Thermal Systems Design Fundamentals of heat transfer Radiative equilibrium Surface properties Non-ideal effects Internal power generation Environmental temperatures Conduction Thermal system components 1 2009 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu
Classical Methods of Heat Transfer Convection Heat transferred to cooler surrounding gas, which creates currents to remove hot gas and supply new cool gas Don t (in general) have surrounding gas or gravity for convective currents Conduction Direct heat transfer between touching components Primary heat flow mechanism internal to vehicle Radiation Heat transferred by infrared radiation Only mechanism for dumping heat external to vehicle 2
Ideal Radiative Heat Transfer Planck s equation gives energy emitted in a specific frequency by a black body as a function of temperature e λb = 2 2πhC 0 λ 5 exp hc 0 1 λkt (Don t worry, we won t actually use this equation for anything ) 3
The Solar Spectrum Ref: V. L. Pisacane and R. C. Moore, Fundamentals of Space Systems Oxford University Press, 1994 4
Ideal Radiative Heat Transfer Planck s equation gives energy emitted in a specific frequency by a black body as a function of temperature 2 2πhC e λb = 0 λ 5 exp hc 0 1 λkt Stefan-Boltzmann equation integrates Planck s equation over entire spectrum P rad =σt 4 σ = 5.67x10 8 5 W m 2 K 4 ( Stefan-Boltzmann Constant )
Thermodynamic Equilibrium First Law of Thermodynamics Q W = du dt heat in -heat out = work done internally Heat in = incident energy absorbed Heat out = radiated energy Work done internally = internal power used (negative work in this sense - adds to total heat in the system) 6
Radiative Equilibrium Temperature Assume a spherical black body of radius r Heat in due to intercepted solar flux Q in = I s π r 2 Heat out due to radiation (from total surface area) Q out = 4π r 2 σt 4 For equilibrium, set equal I s π r 2 = 4π r 2 σt 4 I s = 4σT 4 1 AU: I s =1394 W/m 2 ; T eq =280 K T eq = I s 4σ 7 1 4
Effect of Distance on Equilibrium Temp Mercury Venus Earth Mars Asteroids Jupiter Saturn Uranus Neptune Pluto 8
Shape and Radiative Equilibrium A shape absorbs energy only via illuminated faces A shape radiates energy via all surface area Basic assumption made is that black bodies are intrinsically isothermal (perfect and instantaneous conduction of heat internally to all faces) 9
Effect of Shape on Black Body Temps 10
Incident Radiation on Non-Ideal Bodies Kirchkoff s Law for total incident energy flux on solid bodies: Q Incident = Q absorbed + Q reflected + Q transmitted Q absorbed Q Incident + Q reflected Q Incident + Q transmitted Q Incident =1 α Q absorbed ; Q Incident where ρ Q reflected ; Q Incident α =absorptance (or absorptivity) ρ =reflectance (or reflectivity) τ =transmittance (or transmissivity) τ Q transmitted Q Incident 11
Non-Ideal Radiative Equilibrium Temp Assume a spherical black body of radius r Heat in due to intercepted solar flux Q in = I s απ r 2 Heat out due to radiation (from total surface area) Q out = 4π r 2 εσt 4 For equilibrium, set equal (ε = emissivity - efficiency of surface at radiating heat) I s απr 2 = 4π r 2 εσt 4 I s = 4 ε α σt 4 T eq = α ε I s 4σ 1 4 12
Effect of Surface Coating on Temperature α = absorptivity ε = emissivity 13
Non-Ideal Radiative Heat Transfer Full form of the Stefan-Boltzmann equation ( ) P rad =εσa T 4 4 T env where T env =environmental temperature (=4 K for space) Also take into account power used internally I s α A s + P int =εσa ( rad T 4 4 T ) env 14
