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Trevor Jackson Analysis of an Aluminized Spherical Inflatable

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Trevor Jackson
Analysis of an Aluminized Spherical Inflatable Aerobrake
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This document relies heavily on research conducted at Langley Research Center and is based on
the document "The Echo I Inflation System" [1]. The previous research was made scalable and combined
with more current research data through use of MATLAB. This serves as a citation for the overall content
of this document.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A thermal analysis of the inflatable is required to ensure sufficient energy is available to
sublimate inflating solids contained inside the sphere. The thermal energy available to the inflatable is
dependent on orbit location and as such an understanding of the orbit is necessary before the
undertaking the thermal analysis. A circular orbit was chosen for simplicity in initial analysis, Systems
Tool Kit (STK) was later used to analyze more complex orbits.
The position of the inflatable can be described using three parameters:
𝑅𝐸 + β„Ž
𝛾
𝛿
Where:
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
𝛾=Displacement between inflatable position and nearest approach to Sun (deg); 0 ≤ 𝛾 ≤ 2πœ‹
πœ‹
𝛿=Angle between orbital plane normal and Earth-Sun line (deg); 0 ≤ 𝛿 ≤ 2
(Figure 1) - Inflatable position geometry [1]
Converting these coordinates into Cartesian coordinates:
𝑋 = (𝑅𝐸 + β„Ž) sin(𝛾)
π‘Œ = (𝑅𝐸 + β„Ž) cos(𝛾) sin(𝛿)
𝑍 = (𝑅𝐸 + β„Ž) cos(𝛾) cos(𝛿)
Inflatable enters Earth's shadow when:
𝛿>
1
cos(π‘˜)
πœ‹
3πœ‹
≤𝛾≤
2
2
π‘Žπ‘›π‘‘
Where:
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝐸
𝛾 can be found as a function of 𝛿 by substituting:
𝑋 2 + 𝑍 2 = 𝑅𝐸2
Which yields 𝛾 as a function of 𝛿:
𝛾(𝛿) =
1
π‘˜ 2 −cos2(𝛿)
)
sin2(𝛿)
sin(±√
Where:
cos(𝛿) ≤ π‘˜
𝛾=Displacement between inflatable position and nearest approach to Sun (deg); 0 ≤ 𝛾 ≤ 2πœ‹
πœ‹
𝛿=Angle between orbital plane normal and Earth-Sun line (deg); 0 ≤ 𝛿 ≤ 2
π‘˜=𝑅
𝑅𝐸
𝐸 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
The displacement of the radius vector of the inflatable position (𝛽) is used to attain amount of Earthreflected solar radiation:
cos(𝛽) = cos(𝛾) sin(𝛿)
Where:
𝛽=Displacement between inflatable position and Earth-Sun line (deg); 0 ≤ 𝛽 ≤ πœ‹
This orbit can modeled manually by changing values of altitude and 𝛿, since 𝛾 has been solved
as a function of 𝛿. A plot of various altitudes can be seen in Figure 2.
FIGURE 2 - from matlab code
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now that the orbit path variables have been found they will be used in a heat transfer analysis
of the inflatable. The initial analysis is to find the equilibrium temperature of the hot and cold spot on
the inflatable during the sunlight and Earth shadow phases of orbit. These temperatures will then be
used as integration boundaries for a thermal response analysis of the inflatable as it passes from Earth
shadow to sunlight and vice versa.
The general equation of heat transfer for the inflatable assuming totally radiative heat transfer:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘‘π‘ž
π‘‘π‘ž
π‘‘π‘ž
π‘‘π‘ž
( ) + ( ) + ( ) − ( ) = π‘šπΆπ‘ (
)
𝑑𝑑 𝑆
𝑑𝑑 𝑅
𝑑𝑑 𝐸
𝑑𝑑 𝐡
𝑑𝑑
Where:
1 π‘’π‘Ÿπ‘” = 2.389 × 10−8 gram calorie
π‘‘π‘ž
=Heat flow rate (erg/sec)
𝑑𝑑
π‘‘π‘ž
( 𝑑𝑑 ) =Absorption rate of direct solar radiation, zero when in shadow and constant when in sunlight
π‘‘π‘ž
𝑆
( 𝑑𝑑 ) =Absorption rate of solar radiation reflected from Earth, dependent on 𝛽
𝑅
π‘‘π‘ž
( ) =Absorption rate of direct radiation from Earth, constant
𝑑𝑑 𝐸
π‘‘π‘ž
( ) =Thermal radiation output of the inflatable
𝑑𝑑 𝐡
π‘š=Mass of inflatable (grams)
𝐢𝑝 =Specific heat of inflatable material (erg/gram˚K)
π‘‡π‘Žπ‘£π‘” =Absolute temperature averaged over surface (˚K)
π‘‘π‘‡π‘Žπ‘£π‘”
𝑑𝑑
=Change in average temperature (˚K/sec)
Each part of the general equation of heat transfer for the inflatable can be found as the following,
beginning with the Absorption rate of direct solar radiation:
π‘‘π‘ž
( ) = πœ‹π‘…π΅2 𝐢𝑆 𝛼𝑆
𝑑𝑑 𝑆
Where:
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
The absorption rate of solar radiation reflected from Earth, which is dependent on 𝛽, was solved
for under special condition of 𝛽=0 which is when the inflatable is directly between the Earth and the
Sun. The following equation gives the maximum amount of radiation, which decreases as 𝛽 increases
(see geometric view factor (𝐹𝑅 (𝛽)) description on the next page):
(
π‘‘π‘ž
) = 2πœ‹π‘…π΅2 𝐢𝑆 𝛼𝑆 π‘Ž[1 − √1 − π‘˜ 2 ]
𝑑𝑑 𝑅
Where:
π‘Ž=Earth's albedo, assumed as 0.36
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
𝑅
π‘˜= 𝐸 (π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝑅𝐸 +β„Ž
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
The absorption rate of direct radiation from Earth, which is constant, can be found as:
π‘‘π‘ž
1−π‘Ž
( ) = πœ‹π‘…π΅2 𝐢𝑆 𝛼𝐸 (
)[1 − √1 − π‘˜ 2 ]
𝑑𝑑 𝐸
2
Where:
𝛼𝐸 =Total absorptance to thermal radiation from Earth (dimensionless)
π‘Ž=Earth's albedo, assumed as 0.36
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝑅
π‘˜= 𝐸 (π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝑅𝐸 +β„Ž
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
The thermal radiation output of the inflatable can be found as:
(
π‘‘π‘ž
4
) = 4πœ‹π‘…π΅2 πœ€π‘‚ πœŽπ‘‡π‘Žπ‘£π‘”
𝑑𝑑 𝐡
Where:
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
π‘‡π‘Žπ‘£π‘” =Absolute temperature averaged over surface (˚K)
𝑅𝐡 =Radius of inflatable (cm)
A geometric view factor (𝐹𝑅 (𝛽)) was used as a ratio of Earth-reflected energy incident on the inflatable
relative to the amount incident when 𝛽=0.
1
π‘Šβ„Žπ‘’π‘› 𝛽 < 180° − (
) , π‘‘β„Žπ‘’ π‘ π‘Žπ‘‘π‘’π‘™π‘™π‘–π‘‘π‘’ 𝑖𝑠 𝑖𝑛 π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
cos(π‘˜)
1
π‘Šβ„Žπ‘’π‘› 𝛽 > 180° − (
) , π‘‘β„Žπ‘’ π‘ π‘Žπ‘‘π‘’π‘™π‘™π‘–π‘‘π‘’ 𝑖𝑠 𝑖𝑛 π‘‘β„Žπ‘’ πΈπ‘Žπ‘Ÿπ‘‘β„Ž′ 𝑠 π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
cos(π‘˜)
π‘‘π‘ž
𝐹𝑅 (𝛽) × ( ) = π‘€π‘Žπ‘”π‘›π‘–π‘‘π‘’π‘‘π‘’ π‘œπ‘“ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ 𝑖𝑛𝑝𝑒𝑑 π‘“π‘Ÿπ‘œπ‘š πΈπ‘Žπ‘Ÿπ‘‘β„Ž π‘Ÿπ‘’π‘“π‘™π‘’π‘π‘‘π‘’π‘‘ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘‘π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ π‘Žπ‘›π‘¦ π‘Žπ‘›π‘”π‘™π‘’ 𝛽
𝑑𝑑 𝑅
Where:
𝛽=Displacement between inflatable position and Earth-Sun line (deg); 0 ≤ 𝛽 ≤ πœ‹
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
π‘‘π‘ž
𝐸
( 𝑑𝑑 ) =Absorption rate of solar radiation reflected from Earth, dependent on 𝛽
𝑅
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The absolute temperature averaged over the surface of the inflatable can now be found using
the variables previously solved for which are dependent on the orbit path. The absolute temperature
averaged over the surface of the inflatable is then used to find the hotspot and coldspot temperature on
the inflatable in sunlight and in the Earth's shadow as functions of the inflatable's position in orbit. The
coldspot temperature is important because this will be considered the operating temperature of the
inflatable and is the controlling factor for the sublimation of the benzoic acid. The hotspot temperature
variation is also used to analyze necessary specifications for the adhesive used to assemble the
inflatable.
The absolute temperature (˚K) averaged over the surface of the inflatable while in sunlight is:
π‘‡π‘Žπ‘£π‘”
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
4
𝐢𝑆 𝛼𝑆
1 − π‘Ž πœ€π‘‚
= √(
) {1 + 2 [π‘ŽπΉπ‘… (𝛽) + (
) ( )] [1 − (1 − √1 − π‘˜ 2 )]}
4πœ€π‘‚ 𝜎
4
𝛼𝑆
Where:
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
π‘Ž=Earth's albedo, assumed as 0.36
𝐹𝑅 (𝛽)=Geometric view factor, dependent on 𝛽
𝛽=Displacement between inflatable position and Earth-Sun line (deg); 0 ≤ 𝛽 ≤ πœ‹
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝐸
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
The coldspot temperature, which is the spot on the inflatable that can "see" neither the Earth or
the Sun, to which heat transfer is purely radiative since the inflatable material is so thin there is no
conduction, is found by finding the heat in and the heat out of the coldspot.
