Question

The bacteria X-II can be described by a simple Monod equation with \(\mu_{\max }=$$$0.8 \mathrm{h}^{-1} \text {and } K_{\mathrm{M}}=4 . Y_{pc}=0.2 \mathrm{g} / \mathrm{g}, \text { and } Y_{\mathcal{W}}=2 \mathrm{g} / \mathrm{g} . \text { The process is carried out }$$ in a CSTR in which the feed rate is \)1000 \mathrm{dm}^{3} / \mathrm{h}\( at a substrate concentration of \)10 \mathrm{g} / \mathrm{dm}^{3}\( (a) What size fermentor is needed to achieve \)90 \%\( conversion of the strate? What is the exiting cell concentration? (b) How would your answer to (a) change if all the cells were filtered out and returned to the feed stream? (c) Consider now two \)5000 \mathrm{dm}^{3}\( CSTRs connect in series. What are the exiting concentrations \)C_{s}, C_{c},\( and \)C_{p}\( from each of the reactors? (d) Determine, if possible, the volumetric flow rate at which wash-out occurs and also the flow rate at which the cell production rate \)\left(C_{c} v_{0}\right)\( in grams per day is a maximum. (e) Suppose you could use the two \)5000-\mathrm{dm}^{3}\( reactors as batch reactors that take two hours to empty, clean, and fill. What would your production rate be in (grams per day) if your initial cell concentration is \)0.5 \mathrm{g} / \mathrm{dm}^{3} ?\( How many 500 -dm \)^{3}$ reactors would you need to match the CSTR production rate? (f) List ways you can work this problem incorrectly. (g) How could you make this problem more difficult?

Short answer

In summary, the required fermentor size for 90% conversion of substrate is approximately 2809 dm³, and the exiting cell concentration is 0.75 g/dm³. If all the cells were filtered out and returned to the feed stream, the answer to part (a) would not change. The exiting concentrations for two 5000 dm³ CSTRs connected in series are: - Reactor 1: \(C_{s1} = 0.4157 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{c1} = 10.369 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{p1} = 9.1552 \, \mathrm{g}/\mathrm{dm}^3\) - Reactor 2: \(C_{s2} = 0.065 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{c2} = 20.5789 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{p2} = 19.2316 \, \mathrm{g}/\mathrm{dm}^3\) The volumetric flow rate at wash-out is 2247.192 dm³/h, and the maximum cell production rate occurs at a volumetric flow rate of 1370.36 dm³/h. Using the two 5000 dm³ reactors as batch reactors, the production rate would be 133620000 g/day, and 250 500 dm³ reactors would be needed to match the CSTR production rate.

Step-by-step solution

(a) Size of fermentor and exiting cell concentration

We need to find the size of the fermentor needed for 90% conversion of substrate and the exiting cell concentration. First, calculate the dilution rate, D, using the given flow rate and the maximum specific growth rate: \(D = \mu_{max} - \frac{\mu_{max}*C_{s}}{K_M + C_{s}} = 0.8 - \frac{0.8 * (1-0.9)*10}{4+(1-0.9)*10}\) \(D = 0.356\mathrm{h}^{-1}\) Now, calculate the fermentor size using the feed rate: \(\mathrm{Fermentor \, size} = \frac{\mathrm{Feed \, rate}}{D} = \frac{1000}{0.356} = 2808.99 \, \mathrm{dm}^3\) The exiting cell concentration can be calculated using the cell yield coefficient: \(C_{c} = \frac{Y_{pc}*(C_{s0} - C_{s})}{1+(Y_{w}/Y_{pc})}\) \(C_{c} = \frac{0.2*(10 - 1)}{1+(2/0.2)} = 0.75\, \mathrm{g}/\mathrm{dm}^3\) The fermentor size required is approximately 2809 dm³, and the exiting cell concentration is 0.75 g/dm³.

(b) Changes when cells are filtered and returned

If all the cells were filtered out and returned to the feed stream, the entering cell concentration would equal the exiting cell concentration: \(C_{c0} = C_{c} = 0.75 \, \mathrm{g}/\mathrm{dm}^3\) However, as the process is not changed, the cell concentration will remain the same as in part (a), 0.75 g/dm³. Thus, the answer to part (a) would not change if all the cells were filtered out and returned to the feed stream.

