A temperature control system

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A temperature control system

for the Alfvén Laboratory Balloon ExpeRimenT

MAGNUS SÖDERQUIST

Master’s Thesis at Space and Plasma Physics, School of Electrical Engineering, KTH, Stockholm, Sweden

Supervisor: Nickolay Ivchenko Examiner: Lars Blomberg

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Abstract

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Introduction

The polarized Gamma-ray Observer (PoGOLite) is a collaboration between Swe-den, Japan, USA, and France. The balloon-borne experiment will measure the polarization of soft gamma rays in the energy range 25 keV-80 keV, which are ex-pected from a variety of sources including rotation-powered pulsars, accreting black holes and neutrons stars, and jet-dominated active galaxies [1]. The aim is observe the northern sky sources including Crab Nebula and Cygnus X-1. Measurement of soft gamma rays polarization at these energy levels, where non-thermal processes are likely to produce high levels of polarization, have never been performed before. Auroral emissions will enhance the background level and to characterize the back-ground level an auroral diagnostics package will be used [2]. The measurement will take place at an altitude of 40 km and the balloon will fly for 24 h after its launch from Esrange, northern Sweden in August 2011.

1.1 The Gondola

The gondola will be attached to the balloon and the polarimeter telescope pressure vessel is mounted in the middle of the structural frame, see figure 1.1. At the top is a flywheel and an azimuth rotator and on the bottom of the frame is the altitude control, electronics, and power supply. The ballast hangs underneath. ALBERT will be mounted on the left side. The dimensions of the base is 3 x 3 m and the height is 4 m.

1.2 ALBERT- Alfvén Laboratory Balloon ExpeRimenT

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CHAPTER 1. INTRODUCTION

ALBERT

Figure 1.1. A sketch of the gondola

pass-frequency is temperature dependent and to get reliable measurements it is therefore crucial to keep a stable temperature.

1.3 The layout of ALBERT

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1.3. THE LAYOUT OF ALBERT

Figure 1.2. The optical equipment for auroral diagnostics

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CHAPTER 1. INTRODUCTION

1.4 Aim of thesis

The aim of this thesis is to design and analyze a temperature control system for ALBERT. Such a system is needed for most technical applications working within narrow temperature bounds and ALBERT is no exception. Since the accuracy of the measurement performed by ALBERT are highly dependent on the temperature stability it is crucial that the temperature control system is working properly. The aims were broken down into the following partial goals

- Analysis of the global thermal environment of ALBERT

- Development of numerical/analytical model

- Assements how design parameters affect temperature stability and heating/cooling load

- Assessment of the heating/cooling device behavior

- Modeling and selection of fans for the PMT and the heating/cooling device

- Selection of final design of temperature control system

- Mechanical design of thermal control system

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Chapter 2

Theory

This chapter will cover the basic theory behind the calculations made in this the-sis. The chapter is divided in to three parts covering heat transfer, pressure loss calculations, and control theory.

2.1 Heat transfer

The section will cover basic heat transfer by conduction, convection, and radiation.

2.1.1 Conduction

In all liquids and solids where a temperature gradient exists conduction will take place. The general equation for conduction in three dimensions is the Fourier’s law, which applies for the Cartesian system.

∂ ∂x kx ρcp ∂T ∂x ! + ∂ ∂y ky ρcp ∂T ∂y ! + ∂ ∂z kz ρcp ∂T ∂z ! + ˙q = ∂T ∂τ (2.1) where

ki is the thermal conductivity in each direction [W/m K],

ρ the density [kg/m3_],

cp the specific heat of the material [kJ/kg K],

T the temperature [K],

˙q the internal heat generation [W/m3_] and τ the time [s].

For steady-state and isotropic materials (2.1) can be reduced to

k ∂ 2_T ∂2_x + ∂2_T ∂2_y + ∂2_T ∂2_z + ˙q ρ · c ! = 0 (2.2)

In many cases the problems can be simplified to one dimension and the parameter of interest is the heat flow Q, in which case the Fourier’s law then becomes

Q = −kA∂T

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CHAPTER 2. THEORY

where

A is the area where the heat transfer takes place [m2].

Where the temperature distribution through the body is not of interest (2.3) can be reduced to

Q = −kAT1−T2

∆x (2.4)

2.1.2 Convection

Convection is a phenomenon that occurs in moving liquids and gases. The cause of the motion or flow could be either difference in density due to temperature gradients (natural convection) or a forced motion by a fan or a pump (forced convection). For the overall effect of convection, the Newton’s law of cooling is used

Q = hA∆T (2.5)

where

h is the heat transfer coefficient [W/m2 _K].

To use this equation the fluid regime must first be characterized, turbulent or lami-nar, and then determine the heat transfer coefficient. This is done by using dimen-sionless numbers.

2.1.3 Dimensionless numbers

Dimensionless numbers are a powerful tool to make heat transfer correlations simple and general. There are plenty of dimensionless numbers and some that are used in this study are presented below.

Reynolds number

The Reynolds number Re is one of the most used dimensionless numbers and it provides information about the boundary layer. For flow between parallel plates the flow regime is fully laminar for Re below 2500 and fully turbulent for Re above 7000. In between is the transition regime which is usually categorized as turbulent [7].

Re = wL

ν (2.6)

where

w is the flow speed [m/s], L the characteristic length [m],

and ν the kinematic viscosity [m2/s].

Prandtl number

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2.1. HEAT TRANSFER

boundary layers [9]. Since the Prandtl number for a fluid only depends on temper-ature (cp and k is temperature dependent) the value is tabulated in the literature [6] or can be calculated by

P r = cp·µ

k (2.7)

where

µ the dynamic viscosity [Ns/m2_].

Gratz number

The Gratz number gives information about whether the laminar flow is fully devel-oped, meaning that the velocity profile is in a steady-state. The number is used for calculations in channels and ducts.

Gz = Re · P rL

de

(2.8)

where

L is the characteristic length [m],

and de the hydraulic diameter [m].

