Cooling System Last update: 2019-08-13
The specific heat of water is around 4200 Joules per Kg per
Kelvin. (you need 4200 Watt-seconds to heat 1 liter of water 1
Let’s say our laser tube dissipates 420 Watt and we have a buffer
vessel with 10 liter water. Now we can operate the laser for 100
seconds before the water becomes 1 °C warmer.
We implemented a buffer vessel from which the water is pumped into
the laser tube, then passes the chiller, then the flow meter, and
back into the buffer vessel. The flow meter is at the end of the
circuit, so a loose hose anywhere in between will be detected.
Temperature measurement devices (DS1820) are placed at the tube’s
input, output and at the chiller’s output.
The chiller can be of various principle. One option is a heat
exchanger with tap water on the other side. After streaming some
time tapwater will arrive at some 10 - 12 °C (at least in our
premises). I did an experiment with a heat exchanger from an old
central heating (gas) unit where it was used for hot tapwater for
the shower etc. In the tapwater circuit was an electric valve from
a dismantled laundry machine. When the temperature on the output
of the heat exchanger was to high the valve was opened shortly.
Another option is a Peltier based chiller. I made one, but due to bad
Peltiers it did not work.
Or a compressor based one. A (second-hand) household refrigerator
or freezer may also do a great job in providing a large amount of
cold water. But do not pump it directly into the Lasertube, the
temperature shock may destroy the tube.
F.t.t.b we will use a compressor based chiller, the type used to
provide small amounts of cool drinking water.
In our situation the chance of temperatures below zero °C is very
low, so we did not use anti-freeze or take other measures to
prevent the laser tube to get damaged by frozen water.
We should not cool the system to such a low temperature that moist
from the surrounding air condenses on any part of the system,
especilay not on the output window of the laser tube. How far can
we cool down given the room temperature and the relative humidity?
Well there is a formula [found on wikipedia] giving the relation:
It is a two step sum:
γ (gamma) = a * T / (b +T) + ln (Rh / 100) a
= 17.27, b = 237.7 °C, T = room temperature, Rh = Relative
Humidity in %, ln = natural logarithm.
Tdew = b * γ / (a - γ) [°C]
We implemented a combined temperature and humidity sensor in the
electronics compartment, send these data to the LaserControl
program on the Pi and calculate how far the system may be cooled
down without the risk of condensation.