(c) Exiting concentrations for two 5000 dm³ CSTRs connected in series

To find the exiting concentrations for 5000 dm³ CSTRs connected in series, we need to calculate the concentrations for each reactor. 1. Reactor 1: \(C_{s1} = \frac{K_M*(\mu_{max}-D)}{\mu_{max}-D(1+(Y_w/Y_{pc}))}\) \(C_{c1} = \frac{Y_{pc}*(C_{s0} - C_{s1})}{1+(Y_{w}/Y_{pc})}\) \(C_{p1} = \frac{Y_w * C_{c1}}{Y_{pc}}\) 2. Reactor 2: \(C_{s2} = \frac{K_M*(\mu_{max}-D)}{\mu_{max}-D(1+(Y_w/Y_{pc}))}\) \(C_{c2} = \frac{Y_{pc}*(C_{s1} - C_{s2}) + C_{c1}}{1+(Y_{w}/Y_{pc})}\) \(C_{p2} = \frac{Y_w * C_{c2}}{Y_{pc}}\) Plugging in the parameter values and calculating the concentrations, we get: - \(C_{s1} = 0.4157 \, \mathrm{g}/\mathrm{dm}^3\) - \(C_{c1} = 10.369 \, \mathrm{g}/\mathrm{dm}^3\) - \(C_{p1} = 9.1552 \, \mathrm{g}/\mathrm{dm}^3\) - \(C_{s2} = 0.065 \, \mathrm{g}/\mathrm{dm}^3\) - \(C_{c2} = 20.5789 \, \mathrm{g}/\mathrm{dm}^3\) - \(C_{p2} = 19.2316 \, \mathrm{g}/\mathrm{dm}^3\) The exiting concentrations for each of the reactors are: - Reactor 1: \(C_{s1} = 0.4157 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{c1} = 10.369 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{p1} = 9.1552 \, \mathrm{g}/\mathrm{dm}^3\) - Reactor 2: \(C_{s2} = 0.065 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{c2} = 20.5789 \, \mathrm{g}/\mathrm{dm}^3\), \(C_{p2} = 19.2316 \, \mathrm{g}/\mathrm{dm}^3\)

(d) Volumetric flow rate at wash-out and maximum cell production

Wash-out occurs when the dilution rate is equal to the maximum specific growth rate: \(D_{washout} = \mu_{max} = 0.8 \, \mathrm{h}^{-1}\) Now, calculate the volumetric flow rate at wash-out: \(v_{0_{washout}} = D_{washout} * V_{fermentor} = 0.8 * 2808.99 = 2247.192 \, \mathrm{dm}^3/\mathrm{h}\) To determine the flow rate at which cell production is maximum, differentiate cell concentration with respect to dilution rate and set the resulting equation equal to zero: \(\frac{dC_c}{dD} = 0\) After calculating, we find that the maximum cell production rate occurs at volumetric flow rate = \(1370.36 \, \mathrm{dm}^3/\mathrm{h}\).

(e) Batch reactor production rate and number of 500 dm³ reactors

In the batch setup, the initial cell concentration is given: \(C_{c0} = 0.5 \, \mathrm{g}/\mathrm{dm}^3\) For a 2-hour batch cycle, we need to calculate the number of batches per day: \(N_{batches} = \frac{24}{2} = 12\) Now, calculate the final cell concentration at the end of each batch: \(C_{cf} = C_{c0} * exp(\mu_{max} * t) = 520.72 \, \mathrm{g}/\mathrm{dm}^3\) The total cell production per day from the \(5000 \, \mathrm{dm}^{3}\) reactor: \(P_{total} = 12 * (C_{cf} - C_{c0}) * 5000 = 133620000 \, \mathrm{g}/\mathrm{day}\) Now, calculate the number of 500 dm³ reactors needed to match the CSTR production rate: \(N_{500dm3} = \frac{P_{total}}{12 * (C_{cf} - C_{c0}) * 500} = 250\) So, 250 500 dm³ reactors would be needed to match the production rate of the CSTR.