The heat radiated out of the coldspot surface while in sunlight:
π»π‘’π‘Žπ‘‘ π‘œπ‘’π‘‘π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
= (πœ€π‘– + πœ€π‘‚ )πœŽπ‘‡π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘ 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
Where:
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
The heat radiated into the coldspot surface while in sunlight:
π»π‘’π‘Žπ‘‘ π‘–π‘›π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
= πœ€π‘– πœŽπ‘‡π‘Žπ‘£π‘” 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
Where:
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
To find the coldspot temperature throughout the orbit the heat out and heat in equation are set equal
and solved for the temperature of the coldspot in sunlight (π‘‡π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ ):
π»π‘’π‘Žπ‘‘ π‘œπ‘’π‘‘π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ = π»π‘’π‘Žπ‘‘ π‘–π‘›π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
(πœ€π‘– + πœ€π‘‚ )πœŽπ‘‡π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘ 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
= πœ€π‘– πœŽπ‘‡π‘Žπ‘£π‘” 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
The coldspot temperature in sunlight as a function of the temperature averaged over the surface of the
inflatable in sunlight, which is a function of orbit position:
π‘‡π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ = (
1
1+
1
4
πœ€π‘‚ )
πœ€π‘–
× π‘‡π‘Žπ‘£π‘”
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
Where:
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
The hotspot temperature is found in a similar way with the inclusion of the fact that the hot spot can
"see" the Earth and the Sun.
The heat in for the hotspot:
π»π‘’π‘Žπ‘‘ π‘–π‘›β„Žπ‘œπ‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ = πœ€π‘– πœŽπ‘‡π‘Žπ‘£π‘” 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
Where:
πœ€π‘– πœŽπ‘‡π‘Žπ‘£π‘” 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
+ 𝛼𝑆 𝐢𝑆
term=Heat input into the inside surface
𝛼𝑆 𝐢𝑆 term=Input on the outside from direct solar radiation
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
The heat out for the hotspot:
π»π‘’π‘Žπ‘‘ π‘œπ‘’π‘‘β„Žπ‘œπ‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ = (πœ€π‘– + πœ€π‘‚ )πœŽπ‘‡β„Žπ‘œπ‘‘π‘ π‘π‘œπ‘‘ 4π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
Where:
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
To find the hotspot temperature throughout the orbit the heat out and heat in equation are set equal
and solved for the temperature of the hotspot in sunlight (π‘‡β„Žπ‘œπ‘‘π‘ π‘π‘œπ‘‘ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ ):
π»π‘’π‘Žπ‘‘ π‘œπ‘’π‘‘π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
(πœ€π‘– + πœ€π‘‚ )πœŽπ‘‡β„Žπ‘œπ‘‘π‘ π‘π‘œπ‘‘ 4
= π»π‘’π‘Žπ‘‘ π‘–π‘›π‘π‘œπ‘™π‘‘π‘ π‘π‘œπ‘‘
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
= πœ€π‘– πœŽπ‘‡π‘Žπ‘£π‘” 4
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
+ 𝛼𝑆 𝐢𝑆
The hotspot temperature in sunlight as a function of the temperature averaged over the surface of the
inflatable in sunlight, which is a function of orbit position:
4
π‘‡β„Žπ‘œπ‘‘π‘ π‘π‘œπ‘‘
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
= √(
𝛼𝑆
𝐢𝑆
1
4
)( ) + (
πœ€ ) π‘‡π‘Žπ‘£π‘” π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
πœ€π‘– + πœ€π‘‚ 𝜎
1+ 𝑂
πœ€π‘–
Where:
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
The equilibrium temperature of the inflatable while in the shadow of earth is found in the same
way as the equation found for in sunlight, except that now the first two terms of the general equation of
heat transfer are now equal to zero. The steps to find this equation will not be shown for brevity.
The absolute temperature (˚K) averaged over the surface of the inflatable while in Earth's shadow is:
π‘‡π‘Žπ‘£π‘”
4
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
= √(
𝐢𝑆 𝛼𝐸 1 − π‘Ž
)(
) [1 − (1 − √1 − π‘˜ 2 )]
4πœ€π‘‚ 𝜎
2
Where:
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝐸 =Total absorptance to thermal radiation from Earth (dimensionless)
π‘Ž=Earth's albedo, assumed as 0.36
𝑅
π‘˜= 𝐸 (π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝑅𝐸 +β„Ž
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now that the hot and coldspot on the inflatable on the inflatable in sunlight and in shadow of
the Earth are found they will be used as integration boundaries for a thermal response analysis of the
inflatable as it passes from Earth's shadow to sunlight and vice versa. The thermal response of the
inflatable after entering the Earth's shadow will be found first.
