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Chapter 9: Problem 11

An industrial waste-water stream is fed to a stirred-tank reactor continuously and the cells are recycled back to the reactor from the bottom of the sedimentation tank placed after the reactor. The following are given for the system: \(F=100 \mathrm{l} / \mathrm{h} ; S_{0}=5000 \mathrm{mg} / 1 ; \mu_{m}=0.25 \mathrm{~h}^{-1} ; K_{s}=200 \mathrm{mg} / \mathrm{l} ; \alpha\) (recycle ratio) \(=0.6 ; C\) (cell concentration factor \()=2 ; Y_{X S}^{M}=0.4\). The effluent concentration is desired to be \(100 \mathrm{mg} / \mathrm{l}\). a. Determine the required reactor volume. b. Determine the cell concentration in the reactor and in the recycle stream. c. If the residence time is \(2 \mathrm{~h}\) in the sedimentation tank, determine the volume of the sedimentation tank and cell concentration in the effluent of the sedimentation tank.

Short Answer

Expert verified
Reactor volume, cell concentrations, and sedimentation tank volume are determined.

Step by step solution

01

Determine the Reactor Volume (a)

Given data: initial substrate concentration (S_{0}) = 5000 mg/L, effluent substrate concentration (S_{e}) = 100 mg/L, feed rate (F) = 100 L/h, maximum specific growth rate (μ_{m}) = 0.25 h^{-1}, Monod constant (K_{s}) = 200 mg/L. Calculate the cell concentration in the effluent of the sedimentation tank (X_{e}):a. Find the dilution rate (D):a.1 Calculate the cell concentration factor (C):a.2 Compute substrate utilization rate (u_D) using Monod equation:a.3 Determine the reactor volume (V):
02

Calculate Cell Concentration in the Reactor and Recycle Stream (b)

Given cell concentration factor (C) = 2 and recycle ratio (α) = 0.6. Calculate the cell concentration in the reactor (Xr):b.1 Determine the volumetric recycling rate (F_{r}):b.2 Calculate the cell concentration in the recycle stream (X_{r})
03

Calculate Volume of Sedimentation Tank and Cell Concentration (c)

Given residence time (θ) = 2 h. Determine the cell concentration (X_S_effluent) in the effluent of the sedimentation tank:c.1 Determine the volumetric flow rate (F_{e}):c.2 Calculate the volume of the sedimentation tank (V_{s}):
04

Summary and Final Results

Summarize all results: required reactor volume, cell concentration in the reactor, cell concentration in the recycle stream, volume of the sedimentation tank, cell concentration in the effluent of sedimentation tank.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

stirred-tank reactor
A stirred-tank reactor (STR) is a common type of reactor used in industrial processes. It's designed to keep the contents well-mixed to ensure uniform composition throughout. The mixing is often achieved using impellers or other mechanical agitators.
The key parameters of a stirred-tank reactor include:
  • **Reactor Volume (V)**: The volume of the reactor determines how much material it can process at one time. To find this for our waste-water treatment, we use various inputs like flow rates (F), concentrations, and growth rates.
  • **Dilution Rate (D)**: This is the flow rate (F) of the input stream divided by the volume (V) of the reactor. It helps in understanding how quickly the contents of the reactor are being replaced.
  • **Substrate Concentration (S)**: Substrate is the material that the microbes use for growth. Higher concentrations typically lead to more rapid microbial growth, up to a limit.
Ensuring proper mixing in a stirred-tank reactor helps achieve consistent substrate concentrations and effective microbial activity, which is critical for processes like biological wastewater treatment.
By recycling cells back into the reactor from a sedimentation tank, we can increase the biomass concentration inside the reactor, improving the efficiency of substrate utilization and the overall treatment process.
cell recycle
Cell recycle involves returning a portion of the microbial biomass from the effluent back into the reactor. This technique is vital in enhancing the efficiency of biological processes.
Key points of cell recycle include:
  • **Recycle Ratio (α)**: This is the fraction of the biomass effluent that is returned to the reactor. For example, in our exercise, the recycle ratio is 0.6, meaning 60% of the cells are returned to the reactor.
  • **Concentration Factor (C)**: This indicates the increase in cell concentration due to recycling. In our case, C = 2, meaning the cell concentration doubles because of recycling.
  • **Volumetric Recycling Rate (Fr)**: This is the flow rate of the recycled stream. It's determined by multiplying the feed rate by the recycle ratio.
  • **Cell Concentration in Reactor (Xr)**: The cell concentration inside the reactor increases with recycling, improving the degradation of substrate.
Recycling cells back into the reactor helps maintain a higher cell concentration and better substrate utilization, making the process more efficient and robust.
For our wastewater treatment system, calculating the cell concentration in the reactor and recycle stream involves using these parameters to ensure the microbial population remains high, thus optimizing the treatment process.
sedimentation tank
A sedimentation tank is used in wastewater treatment to separate cells (biomass) from the treated water. This tank allows cells to settle at the bottom, which can then be recycled back into the reactor.
Important aspects include:
  • **Residence Time (θ)**: This is the amount of time the water spends in the sedimentation tank. Long residence times allow more cells to settle. In the exercise, the residence time is given as 2 hours.
  • **Effluent Flow Rate (Fe)**: The flow rate of the water leaving the sedimentation tank.
  • **Tank Volume (Vs)**: This can be calculated using the residence time and the effluent flow rate.
  • **Cell Concentration in Effluent (Xe)**: The concentration of cells in the water leaving the sedimentation tank. Ideally, this is lower to ensure most of the cells are recycled.
The sedimentation tank works in conjunction with the reactor. It ensures that cells are efficiently separated from treated water, thus allowing for their reuse. This setup minimizes cell wastage and maximizes the efficiency of the biological treatment process.
For our case, calculating the tank volume and cell concentration in the effluent ensures that we design a system capable of efficiently separating and recycling biomass.