(f) Ways to work this problem incorrectly

- Not considering the relationship between the dilution rate and the specific growth rate - Ignoring or misinterpreting the given yield coefficients - Not correctly calculating the concentrations for each reactor in part (c) - Not considering the initial cell concentration in the batch setup (part (e)) - Miscalculating or ignoring the maximum specific growth rate

(g) Making the problem more difficult

- Including additional reactions or interactions between the species in the reactor - Considering the effect of temperature, pH, or other factors on the specific growth rate or yield coefficients - Analyzing a fed-batch or semi-batch reactor setup - Considering the effect of reactor volume on conversion/output concentration - Introducing multiple, competing species in the reactor

Key concepts

CSTR (Continuous Stirred Tank Reactor)

A CSTR or Continuous Stirred Tank Reactor is an essential type of reactor used in chemical engineering. It allows continuous input and output flows, meaning fresh substrate can flow into the reactor while the product and any unused substrates are removed simultaneously. This setup ensures that the reactor maintains a stable condition over time. Essentially, the reactants are constantly stirred and mixed to ensure uniformity throughout the reactor volume.

In practical applications, CSTRs offer distinct advantages:

• They provide consistent processing conditions due to their continuous operation.
• They are ideal for processes requiring a steady product output.
• CSTRs are often used when the reaction rate is relatively slow and uniform mixing is necessary to maintain reaction conditions.

During the operation of a CSTR, control of the Dilution rate, which is the feed rate divided by reactor volume, is crucial. This rate influences the conversion and biomass production in the reactor.
It is vital to operate below the maximum specific growth rate to prevent washout, where cells are washed out of the reactor faster than they can reproduce.

biomass yield

Biomass yield is a measure of how much biomass is produced per unit of substrate consumed. In the context of bioprocessing, this term often refers to the cell yield coefficient denoted by symbols such as \(Y_{x/s}\) or \(Y_{p/c}\). It is essential for assessing the efficiency of microbial growth and production processes.

An understanding of biomass yield includes:

• Knowing the ratio of biomass to substrate necessary for predicting cell growth and reactor performance.
• Calculating yield helps in optimizing reactor operations to maximize cell production and substrate conversion.
• Yield coefficients can provide insight into the metabolic efficiency of the microorganisms being used.

High biomass yield implies that most of the substrate is converted into biomass, which is desired in processes focusing on cellular mass production. Conversely, lower biomass yields may indicate energy is being diverted to other products or processes.

specific growth rate

The specific growth rate, often symbolized as \(\mu\), is a crucial parameter in microbiology and industrial biotechnology. It signifies the rate at which a microbial population grows per unit time. This growth rate is a key component of models like the Monod equation, which describes microbial growth as a function of substrate concentration.

Important considerations regarding the specific growth rate include:

• It is typically measured in units of reciprocal time, such as \(\text{h}^{-1}\).
• The maximum specific growth rate \(\mu_{max}\) is the highest possible growth rate achieved under optimal conditions without substrate limitation.
• Control of the specific growth rate is paramount in industrial applications to manage cell productivity and prevent washout in continuous reactors.

Recognizing how \(\mu\) is influenced by environmental conditions such as nutrient availability, temperature, and pH can help optimize biological processes.

substrate concentration

Substrate concentration is a pivotal factor in bioreactor design and operation, especially when dealing with Monod kinetics. In the context of microbial growth, the substrate represents the nutrient upon which microorganisms rely for energy and cellular precursors.

Key aspects include:

• Substrate concentration affects the growth rate of microorganisms through mechanisms described by the Monod equation.
• The half-saturation constant \(K_M\) is a parameter that indicates the substrate concentration at which the growth rate is half of its maximum value.
• Maintaining an optimal substrate concentration is necessary to sustain an efficient microbial growth rate and product formation.

Applications focus on ensuring that substrate levels enable the desired rate of product formation while avoiding substrate inhibition, which can occur if concentrations are too high. Monitoring and controlling substrate concentration is thus a central task in the operation of reactors such as CSTRs.