英國 Labplant 噴霧干燥儀在奶粉中的應(yīng)用
Labplant spray dryer tests
The milk used was reconstituted in the following way:
-200g milk powder
-1.7L of tap water
giving 2L of milk with a measured density of 1.045 at 21’C.
We used a fixed flow, whatever the experiment ; pump flow set at 5, corresponding to
13.5mL/min.
Varying the injection temperature of the product
We did a first test with an injection temperature of 130’C and then a second test at 140’C.
We saw that spray drying was achieved, apparently, comfortably at these two
temperatures.Effectively no liquid ran along the walls of the main spray chamber, even at
130’C. This meant that we could work at 140’C or 130’C given the stipulated flow.
In theory it is preferable to work at 140’C, because the higher the temperature the better
the yield. We will try to prove this through our experiments.
Varying the compressed air ratio / feed flow
We worked with a flow set at 5 (13.5mL/min) and compressed air set at 3 bars
(constant air inlet valve opening).
In theory to increase the size of the agglomerate, it is necessary to favour the agglomeration
mechanism over the drying process. One of the possible means is to decrease the spraying
rate. In the case of this equipment, to decrease the spraying rate you can either decrease the
flow of compressed air through the injection nozzle (while keeping a constant pressure) or
you can decrease the pressure of the compressed air (while keeping a constant flow).
Therefore we tried two tests with constant air and liquid flows, varying the pressure from 2
to 3 bars.We observed the look of the powders we obtained ; it was difficult to decide just
with the naked eye, an additional granulometric(?) study would be necessary, but it did seem
that the powder obtained with 3 bars of pressure was effectively finer than that obtained with
2 bars.
Research into the effective operational limits of the spray dryer
We retained the same solution of reconstituted milk.
At a given flow and pressure of air, we increased the flow of liquid from level 5
(13.5mL/min) to level 10 (28.8mL/min). We very quickly saw that the formation of the
spray in the atomisation tube was not good : in effect the quantity of liquid going through
the tube was too much and could not be vaporised on exiting the tube. This was why we had
some liquid that ran out of the tube, ran along the walls of the spray chamber, of the fan
chamber (cyclone?) and even in the recuperation chamber. Under these conditions the yield
of finished product would be bad.
QUANTITATIVE STUDY
The experiments carried out and the experiment details are given below.
Experiment 1 : starting from 100g/L of reconstituted milk
Amount of milk powder | 200g | |
Amount of water | 1700g | |
Volume of milk | 2L | |
Density of milk | 1.045g/mL | |
Humidity of milk | 89.47 % mas | |
Injection temp (??) | 130’C | |
Injection flow | 13.5mL/min | |
Working time | 40 min | |
Compressed air pressure | 3 bars | |
Humidity of labo | 21.8 %HR | 6g vapour / m3 air |
Ventilator flow | 70 m3/h | |
Gas exit temp | 77’C | |
Air exit humidity | 18.8 %HR | 21.3g vapour / m3 air |
Bottle size | 339g | |
Bottle + wet milk | 391.9 | |
Bottle + dry milk | 390 |
From the experiment details we calculated the following:
-humidity of the milk : 100 x water mass (water mass + powder mass)
-numerical application : % humidity of the milk = 100 x 1700/(1700+200) = approx 89.5%
-the mass of the wet milk we collected = 391.9 – 339 = 52.9g
-the mass of the dry matter we collected = 390 – 339 = 51g
-humidity of the solid = 100 x (52.9 – 51)/52.9 = approx 3.6%
Materials ‘balance sheet’ of the dry milk over the life of the experiment:
-at the start : dry matter is the result of the solution to be tested
-at the exit : dry matter of the solid that was obtained
Numerical application
a) at the start : 13.5mL/min x 1.045 g/mL x 40 min x (100-89.47)/100 = approx 59.4g
b) at the exit : 51g
c) solid yield = 100 x 51 / 59.4 = approx 85.9%
Materials ‘balance sheet’ of the water over the life of the experiment
b) at the start : (13.5mL/min x 1.045 g/mL x 40 min x 89.47 / 100) + 70 m3/h x 6 g/m3 x40/60 = 784.8 approx of water
c) at the exit : (52.9g x 3.6 /100) + (70m3/h x 21.3 g/m3 x 40/60) = approx 995.9
d) water yield = 100 x 995.9 / 784.8 = approx 127%