Finding Charge Density of Solution and then Amps

In summary, the conversation discusses the calculations and errors involved in designing a MHD generator using a water-based solution with various ions. The correct equation for determining the current is determined to be I = q * n * v * A, where q is charge, n is charge density, v is velocity, and A is area. The final calculation using 1 milligram of chloride ion per liter results in a current of 143.29 amps, which may still seem high.
  • #1
Shelnutt2
57
0
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.

Thanks
 
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  • #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.
 
  • #3
for sharing your calculations and thought process. It looks like you are on the right track with your calculations, but there may be a few errors in your calculations and units.

First, let's address the charge density of your solution. You are correct in multiplying the molar concentration by the volume conversion factor to get the charge density in units of C/m^3. However, your final answer of 5,433,336,000 C/m^3 is incorrect. It should be 5,630,400 C/m^3. This is because you also need to account for the charge of each ion species. In this case, Na+ and K+ both have a charge of +1, while Ca^2+ has a charge of +2. So the total charge density would be (272 + 160 + 250) * (9.65 x 10^4 C/mol) * (56.304 mol/m^3) = 5,630,400 C/m^3.

Next, let's address your calculation for the current (I). The equation you provided, I = n * v * A, is correct. However, you have used the wrong value for the charge density (n). Using the corrected value of 5,630,400 C/m^3, and substituting in the given values for velocity (v) and area (A), we get I = (5,630,400 C/m^3) * (0.00657 m^3/s) * (1 m^2) = 36.97 amps. This is still a high current, but it is closer to the correct value.

It is important to always double check your units and calculations to ensure accuracy. Also, make sure you are using the correct values for charge density and taking into account the charge of each ion species. Best of luck with your MHD generator project!
 

Related to Finding Charge Density of Solution and then Amps

1. What is charge density and why is it important in scientific research?

Charge density refers to the amount of electric charge per unit volume of a material or solution. It is an important concept in scientific research because it helps us understand the behavior of electrically charged particles and their interactions with each other.

2. How do you calculate the charge density of a solution?

The charge density of a solution can be calculated by dividing the total charge of the solution by its volume. This can be expressed using the formula: ρ = Q/V, where ρ is the charge density, Q is the total charge, and V is the volume of the solution.

3. What is the unit of measurement for charge density?

The unit of measurement for charge density is coulombs per cubic meter (C/m³). This is because charge is measured in coulombs (C) and volume is measured in cubic meters (m³).

4. How is charge density related to amps?

Charge density and amps (amperes) are related because amps represent the rate of flow of electric charge. In other words, amps measure the amount of charge passing through a given point in a circuit per unit time. Therefore, the higher the charge density, the higher the amps in a solution.

5. How can charge density be manipulated or controlled in a solution?

Charge density can be manipulated or controlled by adjusting the amount or concentration of charged particles present in the solution. This can be done by adding or removing ions or molecules with a specific charge, or by changing the temperature or pH of the solution, which can affect the dissociation of charged particles.

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