Nusselt number

The number is a way to describe the temperature profile in the boundary layer close to the surface and is based on similarities of the temperature fields and velocity fields [6]. It is used to find the heat transfer coefficient h and is defined as

N u = h · d

k (2.9)

Approximations of the Nusselt number can be found in the literature for specific cases and the characteristic length is dependent on geometry [6]. An example of an approximation that is valid for laminar flow in short circular tubes is

N u = 1.62 _{Re P r d} L (2.10) 2.1.4 Radiation Radiation properties

The incident radiation on a surface can either be absorbed α, reflected ρ, transmitted

τ , or a combination of these three. Each of these is between zero and one and the

sum must be unity.

ρ + α + τ = 1 (2.11)

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CHAPTER 2. THEORY

Black body

Black bodies are idealized bodies that absorb all radiation incident upon them. The absorption of all light is the reason for the name, black body, it appears black to the eye. Energy emitted from a black body, Eb, at a temperature T is given by the Stefan-Boltzmann law

Eb = σT4 (2.12)

where

σ is Stefan-Boltzmann’s constant, 5.669 ·10−8 [W/m2 K4].

Gray body

The definition of a gray body is that the monochromatic emissivity is independent of the wavelength. The monochormatic emissivity is defined as the ratio between the emitted power of the body for each wavelength to the emitted power of a black body at the same wavelength and temperature by

ελ =

Eλ

Ebλ

(2.13)

where

Eλ is the energy emitted by a gray body at a certain wavelength [W],

Ebλ the energy emitted by a blackbody at a certain wavelength [W],

and ελ is the emissivity at a certain wavelength . Thus the gray body definition result in

ε = ελ (2.14)

where

ε is the emissivity for all wavelengths.

Real body

For the real body the monochromatic emissivity is dependent on not only the wave-length but also the temperature. The gray body assumption is usually made when making analysis even though this means that the emissivity may differ considerably from the idealized cases with the black and the gray body. In most real cases, where the radiant environment is not 0 K, the surrounding is radiating to the body, equation (2.12) becomes

Q = Aσε T₁4−α T₂4 (2.15)

where

T1 is the temperature of the body [K]

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2.2. FANS AND PRESSURE LOSS

View factors

The definition of the view factor is the fraction of energy leaving one surface which reaches another surface. View factors are used when calculating the radiative ex-change between two bodies and is a result of the geometry, distance, and angles between the bodies. For simple cases the view factors are listed but in most real cases the geometries are complex and the calculations become quite complicated. When introducing the view factor the net heat exchange becomes

Q = F12Aσ

ε T₁4−α T₂4 (2.16)

where

F12 is the view factor from body 1 to body 2.

2.1.5 Thermoelectrics

A thermoelectric element consists of a closed loop with as junction between dissim-ilar conductors. If one junction is heated a current flow through the loop. This is called the Seebeck effect. If a current is led through the junction a temperature gradient arises between the two conductors. The effect is called the Peltier effect [5]. Themoeclectrics are used in a variety of commercial products, e.g., refrigeration, lab equipment, and cooling of optics.

2.2 Fans and pressure loss

In all systems where there is an air motion there will be variations in pressure. Most of these pressure variations can be characterized as losses. The losses can be divided into two parts; friction losses and dynamic losses. The total loss coefficient

ζtot is the sum of these

ζtot = ζf+ ζd (2.17)

where

ζf is the friction loss coefficient and ζd is the dynamic loss coefficient.

The total pressure loss ∆ptot is calculated from

∆ptot= ζtotρw

2

2 (2.18)

where

ρ is the fluid density [kg/m3]

and w is the fluid velocity [m/s].

2.2.1 Friction losses

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CHAPTER 2. THEORY

and decreasing hydraulic diameter de.

ζf = f ·

L de

(2.19)

The friction factor f is dependent on whether the flow is turbulent or laminar. For the fully developed laminar case the friction factor is only dependent on the geometry and not the roughness of the walls. The friction factor is then defined as

f = C

Re (2.20)

where

the constant value C is 53.3 for a triangular duct, 64 for a circular pipe, 56.9 for square duct and 96 for flow between parallel plates [6] .

For most real applications, where the flow channel is relatively short, fully devel-oped laminar flow is unusual. Since the turbulent flow is dependent on the surface roughness the friction factor has to be determined from case to case. On relation that is approximately valid for fully developed turbulent flow in smooth circular channels is

f = 0.316

Re0.25 (2.21)

Another way to calculation the friction factor is to use the Reynolds analogy which is valid for both laminar and turbulent flow. The analogy expresses the relation between heat transfer and friction loss [7].

f = 8 · St

Λapp (2.22)

where

Λapp is the analogy number,

and St is the Stanton number which is defined as St = N um

ReP r

Λapp is used to indicate that the equation also is valid for developing flow. The analogy number for fully developed flow between parallel plates is 0.9 and there are a number of relations for other geometries and developing flow.

2.2.2 Dynamic losses

Losses that occur when the flow is changing direction are called dynamic losses. If the change is sudden, vortices will form and cause larger losses. Estimates of the dynamic loss coefficient ζ are listed for different flow change cases [7].

2.2.3 Fans and system curves

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2.3. CONTROL THEORY

rise and maximum pressure increase when the flow rate is zero. The correlation between flow rate and pressure increase depend on the fan type. The three main type of fans that can be distinguished are axial, radial, and cross blow. Axial fans are the most common type in cooling of electronics. Radial and cross blow types are used in larger systems, e.g., ventilation.

System curves, i.e., the pressure drop in the system as a function of the flow rate, works the other way around. The pressure drop increases with increasing flow rate and can in a simplified manner be described by

∆p = C · ˙Vn (2.23)

where

C is a constant,

˙

V is the volume flow [m3/s]

and the exponent, n, is theoretically one for pure laminar flow and close to two for pure turbulent flow. By combining the system curve with a fan curve one can gain the operating point which is the intersection point between these two. If the fan gives the desired volume flow at the operating point the fan selection is done other wise this iterative process has to continue until the flow requirements are fulfilled.

