Thread: mass of water, word problem

A mass of 100 kg of filter cake contains 55 wt% solids and 45 wt% solution, where the solution consists of 32 wt% H3PO4, 7.5% w/w H2SO4 , the remainder being water. The filter cake is reslurried with water and subjected to a second filtration. The second filter cake contains 60 wt% solids and 40 wt% liquid, and all of the solids from the original filter cake are present in the new cake. If 5.7% of the phosphoric acid present in the original filter cake is retained in the liquid trapped in the second filter cake, calculate the mass of water used to reslurry the filter cake. Give your answer in kg to an accuracy of three significant figures.

In the first filter cake there is 55% solids,

solid in first filter cake = 0.55 x 100 = 55kg
Liquid in first filter cake = 0.45 x 100 = 45kg

so solution composition in first filter cake =
H3PO4 - 0.32 X 45 = 14.4 Kg
H2SO4- 0.075 X 45 = 3.375 kg
H20- 0.605 x 45 = 27.225 kg

where would I go from here?

Are we to assume, since 5.7% of the phosphoric acid is retained, that, in fact, 5.7% of the original liquid is retained? If not, I see no way to do this. If so, since you know there were 45kg of liquid, there must be .057(45)= 25.65 kg of the original liquid in the second cake. You know the mass of solids in the second cake and their percent so you can calculate the mass of the second cake. You know the percentage of the second cake that is water so you can calculate the mass of the water in the cake and, subtracting of the mass of water retained, calculate the mass of "new" water in the cake.

Originally Posted by HallsofIvy

Are we to assume, since 5.7% of the phosphoric acid is retained, that, in fact, 5.7% of the original liquid is retained? If not, I see no way to do this. If so, since you know there were 45kg of liquid, there must be .057(45)= 25.65 kg of the original liquid in the second cake. You know the mass of solids in the second cake and their percent so you can calculate the mass of the second cake. You know the percentage of the second cake that is water so you can calculate the mass of the water in the cake and, subtracting of the mass of water retained, calculate the mass of "new" water in the cake.

so the mass of the second cake ; 55kg = 0.60x

x = 91.7 kg?

Is this the mass of the second filter cake?

How do you know the percentage of water in the second cake?

Yes, this is the total mass of the second cake.

I will keep the assumption we made about the percentage of liquid retained.

You know the mass of solution in the second cake brought from the first cake that is 25.65 kg.

You can find the mass of solution in the second cake using the percentage of solution in the second cake.

Mass of solution = (55/60%) x 40% = 36.67 kg

The 'extra' mass of solution is from the water added to the second cake.

Originally Posted by Unknown008

Yes, this is the total mass of the second cake.

I will keep the assumption we made about the percentage of liquid retained.

You know the mass of solution in the second cake brought from the first cake that is 25.65 kg.

You can find the mass of solution in the second cake using the percentage of solution in the second cake.

Mass of solution = (55/60%) x 40% = 36.67 kg

The 'extra' mass of solution is from the water added to the second cake.

I am really not understanding this, can you please post a full solution so that I can follow from the beginning to the end.

Thank you.

In the first filter cake there is 55% solids,

Solid in first filter cake = 0.55 x 100 = 55kg
Liquid in first filter cake = 0.45 x 100 = 45kg

In second filter cake there is 60% solids,

Sold in second filter cake = solid in first filter cake = 55 kg.

55 kg represents 60% of mass of second filter cake.
The rest 40% is liquid, and is equal to (55/60%) x 40% = 36.67 kg

5.7% of liquid from first filter cake is retained.
Assuming that this is the same for all the liquid in the first filter cake, it means that 5.7% of 45 kg of liquid is kept.

This is equivalent to = 45 x 0.057 = 2.565 kg
(I just realised that HallsOfIvy did a typo in his post)

So, in the liquid of the second filter cake, we have 2.565 kg of liquid from the first cake.

Since there is a total of 36.67 kg of liquid, the mass of water added must be 36.67-2.565 = 34.1 kg

Thank you, that makes sense. However the book says the correct answer is 598?

598 kg or water!? The least I can say is that it's quite a lot...

Maybe the assumption we made was not good... but the question is definitely strange without that assumption.

Hi Tweety,
I calculate as follows
The second cake contains 36.7kg of fluids 35.7 kg water .8kg phosphoric .2kg sulfuric concentration of phosphoric in fluids = .8/36.7 = .0218 wt fraction
This is the conc of the filtrate
13.6 /.0218 = wt of filtrate 624 kg 624 -13.6 - 3.1 = 607 kg water added
May need refinement


bjh

Can you say where you got 13.6 and 3.1 and what are they exactly?

Hi unknown008,
The first cake contains 14.4 kg of phosphoric acid and 3.3 kg of sulfuric acid. If the reslurry and filtration reduces the former to .8 kg the latter will be reduced in the same proportion or to .2kg. Therefore 13.6 kg of the former and 3.1 kg of the latter must appear in the filtrate.
Reason I said some refinement may be required is that I made no correction for the difference in the water contents in the two cakes



bjh