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Math Help - Word problem help,

  1. #1
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    Word problem help,

    The slurry that leaves the reactor on a plant manufacturing phosphoric acid contains 20% w/w solids together with a solution of phosphoric and sulphuric acids
    In water. The solution consists of 38% w/w H3PO4 , 2.5% w/w H2SO4 , the
    Remainder being water.
    The slurry is filtered and all the solids are removed in the filter cake. The filter
    cake that is produced contains 50% w/w solids and 50% w/w liquid. The filter
    Cake is then re-slurried with water and subjected to a second filtration. The second
    Filter cake contains 55% w/w solids and 45% w/w liquid. In this case also, all the
    Solids are present in the filter cake.
    If 1% of the phosphoric acid present in the initial slurry is retained in the liquid
    Trapped in the second filter cake then, using a basis of 1000 kg initial slurry,
    Calculate:
    (a) The composition of the first filter cake,
    (b) The composition of the second fillter cake,
    (c) The amount of water used to reslurry the first filter cake.
    I have drawn a diagram, and I still not sure how to calculate the composition of the first filtration,

    From the question it says that the slurry contains 20% solids, so of the 1000kg, 200kg is solid and the rest must be liquid, of which 38% is H3PO4 and 2.5% is H2SO4. Which means 59.9% must be water.

    Not sure how to go about this, any help appreciated
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  2. #2
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    of the 1000kg, 200kg is solid and the rest must be liquid, of which 38% is H3PO4 and 2.5% is H2SO4. Which means 59.9% must be water.
    It's 59.5% that is the percentage of water.

    First, you can calculate the mass of H3PO4 in the original 1000kg slurry and thus find 1% of the phosphoric acid present in the initial slurry. This will be used later.

    It is not clear from the problem text in what proportions the components of the solution are retained by the filtration. I mean, there is 2.5% w/w H2SO4 in the solution, but filtration theoretically may retain all H2SO4 and some water, but none of H3PO4. I don't think it is possible to calculate anything without this information.

    Most likely, however, filtration preserves the ratio of the solution components. Since the solids account for 50% of the first cake, the cake's mass is 400kg, and the solution's mass is also 200kg. Using the proportions above, you can find the mass of each solution component.

    After that let x kg of water has been added. Then the concentrations of the three components of the liquid solution in the new slurry can be expressed through x.

    In the second cake, the same 200kg of solids constitute 55%, so you know the mass of the cake and the mass of its liquid part. Therefore, using the concentration of H3PO4 you found, you can calculate the mass of H3PO4 as an expression of x. Finally, it is equated to the 1% of the H3PO4in the initial slurry that you found in the beginning to form an equation for x.
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  3. #3
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    Wink

    Quote Originally Posted by emakarov View Post
    It's 59.5% that is the percentage of water.

    First, you can calculate the mass of H3PO4 in the original 1000kg slurry and thus find 1% of the phosphoric acid present in the initial slurry. This will be used later.

    It is not clear from the problem text in what proportions the components of the solution are retained by the filtration. I mean, there is 2.5% w/w H2SO4 in the solution, but filtration theoretically may retain all H2SO4 and some water, but none of H3PO4. I don't think it is possible to calculate anything without this information.

    Most likely, however, filtration preserves the ratio of the solution components. Since the solids account for 50% of the first cake, the cake's mass is 400kg, and the solution's mass is also 200kg. Using the proportions above, you can find the mass of each solution component.

    After that let x kg of water has been added. Then the concentrations of the three components of the liquid solution in the new slurry can be expressed through x.

    In the second cake, the same 200kg of solids constitute 55%, so you know the mass of the cake and the mass of its liquid part. Therefore, using the concentration of H3PO4 you found, you can calculate the mass of H3PO4 as an expression of x. Finally, it is equated to the 1% of the H3PO4in the initial slurry that you found in the beginning to form an equation for x.
    Thank you,

    I also quite confused with the wording of the problem, because when it says all of the solids is removed, does that mean, the filtration contains 800kg of the liquid, which is than separated into 50% solid 50% liquid? which would mean, there is still the original 200kg of solids that is not in the filter.

    before filtration:
    mass of solid = 0.20 x 1000 = 200kg
    mass of liquid = 0.80x1000= 800kg

    of that 800kg liquid:
    304kg H3PO4
    20kg-H2S04
    479.2kg-H20

    right, now how do I work out the proportions that is in the first filtration?