2.3 Control theory

A quick introduction to the basics of control theory is presented below, covering open, close loop, and feed forward controllers. When designing a controller for a system, a simplified model of the system is used. For a dynamic system i.e. a system with an inertia, the output y of the system is not only a function of the current input u but also a function of previous inputs

h (t) = H (u (τ ) , τ < T ) (2.24)

Calculations are made in the frequency domain by using the Laplace transform which, for a function h (t), is defined as

H (s) =

Z ∞ 0 e

−st_h(t)dt _(2.25)

where

s = σ + iω, σ and ω real numbers.

2.3.1 Open loop control

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CHAPTER 2. THEORY

2.3.2 Closed loop control

A closed loop controller regulates a system based on the error e between the output

y of the system and the a desired output of the system r, i.e., the reference signal.

The system’s transfer function is denoted G(s) and for a closed loop system with a regulator denoted F(s) the relation between the reference signal R(s) and the output Y(s) becomes

Y (s) = F (s) G (s)

1 + F (s) G (s)R (s) (2.26) There are three common types of regulators. The simplest regulator regulates the system with a input signal that is proportional to the error. This regulator is denoted as the P-regulator and is described by

F (s) = Kp (2.27)

Another way is to integrate the error

F (s) = Ki

s (2.28)

The third way is to regulate on the derivative of the error according to

F (s) = Kds (2.29)

This is an ideal regulator. In reality it is used with a low-pass filter to get the regulator itself stable and then becomes

F (s) = Kds

1 + N s (2.30)

where

N is the filter coefficient.

Combinations of these are usually used. The most common ones are PI, PD and PID.

2.3.3 Feed forward control

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Chapter 3

Preliminary design and calculations

3.1 Orientation and direction

The main target of the experiment is to observe the Crab nebula and the Cygnus X-1, the position of these two and the sun can be viewed in table 3.1. To do the thermal analysis the heat load from the Sun, the Albedo and Earths IR has to be estimated. The heat load from the sun is estimated by calculating the solar angle and the rotational angle, where the solar angle is the angle between the aiming direction and the sun and the rotational angle is the angle between the Sun and the vertical axis of ALBERT . The first step is to calculate the elevation and the azimuth, shown in figure 3.1.

When ALBERT is aiming at the Crab nebula the solar angle becomes 69◦ and the rotational angle is displayed in figure 3.1. The result is that ALBERT can at times be exposed to direct sun light. For the second case, aiming at the Cygnus X-1 the solar angle becomes 129◦ and the rotational angle is displayed in figure 3.1. This results in that when aiming at Cygnus X-1 ALBERT will be shaded by the gondola.

3.2 Thermal environment

3.2.1 Ascent

During the ascent ALBERT will pass through the troposphere (from 0 to 11 km) and then stop in the stratosphere (from 11 to 50 km) at an altitude of 40 km. The duration of the ascent will be 100-120 min [3] and ALBERT will be subjected to

Coordinates Sun Crab Nebula Cygnus X-1 R.A.: 09h₄₀m₃₆s ₀₅h₃₄m₃₂s ₁₉h₅₈m₂₂s Dec: 13◦ 54´ 47´´ 22◦ 00´ 52´´ 35◦ 12´ 06´´

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CHAPTER 3. PRELIMINARY DESIGN AND CALCULATIONS

Figure 3.1. Left top: Elevation and azimuth to Crab, Right top: Elevation and azimuth to Cygnus, Left bottom: Rotational angle to Sun when aiming at Crab, Right bottom: Rotational angle to Sun when aiming at Cygnus

temperature change by the surrounding air, irradiation of the sun, the albedo and Earth’s IR. This will cause heat or cooling loads that can vary a lot during the ascent depending on the weather and altitude of the clouds. The temperature in the air surrounding ALBERT during the ascent was obtained with a model from NASA [11]. The model is an approximation since there are seasonal variations in therms of thickness, temperature and density. The temperature profile in figure 3.2 was calculated with the following equations

T = 59 − 0.0011h 0.555 0 m ≤ h < 11022 m (3.1) T = −70 0.555 11022m < h < 25105 m (3.2) T = 205.5 + 4.9987 × 10 −4_h 0.555 25105 ≤ h (3.3)

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3.2. THERMAL ENVIRONMENT 0 10 20 30 40 50 −60 −50 −40 −30 −20 −10 0 10 20 Altitude [km] Temperature [degree C]

Temperature profile in the atmosphere

Figure 3.2. Atmospheric temperature profile [11]

for the PMT is −20 − +50◦_{C. The temperature should be stabilized at the time} when the balloon has reached its operating altitude so that the observations can start at once.

3.2.2 Observation

During the observation, at an altitude of 40 km, the air pressure will be about 0.3 % of the atmospheric pressure [11], i.e., almost vacuum. As a results of this the convective heat transfer is reduced dramaticlly and will be neglected. Hence only a radiative heat balance study will be performed. The expected heat load from external sources is the Sun, the Albedo and Earth’s IR.

3.2.3 Internal heat generation

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Appliance Power consumption [W]

PMT 0.5 - 6

Stepper motor 0 - 3.4

Fans 0.5 - 3

Electronics 1 - 2

Table 3.2. Power consumption of different appliances

is tested. This is mainly because of the uncertainties of the duty cycle of the stepper motor and the cooling load of the Peltier element for the PMT.

