The biological treatment of Total Organic Carbon (TOC) in water uses activated carbon adsorption.The adsorption Accords with Freundlich Equation, K=0.007 , n=1.13. In the equation,the unit of qe is mg.mg/C, and the unit of Ce is mg/L.To dispose 10L wastewater with 20mg/l TOC, 1g activated carbon is applied. At equilibrium of adsorption, what is capacity of equilibrium adsorption? (mg.mg/C). Note: (based on material balance for every litre of water 100qe + Ce= 200.)

Question

(NOTE: Based on the material balance, for every litre of water 100qe + Ce =200).

Answer 1

To calculate the capacity of equilibrium adsorption ($qe$) based on the given relationships, we will use the information provided in the material balance equation and the Freundlich equation.

1. Material Balance Equation: $$100qe+Ce=200$$

2. Initial Conditions:

   • Volume of wastewater = 10 L
   • Initial concentration of TOC ($C_i$) = 20 mg/L
   • Amount of activated carbon used = 1 g

3. Final Water Volume: Since we are disposing of 10 L of wastewater, we focus on the concentration in the final volume of 10 L.

4. Equilibrium Concentration ($Ce$): According to the material balance from our equation: $$Ce=200-100qe$$

   Since we are dealing with 1 g of activated carbon and modifying the original concentration based on $qe$, we can multiply $qe$ by the number of grams of carbon to find the equation in terms of total mass adsorbed to the carbon.

5. Instead of focusing on total mg/L in the 10 L volume, we can express everything in terms of 1 L, leading to:
   For 1 g of activated carbon, we need to understand the mass transfer relationship involving the adsorption capacity.

6. Total TOC in 10 L: $$\text{Total TOC}=20\text{ mg/L}\times 10\text{ L}=200\text{ mg}$$

   This is the total mass of TOC that exists before adsorption.

Now, we need to understand that in the presence of $qe$, where $qe$ is the amount adsorbed per unit mass of activated carbon.

7. Assuming no other breakdown or reaction, we can consider: At equilibrium and with the given relationships in the material balance for 1 L of water:

   For 1 g of activated carbon: $$100qe+Ce=200$$

   Substituting $Ce=20-10qe$ into this balance will yield a straightforward equation to use:

   • For every 1 g of activated carbon (or 1000 mg), the equation results into adsorption related to the mass of carbon and its active sites.

To determine $qe$:

1. Substitute various $qe$ back to see if $Ce$ stays zero; because $$Ce$$ simply reduces the TOC.

2. For equilibrium: $$(100\times qe)+Ce=200$$ Equating with the total throughput flowing through.

Rearranging for computation:

1. Solve from: $$Ce=200-100qe$$

2. $qe$ can be seen simply as: With empirical observations, let's just push through for: Assume TOC goes down to the limits of:
   Iterate: targeting a $Ce=0$ which yields the maximum capacity cleared up.

3. Test for practical conservations. (Final estimation ultimately rounds normally yielding to exact specifics.)

Hence solving iteratively finally leads up, yielding: $$100qe+0=200⟹qe=2.0\text{ mg.mg/C}$$

Thus conclude stability at equilibrium: Conclusion: Based on material balance, calculation shows $qe=2.0$ mg/mg of carbon.