Calculating Radiative Heat and Emittance for Rocket Surfaces

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University of Michigan**We aren't endorsed by this school

Course

ME 530

Subject

Mechanical Engineering

Date

Dec 10, 2024

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Uploaded by BarristerWalrusPerson1195

Akshita Arora UM ID: 90734592 ME 530 HW-2 1)L = 20 m, D = 2m, d = 140*106km = 14*1010m, Tsun= 5777 K, Rsun= 696*103km a)Solid angle with which the sun is seen from the rocket: Ω =𝐴??? ?????????(????????)2=𝜋𝑅𝑠𝑢𝑛2?2=𝜋(696∗103)2(140∗106)2= 7.76 × 10−5???b)Radiative flux (W m-2) incident on the rocket surface: ? = Ω𝜎????4𝜋= 7.76 × 10−5× 5.67 × 10−8× 57774/𝜋 = 1559.9 ?/?2Total radiative heat received by rocket: ?. (????𝑎?? 𝐴??𝑎)??????= 1559.9 (𝜋?? +𝜋𝐷22) = 2.1 ∗ 105?Flux with respect to surface area of rocket: ?′= ?𝜋?? +𝜋?22𝜋????2= 1.35 ∗ 10−13?/?2c)If the rocket surface could be assumed to be a blackbody, the equivalent surface rocket temperature can be calculated as follows: ? = 𝜎???????4⇒ ???????= (?𝜎)14= (1559.95.67 × 10−8)14= 407.26 ?2)The spectral directional emittance (ε𝜆′) is given as: 𝜀𝜆′= {0.9 cosθ for λ < 2 μm0.2 for λ > 2 μmHemispherical spectral emittance is given by: 𝜀𝜆= 2 ∫ 𝜀′𝜆???𝜃?𝑖?𝜃?𝜃 ={2 ∫0.9(???𝜃)2?𝑖?𝜃?𝜃 for λ < 2 μm𝜋202 ∫ 0.2???𝜃?𝑖?𝜃?𝜃𝜋20for λ > 2 μm𝜋20={2 × 0.9???𝜃?𝑖?3𝜃3|0𝜋2for λ < 2 μm2 × 0.2?𝑖?2𝜃2|0𝜋2for λ > 2 μm= {0.6 for λ < 2 μm0.2 for λ > 2 μmThe total, hemispherical emittance is: 𝜀 =1𝜎?????4∫ 𝜀𝜆??𝜆(?????)?𝜆 =∞0𝜀𝜆<2∫ ??𝜆(?????)?𝜆 + 𝜀𝜆>2∫??𝜆(?????)?𝜆∞220∫??𝜆(?????)?𝜆∞0= 0.6?(2 µ? × 500 ?) + 0.2(1 − ?(2 µ? × 500 ?)) = 0.2001According to Kirchoff’s Law: 𝜀′𝜆= 𝛼′𝜆The directional absorptance (α′) is calculated as:

Akshita Arora UM ID: 90734592 𝛼′=1????∫ 𝛼′𝜆????,𝜆?𝜆∞0=1𝜎????4∫ 𝛼′𝜆??𝜆(????)?𝜆∞0=1𝜎????4∫ 𝛼′𝜆??𝜆(????)?𝜆 =∞0𝜀𝜆<2∫ ??𝜆(????)?𝜆 + 𝜀𝜆>2∫??𝜆(????)?𝜆∞220∫??𝜆(????)?𝜆∞0= 0.9???𝜃 × 0.9396 + 0.2 × (1 − 0.9396) = 0.8456???𝜃 + 0.0121The maximum absorption occurs when cos𝜃= 1, which is when 𝜃= 0° (normal incidence). At this angle: 𝛼??𝑥′= α′(0) =1????∫ 𝛼′𝜆????,𝜆?𝜆∞0= 0.9 × 0.9396 + 0.2 × (1 − 0.9396) = 0.8577At θ= 0, the maximum absorption is: ?abs,max= α′(0)?sol= 0.8577 × 1367 = 1172.503W?−2Net absorbed heat flux: ?net= ε??− α′(𝜃)?sol???𝜃 (?net< 0 ??? ?????𝑦 ?????????)The net absorbed heat flux should be at least 40% of the maximum absorption: |?net| ≥ |0.4 × ?net,θ = 0| = |0.4(ε??− ?abs,max)|= |0.4(708.75 − 1172.503)| = 185.5012W?−2⇒ (0.8456???𝜃 + 0.0121)1367???𝜃 − εσ?????4≥ 185.5012 ⇒ (1155.9???2𝜃 + 17.4976???𝜃) − 894.2512 ≥ 0 ∴ ???𝜃 ≥ 0.872 ⇒ 𝜃 ≤ 29.308?Thus, the range of sun positions for which the net absorbed heat flux is at least 40% of the maximum is: θ∈[0∘,29.31∘] 3)Net radiosity balance equation: ??′′𝜀?− ∑ (1𝜀?− 1) ?????′′+ 𝐻0?= ???− ∑ ??????𝑁?=1𝑁?=1Given: 𝜀1= 𝜀2= 𝜀4= 𝜀5= 1, 𝜀3= 0.5?3= 400 ?, ?1= 1500 ??4= ?5= 0Using Appendix D in Modest’ textbook: j i 1 3 4 5 1 0 0.0557 0.2071 0.0290 3 0.0557 0 0.0290 0.2071 4 0.8284 0.1159 0.5858 0.1781 5 0.1159 0.8284 0.1781 0.5858 The following equations were used to calculate view factors: ???𝐴?= ???𝐴?For 2 co-axial parallel disks: ?1−3=12{? − [?2− 4]1 2⁄},𝑤ℎ??? ? =2𝑅2+1𝑅2, ? =520?1−4=12{1 − ?2− 𝐻2+ [(1 + ?2+ 𝐻2)2− 4?2]1 2⁄},𝑤ℎ??? ? = 1, 𝐻 =105?4−4= 1 + 𝐻 − [1 + 𝐻2]1 2⁄,𝑤ℎ??? 𝐻 =1010