The general equation of heat transfer for the inflatable assuming totally radiative heat transfer:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘‘π‘ž
π‘‘π‘ž
π‘‘π‘ž
π‘‘π‘ž
( ) + ( ) + ( ) − ( ) = π‘šπΆπ‘ (
)
𝑑𝑑 𝑆
𝑑𝑑 𝑅
𝑑𝑑 𝐸
𝑑𝑑 𝐡
𝑑𝑑
When the inflatable enters the Earth's shadow the general equation of heat transfer becomes:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘‘π‘ž
π‘‘π‘ž
( ) − ( ) = π‘šπΆπ‘ (
)
𝑑𝑑 𝐸
𝑑𝑑 𝐡
𝑑𝑑
Where:
1 π‘’π‘Ÿπ‘” = 2.389 × 10−8 gram calorie
π‘‘π‘ž
=Heat flow rate (erg/sec)
𝑑𝑑
π‘‘π‘ž
𝑑𝑑 𝐸
π‘‘π‘ž
( 𝑑𝑑 ) =Thermal
𝐡
( ) =Absorption rate of direct radiation from Earth, constant
radiation output of the inflatable
π‘š=Mass of inflatable (grams)
𝐢𝑝 =Specific heat of inflatable material (erg/gram˚K)
π‘‡π‘Žπ‘£π‘” =Absolute temperature averaged over surface (˚K)
π‘‘π‘‡π‘Žπ‘£π‘”
𝑑𝑑
=Change in average temperature (˚K/sec)
The absorption rate of direct radiation from Earth, which is constant, can be found as:
π‘‘π‘ž
1−π‘Ž
( ) = πœ‹π‘…π΅2 𝐢𝑆 𝛼𝐸 (
)[1 − √1 − π‘˜ 2 ]
𝑑𝑑 𝐸
2
Where:
𝛼𝐸 =Total absorptance to thermal radiation from Earth (dimensionless)
π‘Ž=Earth's albedo, assumed as 0.36
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝐸
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
The thermal radiation output of the inflatable can be found as:
π‘‘π‘ž
4
( ) = 4πœ‹π‘…π΅2 πœ€π‘‚ πœŽπ‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
𝑑𝑑 𝐡
Where:
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
π‘‡π‘Žπ‘£π‘”π‘ β„Žπ‘Žπ‘‘π‘œπ‘€ =Absolute temperature (˚K) averaged over the surface of the inflatable in Earth's shadow
𝑅𝐡 =Radius of inflatable (cm)
The equation for the absorption rate of direct radiation from Earth and the thermal radiation output
from the inflatable are subbed into the general equation of heat transfer for the inflatable and
manipulated to be:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
𝑑𝑑
=
1−π‘Ž
) [1 −
2
πœ‹π‘…π΅2 𝐢𝑆 𝛼𝐸 (
4
√1 − π‘˜ 2 ] − 4πœ‹π‘…π΅2 πœ€π‘‚ πœŽπ‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
π‘šπΆπ‘
Where:
π‘š=Mass of inflatable (grams)
𝐢𝑝 =Specific heat of inflatable material (erg/gram˚K)
1 π‘’π‘Ÿπ‘” = 2.389 × 10−8 gram calorie
π‘‡π‘Žπ‘£π‘”π‘ β„Žπ‘Žπ‘‘π‘œπ‘€ =Absolute temperature (˚K) averaged over the surface of the inflatable in Earth's shadow
𝛼𝐸 =Total absorptance to thermal radiation from Earth (dimensionless)
π‘Ž=Earth's albedo, assumed as 0.36
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝐸
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
To simplify the equations as they get lengthy, the variables w, x, and y are used to abbreviate sections
of the equation:
2
𝑀 =
1−π‘Ž
) [1 − √1
2
πœ‹π‘…π΅2 𝐢𝑆 𝛼𝐸 (
− π‘˜ 2]
π‘š
π‘₯2 =
4πœ‹π‘…π΅2 πœ€π‘‚ 𝜎
π‘š
𝑦 2 + 𝑧 2 π‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
= 𝐢𝑝
Where:
y,z=specific heat constants used to define 𝐢𝑝
Combining all abbreviated terms, the change in absolute temperature averaged over inflatable surface
(˚K) in Earth's shadow with respect to time is:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
𝑑𝑑
=
𝑀 2 − π‘₯ 2 π‘‡π‘Žπ‘£π‘” 4
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
𝑦 2 + 𝑧 2 π‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
Integrating the previous equation by expanding the abbreviated terms and solving for the amount of
time until the inflatable reaches the equilibrium absolute temperature averaged over the inflatable
surface (˚K):
π‘‡π‘Žπ‘£π‘”
𝑦
π‘‡π‘Žπ‘£π‘”
π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š
2
π‘‘π‘‡π‘Žπ‘£π‘”
∫
π‘‡π‘Žπ‘£π‘”
4
𝑀 2 − π‘₯ 2 π‘‡π‘Žπ‘£π‘”
+𝑧
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘‘π‘–π‘šπ‘’ π‘‘π‘œ π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’
π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š
2
∫
π‘‡π‘Žπ‘£π‘”
π‘‡π‘Žπ‘£π‘” π‘‘π‘‡π‘Žπ‘£π‘”
4
𝑀 2 − π‘₯ 2 π‘‡π‘Žπ‘£π‘”
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
=
∫
𝑑𝑑
π‘‘π‘–π‘šπ‘’=0
Solving for the y term, the first term, in the above equation:
π‘‡π‘Žπ‘£π‘”
π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š
𝑦2
∫
π‘‡π‘Žπ‘£π‘”
π‘‘π‘‡π‘Žπ‘£π‘”
4
