document concerning the heating problem

LUXPAK report # 2
Heating problems for LUXPAK
Francis Massen, Claude Baumann
version 1.1
This is a general report concerning the heating problems for the LUXPAK project.
It covers two main chapters:
1. general heating problems of the Luxpak container
2. air flow heating for the ozone sensor
The preliminary results are:
1. The ozone heater will deliver enough heat to maintain the Luxpak inner
temperature well above 0°C
2. The ozone heater will possibly fail to maintain a minimum temperature of 0°C to
the air flow through the ozone sensor when outside temperatures reach -50°C.
1. General heating problems of the Luxpak container
The ascending balloon will encounter quite different temperature zones. In a very
approximate manner, they are:
height (km) approx. density (kg/m3)
approx. temperature (K) and (°C)
(launch-pad) 20
The main question is: what will be the needed electrical heating power inside the
container to keep it's temperature at least at 0°C (273 K) ?
1.1. Simple thermal conduction computation
Let us assume a steady state situation with outside air at 220 K and inside air at 273 K
The thermal conductivity λ of the styropore is about 0.032 (W/(m*K), the thickness e
of the box e = 2.9 cm, and the approx. outer dimensions 300mm*250mm*153mm.Ths
gives an overall area A = 0.32 m2.
The general relation of the thermal conduction is:
Q  * A * t 
* A * t
[eq. 1]
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Rth is the thermal resistance, which can be written as:
Rth = αo + e/λ + αi
The α's correspond to the "bridging" resistances at the outside and the interior.
They depend on a lot of other parameters, as for instance air flow velocity, air density
and viscosity etc.
As an example, Recknagel [1] gives for αi the expression:
i  (4.2* w0.75 ) / d 0.25
for a pipe of diameter d transporting an air flow of normalized velocity w and
suspended in the open.
We will take conservative, probably too low values as used in traditional thermal
isolation calculations for housings [2] (αo is probably much too low):
αo = 15
αi = 8
So Rth = 15 + 0.029/0.032 + 8 ~ 24 [m2*K/W], what gives a thermal conductance of
k = 1/Rth = 0.04 [W/(m2*K)]
Q k * A * t  0.04*0.32*(273  220)  0.68 [W]
So in the worst case we would need to compensate by 0.68 Watt internal heating the
loss of internal heat through the styropore to the outside, when the inside temperature
of the box should be held at 0°C and the outside temperature is -50°C..
1.2. Experimental cooling
Let us compare this theoretical result with practical experiments.
The container was put with an included RCX and other material into a freezer. The
toal mass of the material was about 1 kg, representative for the planned payload.
Outside temperature = 26 °C, freezer = -18°C.
As can be seen from fig.1 (and as established by Fourier), the heat loss is exponential:
a raw, but reasonable approximation is to assume that the cooling is more or less
linear during the first half of the time needed to reach equilibrium.
So we have this situation:
Start temp.
Low temp.
Time to
reach 6°C
6000 s
Rate of
0.003 °C/s
We may assume that the initial rate of cooling is proportional to the difference
between inside and outside temperature; if the temperature of the freezer had been
-40°C, and so the Hi-Lo temperature difference between inside/outside 66°C instead
of 44°C, we would assume a cooling rate of 66/44*0.003 = 0.0045 °C/s
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The experimental cooling rate allows us to compute in a very simple way the needed
electrical power to keep the internal temperature constant (or at least above 0°C), as
the electrical heating power = loss of thermal energy per second
fig.1. Cooling of the Luxpak container (exp. done at the Konvikt)
Internal container temp. = 0°C
Container is at altitude 10km: outside temp. = -44°C
Temperature difference is 44°C
Rate of cooling = 44/44*0.003 = 0.003 °C/s
Internal mass of payload = 1 kg
Approximate mean heat capacity of payload c = 1000 [J/(kg*K)]
Heat loss per second = 1000*1*0.003 = 3 Watt
The power of the ozone air heater is about 12*0.6 = 7.2 Watt. Practically all this
power will be kept inside the container, as that needed to heat the air flow to the
ozone sensor is miniscule.
