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CaCO3 breakdown, please help

  1. Feb 28, 2008 #1
    in order for a well to supply 4.5 million litres of drinking water per day, at a minimum total hardness of 60mg l-1 as Ca, calculate the daily throughput of the reverse osmosis plant assuming that 30% of the feed to the plant comes out as reject.

    2. Relevant equations

    a table provides the following info:
    total hardness (as mgl-1 CaCO3) at feed is 660 but at permeate it is 4

    3. The attempt at a solution

    i suspect i must break down CaCO3 into its constituents however i cant remember how to do this. Iwould be very grateful for a pointer in the right direction.[/QUOTE]
  2. jcsd
  3. Feb 28, 2008 #2


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    Your suspicion is half correct if the problem is correctly restated, two different hardness scales. MW Ca+2 is 40, and MW CO3-2 is 60. If you've been able to track the "N" relocations to this point, this is not a chemistry problem, nor an engineering problem, but a math problem; it is best paraphrased as, "Can you set up simultaneous equations?" Pointer enough?
  4. Feb 28, 2008 #3
    thanks for replying, im sorry but i dont understand your answer,
    what is "N" reocations? and what is MW? also how do you get 40 and 60 respectively?
  5. Feb 28, 2008 #4
    please excuse my idioticity, i take it MW is molecular weight so;

    Ca is 40
    CO3 is 1 part C and 3 parts oxygen which is 12 + (16x3) = 60

    again though i dont understand the term "n" relocations? do you mean number?
  6. Feb 28, 2008 #5


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    "Relocations?" This thread has been moved at least once, and I'm thinking more like two or three times --- people post questions, the questions get moved to a more appropriate area, and another moderator moves it again --- "If you've been able to track...." may be translated as, "If you're still able to find the thread ...."

    MW = molecular weight, yes.
  7. Feb 28, 2008 #6


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    First thing you need to do is express the total hardness of the feed in terms of Ca+2 rather than CaCO3.
  8. Mar 3, 2008 #7
    breaking down CaCO3 into its relevent atomic mass;
    CACO3 is 1 part C with an atomic weight of 40 and 3 parts oxygen whose atomic weight is 12 + (16x3) = 60

    Total hardness (as mg l-1 CaCO3)
    660 – 4 = 656
    656 / 660 x 100 = 99.4% total removed by reverse osmosis plant.

    Total hardness of feed as Ca is 660 x 0.4 = 264 mg l-1
    Total hardness of Ca at permeate is = 4 x 0.4= 1.6 mg l-1

    So to achieve 60 mg l-1, the total hardness as CaCO3 mg l-1 must not be more than;
    60 / 0.4 = 150 mg l-1

    i still cant figure out how to translate this into calculating the daily throughput.
  9. Mar 3, 2008 #8


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    The plant will produce water at 4 ppm (expressed as CaCO3). Convert this to ppm as Ca+2. This water will be blended with water at 660 ppm (expressed as ppm CaCO3) to achieve a final mix of around 60 ppm (expressed as ppm Ca+2). You are asked to find the blend ratio of these two water streams. How much 4 ppm water is required to treat 4.5 million liters down to a level of 60 ppm?
  10. Mar 4, 2008 #9
    i really do appreciate all the help and direction Chemistree.......

    so I need to calculate the amount of treated 4ppm water it will take to dilute the 4.5 million litres at 660 CaC03 down to just 60 ppm, i cant wait to get home to get working on this.
  11. Mar 5, 2008 #10


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    Let me rephrase that... How much 4 ppm water will you use to produce 4.5 million gallons of water at 60 ppm and how much 660 ppm water will you use in the process? Don't forget that the 4 ppm plant is only 70% efficient and wastes or throws away 30% of it's output as a stream much higher than 660 ppm Ca..
  12. Mar 6, 2008 #11
    You need to calculate the hardness of the elements involved using the concentrations that are given to you.
    Then you need to multiply by 4.5 (amount of water, leave out the zeros but remember you are working in millions)
    6.6 caco3 (hardness) multiplied by X (?) (not known from the well).
    You should have 150 hardness for the new permeate, multiply this by 4.5
    Now work on the next equation, 315 (70% of 4.5).
    You should now be able to write, the original amount (X), the 0.7x + hardness value.
    This will now result in a simultaneaous equation.
    To learn how to solve these equations go to this web: http://stream.port.ac.uk/ [Broken]
    Last edited by a moderator: May 3, 2017
  13. Mar 12, 2008 #12
    RWW63 have you worked it out ??

    Well, look at this!

    0.7x + y = 4.5
    0.7(4) + y(660) = 4.5(150)

    working out this simultaneaous equation will give you:

    x = 4.99 million litres
    and y = 1.0 ML

    BYE !!
    Last edited: Mar 12, 2008
  14. Mar 15, 2008 #13
    sorry guys i am still struggling to grasp this, Chemistree do you concur with hairypalms calcualtions?
  15. Mar 16, 2008 #14
    From what I can gather after many hours going in circles and after speaking with yourelf;
    Total hardness is 4.5 million litres @ 150 CaCO3 mg l-1
    150 = (0.7 feed x 4) + (Raw x 660)
    Multiply the CaCO3 values by 4.5;
    675 = (0.7 feed x 18) + (Raw x 2970)
    70% of the feed is 462, enter info into equation;
    6,750,000 = (462 x 18) + (Raw x 2970)
    Raw x 2970 = 6,750,000 – 8316
    Raw x 2970 = 6741684
    6741684 / 2970 = Raw
    Raw = 2270

    this is about as far as i can get, can anyone point out where im going wrong? if im on the right track even?
  16. Mar 17, 2008 #15


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    gallons from RO plant = X
    gallons from well = Y

    X+Y = 4.5X10^6 gallons

    Use the weighted average formula to find the relationship between the ratio of RO water to well water (be sure of your units of Ca of course). Solve for 'X'. I haven't done the problem (and never will) so I can't comment on Hairy's answer but if you show me a well worked solution I will find the time and effort to carefully evaluate it for you.
  17. Mar 31, 2008 #16
    Guys I just wanted to say thanks for all the help, I got there in the end, once again thanks for the help and the patients.
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