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Finding Charge Density of Solution and then Amps

  1. Jan 29, 2009 #1
    Alright, I'm working on an idea for a MHD generator, and I have a water based solution. That is
    Cl- 6250 mg/L or 176 mmol/L
    SO4^2- 6250 mg/L or 65 mmol/L
    Na+ 6250 mg/L or 272 mmol/L
    K+ 6250 mg/L or 160 mmol/L
    Ca^2+ 5000 mg/L or 125 mmol/L

    If I'm doing my math right, that means my solution (including the water) is 56.304 mol/L correct?
    So if that is 56.304 mol/liter then I just multiple by 1000 L / m^3 to get 56,304 mol / m^3 correct?

    Now with this 56,304 mol / m^3, I can multiple by the Faraday Constant (9.65 x 10^4 C / mol) to get (9.65 x 10^4) * 56,304 = 5,433,336,000 C / m^3 correct?

    Then I can use the equation of I = delta Q / delta A = n * v * A . Now v * A = flow, and my flow is .00657 m^3/s
    So it's just I = 5,433,336,000 * .00657 = 35,697,017.52 amps? That seems very very wrong (it's very high).

    Now one thing I found said it's I = q * n *v * A, where q is charge, n is charge density, v is velocity and A is area. That doesn't make sense to me though because q is in Coulombs, n is C / m^3, v is m/s and A is m^2, so when multiplied you'd get I (amps) = c^2 / s. So I'm pretty sure it's just n * v * A, but man that is a lot of amps.

  2. jcsd
  3. Jan 30, 2009 #2
    So it seems one of my errors was in using Faraday's constant. That assumed that all electrons counted, but in reality on the valence electrons are what matter in terms of electricity.

    Now redoing my calculations, and to make it simple I'll only use 1 Chloride ion.

    A Chloride ion has 8 valence electrons. Now we must determine the % of the mass that these 8 valence electrons account for. For these we do weight of 8 electrons divided by the weight of the total mass. We get 8 * 9.1093818E^-31 / (5.887E^-26) = .000124 .

    Now using 1 mg/l of Chloride we know that the percent of the mass of electrons is 1 mg/l * .000124 . That leaves us with .000124 mg / l . To convert this to mg / l we multiple by 1 kg / 1000000 mg = 1.24E^-10 Divide this by the mass of an electron and we get 1.24E^-10 / (9.1093818E^-31) = 1.3612E^20 electrons / liter.

    With this number we can then plug into out equation for I.

    I = q n v * A or I = (1.60217646 * 10^-19) * (1.3612 * 10^20) * (6.57) = 143.29 Amps.

    143.29 amps, based on 1 milligram of chloride ion per liter. Still seems really high. Maybe my flow is a lot higher than I am thinking. 6.57 liters per second is about 1.736 gallons per second, still though.
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