Conclusion: there is no need for additional heaters; the ozone heater will do the job!
It should be noted that the results from theory and experiment are remarkably close;
remember that the theoretical calculation probably errs on the high side!
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For security measures and to avoid overheating, the RCX should check the internal
temperature of the box and shut of the ozone heater if it exceeds a threshold of for
instance 50°C
The whole problem could be tackled in a more fine-grained manner, just to get a
feeling for the needed heating power at the different altitudes.
Let us divide the ascent into portions of 5 km, and repeat the previous computations
for each level, starting with an initial temperature of 20°C and assuming 0°C as a
minimum box temperature from 10k on:
height (km)
approx. temp.
cooling rate
heating power [W]
(K) and (°C)
to keep box inside at 0°C
The results do not change much to the conclusion found above: we need approx.
between 2.3 and 4.1 Watt electrical heating power to maintain the Luxpak above 0°C.
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2. Heating air flowing through a hot cyclindrical pipe
2.1. The problem
The Luxpak O3 sensor does not operate on air temperatures below 0°C. As the
stratospheric temperature reaches -50 to -60°C, the air flow must be heated. This
should be a gentle operation, avoiding O3 destruction by fins or other sharp edges in
the tubings.
2.2. The device:
air flow ( v = 5-6 m/s)
length l = 20cm
teflon tube
6mm outer diam.
4mm inner diam.
ceramic heating resistors
(6 , 12V, I = 0.6A)
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2.3. Theoretical calculations
The main problem is: what will be the temperature gain of the cold air flowing
through the heater?
2.3.1. Air heater
A solution has been found in Recknagel [1], page 1004.
The formula for an air heater (where air flows around a heated tube) is given as:
tout  tin
ttube  tin
where Φ is given by a diagram (it depends on velocity of the air stream and the
geometry of the heat exchanger).
For v = 5 m/s, a suggested value is Φ = 0.2
If the temperature of the tube is 80°C, this would give for tin = -50° an tout of -24°C.
The table gives several possible situations:
tout if
These calculations assume a more or less "normal" air density; if the air density is
lower, one should expect higher outlet temperatures. So it might possible to get outlet
tempartures close t 0°C even if outside temperatures are -50°C.
A second argument is that in our case the situation is reversed: air flows through the
tube, and the heating depends essentially on how fast the heat flows from the hot
walls of the tube to the inner air. We will try a second calculation using heat
conduction into the tube.
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2.3.2. Air flowing through a hot pipe: first theoretical calculation
Recknagel gives the approximate calculations for a pipe suspended in a room of
given temperature; hot air streams with a given velocity through the pipe. By how
much this air will it cool?
Again, our situation is the reverse of this, but the formulas are independant of the
direction of the thermal flow:
Basic equation:
heat transfered from hot wall = heat gain by the air inside:
Q  k * A* 
 L*c * t
[eq. 5]
where Q' = heat loss in W
k = "k" factor
A = inner surface of pipe
δm = estimated difference between air temperature in pipe and outside
L = mass flow [kg/s] = ρ*S*v
with ρ = air density
S= inner section of pipe
v = velocity of air
If we rewrite this equation, we get:
A 1
* m
S  *v
t  k * 200*
 *v m
t k*
[eq. 6]
A/S = (2*π*r*length)/( π *r2) = 2*length/r = 2* 0.2/0.002 = 200
k is computed as shown in the first part of this report (eq. 2 and 3); we can safely
estimate k between 0.02 and 0.04.
Let us assume δm = 0 -(-50) = 50 °C ( 0 = supposed temperature of 0°C reached in
the tube, -50°C = minimal outside temperature)
With ρ = 1 kg/m3 and v = 5 m/s we get a heating of:
Δt = 0.02*200*1/(1*5)*50 = 40 °C
This means the temperature of the air leaving the pipe would be -50 + 40 = -10°C
If we take k = 0.04, the result is an improbable Δt= 80 °C and tout = + 30 °C
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The crucial parameter is the estimation of δm. Could δm be also taken as (80+0)/2 -(50) = 90 °C, if we replace the expected minimum inside temperature of 0°C by the
mean between the pipe temperature (80°C) and this 0°C ?