3.3 Radiative heat balance

The first step of the thermal analysis is to make a radiative heat balance for AL-BERT in space, perpendicular to the Sun and Earth, with different surface coatings and without the influence from the PoGOLite. This means that the illuminated area for this case is the same for the Sun, the Albedo, and Earth’s IR. The emis-sivity ε and the absorptivity αs for the different coatings can be found in Appendix D. The values of the Solar constant S, the Albedo factor f , and the Earth’s IR E is tabulated in Appendix B. These values are the normal values and the variation can be rather large. The radiative losses Qlosstospace are equal to the indenced ir-radiation from the Sun QSolar, Albedo QAlbedo and Earths IR QEarthsIR according to

Qlosstospace= QSun+ QAlbedo+ QEarthsIR (3.4)

which can be expanded to

ε σ AeT4 = S · αsAill+ S · f · αsAA+ ε · EAE (3.5)

where

E is the heat flux from Earth’s IR [W/m2_],

Ae is area that is emitting heat [W/m2],

AA is the area illuminated by the Albedo [m2],

AE is the area illuminated by the Earth’s IR [m2], and Aill is the area illuminated by the Sun [m2].

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3.4. AIR CIRCULATION

Case Solar irradiation Shaded Coating Temperature [K] Temperature [K]

White paint 242 207

Black paint 332 251

Polished aluminium 416 303

Table 3.3. Temperature of ALBERT with different coatings

3.3.1 Thermal environment on the gondola

The temperature variation between when ALBERT is shaded by the gondola and when it is in direct sunlight is rather large. When the orientation of ALBERT in relation to the gondola and the sun was evaluated it was found that ALBERT was shaded by the gondola most of the time so total shading of ALBERT was a straight forward decision to reduce temperature variations. This will be done by one plate on each side of ALBERT. The temperature of the gondola is unknown and to further reduce the impact from the gondola on ALBERT the backside of ALBERT will be insulated by thick insulation. The backside of ALBERT is from now on considered as ideally insulated.

3.4 Air circulation

To select a suitable fan for the air circulation the system curve has to be matched to an appropriate fan curve so that a satisfactory operating point is achieved. To obtain the system curve the pressure drop through the heat sink and the pressure drop in the circulation loop have to be calculated. The first step in the design of the air circulation system is to estimate the pressure loss due to the circulation of the air as this is the pressure that the circulation fan has to overcome. The system curve is then calculated by equation (3.6). A principal sketch of the air flow is shown in figure 3.3. The heat transfer coefficient from the air to the wall and from the air to the heat sink will also be calculated in this section. The heat transfer coefficient between air and bench is assumed to be the same as between the air and the wall.

∆ptotal= ∆pairloop+ ∆pheatsink (3.6)

3.4.1 Air circulation loop

A one dimensional model was used to calculate the pressure drop in the air circu-lation loop. The model consists of a duct loop with area Aduct, four branches and a contraction with the area Af an. The total pressure loss ∆p can be calculated by summing pressure losses according to

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Air circ loop

Heat sink Optical setup Pressure vessel Peltier element Fan

Figure 3.3. Principal sketch of the air flow in circulation loop and the heat sink.

The pressure losses from the fan is due to the rapid area change after the fan and is expressed by

∆pf an = ζf anρ

w2_{f an}

2 (3.8)

wf an is the air speed at the fan outlet[m/s].

By assuming that the flow is uniform wf an can be calculated from

wf an= wduct

Aduct

Af an

(3.9)

where

wduct is the speed of the air flow in the duct [m/s].

The branch loss coefficient ζbranch is assumed to be 1.5 and the pressure loss due to the four branches is calculated by

∆pbranch= 4ζbranchρ

w2

duct

2 (3.10)

The friction loss is calculated by assuming that the loop consists of two ducts with the length L because the boundary layers will probably be disturbed by the u-turn in the front. The hydraulic diameter de is the area of the air flow and the constant

C is assumed to be 80 since the geometry of the duct is somewhere between a pipe

and parallel plates. The pressure loss due to friction can be calculated by

∆pf riction = 2ζf riction·ρ ·w

2 duct

2 (3.11)

where ζf riction is calculated from

ζf riction= f ·

L de

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3.4. AIR CIRCULATION 0 0.5 1 1.5 2 0 5 10 15 20 25 Velocity [m/s]

Pressure drop [Pa]

0 0.5 1 1.5 2 0 1 2 3 4 5 6 7 8 9 Velocity [m/s]

Heat transfer coefficient [W/m

2 K]

Figure 3.4. Left: Pressure drop in the air circulation loop as a function of air flow speed, Right: Heat transfer coefficient from air to wall in the pressure vessel as a function of air velocity

f from f = C Re (3.13) and Re from Re = w · L ν (3.14)

The heat transfer coefficient can be calculated from (3.18) when combining the Gratz number (3.16) and the Nusselt number (3.17) [7]. In most cases it is con-venient to use the inlet temperature difference rather then the mean logarithmic temperature which involves the outlet temperature. The transition between N um and N u0 can be made according to

N u0= Gz 4 1 − e−4N umGz (3.15) Gz = Re · P rL de (3.16) where

L is the characteristic length [m],

and de the hydraulic diameter [m].

N um= C1+ 0.0298Gz 1.37

1 + 0.0438Gz0.87 (3.17)

where

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h = N u · k

de

(3.18)

The heat flux from the wall to the fluid is assumed to be constant. The cross section of the duct is somewhere between a pipe and a rectangle so the constant C1 is assumed to be 4.

The heat transfer coefficient h calculated from (3.18) is increasing rapidly when the air speed is increasing, see figure 3.4.

3.4.2 Heat sink

Since the Peltier element is working against the heat sink and the efficiency of the element is highly dependent on the temperature differential it is important that the heat sink has a low thermal resistance. Less than 1 K/W is desired to not compromise the efficiency of the Peltier element. This means that the heat sink needs a large heat exchange area and should be made of a material with good thermal conductivity. The construction is usually a plate with several fins made out of Aluminium or Copper. Calculations where made to ensure that the selected heat sink meet the requirements and to obtain the pressure loss through it.

The flow between the fins can be assumed to be spatial, i.e., flow between plates with narrow gap in between, and therefore the Reynolds number Re can be calculated by

Re = 120 w s (3.19)

where

w is the air velocity [m/s],

and s is the wall to wall distance [mm].