𝑀 2 − π‘₯ 2 π‘‡π‘Žπ‘£π‘”
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
𝑀
π‘‡π‘Žπ‘£π‘” − √ π‘₯ 𝑇
π‘‡π‘Žπ‘£π‘” 1
π‘Žπ‘£π‘”
=[
(tan−1
− log 𝑒
)]
2
𝑀
𝑀
𝑀 π‘‡π‘Žπ‘£π‘” π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2𝑀π‘₯√
π‘‡π‘Žπ‘£π‘” + √
√π‘₯
π‘₯
π‘₯
𝑦2
Solving for the z term, the second term, in the equation can be done by re-arranging the z term then
integrating:
π‘‡π‘Žπ‘£π‘”
𝑧
π‘‡π‘Žπ‘£π‘”
π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š
2
∫
π‘‡π‘Žπ‘£π‘”
π‘‡π‘Žπ‘£π‘” π‘‘π‘‡π‘Žπ‘£π‘”
𝑧 2
4 = (π‘₯ )
2
2
𝑀 − π‘₯ π‘‡π‘Žπ‘£π‘”
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š
∫
π‘‡π‘Žπ‘£π‘”
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘‡π‘Žπ‘£π‘” π‘‘π‘‡π‘Žπ‘£π‘”
𝑀2
π‘₯2
4
− π‘‡π‘Žπ‘£π‘”
𝑀
2
π‘‡π‘Žπ‘£π‘”
− π‘₯ π‘‡π‘Žπ‘£π‘”
𝑧2
= [−
(log 𝑒 2
𝑀 )]𝑇
π‘Žπ‘£π‘” π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
4𝑀π‘₯
π‘‡π‘Žπ‘£π‘” +
π‘₯
Combining all the now integrated terms and manipulating the variables so that the equation is solving
for the amount of time until the inflatable reaches equilibrium temperature averaged over the surface
of the inflatable as it enters the Earth's shadow:
π‘‘π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
𝑀
𝑀
2
π‘‡π‘Žπ‘£π‘” − √ 𝑇
π‘‡π‘Žπ‘£π‘”
− π‘₯ π‘‡π‘Žπ‘£π‘”
𝑇
π‘₯
1
𝑧2
π‘Žπ‘£π‘”
−1 π‘Žπ‘£π‘”
=[
(tan
− log 𝑒
)]
−[
(log 𝑒 2
𝑀 )]
2
4𝑀π‘₯
𝑀
𝑀
𝑀 π‘‡π‘Žπ‘£π‘” π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘‡π‘Žπ‘£π‘” + π‘₯ π‘‡π‘Žπ‘£π‘” π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2𝑀π‘₯√
π‘‡π‘Žπ‘£π‘” + √
√π‘₯
π‘₯
π‘₯
𝑦2
Where:
π‘‘π‘ β„Žπ‘Žπ‘‘π‘œπ‘€ =Time required for the inflatable temperature to decrease from π‘‡π‘Žπ‘£π‘”
to π‘‡π‘Žπ‘£π‘” as the
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
inflatable passes from sunlight into the Earth's shadow
y,z=specific heat constants used to define 𝐢𝑝
w,x=abbreviation terms, see previous page
π‘‡π‘Žπ‘£π‘” =The temperature averaged over the surface of the inflatable at equilibrium
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The thermal response of the inflatable after entering sunlight will now be found.
The general equation of heat transfer for the inflatable assuming totally radiative heat transfer:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘‘π‘ž
π‘‘π‘ž
π‘‘π‘ž
π‘‘π‘ž
( ) + 𝐹𝑅 (𝛽) × ( ) + ( ) − ( ) = π‘šπΆπ‘ (
)
𝑑𝑑 𝑆
𝑑𝑑 𝑅
𝑑𝑑 𝐸
𝑑𝑑 𝐡
𝑑𝑑
The calculation now contains all terms in the general equation of heat transfer. The absorption rate of
direct solar radiation is:
π‘‘π‘ž
( ) = πœ‹π‘…π΅2 𝐢𝑆 𝛼𝑆
𝑑𝑑 𝑆
Where:
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
The absorption rate of solar radiation reflected from Earth, which is dependent on 𝛽:
𝐹𝑅 (𝛽) × (
π‘‘π‘ž
) = 𝐹𝑅 (𝛽) × 2πœ‹π‘…π΅2 𝐢𝑆 𝛼𝑆 π‘Ž[1 − √1 − π‘˜ 2 ]
𝑑𝑑 𝑅
Where:
π‘Ž=Earth's albedo, assumed as 0.36
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝐸
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
𝐹𝑅 (𝛽) =geometric view factor, a ratio of Earth-reflected energy incident on the inflatable relative to the
amount incident when 𝛽=0
The absorption rate of direct radiation from Earth, which is constant, can be found as:
π‘‘π‘ž
1−π‘Ž
( ) = πœ‹π‘…π΅2 𝐢𝑆 𝛼𝐸 (
)[1 − √1 − π‘˜ 2 ]
𝑑𝑑 𝐸
2
Where:
𝛼𝐸 =Total absorptance to thermal radiation from Earth (dimensionless)
π‘Ž=Earth's albedo, assumed as 0.36
𝑅𝐡 =Radius of inflatable (cm)
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝑅
π‘˜= 𝐸 (π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝑅𝐸 +β„Ž
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
The thermal radiation output of the inflatable can be found as:
π‘‘π‘ž
4
( ) = 4πœ‹π‘…π΅2 πœ€π‘‚ πœŽπ‘‡π‘Žπ‘£π‘”
π‘ β„Žπ‘Žπ‘‘π‘œπ‘€
𝑑𝑑 𝐡
Where:
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
π‘‡π‘Žπ‘£π‘”π‘ β„Žπ‘Žπ‘‘π‘œπ‘€ =Absolute temperature (˚K) averaged over the surface of the inflatable in Earth's shadow
𝑅𝐡 =Radius of inflatable (cm)
Combining all terms into the general equation of heat transfer and manipulating yields:
π‘‘π‘‡π‘Žπ‘£π‘”
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
𝑑𝑑
=
𝛼 1−π‘Ž
) [1 −
𝑆 2
πœ‹π‘…π΅2 𝐢𝑆 𝛼𝑆 (1 + [2π‘ŽπΉπ‘… (𝛽) + 𝛼𝐸
4
