Question

The hydrolysis rate coefficient of the biodegradable fraction of a feedstock is \(0.05 \mathrm{~h}^{-1}\). What retention time is required to obtain \(80 \%\) degradation of the biodegradable fraction of the substrate in a continuous stirred tank reactor (CSTR)? What happens to the retention time required if we place two CSTRs in series or in parallel?

Short answer

#tag_title# Step 3: Solve for the retention time in a single CSTR #tag_content# By calculating the values in the formula, we find the retention time for a single CSTR. $$ \text{retention time} = \frac{-\ln(1 - 0.8)}{0.05} = \frac{-\ln(0.2)}{0.05} \approx 3.22 \mathrm{~h} $$ So, the retention time for 80% degradation in a single CSTR is approximately 3.22 hours. #tag_title# Step 4: Calculate the retention time for two CSTRs in series and parallel #tag_content# In our scenario, we have two CSTRs arranged in either series or parallel. When arranged in series, the substrate flows sequentially from one CSTR to the other. In parallel, the substrate is split between two identical reactors operating simultaneously. The retention time for two CSTRs in series is the sum of retention times for each individual CSTR, while in parallel, it is the same as for a single CSTR. For two CSTRs in series: $$ \text{retention time (series)} = 3.22 + 3.22 = 6.44 \mathrm{~h} $$ For two CSTRs in parallel: $$ \text{retention time (parallel)} = 3.22 \mathrm{~h} $$ #tag_title# Conclusion #tag_content# The retention time needed for 80% degradation of the biodegradable fraction in a single CSTR is approximately 3.22 hours. For two CSTRs in series, the retention time is approximately 6.44 hours, while for two CSTRs in parallel, the retention time remains at 3.22 hours.

Step-by-step solution

Understand the CSTR process and components

In a continuous stirred tank reactor (CSTR), the substrate is continuously mixed while undergoing a reaction. The hydrolysis process is characterized by its rate coefficient, which shows how quickly the biodegradable fraction breaks down. Our goal is to determine how long the substrate must remain in the CSTR to achieve 80% degradation.

Calculate the retention time for 80% degradation in a single CSTR

To find the retention time, we'll use the formula: $$ \text{retention time} = \frac{-\ln(1 - \text{degradation fraction})}{\text{hydrolysis rate coefficient}} $$ Here, the degradation fraction is \(80\% = 0.8\), and the hydrolysis rate coefficient is \(0.05 \mathrm{~h}^{-1}\). Plug these values into the formula to find the retention time for a single CSTR. $$ \text{retention time} = \frac{-\ln(1 - 0.8)}{0.05} $$

Key concepts

Hydrolysis Rate Coefficient

The hydrolysis rate coefficient is a crucial parameter in evaluating the treatment process in a reactor. It represents the speed at which hydrolysis occurs, which is an essential step in the degradation of complex organic substances into simpler, more biodegradable forms. In simpler terms, it tells us how fast the biodegradable components of a feedstock break apart in the presence of water. This coefficient is typically expressed in reciprocal time units (e.g., inverse hours, \( \mathrm{h}^{-1} \)).

To visualize its impact, imagine the coefficient as a 'speed setting' for the degradation. A higher coefficient means the substrate will degrade faster, requiring less time for the process to complete. When we talk about the calculation of retention time in a Continuous Stirred Tank Reactor (CSTR), the hydrolysis rate coefficient directly influences the duration needed for a specific percentage of degradation. Essential factors that can affect this rate include temperature, pH, and the nature of the substrate itself. Understanding how to adjust and control this rate is vital for the efficient operation of wastewater treatment or chemical reaction processes.

Biodegradable Fraction Degradation

The process of biodegradable fraction degradation involves breaking down organic matter into its constituent elements - typically carbon dioxide, water, and biomass - through biological activity. In the context of a CSTR, we are primarily interested in the percentage of the feedstock that is biodegradable and the efficiency with which it can be degraded. This percentage is expressed as a fraction (e.g., 80% \( \frac{4}{5} \)).

To determine the required retention time for a certain degradation level, we need to understand the percentage of the biodegradable fraction we aim to degrade. In our exercise, we are tasked with achieving 80% degradation. Different levels of degradation are suitable for various purposes. For example, less degradation might be acceptable if the effluent is discharged into a robust ecosystem that can handle some organic matter. However, for discharges into sensitive environments or those undergoing further treatment, a higher level of degradation may be necessary. The retention time must be calculated accordingly to ensure the right balance between treatment time, reactor capacity, and process efficiency.

Continuous Stirred Tank Reactor (CSTR)

A Continuous Stirred Tank Reactor (CSTR) is a common piece of equipment in chemical engineering and environmental applications. A CSTR provides a mixing action that ensures a uniform concentration throughout the reactor. This homogeneous environment allows for the continuous input of reactants (the feedstock) and the continuous output of products (e.g., treated wastewater), which happens simultaneously.

In the context of our exercise, retention time in a CSTR refers to the amount of time a particular volume of substrate remains in the reactor to achieve a certain level of degradation. It is a critical factor that determines the size and number of reactors needed for a process. If the retention time is too short, the substrate will not degrade sufficiently. If it is too long, it suggests an overdesign that could be more costly and inefficient.

Effects of Reactor Configuration

When considering placing two CSTRs in series, the overall retention time will be the sum of the time spent in each reactor. In contrast, placing two CSTRs in parallel does not change the individual retention time but doubles the system's capacity, allowing twice the amount of substrate to be processed in the same amount of time.