that there is no internal heat generation.

3

Represents radiation, which is to the fourth power in terms of temperature.

Represents the heat that is entering the solid.

Figure 2. Space shuttle descending through Earth’s atmosphere. (Ref. 2)

My task was to create a code that modeled a one-layer substance of exposed insulation covering a thin piece of aluminum. Everything started at 20 degrees Celsius, and the back-face boundary condition was adiabatic. This means that there is no heat loss from the back side, which is a conservative assumption. A varying heat flux was created as a series of linear piece-wise functions of time to model the effect of a lift body that is experiencing reentry into Earth’s atmosphere as depicted in Figure 2. Also, since the thermal conductivity of the insulation is dependent upon atmospheric pressure to a significant degree, a variable was created to model the increase of pressure over time. For the outer facing piece of insulation, the equation had to let the substance radiate to another distant body of matter, and this was also dependent upon the temperature of said distant body of matter.

This section calculates all of the values necessary for the insulation material ( if i >= 0 and i <= M2-2:).:

T avg is used to calculate thermal conductivity and specific heat below:

T avg=T i+1t +T i

t

2

The thermal conductivity of the insulation is modeled by the following equation:

K it=1.064∗10−2−3.636∗10−6∗T avg+5.098∗10−8∗Tavg

2 +1.0242∗10−11∗T avg3 +KgasC1

4

The windward side of a lift body experiences most of the heat during reentry to Earth’s atmosphere.

The first four terms represent the thermal conductivity of the insulation at vacuum (Ref. 3), and the trend that this equation creates is shown in Figure 3. K varies with temperature and pressure due to the large amount of void space between the fibers. This allows a large amount of gas conduction to take

place, which varies with altitude. At high altitude, where there is near zero gas molecules, there is no gas conduction. At sea level (P = 101,325 Pa) there is full blown gas conduction.

Figure 3. Thermal conductivity of insulation versus temperature in vacuum.

The variation of gas conductivity for air with temperature is shown below and in Figure 4:

K gas=−1.8763∗10−3+1.218∗10−4∗T avg−1.328∗10−7∗T avg2 +1.479∗10−10∗T avg

3 −8.58∗10−14∗T avg4 +1.705∗10−17∗T avg

5 +9.527∗10−22∗T avg6

Figure 4. Thermal conductivity of air versus temperature.

5

200 300 400 500 600 700 800 900 1000 11000

0.02

0.04

0.06

0.08

0.1

0.12

Temperature (K)

Ther

mal

Con

ducti

vity

250 415 580 745 910 1075 1240 1405 1570 1735 19000

0.05

0.1

0.15

0.2

0.25

0.3

Temperature (K)

Ther

mal

Cond

uctiv

ity (W

/(mK)

)

100 450 800 1150 15000

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

Gas Conduction Versus Temperature

Temperature (K)

Gas C

ondu

ction

(W/(

mK)

)

The equation for the specific heat of the insulation is shown below and in Figure 5:

c it=1216(1−e3.233∗10−3∗Tavg)

The if statement designates every material space as insulation except for the final space which has been defined below as aluminum.T avgwas calculated as an average temperature between two adjacent nodes.K gas is calculated here as a function of temperature. The e, that is used to calculate the specific heat of the insulation, simply stands for Euler’s number.

Figure 5. Specific heat of Insulation versus temperature.

The overarching loop of my program also contains parameters for heat flux and the C1-value that represents the coefficient to be multiplied by thermal conductivity with respect to pressure (K gas). The variation of C1 with time is shown in Figure 6. Code 1 in the Appendix was used to create C1 data shown in Figure 6.

6

250 350 450 550 650 750 850 950 1050600

700

800

900

1000

1100

1200

Temperature (K)

Spec

ific H

eat

0 200 400 600 800 1000 1200 1400 1600 1800 20000

200

400

600

800

1000

1200

1400

Specific Heat of Insulation Versus Temperature

Temperature (K)

Spec

iific H

eat (

J/(K

gK))

Figure 6. Variation of C1 versus time forK gas.

