LSU 06/21/2004Thermal Issues1 Payload Thermal Issues & Calculations Ballooning Unit, Lecture 3

LSU 06/21/2004Thermal Issues2 Thermal Requirements All payload components can function properly only within particular temperature ranges Operating temperature range (narrowest) –In this temperature range the component will perform to within specified parameters Non-Operation temperature range (wider) –Component will not perform within specs, but will do so when returned to operating temperature range Survival temperature range (widest) –If this range is exceeded component will never return to proper operation Thermal requirements constitute specifying these ranges for all components

LSU 06/21/2004Thermal Issues3 Thermal Control Plan Systems and procedures for satisfying the thermal requirements Show that thermal system (i.e. heaters, insulation, surface treatment) is sufficient to avoid excursions beyond survival temperature range Show critical components remain mostly in the operating temperature range Specify mitigation procedures if temperature moves to non-operating range (e.g. turn on heaters)

LSU 06/21/2004Thermal Issues4 Determining Temperature Ranges Start with OEM (original equipment manufacturer) datasheet on product –Datasheets usually specify only operating temperature range –Definition of “operating” may vary from manufacturer to manufacturer for similar components Look for information on how operating parameters change as a function of temperature –Your operating requirement may be more stringent than the manufacturer Find similar products and verify that temperature ranges are similar Search for papers reporting results from performance testing of product Call manufacturer and request specific information

LSU 06/21/2004Thermal Issues5 Survival Temperature Range Survival temperature range will be the most difficult to quantify Range limits may be due to different effects –Softening or loss of temper –Differential coefficients of expansion can lead to excessive shear Contact manufacturer and ask for their measurements or opinion Estimate from ranges reports for similar products Measure using thermal chamber

LSU 06/21/2004Thermal Issues6 Heat Transfer The payload will gain or lose heat energy through three fundamental heat transfer mechanisms Convection is the process by which heat is transferred by the mass movement of molecules (i.e. generally a fluid of some sort) from one place to another. Conduction is the process by which heat is transferred by the collision of “hot” fast moving molecules with “cold” slow moving molecules, speeding (heating) these slow molecules up. Radiation is the process by which heat is transferred by the emission and absorption of electromagnetic waves.

LSU 06/21/2004Thermal Issues7 Convection Requires a temperature difference and a working fluid to transfer energy Q conv = h A ( T 1 – T 2 ) The temperature of the surface is T 1 and the temperature of the fluid is T 2 in K o The surface area exposed to convection is given by A in m 2 The coefficient h depends on the properties of the fluid. –5 to 6 W/(m 2 K o ) for normal pressure & calm winds –0.4 W/(m 2 K o ) or so for low pressure In the space environment, where air pressure is at a minimum, convection heat transfer is not very important.

LSU 06/21/2004Thermal Issues8 Conduction Requires a temperature gradient (dT/dx) and some kind of material to convey the energy Q cond = k A ( dT / dx ) The surface area exposed to conduction is given by A in m 2 The coefficient k is the thermal conductivity of the material. –0.01 W/(m 2 K o ) for styrofoam –0.04 W/(m 2 K o ) for rock wool, cork, fiberglass –205 W/(m 2 K o ) for aluminium Need to integrate the gradient over the geometry of the conductor. –Q = k A ((T 1 -T 2 )/L) for a rod of area A and length L –Q = 4  k r (r + x) ((T 1 -T 2 )/x) for a spherical shell of radius r and thickness x

LSU 06/21/2004Thermal Issues9 Radiation Requires a temperature difference between two bodies, but no matter is needed to transfer the heat Q rad =   A ( T 1 4 – T 2 4 ) The Stefan-Boltzmann constant, , value is 5.67 x W/m 2 K 4 The surface area involved in radiative heat transfer is given by A in m 2 The coefficient  is the emissivity of the material. –Varies from 0 to 1 –Equal to the aborptivity (  )at the same wavelength –A good emitter is also a good absorper –A good reflector is a bad emitter In the space environment, radiation will be the dominant heat transfer mechanism between the payload & environment

LSU 06/21/2004Thermal Issues10 Emissivity & Absorptivity 1 Kirchoff’s Law of Thermal Radiation: At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity A material with high reflectivity (e.g. silver) would have a low absorptivity AND a low emissivity –Vacuum bottles are “silver” coated to stop radiative emission –Survival “space” blankets use the same principle Kirchoff’s Law requires an integral over all wavelengths Thus, some materials are described as having different absorptivity and emissivity value.