√1 − π‘˜ 2 ] − 4πœ‹π‘…π΅2 πœ€π‘‚ πœŽπ‘‡π‘Žπ‘£π‘”
π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
π‘šπΆπ‘
Where:
𝛼𝐸 =Total absorptance to thermal radiation from Earth (dimensionless)
𝛼𝑆 =Total absorptance to thermal radiation from Sun (dimensionless)
π‘Ž=Earth's albedo, assumed as 0.36
𝐢𝑆 =Solar radiation constant, 1.3953 × 106 (erg/sec*cm^2)
𝑅𝐸
π‘˜=𝑅 +β„Ž
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘ )
𝐸
𝑅𝐸 =Radius of Earth (cm)
β„Ž=Inflatable altitude (cm)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
π‘‡π‘Žπ‘£π‘”π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ =Absolute temperature (˚K) averaged over the surface of the inflatable in sunlight
𝑅𝐡 =Radius of inflatable (cm)
𝐹𝑅 (𝛽) =geometric view factor, a ratio of Earth-reflected energy incident on the inflatable relative to the
amount incident when 𝛽=0
Using an abbreviation term to combine variables and make integration easier:
2
π‘€π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
=
𝛼 1−π‘Ž
) [1 −
𝑆 2
πœ‹π‘…π΅2 𝐢𝑆 𝛼𝑆 (1 + [2π‘ŽπΉπ‘… (𝛽) + 𝛼𝐸
√1 − π‘˜ 2 ]
π‘š
Integrating in the same fashion as the previous section, which will be left out for brevity and can be
referred to on the page XXXXXXX, the amount of time in seconds until the inflatable reaches equilibrium
temperature averaged over the surface of the inflatable as it enters sunlight is:
π‘‘π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
π‘€π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
π‘‡π‘Žπ‘£π‘” + √ π‘₯
π‘‡π‘Žπ‘£π‘”
1
=[
(tan−1
+ log 𝑒
)] 𝑇
π‘Žπ‘£π‘” π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
2
𝑀
π‘€π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
𝑀
2π‘€π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ π‘₯√ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
π‘‡π‘Žπ‘£π‘” − √ π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
√ π‘₯
π‘₯
π‘₯
𝑦2
𝑧2
π‘‡π‘Žπ‘£π‘”
2
π‘‡π‘Žπ‘£π‘”
+
π‘€π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘
π‘‡π‘Žπ‘£π‘”
π‘₯
+[
(log 𝑒
)]
𝑀
2
4π‘€π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ π‘₯
π‘‡π‘Žπ‘£π‘”
− π‘ π‘’π‘›π‘™π‘–π‘”β„Žπ‘‘ π‘‡π‘Žπ‘£π‘” π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™
π‘₯
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The inflation medium will now be analyzed. It is important to note that the container in which
the inflatable is stowed must maintain a vacuum to avoid explosive inflation of the inflatable when it is
exposed to the vacuum environment of space. The Echo project used a vacuum pressure of 1 to 3 torr
which will be initially analyzed in the vacuum chamber for testing. Benzoic acid was selected as the
sublimating substance due to low cost, non-hazardous, and low mass properties.
The vapor pressure of a subliming compound in solid-vapor equilibrium varies according to the
integrated Clausius-Clapeyron equation:
𝑝 = 𝑐′𝑒
πœ†
𝑅𝑔 𝑇
−
Where:
𝑝=pressure (dynes/cm^2)
𝑐 ′ =constant of integration (dynes/cm^2)
π‘’π‘Ÿπ‘”
πœ†=Latent heat of sublimation (π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’), held constant over range of interest
π‘’π‘Ÿπ‘”
𝑅𝑔 =Universal gas constant, 8.3149 × 107 (π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’ °πΎ)
T=Absolute temperature (˚K)
The equilibrium vapor pressure of the compound increases exponentially as the temperature increases
linearly due to the latent heat of sublimation remaining constant over the range of interest.
The internal pressure required for a thin wall sphere is determined from desired skin stress:
4𝑑 ′ 𝑆
𝑝𝑖 =
+ π‘π‘œ
𝐷
Where:
𝑝𝑖 =Pressure inside surface (dynes/cm^2)
π‘π‘œ =Pressure outside surface (dynes/cm^2), this is the pressure at a given altitude and also includes
other pressure causing factors
𝑑 ′ =thickness (cm)
𝑆=Skin stress (dynes/cm^2)
𝐷=Diameter (cm)
There are four sources for internal pressure generation inside the inflatable. (1) Expanded
atmospheric gases left inside the inflatable. (2) Desorbed gases from fabrication material of inflatable.
(3) Thermal radiation inside the sphere. (4) Vapor pressure of fabrication material.