For the first five minutes, C1 would equal zero, so the coefficient would cause the value of Kgas at any given time before 300 seconds to be negligible. Between five minutes and 20 minutes, the coefficient rises at a constant slope to become a solid value of one. Thus, for any time past 1200 seconds, the full computed value of Kgas would be added to the thermal conductivity of the insulation. Kgas is never added to the aluminum since the aluminum is a solid material and is not porous. Aluminum has no void space, so its thermal conductivity is not a function of pressure. In terms of the shuttle, the first 5 minutes represents the time spent in high altitude upon entering the atmosphere. The next fifteen minutes models the descent of the shuttle from high atmosphere to ground level. The following ten minutes represents the space shuttle resting at its landing sight.

7

0 300 600 900 1200 1500 18000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

C1

0 200 400 600 800 1000 1200 1400 1600 18000

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

Time (s)

Hea

t Flu

x (W

/m^2

)

Figure 7. Heat flux versus time.

With respect to heat flux, q begins at zero and rises to the designated value of maximum heat flux (qmax

¿ ) in the first five minutes. In the case of my program, qmax¿ was 200,000 W/m^2. Between five

minutes and fifteen minutes, heat flux remains at the designated value ofqmax¿ . For the following five

minutes, q decreases back down to a value of zero. Then, for any time after twenty minutes, heat flux remains at a value of zero. This is a generic reentry profile for a lifting body such as a space shuttle, and this is illustrated in Figure 7. Code 2 in the appendix created the data used for figure 7. For the more exterior nodes, heats radiates out; however, heat continues to gradually flow from the central nodes to the interior nodes. This would eventually stop, and all of the heat would exit the system after a certain period of time, but the time array is only set for a length of 1800 seconds. In terms of the shuttle, it is experiencing a gradual increase in heating as it descends through the atmosphere for the first five minutes. Then, for the following fifteen minutes, heat flux remains at qmax

¿ since the shuttle is experiencing maximum heating as it is descending. It then cools off as it slows down in the low-Earth atmosphere for five minutes. The shuttle reaches its landing point at twenty minutes and no longer receives any heat flux.

Now, I am going to list the energy balance equations in terms of non-constant properties:

Top boundary condition:

T 0t+∆t=T0

t + 2∆t∆ x0 ρ0 c0

¿

The T 0t+∆trepresents the new temperature of the original temperature array. The T 0

t signifies the original temperature at the same nodal location.∆ t stands for time increment while ∆X stands for spatial distance between nodes. K 0 represents the thermal conductivity. ρ0stands for density, and c0

stand for specific heat. The q} ¿represents heat flux while the εσ represents the emissivity multiplied by the Stefan Boltzmann constant. This number is multiplied by the current temperature to the fourth power of the original array subtracted by the temperature of the distant body to the fourth power which the outer insulation is radiating heat to, in this case the surface of the Earth. So if the far away body is at the same temperature as the outer insulation, there is no net radiation heat exchange.

Internal Nodes:

For i in range (1, M2):

T it+∆t=T i

t+ 2∆ t∆ xi−1 ρi−1c i−1+∆ x i ρic i

(K i−1

∆x i−1(T i−1

t −T it )+ K iDx i

(T i+1t −T i

t ))

This for loop covers all internal nodes of the temperature array. In this case there is no radiation or heat flux from the environment since it is covered by the material itself on both sides. For internal nodes, the density, specific heat, spatial increment, and thermal conductivity have to be taken into account for both the previous half spatial position and the next spatial position because one node could

8 i-1

be resting between two different materials with different properties. This concept is shown below in Figure 8.

Figure 8. Illustration of nodal placements and variable ranges.

Bottom boundary condition:

TM 2t+∆t=TM 2

t + 2∆ t∆xM 2−1 ρM 2−1 cM 2−1

(KM 2−1

∆xM 2−1(TM 2−1

t −TM 2t ))

For the final node (M2) in any situation of this adiabatic set condition, the heat can no longer flow forward; however the heat can travel back out of the substance if the exterior nodes become cooler than the interior nodes. This case doesn’t have to take into account the possibility of different properties since the final node only touches one type of material.