The hydraulic diameter de for a heat sink is calculated from

de = 4

s H

s + 2H (3.20)

The friction factor can be calculated from

f = 8 · St

Λapp

(3.21)

where Λapp is calculated from

Λapp= Λ − (Λ − 0.5)1 − e−0.008Gz (3.22) and Λ from Λ = 0.9 _H/s H/s + 11 0.19 (3.23)

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3.4. AIR CIRCULATION 0 0.5 1 1.5 2 2.5 3 0 5 10 15 20 25 Velocity [m/s]

Pressure drop [Pa]

0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 3.5 Velocity [m/s] Thermal resistance [K/W]

Figure 3.5. Left: Pressure drop in the heat sink as a function of air flow speed, Right: Thermal resistance of the heat sink as a function of air velocity

∆pheatsink = ζin + f L de ζout ρw 2 2 (3.24)

where ζin is calculated from

ζin= 0.42 1 − _s t 2! (3.25) where

s is the spacing between the fins [mm]

and t is the thickness of the fins [mm]. and ζin from

ζout = 1 −

_s

t

2!2

(3.26)

Figure 3.5 shows that the pressure drop is increasing with increasing air velocity. The heat sink used for the initial Peltier test was, after calculation, found to match the thermal requirements at air velocities higher than 0.4 m/s. The heat sink also meets the dimensional requirements.

The heat sink is assumed to be isothermal and the Nusselt number can be calculated by equation (3.17) where C1 is dependent on the shape of the heat sink according to C1= 2.98 + 4.56 1 − eH/s−18 (3.27)

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CHAPTER 3. PRELIMINARY DESIGN AND CALCULATIONS 0 0.05 0.1 0.15 0.2 0.25 0.3 0 5 10 15 20 25 30 35 40 45 50

Volume flow [m3/min]

Pressure drop [Pa]

Figure 3.6. Pressure drop in the air circulation loop with heat sink as a function of the volume flow

3.4.3 Fan

The fan was selected by matching the system curve to a suitable fan curve. The sys-tem curve was calculated by the equation below and is shown in figure 3.6. The fan curve of the selected fan is shown in figure 3.7 which is the San Ace 109P0512A702 [14]. The intersections of the fan curve and the system curve is the operating point at the different voltage input to the fan. When the fan is operating at 12 V the velocity through the heat sink becomes 2 m/s and the thermal resistance becomes 0.19 K/W and when the fan is operating at 10 V the velocity through the heat sink becomes 1.6 m/s and the thermal resistance is 0.23 K/W. The fan fulfills the thermal requirements at 10 V.

3.5 Thermal model

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3.5. THERMAL MODEL

System curve

Figure 3.7. The operation point at different voltages i.e. the intersection between the fan curve and the system curve.

Bench

Air

Heat sink Insulation

Pressure vessel

Insulation

Space Earth Albedo

Figure 3.8. A principal sketch of the thermal model

heating the surface on the outside of the second insulation, denoted as AA in the equations below, while heat is radiating to cold space and Earth from the same area, denoted Ae. Heat is gained or dissipated from the outside of the second insulation to Earth depending on the temperature of it.

The heat transfer from or to the outside of the second layer of insulation was calculated by

Qloss= QAlbedo+ QEarthsIR+ Qlosstospace (3.28)

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QAlbedo = Sf αsAA (3.29)

QEarthsIR from

QEarthsIR= εσ (1 − F12) Ae

T_Earth4 ′_sIR−T_out4

(3.30) and Qlosstospace from

Qlosstospace = εσF12Ae

T_space4 −T_out4 (3.31)

The heat transfer between the nodes inside the vessel was calculated by

Qloss=

Tout−Tin

Rtot

(3.32)

The thermal resistance for each node can be found in Appendix E.

The view factor F12 was varied between 0.2 and 0.9 and the inside insulation thickness t1 was varied between 1-10 mm in the simulation. The outside insulation thickness was fixed to 10 mm. The insulation material used in the calculation is polystyrene and can be found in Appendix C. Two different coatings were simulated; white paint and polished aluminium. If ALBERT is painted with white paint then there will be a heat loss and with polished aluminium there will be a heat gain. The white paint results in greater variations in the heat flow, see figure 3.9. An increase in insulation thickness results in less difference in the heat loss for each view factor independent of the coating. If there is no albedo then the heat loss with the white paint does not change significantly but for the case with polished aluminium the heat gain turns into a heat loss, see figure 3.9.

3.6 Testing of Peltier element

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3.6. TESTING OF PELTIER ELEMENT 200 220 240 260 280 300 320 −40 −30 −20 −10 0 10 20

Outside wall temperature [K]

Heatflow

[W

]

Q

s pace+Qalbedo+QE arths IR−view factor dependent Q_{los s}−insulation thickness dependent

200 220 240 260 280 300 320 −40 −30 −20 −10 0 10 20

Heatflow

[W

]

Q

s pace+QE arths IR−view factor dependent Q_out−insulation thickness dependent

230 240 250 260 270 280 290 300 310 320 330 −20 −15 −10 −5 0 5 10 15 20

Heatflow

[W

]

Q_{s pace}+Q_albedo+Q_{E arths IR}−view factor dependent Q_{los s}−insulation thickness dependent

230 240 250 260 270 280 290 300 310 320 330 −20 −15 −10 −5 0 5 10 15 20

Heatflow

[W

]

Q_{s pace}+Q_{E arths IR}−view factor dependent Q_out−insulation thickness dependent

Figure 3.9. Left top: White paint with albedo, Right top: White paint without albedo, Left bottom: Polished aluminium with albedo, Right bottom: Polished alu-minium without albedo

element, first cooling it, then switching polarity and heating it. Measurements were also performed when the aluminium piece only was cooled and heated by the surrounding. This was done repeatedly with different magnitude of the current.