In that case we would get Δt = 72, and an inside temperature of +22°C.
In my opinion, the heating by the short tube will not be able to maintain a minimum
temperature 0°C if the velocity of the air through the pipe is as high as 5 m/s
This caution is suggested by the first experimental results discussed in the next
2.3.2. Air flowing through a hot pipe: second theoretical calculation
Let us take a different approach, as suggested by Stöcker ([4]).
The thermal heat flow through a pipe of outer and inner dimaeters do and di is given
2*  * l
*  *(Tout  Tin ) [Watt]
ln( )
[eq. 7]
This thermal power will be taken up (in the best of all worlds!) by the mass flowing
through the tube in one second, i.e. L = ρ*S*v.
The power balance is:
2*  * l
*  *(Tout  Tin )  c * L *(Tin  Tatmosphere )
ln( )
[eq. 8]
Here we have as a single unknown Tin . Let us take the following parameters,
corresponding to our device:
do = 6mm
di = 4mm
l = 0.2 m
λ = 0.25 for teflon (PTFE) [W/(m*K)]
Tout = 80°C = 353K for the heater temperature = outer tube temperature
Tatmosphere = -50°C = 223 K
c = 1000 [J/(kg*K)]
L = ρ*S*v = 1 *(1.26*10-5)*5 =6.28*10-5 [kg/s]
We get: Tin = 344 K = 71 °C !!!
If we assume that only 10% of the heat flow really is transferred to the air, we replace
λ by 0.025; the resulting Tin = 294.6K = 21 °C
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So there is no definitive conclusion from these theoretical calculations!
Everything from about -10°C to + 21°C seems a possible result for the heating!
2.4. First experimental results for the heat flow through the heating pipe
The following figure shows both the temperature of the heater and that of the
outflowing air. The well insulated heater was suspended at room temperature (about
25°C). Air was sucked in from a freezer at -15 °C. It should be noted that for reasons
unknown, the measured outlet temp. never was less than 20 °C ( room temp. was 25
The dt given by a heater at 90°C is about 10°C. The second rise consists first at air
from the freezer being heated, than at the change of slope normal room air was
sucked in.
The heater rises from 25 to 95 °C is about 1700 seconds ( 28 minutes); when shut off,
the inital cooling is quite slow, about 25°C in 1000 seconds (17 minutes).
The conclusion from this is that the heater should be switched on at least 15 minutes
before launch! If the RCX controls the inner temperature, and is able to toggle the
heater, all equipement should be switched on 1 hour before launch, as this gives the
ozone HMOS sensor time to heat up.
The great thermal inertia of the ceramic heaters is the cause of the slow responses;
this should be considered as an advantage!
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The mass flow through the pipe in this first experiment is unknown. Probably the
volume flow is less than 1 liter/min, which amounts to a velocity not exceeding 1.3
m/s (mass flow 1.26 g/min or 2.1E-5 kg/s)
The experiment gives us a temperature rise of approx. 10 °C for a steady state
situation, much less than the result of the theoretical calculation at chapter 3.2
The experiment will be repeated with the following changes:
- usage of an electrical thermometer of low thermal inertia
- replacement of the pump by a syringue, so that air flow can be better estimated
(alternative: try to measure the air flow delivered by the membrane pump)
- focus on air flowing through the heater which has a stable temperature of 80°C
- do not try to use cold air from a freezer; dt should be more or less independant of
intial air temperature.
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Recknagel et al: Taschenbuch f. Heizung + Klimatechnik. Oldenbourg 92/93.
Page 1060
Wolfgang Polifke / Jan Kopitz : Wärmeübertragung. Grundlagen, analytische
und numerische Methode.. Pearson Studium
Stöcker: Toute la Physique. Dunod/ Harri Feutsch, 1998. Page 763
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