All of these equations are contained within a larger for loop that runs for the designated time in seconds divided by the time step. For instance, in my code, the designed actual time length went to a full thirty minutes, or 1800 seconds. At a time step of 0.05 seconds per iteration in the overarching loop, the overarching loop iterated thirty-six-thousand times. Pt.3 in the appendix describes the range for loops.

The following section calculates all of the values necessary for the aluminum material ( if i == M2-1:):

T avg=T i+1t +T i

t

2

The equation for the thermal conductivity of aluminum is shown below (this K value does not involve gas conduction since there is no void space in the aluminum.) The data created from the equation is shown in Figure 9:

K it=70.846+0.369∗T avg−3.1∗10−4∗T avg

2

9

ρi i

i+1

ρi−1

Figure 9. Thermal conductivity of aluminum versus temperature.

The equation for the specific heat of aluminum is shown below, and the data created from the equation is shown in Figure 10:

c it=646.64+0.6585∗T avg

10

290 300 310 320 330 340 350 360146

148

150

152

154

156

158

160

162

164

Temperature (K)

Ther

mal

Con

ducti

vty

Figure 10. Specific Heat of aluminum versus temperature.

The if statement tells the code, that if the final node is being iterated, then use the specific heat and thermal conductivity values for aluminum.

Figure 11. Rigid ceramic tile used on shuttles. (Ref. 4)

11

This insulation is connected to the thin sheet of aluminum via the adhesive.

290 300 310 320 330 340 350 360820

830

840

850

860

870

880

890

900

Temperature (K)

Spec

ific H

eat

Figure 11 shows an illustration of the rigid ceramic tile that is used on the space shuttles which my program models. I then compared my Python program to my mentor’s Fortran code to make sure that my program was working correctly and efficiently. Although it took a lot of time, I eventually worked out all of my many minor errors to produce a working code.

Figure 12 below illustrates the values of temperature computed every five minutes at each of the forty-two nodes that were used in this model. At time equals zero, all temperatures are at room temperature. Then at time equals 300 seconds, the exterior of the insulation is at around 1440 K. By 1200 seconds, the surface is cooler than the middle of the insulation. At the end of the full 1800 second run, the aluminum at the back of the insulation is actually warmer than the surface of the insulation. Figure 13 shows the temperatures of certain points through the material with respect to time. The program can use a larger number of nodes to decrease the amount of error that would be involved when comparing a real life situation to the results of the program. As the number of nodes increases, the delta x will decrease, so one can basically model a material at every point in position for a one dimensional problem. For the purpose of computation and comparison with my mentor, the nodal count was kept at a number of 42 nodes to save time in running iterations through the program. The more nodes that the program needs to calculate temperatures for each time step, the longer the program takes to complete its run. I performed convergence studies to see how many nodes would be required to get accurate results. By doubling the amount of nodes used and subtracting the difference between temperatures at the same spatial locations, I was able to compare the solutions. The result of this process concluded that forty-two nodes would be sufficient enough to give accurate results. A time increment of 0.05 had to be used to prevent the program from reaching overflow and printing infinities. The difference at each point in location is less than one degree kelvin when forty-two and eighty-two nodes were compared. In a case where the computed values of temperature were more important, the nodal count could be increased; however, if delta x gets too small, the results could begin to diverge. In order to combat this effect of increasing the nodal count, the time step would need to be decreased. This process could now take a sizable amount of time once the time step is decreased further.

Figure 12. Temperature versus thickness across the multi-layer material.

12

Figure 13. Temperature versus time at various locations through the material.

After the program itself was completed, I then calculated the minimum thickness of insulation that would need to be used to keep the aluminum under the desired temperature of 350 degrees Fahrenheit; however, my program operates in Kelvin, so the actual max number that the aluminum could reach was about 449.8 degrees Kelvin. This temperature was chosen as the maximum temperature to be experienced by the aluminum since higher temperatures would cause the material strength of the aluminum to diminish after repetitive exposure. By running a high insulation thickness and a low insulation thickness to produce final aluminum temperatures above and below the desired max temperature, I was able to use linear interpolation to determine the minimum thickness of insulation necessary. This value was computed to be approximately five centimeters. However, the thickness of the heat shield encasing the shuttles vary with their length from the nose since the heating rate varies across the length of the windward face of the shuttle. Typically, the heat shield is thickest at the nose and thinnest at the base. The heat flux of 200000 W/cm2 represented the exposure close to the nose. To model the generic thickness along the rest of the shuttle’s windward face, values of 34, 12, 720,∧1

