with an sg of 1.2 kg/m^3 , 0.2 grams is off product so 0.2 * 40m^3/h(flow rate) = 8 t/h(dry product), that's what i would think is the correct answer, but i have a person that doesn't agree with me, so im merely looking for an alternative solution or a is this correct.
OK Let me understand this, I'll assume metric tonnes.
You have 40 tons an hour of a wet product with an SG of 1.2
For every 1.2kg you have 1kg of water and 0.2kg of product,
The ratio is 1/6 and 5/6 or 16.66% and 83.33%
So each ton of wet product has 833.33 KG of water and 166.66 kg of product,
40 ton gives 40 X 166.66 kg gives 6.666 tons of product
and 40 X 833.33 of water gives 33.333 tons.
Or you could just transpose the formula and say (40-(40*(1/1.2)
Just like Kev77
I'm interested to know the figure that the person you disagree with has?
Chris
Find one of these unknown variables, and then it will be possible to solve your problem. Someone at your company should have one or both. If you are already removing the water, then the SG of the dry product should already be known. Weigh a known volume of the dry product, and then divide by the equivalent weight of that same volume of water. Now you can use this SG-dry ratio to find the amount of water in the wet product. Subtract the water and you are left with the dry tons of product.You are missing one key piece of information before you can answer your question. You either need the specific gravity of the dry product, or the percent by volume of water for the mixture.