The results were exported to MATLAB and then processed. Figure 3.12 shows an example of a temperature logging at a current of 0.5 A. Figure 3.12 shows the absolute value of the voltage, current and power. The voltage is increasing with increasing temperature when heating and with decreasing temperature when cooling and reaches maximum when the Peltier element is operating at its temperature extremes. The temperature derivative was calculated for each case. The derivative for the case with cooling and heating from the surrounding was subtracted from the other cases to get temperature derivative of the Peltier element without any heat losses. By doing this the heat flow Q caused by the Peltier element can be obtained according to

Q = mc∂T

∂t (3.33)

where m is the mass [kg],

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Figure 3.10. The test setup used for testing the peltier element

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3.6. TESTING OF PELTIER ELEMENT 0 500 1000 1500 2000 2500 3000 10 15 20 25 30 35 40 Time [s] Temperature [C] 0 500 1000 1500 2000 2500 3000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Time [s]

Current [A] Voltage [V] Power [W]

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CHAPTER 3. PRELIMINARY DESIGN AND CALCULATIONS −30 −20 −10 0 10 20 30 40 50 60 −6 −4 −2 0 2 4 6 8 Delta T [C] Power [w] 0.5A 0.5A 1A 1A 2A 2A 3A 3A 10 15 20 25 30 35 40 0 0.5 1 1.5 2 2.5 3

Temperature difference over the Peltier element [K]

Thermal conductivity k [W/K m]

Figure 3.13. Left: The power consumption as a function of current and temper-ature, Right: Thermal conductivity of the peltier element for different temperature differences

and∂T

∂t is the time derivative of the temperature [K/s].

The energy dissipated or absorbed from the Peltier element, shown in figure 3.13, is the energy when there are no losses to the surrounding and no heat losses through the Peltier element. The energy dissipated from the Peltier element at a certain ∆T is higher than the energy absorbed at the same ∆T , see figure 3.13. This is because the energy that is put in to the Peltier element is dissipated on the hot side.

To be able to predict the energy that the Peltier element can move when losses occur through the Peltier element an additional test was performed. The aim of the test was to evaluate the thermal conductivity of the element. The test setup was similar to the setup used to test the performance of the Peltier element but in this test the element was used as a passive component. The top of the aluminium piece was heated by a resistive wire, see figure 3.14, and insulated with mineral wool. The temperature was measured on both side of the Peltier element at steady-state and the thermal conductivity was calculated from:

k = Q t

A ∆T (3.34)

where

A is the area of the Peltier element [m2]

and t is the thickness of the element[m].

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3.6. TESTING OF PELTIER ELEMENT

Figure 3.14. The test setup used for testing the heat transfer through the Peltier element

is defined as the cooling or heating power divided by the power input is shown in figure 3.15. The COP of the Peltier element decreases with increasing temperature difference and increasing current.

Thermal model with Peltier element

By implementing the results from the Peltier tests in the thermal model it is possible to make rough assessments of the size of the Peltier element that is needed and also which temperature range the element has to operate in. The thermal resistance for the Peltier element can be found in Appendix E. The results were used to select another Peltier element for a second test, see section 4.4.

Test of the PMT thermal properties

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CHAPTER 3. PRELIMINARY DESIGN AND CALCULATIONS −30 −20 −10 0 10 20 30 40 50 60 −10 −8 −6 −4 −2 0 2 4 6 Delta T [C]

Power with heat loss through peltier [W]

0.5A 0.5A 1A 1A 2A 2A 3A 3A −30 −20 −10 0 10 20 30 40 50 60 −20 −15 −10 −5 0 5 10 15 20 Delta T [C]

Coefficient of performance COP [−]

0.5A 0.5A 1A 1A 2A 2A 3A 3A

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3.6. TESTING OF PELTIER ELEMENT 0 50 100 150 200 250 300 350 −10 −5 0 5 10 15 20 25 Time [s] Temperature [C]

Figure 3.16. The temperature of the PMT sensor when the Peltier element is switched off

temperature results in lower temperature difference between the PMT sensor and air and should result in lower power consumption.

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CHAPTER 3. PRELIMINARY DESIGN AND CALCULATIONS 0.8 1 1.2 1.4 1.6 1.8 2 −12 −10 −8 −6 −4 −2 0 2 4 Current [A] Temperature [C] 0.8 1 1.2 1.4 1.6 1.8 2 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Current [A] Power [W]

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Chapter 4

Final design and calculations

4.1 Air circulation

To ensure that the cooling of the PMT is sufficient one fan is mounted at the PMT blowing air into the heat sink of the PMT. This fan is the one supplied by the PMT manufacturer. The fan selected in the section 3.4.3 is mounted in the back and will cool the heat sink of the Peltier element. The fans run in parallel and push the air forward on one side of the optical setup and drawing it back on the other side, see figure 4.1

4.2 Insulation

The insulation material chosen for the final design is the Aerogel Spaceloft, see Appendix C for properties. It has excellent thermal properties and is manufactured as blankets available in 5 and 10 mm thickness. The aim was to fit 10 plus 5 mm on the inside but this was not possible everywhere since some of the optical equipment is located closer than 15 mm to the inside of the pressure vessel. Therefore simulations were also made with thinner insulation to show how the insulation thickness affects the heating load. On the outside of the pressure vessel 5 and 10 mm insulation thickness is simulated. For the final design the 5 mm insulation was selected.

4.3 Heating and cooling

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CHAPTER 4. FINAL DESIGN AND CALCULATIONS

Figure 4.1. The flow path of the circulated air

Peltier element is switched off the circulation fan should also be switched off or the fan speed should be reduced to a minimum. This is to reduce the losses through the Peltier element when it is not used. The fan, cooling the Peltier element of the PMT, is then taking care of the air circulation. The internal heat generation was assumed to be 8 W in the final calculations.

4.4 Test of selected Peltier element

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4.4. TEST OF SELECTED PELTIER ELEMENT −30 −20 −10 0 10 20 30 40 −20 −15 −10 −5 0 5 10 15 20 Delta T [C]

Coefficient of performance COP [−]

0.5A 0.5A 1A 1A 1.5A 1.5A 2A 2A 10 15 20 25 30 35 40 0 0.5 1 1.5 2 2.5 3

Temperature difference over the Peltier element [K]

Thermal conductivity k [W/K m]

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Insulation Space Albedo Earth

Shield Shield

Figure 4.3. A principle sketch of the thermal model

4.5 Thermal model

The thermal model was revised in order to improve the accuracy and to model the transients during the flight.