Example: AERCam/SPRINT 30 cm diameter sphere α=0.2; ε=0.8 P int =200W T env =280 K (cargo bay below; Earth above) Analysis cases: Free space w/o sun Free space w/sun Earth orbit w/o sun Earth orbit w/sun 15
AERCam/SPRINT Analysis (Free A s =0.0707 m 2 ; A rad =0.2827 m 2 Free space, no sun 200W P int = εσa rad T 4 T = 0.8 5.67 10 8 W m 2 K 4 0.2827m 2 ( ) 1 4 = 354 K 16
AERCam/SPRINT Analysis (Free A s =0.0707 m 2 ; A rad =0.2827 m 2 Free space with sun I s α A s + P int = εσa rad T 4 T = I sα A s + P int εσa rad 1 4 = 362 K 17
AERCam/SPRINT Analysis (LEO T env =280 K LEO cargo bay, no sun P int =εσa rad T 4 4 ( T env ) T = LEO cargo bay with sun 200W 0.8 5.67 10 8 W 0.2827m 2 m 2 K 4 ( ) + (280 K)4 1 4 = 384 K I s α A s + P int =εσa rad T 4 4 T env ( ) T = I sα A s + P int εσa rad 4 + T env 14 = 391 K 18
Radiative Insulation I s T insulation T wall Thin sheet (mylar/kapton with surface coatings) used to isolate panel from solar flux Panel reaches equilibrium with radiation from sheet and from itself reflected from sheet Sheet reaches equilibrium with radiation from sun and panel, and from itself reflected off panel 19
Multi-Layer Insulation (MLI) I s Multiple insulation layers to cut down on radiative transfer Gets computationally intensive quickly Highly effective means of insulation Biggest problem is existence of conductive leak paths (physical connections to insulated components) 20
Emissivity Variation with MLI Layers Ref: D. G. Gilmore, ed., Spacecraft Thermal Control Handbook AIAA, 2002 21
MLI Thermal Conductivity Ref: D. G. Gilmore, ed., Spacecraft Thermal Control Handbook AIAA, 2002 22
Effect of Ambient Pressure on MLI Ref: D. G. Gilmore, ed., Spacecraft Thermal Control Handbook AIAA, 2002 23
1D Conduction Basic law of one-dimensional heat conduction (Fourier 1822) Q = KA dt dx where K=thermal conductivity (W/m K) A=area dt/dx=thermal gradient 24
3D Conduction General differential equation for heat flow in a solid 2 T r,t ( ) + g(r r,t) K where g(r,t)=internally generated heat ρ=density (kg/m 3 ) c=specific heat (J/kg K) K/ρc=thermal diffusivity = ρc K T( r,t) t 25
Simple Analytical Conduction Model Heat flowing from (i-1) into (i) T i-1 T i T i+1 Q in = KA T T i i 1 Δx Heat flowing from (i) into (i+1) Q out = KA T T i+1 i Δx Heat remaining in cell 26 Q out Q in = ρc K T i ( j +1) T i ( j) Δt
Finite Difference Formulation Time-marching solution T n+1 i = T n i + d(t n i+1 2T n i + T n i 1) where d = α t x 2 α = For solution stability, k ρc v = thermal diffusivity t < x2 2α 27
Heat Pipe Schematic 28
Shuttle Thermal Control Components 29
Shuttle Thermal Control System Schematic 30
ISS Radiator Assembly 31
Case Study: ECLIPSE Thermal Analysis 32 Developed by UMd SSL for NASA ESMD Minimum functional habitat element for lunar outpost Radiator area - upper dome and six upper cylindrical panels
ECLIPSE Heat Sources Solar heat load (modeling habitat as right circular cylinder) A illuminated = ld sin β + 1 4 πd2 cos β Q solar = A illuminated αi s Electrical power load = 4191 W Metabolic work load (4 crew) = 464 W 33
Thermal Modeling for Lunar Surface Assume upper dome radiates only to deep space Assume side panels radiate half to deep space and half to lunar surface Assume (conservatively) that lunar surface radiates as a black body Q internal + Q solar = ɛσ ( [A dome T 4rad + n rad A panel Trad 4 1 )] 2 T moon 4 T rad = [ 1 A dome + n rad A panel ( Qinternal + Q solar ɛσ + 1 2 n rada wall T 4 moon )] 1 4 34
ECLIPSE Thermal Results 35