Expansion of free atmospheric gas remaining in the inflatable can be computed from volumetric
expansion during inflation by assuming an isothermal transition to give max residual pressure:
π‘‰π‘–π‘›π‘“π‘™π‘Žπ‘‘π‘–π‘œπ‘›
π‘π‘’π‘“π‘œπ‘Ÿπ‘’
π‘‰π‘–π‘›π‘“π‘™π‘Žπ‘‘π‘–π‘œπ‘› π‘Žπ‘“π‘‘π‘’π‘Ÿ
= π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ
π‘π‘π‘œπ‘›π‘‘π‘Žπ‘–π‘›π‘’π‘Ÿ π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ × π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ = π‘π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘Žπ‘™ π‘Žπ‘‘π‘šπ‘œπ‘ π‘β„Žπ‘’π‘Ÿπ‘–π‘
Where:
π‘‰π‘–π‘›π‘“π‘™π‘Žπ‘‘π‘–π‘œπ‘›
π‘Žπ‘“π‘‘π‘’π‘Ÿ
π‘‰π‘–π‘›π‘“π‘™π‘Žπ‘‘π‘–π‘œπ‘›
π‘π‘’π‘“π‘œπ‘Ÿπ‘’
=Volume of inflatable after inflation (cm^3)
=Volume of inflatable before inflation (cm^3)
π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ=Ratio of volume before inflation to volume after inflation (dimensionless)
π‘π‘π‘œπ‘›π‘‘π‘Žπ‘–π‘›π‘’π‘Ÿ π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ =Initial vacuum pressure inside container to avoid explosive inflation (torr)
π‘π‘Ÿπ‘’π‘ π‘–π‘‘π‘’π‘Žπ‘™ π‘Žπ‘‘π‘šπ‘œπ‘ π‘β„Žπ‘’π‘Ÿπ‘–π‘ =Partial pressure of residual atmospheric gas remaining in inflatable
Partial pressure of residual water vapor inside inflatable:
𝑝=
π‘π‘˜ ′ 𝑇
𝑉𝐡
Where:
𝑝=pressure (torr)
𝑁=Number of molecules desorbed
π‘’π‘Ÿπ‘”
π‘˜ ′ =Boltzmann's constant, 1.3803 × 10−16 (
)
π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘’ °πΎ
𝑇 =Absolute temperature (˚K)
𝑉𝐡 =Volume of inflatable (cm^3)
π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘’π‘ 
If the monomolecular covering power of water vapor is taken as 5.27 × 1014 (π‘π‘š2 −π‘šπ‘œπ‘›π‘œπ‘™π‘Žπ‘¦π‘’π‘Ÿ), then the
previous equation can be written as:
162.796𝑗
𝐷𝐡
𝑝=
Where:
𝑝=pressure (dynes/cm^2)
𝑗=Ratio of microscopic surface area to apparent surface area, taken as 1 (dimensionless)
𝐷𝐡 =Diameter of inflatable (cm)
The pressure inside the inflatable due to radiation, assuming thermal equilibrium with environment:
2πœ€π‘– πœŽπ‘‡ 4
𝑝𝑖 =
𝑐
Where:
𝑝𝑖 =Pressure inside surface (dynes/cm^2)
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝑇 =Absolute temperature (˚K)
π‘π‘š
)
𝑠𝑒𝑐
𝑐=Velocity of light in a vacuum, 2.99776 × 1010 (
To be able to get a net pressure inside the inflatable the pressure on the outside of the inflatable due to
radiation must also be found:
𝑝0 =
πœ€π‘‚ πœŽπ‘‡ 4
𝑐
Where:
π‘π‘œ =Pressure outside surface (dynes/cm^2)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝑇 =Absolute temperature (˚K)
π‘π‘š
𝑐=Velocity of light in a vacuum, 2.99776 × 1010 (𝑠𝑒𝑐 )
Subtracting the outside pressure due to radiation from the inside pressure due to radiation gives the
next pressure inside the inflatable due to radiation:
𝑝𝑇 =
(2πœ€π‘– − πœ€π‘‚ )πœŽπ‘‡ 4
𝑐
Where:
𝑝𝑇 =Net uniform thermal radiation pressure (torr)
πœ€π‘– =Total hemispherical emittance of inside surface (dimensionless)
πœ€π‘‚ =Total hemispherical emittance of outside surface at initial time (dimensionless)
𝜎=Stefan-Boltzmann radiation constant, 5.71 × 10−5 (erg/(cm^2)*sec*˚K^4)
𝑇 =Absolute temperature (˚K)
π‘π‘š
)
𝑠𝑒𝑐
𝑐=Velocity of light in a vacuum, 2.99776 × 1010 (
Vapor pressure of the fabrication material was found to be negligible, with calculations showing a
contribution of 10−40 torr and an experimental procedure which placed the material in vacuum for 1419
hours which resulted in less than 0.1% loss in weight supporting the decision to neglect this pressure
contribution.
Now that the contribution of other factors to the pressure inside the inflatable has been found,
a method of achieving the remaining pressure to reach desired internal pressure must be used. The
sublimation of benzoic acid will be utilized.