4.5.1 Radiative model

The heat loss from the outside wall to space was reformulated to a network model that include the shadings. The model is shown in figure 4.3 and describes the radiative balance between ALBERT, the shadings, Space, and Earth. The network model is based on a radiant-energy balance on each surface and expressed in terms of radiosities Ji [9]. The shadings are insulated and are not subjected to a heat flux which means that the surface is in radiative balance, resulting in Ebi = Ji. Since the Space and Earth temperatures are constant this is modeled as radiation from a black body with the emissivity of one, Ebi= Ji .

Ji−(1 − εi) X

j

FijJj = εiEbi (4.1)

Equation (4.2) is valid for all surfaces that are insulated Ji = Ebi and are in radiant equilibrium. Ji = 1 1 − Fii X j6=i FijJj (4.2)

Equation (4.3) is valid for all surfaces with a specified heat flux.

qi=

Aiεi

1 − εi (Ebi−Ji) (4.3)

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4.5. THERMAL MODEL 200 220 240 260 280 300 320 −40 −30 −20 −10 0 10 20

Heatflow

[W

]

Q

s pace+Qalbedo+QE arths IR−view factor dependent

Q

los s−insulation thickness dependent

Figure 4.4. The heatflow from ALBERT

   1 −2 (1 − ε1) F12 εi F12 1 − F23 0 −1 0 1   ·    J1 J2 Eb1   = (4.4)    (1 − ε1) (F14J4+ F15J5qAlbedo) F24J4+ F25J5 (1−ε1)Q_loss Aillε1 + F12αs1qAlbedo ε1    (4.5)

The result is shown in figure 4.4 and the range of the outside wall temperature that has been estimated will be used in the simulations below. The outside temper-ature can vary between 230K and 275K. The insulation thickness has large impact on the heat loss from the vessel but does not affect the temperature on the outside significantly.

4.5.2 Conductive model

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Appendix E and the data from the test of the Peltier element were used to model the Peltier element.

∂Ti ∂t = 1 mc _∂Q i−1 ∂t + ∂Qi ∂t + ∂Qi+1 ∂t (4.6) ∂Qi−1 ∂t = Ti−Ti−1 Req (4.7)

4.6 Control system

The control system will consist of several regulators controlling different parts and will be PID and PD regulators with feed-forward. A sketch of the system principle is shown in figure 4.5. The PID and PD regulators where simulated and designed using Matlab Simulink. Simulink, as Labview, is a graphical programming language for solving differential equations. It is a powerful tool when simulating dynamic and transient systems that are regulated by control systems.

4.6.1 Temperature control system for the PMT

To model the PMT the thermal resistance and thermal mass had to be estimated. This was done by assuming that the characteristic of the Peltier element was similar to the one tested in section 3.6 and that the size can be scaled linearly. By using (4.8) with ∆T from figure 3.16 and QP eltier from figure 3.13, 3 A, scaled with a factor of 2/3. This gives a thermal resistance of 18 K/W. The thermal mass was es-timated by dividing temperature time derivative, when the PMT was heated by the surrounding, with the power from the Peltier element according to equation (3.33). The estimations of the thermal resistance and mass were the used in the simulation and the model was validated and adjusted with the data from the test shown in 3.16. The result is shown in figure 4.6. A PID controller for the temperature control of the PMT Peltier element was designed in Simulink and the PID parameters are

KP = −7.49 · 10−2, KI = −4.10 · 10−3, KD = −4.87 · 10−1 and N = 1.39 · 10−1.

R = ∆T

QP eltier

(4.8)

4.6.2 Temperature control system for ALBERT

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4.6. CONTROL SYSTEM Insulation Air Bench PD PID PID Heater Heater Peltier Heat exchange Heat exchange Feed forward PID PMT Feed forward Reference Switch T_air T_bench T_PMT Reference Reference

Figure 4.5. A principle sketch of the control system

Q = Ti−Ti−1

R (4.9)

The control system will switch off the Peltier element when the temperature difference between the heat sink and wall exceeds 30 K, since it will be more efficient to heat the air due to the large heat loss through the Peltier element. The regulator parameters are shown in table 4.1.

4.6.3 Results from Simulink simulations

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CHAPTER 4. FINAL DESIGN AND CALCULATIONS 0 50 100 150 200 250 300 350 −10 −5 0 5 10 15 20 25 Time [s] Temperature [C] Experimental data Simulated data

Figure 4.6. Comparison of experimental data with the data from the model.

Bench heater Air heater Peltier element

KP 14.06 5.07 4.48 ·10−2

KI 9.78·10−5 - 2.21·10−4

KD -1.15·103 -16.97 -9.67 ·10−2

N 9.30·10−3 9.30·10−3 1.58·10−3

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4.6. CONTROL SYSTEM 0 1 2 3 4 5 6 7 8 9 x 104 200 220 240 260 280 300 Time [s] Temperature [K] T Bench T Air T wall T outside wall 0 1 2 3 4 5 6 7 8 9 x 104 0 10 20 30 Time [s] Power [W] Bench Heater Peltier Total Air Heater

Figure 4.7. The energy consumption and temperatures in ALBERT with 5mm Spaceloft insulation on the outside and a outside temperature of 210K

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CHAPTER 4. FINAL DESIGN AND CALCULATIONS 0 1 2 3 4 5 6 7 8 9 x 104 200 220 240 260 280 300 Time [s] Temperature [K] T Bench T Air T wall T outside wall 0 1 2 3 4 5 6 7 8 9 x 104 0 10 20 30 Time [s] Power [W] Bench Heater Peltier Total Air Heater