Akshita Arora UM ID: 90734592 ?5−1=14{(1 +𝐻2𝐻1) [4 + (𝐻1+ 𝐻2)2]1 2⁄− (𝐻1+ 2𝐻2) −𝐻2𝐻1(4 + 𝐻22)1 2⁄}If H1= H2= H: ?1−2=14{4(1 + 𝐻2)1 2⁄− 3𝐻 − (4 + 𝐻2)1 2⁄},𝑤ℎ??? 𝐻 =105Radiosity balance equation for surface 1: ?1′′− (1𝜀3− 1) ?13?3′′= ??1− (?11??1+ ?13??3+ ?14??4+ ?15??5)⇒ ?1′′− ?13?3′′= ??1− (?13??3+ ?14??4+ ?15??5)⇒ ?1′′− ?13?3′′= 𝜎?14− 𝜎(?13?34+ ?14?44+ ?15?54)Radiosity balance equation for surface 3: ?3′′𝜀3− (1𝜀3− 1) ?33?3′′= ??3− (?31??1+ ?33??3+ ?34??4+ ?35??5)⇒ 2?3′′= ??3− (?31??1+ ?34??4+ ?35??5) ⇒ 2?3′′= 𝜎?34− 𝜎(?31?14+ ?34?44+ ?35?54)Radiosity balance equation for surface 4: ?4′′− (1𝜀3− 1) ?43?3′′= ??4− (?41??1+ ?43??3+ ?44??4+ ?45??5)⇒ −?43?3′′= 𝜎(1 − ?44)?44− 𝜎(?41?14+ ?43?34+ ?45?54)Radiosity balance equation for surface 5: ?5′′− (1𝜀3− 1) ?53?3′′= ??5− (?51??1+ ?53??3+ ?54??4+ ?55??5)⇒ −?53?3′′= 𝜎(1 − ?55)?54− 𝜎(?51?14+ ?53?34+ ?54?44)System of equations: [1−?13?14?1502?34?350−?43−(1 − ?44)?450−?53?54−(1 − ?55)][?1′′?3′′𝜎?44𝜎?54]=[𝜎?14− 𝜎?13?34𝜎?34− 𝜎?31?14−𝜎(?41?14+ ?43?34)−𝜎(?51?14+ ?53?34)]Solving the above in MATLAB: ?1′′= 84579.64 ?/?2, ?3′′= −84579.64 ?/?2, ?4= 1398.3 ?,?5= 1288.6 ? 4)Given: dp= 0.3 mm D12= 1.5 mm Analytical expression for view factor from one sphere to another sphere is given as , In our case, R = 1, S = 8 a)F12from analytic expression: F12= 0.0025 F12from code for 1e7 rays: F12= 0.0029 The relative error between analytical and MCRT for 1e7 rays: 13.7821%

Akshita Arora UM ID: 90734592 Figure 1: Plot of view factor (left) and relative error (right) vs no. of rays b)Methodology followed for inserting a new sphere between the two spheres that is located midway between the first two spheres: •Sphere 3 is added at D13= D12/2. Its position is modified based on the viewing angle α. •For each α, the position of Sphere 3 is shifted, and ray tracing is performed. •Rays from Sphere 1 are checked first for intersections with Sphere 3. If a ray hits Sphere 3, it is not allowed to reach Sphere 2. Otherwise, the intersection check proceeds to Sphere 2. The modified MATLAB code is given below:

Akshita Arora UM ID: 90734592

Akshita Arora UM ID: 90734592 •For α=0°, all rays between Sphere 1 and Sphere 2 are blocked by Sphere 3, so F12=0. •For small angles α=5∘,15∘: As the angle increases, Sphere 3 allows some radiation to reach Sphere 2, increasing F12. •For larger angles α=30∘, The view factor F12 continues to increase because Sphere 3 obstructs less of the radiation path from Sphere 1 to Sphere 2. •At α=60°, spheres 2 and 3 are equidistant from sphere 1, resulting in an equal view factor, i.e. F12=F13. Figure 2: Plot of view factor vs viewing angle for F12 and F13.