    The correct answers for part a is: 200kg solids, 76kg H3PO4, 5kg H2S04, and 119kg of H20

    Is it half of 800kg is liquid and half solid?

    800 x 0.5 = 400kg than work out the percentages of acid in that amount of liquid, which does not give the correct answer.


    How did you work out that the cake's mass is 400kg and that the solution's mass is 200kg?
    Last edited by Tweety; October 27th 2010 at 01:42 PM.
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  4. #4
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    I also quite confused with the wording of the problem, because when it says all of the solids is removed, does that mean, the filtration contains 800kg of the liquid, which is than separated into 50% solid 50% liquid?
    The first cake contains all the solids and some of the liquid. Namely, 200kg liquid of the original 800kg (see below).

    of that 800kg liquid:
    304kg H3PO4
    20kg-H2S04
    479.2kg-H20
    Again, there is 59.5% of water, which is 476kg.

    Is it half of 800kg is liquid and half solid?
    No, 800kg is completely liquid.

    How did you work out that the cake's mass is 400kg and that the solution's mass is 200kg?
    Since all of solids (200kg) are in the first cake and they constitute 50% of the cake, the cake is 400kg and its liquid part is 400 - 200 = 200kg.

    right, now how do I work out the proportions that is in the first filtration?
    Of the original 800kg liquid, 200kg is still in the first cake. This is 1/4 of the liquid in the original slurry. If the concentrations of H2S04, H3PO4 and H2O in the liquid part of the slurry and in the liquid part of the cake are the same, you can multiply 304, 20 and 476 by 1/4 to find the component masses of the liquid part of the cake. This is the same as the answer you have.
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  5. #5
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    Since all of solids (200kg) are in the first cake and they constitute 50% of the cake, the cake is 400kg and its liquid part is 400 - 200 = 200kg.
    Sorry, still a little confused, Is that 400kg of liquid and solid? So the mass of the first filter cake is 400kg? How do you know this?

    to work out the amount of solid present in the first filter cake; you could just set up this equation.

    200kg = 0.5S

    solving for s = 400kg

    wont you do the same for the liquid?

    800kg = 0.5L
    L = 1600kg

    I know this does not work, because than the total is more than 1000kg, but as the question says 50% of liquid and solid, why are you not just taking 50% of the original masses?



    Thank you
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  6. #6
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    Actually, I understand now why is it 200kg, liquid and 200kg solid. For the second part, we know that 55% is the orignal 200kg of solid, but from that how would I know the mass of the liquid? is it 0.45 x 800 = 360 ?
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  7. #7
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    Let C2 be the mass of the second cake. We know that 55% of it is 200kg (solids). Therefore, 0.55 * C2 = 200, from where C2 = 200 / 0.55 = 364 kg (approximately). So, the second cake has 364 - 200 = 164 kg of liquid.

    Now suppose that x kg of water was added during re-slurrying. Then the mass of liquid in the second slurry is (200 + x) kg. Since there was 76kg of H3PO4 in the second cake, the concentration of H3PO4 in the liquid part of the second slurry is 76 / (200 + x). H3PO4 has the same concentration in the liquid part of the second cake, so the second cake has 164 * 76 / (200 + x) kg of H3PO4. Finally, we are told that this is 1% of the amount of H3PO4 in the original slurry, or 304kg. From here, x can be found.
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  8. #8
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    Quote Originally Posted by emakarov View Post
    Let C2 be the mass of the second cake. We know that 55% of it is 200kg (solids). Therefore, 0.55 * C2 = 200, from where C2 = 200 / 0.55 = 364 kg (approximately). So, the second cake has 364 - 200 = 164 kg of liquid.

    Now suppose that x kg of water was added during re-slurrying. Then the mass of liquid in the second slurry is (200 + x) kg. Since there was 76kg of H3PO4 in the second cake, the concentration of H3PO4 in the liquid part of the second slurry is 76 / (200 + x). H3PO4 has the same concentration in the liquid part of the second cake, so the second cake has 164 * 76 / (200 + x) kg of H3PO4. Finally, we are told that this is 1% of the amount of H3PO4 in the original slurry, or 304kg. From here, x can be found.



    Hello,

    Can you please post a full solution for this problem, as I am not able to follow through. I don't quite understand, why do you not multiply by the amount that the question has given?

    if you could please start from the beginning.

    Thank you.
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