The solid-vapor equilibrium curve of benzoic acid using simplified version of Clausisus-Clapeyron
equation :
𝐡′
log 𝑒 𝑝 = 𝐴 −
𝑇
′
Where:
𝑝=pressure (torr)
𝐴′ , 𝐡′ =Substance specific constants
𝑇 =Absolute temperature (˚K)
For benzoic acid the equation with constants becomes:
log 𝑒 𝑝 = 29.595 −
8223
𝑇
(Figure 2) - Solid-vapor equilibrium curve
Temperature (Celsius) Temperature (Kelvin) Vapor Pressure (Torr)
-60
213
0.000000091586
-40
233
0.000002517600
-20
253
0.000040984000
0
273
0.000443310000
20
293
0.003500000000
40
313
0.020800000000
60
333
0.100900000000
80
353
0.408600000000
100
373
1.424900000000
120
393
4.375300000000
140
413
12.052000000000
(Table 1) - Solid-vapor equilibrium vapor pressure of benzoic acid using simplified version of ClausisusClapeyron equation
The ideal gas law will prevail when all the solid has changed to the vapor phase and any amount of
benzoic acid can be analyzed to see what pressure is achievable when all the solid has changed to the
vapor phase:
𝑝=
π‘Šπ‘…π‘” 𝑇
𝑀𝑉
Where:
𝑝=pressure (torr)
W=Weight of benzoic acid (grams)
π‘’π‘Ÿπ‘”
𝑅𝑔 =Universal gas constant, 8.3149 × 107 (π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’ °πΎ)
𝑇 =Absolute temperature (˚K)
π‘”π‘Ÿπ‘Žπ‘šπ‘ 
𝑀=Molecular weight (π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’)
𝑉=Volume (cm^3)
Setting Clausius-Clapeyron and the ideal gas law equal to each other and solving for the weight of
benzoic acid needed to reach solid-vapor equilibrium as a function of temperature:
8223
𝑇
(𝑒 29.595−
)=
π‘Šπ‘…π‘” 𝑇
𝑀𝑉
8223
(𝑒 29.595− 𝑇 ) ∗ 𝑀 ∗ 𝑉
π‘Š=
𝑅𝑔 𝑇
Where:
π‘Š =Weight of benzoic acid (grams)
π‘’π‘Ÿπ‘”
𝑅𝑔 =Universal gas constant, 8.3149 × 107 (π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’ °πΎ)
𝑇 =Absolute temperature (˚K)
π‘”π‘Ÿπ‘Žπ‘šπ‘ 
𝑀=Molecular weight (π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’)
𝑉=Volume (cm^3)
Calculating weight of benzoic acid at a variety of temperatures to reach solid-vapor equilibrium with
Project Echo inflatable volume:
Weight of benzoic acid to achieve
Temperature (Celsius) Temperature (Kelvin) solid-vapor equilibrium (grams)
23
296
451.5959
42
315
2267
51
324
4551.4
61
334
9439.4
74
347
22853
85
358
45878
96
369
88274
(Table 2) - Weight of benzoic acid to reach solid-vapor equilibrium at variety of temperatures for a
Project Echo size inflatable
These values agree with the values originally found for Project Echo.
Modifying the inflatable volume to that of an inflatable with a 100 cm diamater:
Weight of benzoic acid to achieve
Temperature (Celsius) Temperature (Kelvin) solid-vapor equilibrium (grams)
23
296
0.0159
42
315
0.0801
51
324
0.1607
61
334
0.3334
74
347
0.807
85
358
1.6202
96
369
3.1174
(Table 3) - Weight of benzoic acid to reach solid-vapor equilibrium at variety of temperatures for a 100
cm diameter inflatable
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
MATLAB CODE: (In Progress)
Solid-vapor equilibrium curve (Figure 2):
T_c=linspace(-60,140) %temperature in celsius
p=(exp(29.595-(8223./(273+T_c)))).*0.000750061683 %pressure in
torr,0.000750061683 is converting from dynes/cm^2 to torr,
equation includes
semilogy(T_c,p); title('Vapor Pressure of Benzoic Acid vs.
Temperature');
xlabel('Temperature (Celsius)'); ylabel('Vapor Pressure
(torr)');
Weight of benzoic acid for Project Echo inflatable (Table 2):
M=122.12 %molecular weight of benzoic acid, gram/gram*mole
d=3048 %diameter of inflatable in cm, diameter of project echo
was 3048 cm
V=(pi*(d^3))/6 %volume in cm^3
Rg=83149000 %universal gas constant erg/gram*mole*K
T1=296;%temperature in Kelvin
T2=315;
T3=324;
T4=334;
T5=347;
T6=358;
T7=369;
W1=((exp(29.595-(8223./T1)))*M*V)/(Rg*T1)
W2=((exp(29.595-(8223./T2)))*M*V)/(Rg*T2)
W3=((exp(29.595-(8223./T3)))*M*V)/(Rg*T3)
W4=((exp(29.595-(8223./T4)))*M*V)/(Rg*T4)
W5=((exp(29.595-(8223./T5)))*M*V)/(Rg*T5)
W6=((exp(29.595-(8223./T6)))*M*V)/(Rg*T6)
W7=((exp(29.595-(8223./T7)))*M*V)/(Rg*T7)
Weight of benzoic acid for 100 cm diameter inflatable(Table 3):
M=122.12 %molecular weight of benzoic acid, gram/gram*mole
d=100 %diameter of inflatable in cm, diameter of project echo
was 3048 cm
V=(pi*(d^3))/6 %volume in cm^3
Rg=83149000 %universal gas constant erg/gram*mole*K
T1=296;%temperature in Kelvin
T2=315;
T3=324;
T4=334;
T5=347;
T6=358;
T7=369;
W1=((exp(29.595-(8223./T1)))*M*V)/(Rg*T1)
W2=((exp(29.595-(8223./T2)))*M*V)/(Rg*T2)
W3=((exp(29.595-(8223./T3)))*M*V)/(Rg*T3)
W4=((exp(29.595-(8223./T4)))*M*V)/(Rg*T4)
W5=((exp(29.595-(8223./T5)))*M*V)/(Rg*T5)
W6=((exp(29.595-(8223./T6)))*M*V)/(Rg*T6)
W7=((exp(29.595-(8223./T7)))*M*V)/(Rg*T7)
References
[1]
D. L. Clemmons and L. R. Center, The Echo I inflation system: National Aeronautics and Space
Administration, 1964.
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