Samson TRUCK MASTER 1700, Инструкция по эксплуатации, Страница: 14 / 36

DOC9091B |
14
Truck Master 1700
3.2 Air cooling with fan cooler
Compression of air inside the liquid ring pump will create heat that is transferred to the water.
Therefore, it can be necessary to install a fan cooler depending on the expected use, the climate etc. The
time it takes to heat up the water also depends on for example ambient temperature, suction pressure,
amount of water and the cooling effect in the truck itself.
The operation temperature will go up until there is a balance between the heat input and the heat output.
So basically, there are only two things that can lower the operation temperature. Reduce the heat input or
increase the heat output.
The amount of water has no or only a little influence on the final operating temperature. The truck itself
will work as a big radiator and if there is a huge amount of water in the liquid separator and thereby good
contact area between the water and the steel tank, it will give a higher cooling effect. This in combination
with low ambient temperature and short time of operation, could mean that the truck can operate without
any additional cooling.
In general, the time it takes to heat up the water can be calculated from the formula below.
t
sec
=
C
p
x
m
x
Δt
Q
t
sec
= Time in seconds
C
p
= Heat capacity of the media. Water= 4,2
Δt = Temperature difference
m =
Mass of the media heating up [Kg]
Q = Heat input in [kW] See specifications, chapter 2.2
5 kW 10 kW 20 kW
Small liquid separator mounted
external from the truck tank
Water content below 100 L
Small liquid separator inside
slurry tank. Located with only
minor contact to the product
Water content 300 L
Normal liquid separator inside
slurry tank with good contact to
the product
Water content 400 L
With a temperature difference on 20°C it is typical to have a radiator affect in a truck on somewhere
between 5 to 20 kW depending on the construction.
The table below shows truck radiator effect at a temperature difference of 20°C.
Example:
We have a tank with 300 litres of water corresponding to 300 kg. The heat input is 30 kW.
How long will it take to heat it up from 20°C to 40°C ?
t
sec
=
4,2
x
300
x
20
= 840 s = 14 min
30
The temperature will continue to go up until the steel construction can absorb the heat and transfer it to
the surroundings.