LSU 06/21/2004Thermal Issues11 Emissivity & Absorptivity 2 Manufacturers define absorption and emission parameters over specific (different) wavelength ranges –Solar Absorptance (  s ): absorptivity for 0.3 to 2.5 micron wavelengths –Normal Emittance (  n ): emissivity for 5 to 50 micron wavelengths The Sun, Earth and deep space are all at different temperatures and, therefore, emit power over different wavelengths –A blackbody at the Sun’s temperature (~6,000 K o ) would emit between about 0.3 and 3 microns and at the Earth’s temperature (~290 K o ) would emit between about 3 and 50 microns For space we want to absorb little of the Sun’s power and transfer much of the payload heat to deep space. Want a material with low  s and high  n. –Sherwin Williams white paint has  s of 0.35 and  n of 0.85

LSU 06/21/2004Thermal Issues12 Steady State Solution In a steady state all heat flows are constant and nothing changes in the system Sum of all heat generators is equal to the sum of all heat losses ( Q in = Q out ) Example flow of heat through a payload box wall –Assume vacuum so no convection –Input heat ( Q in ) generated by electronics flows through wall by conduction and is then radiated to space. Q cond = Q in or kA ( T 1 – T 2 ) / L = Q in (1) Q rad = Q in or  A (T 2 4 – T s 4 ) = Q in (2) –Use eq. 2 to determine T 2 and then use eq. 1 to determine T 1 But real systems are never this simple T1T1 T2T2 TsTs Q in Q cond Q rad

LSU 06/21/2004Thermal Issues13 Balloon Environment is Complex Multiple heat sources –Direct solar input (Q sun ), Sun reflection (Q albedo ), IR from Earth (Q IR ), Experiment power (Q power ) Multiple heat sinks –Radiation to space (Q r,space ), Radiation to Earth (Q r,Earth ), Convection to atmosphere (Q c ) Equation must be solved by iteration to get the external temperature Then conduct heat through insulation to get internal temperature Q sun +Q albedo +Q IR +Q power = Q r,space +Q r,Earth +Q c

LSU 06/21/2004Thermal Issues14 Solar Input Is Very Important Nominal Solar Constant value is 1370 W / m 2 Varies ~2% over year due to Earth orbit eccentricity Much larger variation due to solar inclination angle –Depends upon latitude, time of year & time of day Albedo is reflection of sun from Earth surface or clouds –Fraction of solar input depending on surface conditions under payload ATIC-02 data showing effects of daily variation of sun input

LSU 06/21/2004Thermal Issues15 Other Important Parameters IR radiation from the Earth is absorbed by the payload –Flux in range 160 to 260 W/m 2, over wavelength range ~5 to 50 microns, depending on surface conditions –Radiation is absorbed in proportion of Normal Emittance (  n ) Heat is lost via radiation to Earth and deep space –Earth temperature is 290 K o and deep space is 4 K o There is also convective heat loss to the residual atmosphere –Atmosphere temperature ~260 K o For a 8 cm radius, white painted sphere at 100,000 feet above Palestine, TX on 5/21 at 7 am local time with 1 W interior power: Q sun = 9.5 W, Q albedo = 3.7 W, Q IR = 1.6 W Q r,Earth = 0.1 W, Q r,Space = 15.3 W, Q conv = 0.4 W

LSU 06/21/2004Thermal Issues16 Application for BalloonSat Can probably neglect heat loss due to convection and radiation to Earth –Simplifies the equation you need to solve Need to determine if the solar inclination angle will be important for your payload geometry –e.g. a sphere will absorb about the same solar radiation regardless of time of day and latitude Spend some time convincing yourself that you know values for your payload surface  s and  n and your insulation k. Biggest problem will be to estimate albedo and Earth IR input – Use extremes for albedo and IR to bracket your temperature range

LSU 06/21/2004Thermal Issues17 References “HyperPhysics” web based physics concepts, calculators and examples by Carl R. Nave, Department of Physics & Astronomy, Georgia State University –Home page at –Thermodynamics at –Heat Transfer at –Vacuum Flask at –Thermal Conductivity Table at Table of Solar Absorptance and Normal Emmittance for various materials by Dr. Andrew Marsh and Caroline Raines of Square One research and the Welsh School of Architecture at Cardiff University. – Sun, Moon Altitude, Azimuth table generator from the U.S. Naval Observatory –