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Tips to Quickly Compare Energy Consumption of Sludge Dryers

Methods to quickly compare energy consumption of sludge dryers under different experssions.

When we talk about a method of sludge dewatering or drying, we take energy consumption as an important performance indicator of that method. However, due to different applications and /or individuals’ preferences, a lot of expressions with different terms are used to describe such performance, which caused confusion and sometimes misunderstanding. For example,
1) take 1200 kWh to get 1 ton of dried sludge with 95% d.s;
2) take 400 kWh to remove 1 ton of water
3) take 390 kWh electricity and 2 tons of steam to get 1 ton of dried sludge with 70% d.s
4) take 180 kWh to dry wet sludge from 80% m.c to 10% m.c
How to evaluate the power performance and compare these numbers become a headache. Simple solution is to compare “apple” to “apple”; if it is an “orange”, convert the “orange” to “apple” and then compare. Here I would like to share some clarifications between “apples” and “oranges” and tips of quick conversion among them based on GreenteQ’s experiences.
Firstly let us have a look of some terms ( skip this part if you already know clearly these terms)
1)     dry solid (d.s) or dry matter (d.m)
d.s = d.m = total solids content = suspended solids + dissolved salts.
The total solids content is expressed as a ratio of weights (expressed as a %) obtained before and after the drying process. There are strictly protocol of how to test the total solids content.
2)     water content (w.c) or moisture content (m.c) , expressed as a %
w.c =m.c = 100- d.s= 100- d.m
3)     Dried sludge
the weight of sludge after the dewatering or drying process. Actually it has to come with a certain d.s or m.c specs or else it is an incomplete term to describe the sludge condition.
4)     Wet Sludge
the weight of sludge before the dewatering or drying process. Similar to the term “dried sludge”, it has to come with a certain d.s or m.c specs or else it is an incomplete term to describe the sludge condition.
5)     water removal (w.r)
the weight loss of sludge before and after the drying process.
w.r= Wet Sludge weight – Dried sludge weight
Come back to the examples in the beginning of this article, we shall bear in mind that a statement without initial m.c or d.s and the end m.c or d.s is incomplete (except the expression using water removal), and additional info are required. After we have adequate information and understand difference of these terms, we can easily find out their relationship with simple calculations. I illustrate the conversion methods with some examples.
E.g. 1. wastewater with 99.99% w.c is dewatered to sludge with 80% m.c
before dewatering: d.s=100-99.99=0.01 ; after dewatering, d.s=100-80=20
since the absolute dry solid doesn’t change before and after dewatering, so the ratio of the weight (wastewater) before dewatering to the weight after dewatering (wet sludge) is 1/0.01: 1/20 = 20:0.01=2000:1=10000:5, i.e., the weight before and after is reduced by factor 2000. so we can say 1) 2000 tons of wastewater with 99.99% w.c becomes 1 tons of sludge with 80% w.c after dewatering , or 2) 10000 tons of wastewater with 99.99% w.c is dewatered to 5 tons of sludge with 80% w.c
E.g. 2. sludge with 80% m.c is dried to sludge with 60% m.c
before drying: d.s=100-80=20 ; after drying, d.s=100-60=40
the weight ratio before to after =40:20=2:1, i.e., so we can say “2 tons of wet sludge with 80% m.c becomes 1 tons of dried sludge with 60% m.c.”. Simply use the ratio number of before minus the ratio number of after to get the ratio number of water removal.
the weight before : the water removal: the weight after = 2 : (2-1) :1= 2:1:1
E.g. 3. Sludge with m.c 60% is dried to sludge with m.c 30%
before drying: d.s=100-60=40; after drying, d.s=100-30=70
the weight before : the water removal: the weight after =70: (70-40):40=7:3:4
i.e., 7 tons of sludge with 60% m.c becomes 4 tons of dried sludge with 30% m.c after removing 3 tons of water.
So if somebody says B tons of sludge takes x kWh to dry, firstly you should clarify the conditions of moisture content before (m.c.b) and after drying (m.c.a), then you can quickly determine the ratio by using the above method and convert the different expressions into a same benchmark, which are used to compare the power consumption. Details are illustrated as following
sludge weight before: water removal: sludge weight after
= (100-m.c. a): (m.c.b – m.c.a) :(100-m.c.b)
= b:(b-a):a  ( assuming the simplified ratio of before to after is b: a )
weight before=B tons; weight after = B x a/b tons; water removal= B x (b-a)/b tons
This method is extendable. If you know any of the sludge weight before, sludge water after or water removal, you can calculate the other two.
E.g. 4. 20 tons of water removal, sludge from 70% m.c to 10% m.c
before: water: after=(100-10): (70-10): (100-70)=9:6:3 =3:2:1
wet sludge weight =20/2*3=30 tons, water removal=20 tons, dried sludge weight=20/2*1=10tons
I end this article with the answer to the beginning comparison. First let us collect the additional information to make the statement complete ( assuming as the red font part)
1) take 1200 kWh to get 1 ton of dried sludge with 95% d.s from wet sludge with 80% m.c;
2) take 400 kWh to remove 1 ton of water;
3) take 390 kWh electricity and 2 tons of steam to get 1 ton of dried sludge with 70% d.s from wet sludge with 85% m.c;
4) take 180 kWh to dry 1 ton of wet sludge from 80% m.c to 10% m.c
 
Since the power consumption per ton of wet sludge or dried sludge is related to the sludge initial and end moisture content, we prefer to use power consumption per ton of water removal as an independent performance indicator of sludge dryers. Based on the table above, we can say the sludge dryer under Exp 4 has the lowest power consumption, which is exactly what GreenteQ has achieved.