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4.6. CONTROL SYSTEM 0 1 2 3 4 5 6 7 8 9 x 104 270 280 290 300 Time [s] Temperature [K] T Bench T Air T wall T outside wall 0 1 2 3 4 5 6 7 8 9 x 104 0 5 10 15 20 Time [s] Power [W] Bench Heater Peltier Total Air Heater

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CHAPTER 4. FINAL DESIGN AND CALCULATIONS 0 1 2 3 4 5 6 7 8 9 x 104 275 280 285 290 295 Time [s] Temperature [K] T Bench T Air T wall T outside wall 0 1 2 3 4 5 6 7 8 9 x 104 0 5 10 15 Time [s] Power [W] Bench Heater Peltier Total Air Heater

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4.6. CONTROL SYSTEM 0 1 2 3 4 5 6 7 8 9 x 104 200 220 240 260 280 300 Time [s] Temperature [K] T Bench T Air T wall T outside wall 0 1 2 3 4 5 6 7 8 9 x 104 0 10 20 30 Time [s] Power [W] Bench Heater Peltier Total Air Heater

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Chapter 5

Summary and Conclusions

In this thesis an analysis of the thermal environment of ALBERT has been per-formed, a final design and a control system have been suggested. It was found, in the preliminary design, that there was a risk that ALBERT would be exposed to Sun light and had to be shaded to reduce the temperature range of ALBERT. Due to the low air pressure at observation altitude only a radiative heat balance study was performed. It showed that with polished aluminium finish the heat load varied between -5 and +5 W and for the white paint finish the heat load varied between -10 and -35 W. The white paint was selected since there was no risk of heat gains. The San Ace 109P0512A702 was selected as the circulation fan. A Peltier element was tested and was found to work very efficiently at low current. The model was revised for the final design and calculations and simulations were made in Simulink. A larger Peltier element, the Supercool PE-127-10-13-S, was tested and was found to be suitable. The Aerogel Spaceloft was selected as the insulation material. At 275 K the energy consumption for both the case with 5 mm and 10 mm insulation on the outside becomes 13 W. The difference is the pressure vessel wall tempera-ture. For the case with 10 mm, it keeps increasing during the whole simulation. This could mean that if there is an error in the model of the Peltier element the cooling will not be sufficient with 10 mm insulation. With both thicknesses the temperature and the power limit are kept at 210 K. Therefore the vessel should be insulated with 5 mm.

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CHAPTER 5. SUMMARY AND CONCLUSIONS

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Appendix A

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Magnus Söderquist soderqu@kth.se

KTH-ALBERT-20091118 Version: 1.0

Pressure test of the ALBERT pressure vessel

Introduction

The pressure vessel which will contain the auroral diagnostics package was pressure tested on the 13-18 th of november. During the balloon flight, scheduled for August 2010, the vessel will contain Nitrogen at 1 atm while the outside walls will be subjected to nearly vacuum at an altitude of 40 km. The test was performed to ensure that the vessel is not leaking.

Pressure vessel

The pressure vessel consists of three main parts: the tube, the lid and the bottom. The lid and the bottom seal the tube which a O-ring and is tighten with bolted connections. Two windows are mounted on the lid, sealed with O-rings and attached with flanges. On the lid is also a electrical connector mounted and sealed with a O-ring. The walls of the pressure vessel is 6 mm aluminium. The test setup are shown in figure 1 and the drawings of the pressure vessel is attached as PDF-files. Experimental setup

One of the windows where replaced with a aluminium dummy of the same dimensions to be able to connect the helium supply. The helium supply consists of a helium tube with a low and high pressure manome-ter and regulator valve. The setup is shown in 1.

Method

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Figure 1: The test setup

leaks was performed. The vessel was then depressurized, repressurized again and left for another 24 hours. A last search for leaks was then performed with the leak detector.

Results

The first search for leaks showed no indications of leakage from the pres-sure vessel. There was a small amount of of leakage on the manometer but it was considered to not have a significant effect on the test. After the first pressure cycle the pressure loss was less than 2 % and there no indications of leakage. There where no indcations from the bla of leakage during and after the second pressure cycle. The pressure loss after the second pressure cycle was less than 1 %. There where no signs of deformation of the structure of the vessel.

Conclusion

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Appendix B

Solar data

Solar constant S = 1450 [W/m2_] Albedo factor f = 0.33

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Appendix C

Material properties

Material Aluminium T6 Copper Pure Polystyrene Spaceloft

Density [kg/m3] 2780 8933 20 150

Specific heat [kJ/kg K ] 880 385 -

-Thermal conductivity [W/m K] 186 389 3.55 · 10−2 1.36 · 10−2

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Appendix D

Surface properties

Coating Black paint White paint Polished aluminium

Emissivity 0.84 0.86 0.07

Absorptivity 0.97 0.17 0.13

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Appendix E

Thermal resistance

Rair,bottom= 1 hAbottom (E.1) Rair,lid= 1 hAlid (E.2) Rair,wall1 = 1 hAin (E.3) Rbottom = t1 k1Abottom (E.4) Rlid= t1 k1Abottom (E.5) Rwall1= t1 k1Am1 (E.6) Rlid2= t2 k2Abottom2 (E.7) Rwall2= t2 k2Am2 (E.8) 1 Rintowall = 1 Rair,bottom+ Rbottom + 1 Rair,lid+ Rlid + 1 Rair,wall1+ Rwall1 (E.9) 1 Rwalltoout = 1 Rlid2 + 1 Rwall2 (E.10)

Rtot = Rintowall+ Rwalltoout (E.11)

Amax=

2πroutl

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APPENDIX E. THERMAL RESISTANCE Aill= 2πroutl 2 + πr 2 out (E.13)

Abottom = (rin−t1)2π (E.14)

Abottom2 = (ro+ t2)2π (E.15)

Am1= Aout1−Ain1 lnAout1 Ain1 (E.16) Am2= Aout2 −Ain2 lnAout2 Ain2 (E.17) RP eltier = tP eltier kP eltierAP eltier (E.18)

AP eltier = WP eltier2 (E.19)

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BIBLIOGRAPHY

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A temperature control system