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

LSU 06/21/2004 Thermal Issues 1

Payload Thermal Issues & Calculations

Ballooning Unit, Lecture 3


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


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)


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


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


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.


Convection

• Requires a temperature difference and a working fluid to transfer energy

Qconv = h A ( T1 – T2 )• The temperature of the surface is T1 and the temperature of

the fluid is T2 in Ko

• The surface area exposed to convection is given by A in m2

• The coefficient h depends on the properties of the fluid.– 5 to 6 W/(m2 Ko ) for normal pressure & calm winds– 0.4 W/(m2 Ko ) or so for low pressure

• In the space environment, where air pressure is at a minimum, convection heat transfer is not very important.


Conduction

• Requires a temperature gradient (dT/dx) and some kind of material to convey the energy

Qcond = k A ( dT / dx )• The surface area exposed to conduction is given by A in m2

• The coefficient k is the thermal conductivity of the material.– 0.01 W/(m2 Ko ) for styrofoam– 0.04 W/(m2 Ko ) for rock wool, cork, fiberglass– 205 W/(m2 Ko ) for aluminium

• Need to integrate the gradient over the geometry of the conductor.– Q = k A ((T1-T2)/L) for a rod of area A and length L– Q = 4 k r (r + x) ((T1-T2)/x) for a spherical shell of radius r and

thickness x


Radiation

• Requires a temperature difference between two bodies, but no matter is needed to transfer the heat

Qrad = A ( T14 – T2

4 )

• The Stefan-Boltzmann constant, , value is 5.67 x 10-8 W/m2 K4

• The surface area involved in radiative heat transfer is given by A in m2

• 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


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.


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 Ko) would emit between about 0.3

and 3 microns and at the Earth’s temperature (~290 Ko) 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


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 ( Qin = Qout )

• Example flow of heat through a payload box wall– Assume vacuum so no convection

– Input heat ( Qin ) generated by electronics flows through wall by conduction and is then radiated to space.

Qcond = Qin or kA ( T1 – T2 ) / L = Qin (1)

Qrad = Qin or A (T24 – Ts

4 ) = Qin (2)

– Use eq. 2 to determine T2 and then use eq. 1 to determine T1

• But real systems are never this simple

T1T2

Ts QinQcond

Qrad


Balloon Environment is Complex

• Multiple heat sources– Direct solar input (Qsun), Sun

reflection (Qalbedo), IR from Earth (QIR), Experiment power (Qpower)

• Multiple heat sinks– Radiation to space (Qr,space),

Radiation to Earth (Qr,Earth), Convection to atmosphere (Qc)

• Equation must be solved by iteration to get the external temperature

• Then conduct heat through insulation to get internal temperature

Qsun+Qalbedo+QIR+Qpower = Qr,space+Qr,Earth+Qc


Solar Input Is Very Important

• Nominal Solar Constant value is 1370 W / m2

• 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


Other Important Parameters

• IR radiation from the Earth is absorbed by the payload– Flux in range 160 to 260 W/m2, 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 Ko and deep space is 4 Ko

• There is also convective heat loss to the residual atmosphere– Atmosphere temperature ~260 Ko

• 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:

Qsun = 9.5 W, Qalbedo = 3.7 W, QIR = 1.6 W

Qr,Earth = 0.1 W, Qr,Space = 15.3 W, Qconv = 0.4 W


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


References• “HyperPhysics” web based physics concepts, calculators and examples by Carl

R. Nave, Department of Physics & Astronomy, Georgia State University– Home page at http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#hph

– Thermodynamics at http://hyperphysics.phy-astr.gsu.edu/hbase/heacon.html#heacon

– Heat Transfer at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html#c1

– Vacuum Flask at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/vacfla.html#c1– Thermal Conductivity Table at http://hyperphysics.phy-astr.gsu.edu/hbase/tables/thrcn.html#c1

• 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.

– http://www.squ1.com/index.php?http://www.squ1.com/materials/abs-emmit.html

• Sun, Moon Altitude, Azimuth table generator from the U.S. Naval Observatory– http://aa.usno.navy.mil/data/docs/AltAz.html

http://hyperphysics.phy-astr.gsu.edu/hbase/heacon.html#heacon

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

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