4 of the maximum heat flux experienced by the nose of the shuttle were arbitrarily

chosen.34 of the maximum heat flux produced a minimum thickness of about 4.6 cm. For ½ the max

13

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5290

520

750

980

1210

1440

t=0 t=300 t=600 t=900 t=1200 t=1500 t=1800

Thickness across multi-layer material (Cm)

Tem

pera

ture

(K)

0 200 400 600 800 1000 1200 1400 1600 1800200

300

400

500

600

700

800

900

1000

1100

Surface Middle Back

Time (s)

Tem

pera

ture

(K)

heat flux, a heat flux of 100000 W/cm2 was used. This produced a minimum thickness of 4 cm of

insulation. At 720 the maximum heat flux experienced at the nose of the shuttle, a heat flux of 70000

W/cm2 was used. This value produced a minimum thickness of about 3.3 cm. Finally, at ¼ the maximum heat flux, a heat flux of 50000 W/cm2 was used. This value produced a minimum thickness of about 2.8 cm. Comparing these values in excel produced a somewhat linear line as shown in Figure 14.

Figure 14. Thickness of Insulation versus heat flux.

Conclusion

The purpose of this exercise was to shadow a professional in their area of study with the hope that this would clarify my educational goals and/or allow me to increase my knowledge of my mentor’s area of study. Through this experience, I have had the opportunity to learn the basics of heat transfer and how to use it in order to model realistic situations. Using this knowledge I built a base program of the most simplistic parameters and boundary conditions to model one-dimensional heat transfer. Gradually, under the guidance of my mentor, I added to the complexity of the program until it eventually became able to represent a realistic descent of a lifting body through Earth’s atmosphere. When compared my own results to those of my mentor, they matched exactly, so I was able to acknowledge my code as functional; however, these results have not been compared to a physical thermal test of the insulation. Also, this case study has given me the chance to learn the coding language of Python which will undoubtedly help me as I continue my education with respect to computer science and engineering in the future.

14

4 6 8 10 12 14 16 18 20 220

0.01

0.02

0.03

0.04

0.05

0.06

f(x) = 0.00146019034852547 x + 0.0226866300268096R² = 0.957276530221908

Heat Flux (W/cm2 )

Thick

ness

of I

nsul

ation

(m)

References

1. Kreith, F., and Black, W., Basic Heat Transfer, New York, N.Y, 1980, pp. 140-180.

2. Anon.,”Orbital Thermal Protection System.” NASA/Kennedy Space Center. 1989.

3. Daryabeigi, K., Knutson, J., and Cunnington, G., “Heat Transfer Measurement and Modeling in Rigid High-Temperature Reusable Surface Insulation Tiles”, AIAA 2011-345, January 2011.

4. Anon.,” Orbital maneuvering system/reaction control system low temperature reusable surface insulation tiles (LRSI) replaced with advanced flexible reusable surface insulation (AFRSI) consisting of a sewn composite quilted fabric blanket with same silica tile material sandwiched between outer and inner blanket.” Mar. 1983, p. 7.

15

Appendix:

Code 1 [

if time <= 300:

C1 = 0.

if time > 300 and time <= 1200:

C1 = (1.0/900)*(time - 300)

if time > 1200:

16

C1 = 1.

]

Code 2 [

If time <= 300:

Slope = qmax/300.

q = slope*time

If time > 300 and time <= 900:

q = qmax

If time > 900 and time <= 1200:

Slope = -qmax/ (300)

q = qmax + slope*(time-900)

If time > 1200:

q = 0.0

]

Pt.3

This for loop allows the program to create enough values to satisfy each nodes equation. There needs to be one less value for specific heat and thermal conductivity since these values are calculated “between” the nodes themselves. If there is eleven nodes, there are ten spaces of material. Each space of material is then assigned their according K and Cp values below.

17