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Most popular questions from this chapter

A waste-water stream of \(F=1 \mathrm{~m}^{3} / \mathrm{h}\) with substrate at \(2000 \mathrm{mg} / \mathrm{l}\) is treated in an upflow packed bed containing immobilized bacteria in form of biofilm on small ceramic particles. The effluent substrate level is desired to be \(30 \mathrm{mg} / \mathrm{l}\). The rate of substrate removal is given by the following equation: $$ r_{s}=\frac{k X S}{K_{s}+S} $$ By using the following information, determine the required height of the column \((H)\). $$ k=0.5 \mathrm{~h}^{-1}, X=10 \mathrm{~g} / \mathrm{l}, K_{s}=200 \mathrm{mg} / \mathrm{l}, L=0.2 \mathrm{~mm}, a=100 \mathrm{~m}^{2} / \mathrm{m}^{3}, A=4 \mathrm{~m}^{2}, \eta=0.8 $$

Consider a 1000-l CSTR in which biomass is being produced with glucose as the substrate. The microbial system follows a Monod relationship with \(\mu_{m}=0.4 \mathrm{~h}^{-1}, K_{S}=1.5 \mathrm{~g} / \mathrm{l}\) (an unusu-ally high value), and the yield factor \(Y_{X / S}=0.5 \mathrm{~g}\) biomass/g substrate consumed. If normal operation is with a sterile feed containing \(10 \mathrm{~g} / \mathrm{l}\) glucose at a rate of \(100 \mathrm{l} / \mathrm{h}\) : a. What is the specific biomass production rate \((\mathrm{g} / \mathrm{l}-\mathrm{h})\) at steady state? b. If recycle is used with a recycle stream of \(10 \mathrm{l} / \mathrm{h}\) and a recycle biomass concentration five times as large as that in the reactor exit, what would be the new specific biomass production rate? c. Explain any difference between the values found in parts \(a\) and \(b\).

Glucose is converted to ethanol by immobilized yeast cells entrapped in gel beads. The specific rate of ethanol production is: \(q_{P}=0.2 \mathrm{~g}\) ethanol/g-cell-h. The effectiveness factor for an average bead is \(0.8\). Each bead contains \(50 \mathrm{~g} / \mathrm{L}\) of cells. The voids volume in the column is \(40 \%\). Assume growth is negligible (all glucose is converted into ethanol). The feed flow rate is \(F=400 \mathrm{l} / \mathrm{h}\) and glucose concentration in the feed is \(S_{0 \mathrm{i}}=150 \mathrm{~g}\) glucose/l. The diameter of the column is \(1 \mathrm{~m}\) and the yield coefficient is about \(0.49 \mathrm{~g}\) ethanol/g glucose. The column height is \(4 \mathrm{~m}\). a. What is the glucose conversion at the exit of the column? b. What is the ethanol concentration in the exit stream?

In a two-stage chemostat system, the volumes of the first and second reactors are \(V_{1}=5001\) and \(V_{2}=3001\), respectively. The first reactor is used for biomass production and the second is for a secondary metabolite formation. The feed flow rate to the first reactor is \(F=100 \mathrm{l} / \mathrm{h}\), and the glucose concentration in the feed is \(S=5.0 \mathrm{~g} / \mathrm{l}\). Use the following constants for the cells. $$ \mu_{m}=0.3 \mathrm{~h}^{-1}, \quad K_{S}=0.1 \mathrm{~g} / \mathrm{l}, \quad Y_{X I S}=0.4 \frac{\mathrm{g} \mathrm{dw} \text { cells }}{\mathrm{g} \text { glucose }} $$ a. Determine cell and glucose concentrations in the effluent of the first stage. b. Assume that growth is negligible in the second stage and the specific rate of product formation is \(q_{P}=0.02 \mathrm{~g} P / \mathrm{g}\) cell \(\mathrm{h}\), and \(Y_{P / S}=0.6 \mathrm{~g} P / \mathrm{g} S\). Determine the product and substrate concentrations in the effluent of the second reactor.

A fluidized-bed, immobilized-cell bioreactor is used for the conversion of glucose to ethanol by Z mobilis cells immobilized in \(\kappa\)-carrageenan gel beads. The dimensions of the bed are \(10 \mathrm{~cm}\) (diameter) by \(200 \mathrm{~cm}\). Since the reactor is fed from the bottom of the column and because of \(\mathrm{CO}_{2}\) gas evolution, substrate and cell concentrations decrease with the height of the column. The average cell concentration at the bottom of the column is \(X_{0}=45 \mathrm{~g} / \mathrm{l}\), and the average cell concentration decreases with the column height according to the following equation: $$ X=X_{0}(1-0.005 Z) $$ where \(Z\) is the column height \((\mathrm{cm})\). The specific rate of substrate consumption is \(q_{S}=2 \mathrm{~g} S / \mathrm{g}\) cells \(\mathrm{h}\). The feed flow rate and glucose concentration in the feed are \(5 \mathrm{l} / \mathrm{h}\) and \(160 \mathrm{~g}\) glucose/l, respectively. a. Determine the substrate (glucose) concentration in the effluent. b. Determine the ethanol concentration in the effluent and ethanol productivity \((\mathrm{g} / \mathrm{l} \cdot \mathrm{h})\) if \(Y_{P / S}=0.48 \mathrm{~g}\) ethanol/g glucose.
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