Give step-by-step solution with explanation and final answer:10C Palallcicis O1 a plot tallk aic as I010Ws.
Pilot Parameter Unit
“Tank diameterc®dy lL wm
Tank height (H) 2 m
Loading factor 64 %
Impeller diameter (d) 0.5 m
Agitation speed 300 pm
Fermentation time 180 h
Operation time (cleaning, 0 h
sterilization, erc.)
Gauge pressure of tank 0.3 MPa
Temperature in tank 28 °C
Aeration rate 12 VVM (standard conditions)
Broth viscosity 0.1 Pas
Broth density 1050 kg/m?
Final lipase concentration in broth ~~ 18.18 gL
Lr
{3 t]
Ve
Ww | =
| aa || |
= t]
= — dt
Ee F===F— hy
| — NA
(3) Agitator Power
Four baffles and a two-layer, six-curved-blade turbine agitator are installed to ensure efficient
mixing. To select and procure a suitable drive motor, calculate agitation power (P) and the
agitation power (Pg) with aeration using Michel's formula for this pilot-scale tank. (20 marks)
(4) Scale-Up Calculation
A scale-up from pilot scale to an annual (300 working days) production target of 30,000 kg of lipase
is required.
Calculate the required production-scale bioreactor volume (in m*) and the corresponding scale-up
factor.
Based on the scale-up factor, determine the following for the production tank: main dimensions,
linear velocity of air, rotational speed, and impeller (agitator) power. (40 marks)
Question:
Give step-by-step solution with explanation and final answer:10C Palallcicis O1 a plot tallk aic as I010Ws.
Pilot Parameter Unit
“Tank diameterc®dy lL wm
Tank height (H) 2 m
Loading factor 64 %
Impeller diameter (d) 0.5 m
Agitation speed 300 pm
Fermentation time 180 h
Operation time (cleaning, 0 h
sterilization, erc.)
Gauge pressure of tank 0.3 MPa
Temperature in tank 28 °C
Aeration rate 12 VVM (standard conditions)
Broth viscosity 0.1 Pas
Broth density 1050 kg/m?
Final lipase concentration in broth ~~ 18.18 gL
Lr
{3 t]
Ve
Ww | =
| aa || |
= t]
= — dt
Ee F===F— hy
| — NA
(3) Agitator Power
Four baffles and a two-layer, six-curved-blade turbine agitator are installed to ensure efficient
mixing. To select and procure a suitable drive motor, calculate agitation power (P) and the
agitation power (Pg) with aeration using Michel's formula for this pilot-scale tank. (20 marks)
(4) Scale-Up Calculation
A scale-up from pilot scale to an annual (300 working days) production target of 30,000 kg of lipase
is required.
Calculate the required production-scale bioreactor volume (in m*) and the corresponding scale-up
factor.
Based on the scale-up factor, determine the following for the production tank: main dimensions,
linear velocity of air, rotational speed, and impeller (agitator) power. (40 marks)
This is AI generated Response.
Asked by: vemuri swathi
Created at: 2025-12-18 03:05:24
Final Answer:
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Let's solve the problem step-by-step as per the requirements:
---
# Step 1: **Calculate Agitation Power (P) for Pilot-Scale Tank**
### **Given Data**
- Tank diameter, \( D = 1 \) m
- Tank height, \( H = 2 \) m
- Impeller diameter, \( d = .5 \) m
- Agitation speed, \( N = 300 \) rpm \( = 300/60 = 5 \) s\(^{-1}\)
- Broth viscosity, \( \mu = .1 \) Pa·s
- Broth density, \( \rho = 105 \) kg/m³
### **Assumptions**
- Four baffles, two-layer six-blade turbine (typical for bioreactors)
- Power number (\(N_P\)) for a standard Rushton turbine is about 5 (for fully baffled tanks).
#### **Convert agitation speed to revolutions per second:**
\[
N = \frac{300}{60} = 5 \ \text{s}^{-1}
\]
### **Power Required (No Aeration):**
\[
P = N_P \cdot \rho \cdot N^3 \cdot d^5
\]
Where:
- \(N_P = 5\)
- \(\rho = 105\) kg/m³
- \(N = 5\) s\(^{-1}\)
- \(d = .5\) m
\[
P = 5 \times 105 \times (5)^3 \times (.5)^5
\]
\[
(5)^3 = 125, \quad (.5)^5 = .03125
\]
\[
P = 5 \times 105 \times 125 \times .03125
\]
\[
= 5 \times 105 \times 3.90625
\]
\[
= 5 \times 4091.56 = 20,457.8 \ \text{W} \approx 20.46 \ \text{kW}
\]
---
### **With Aeration (Michel’s Formula):**
Michel’s correlation for power reduction due to aeration:
\[
P_g = P \left[1 - .15 \left(\frac{Q}{N d^3}\right)^{.56}\right]
\]
Where:
- \(Q\) = volumetric flow rate of air (m³/s)
#### **Calculate \(Q\) from VVM:**
VVM = 1.2 (volume of air per volume of liquid per minute)
- Working volume (\(V\)) = Loading factor \(\times\) Tank volume
Loading factor = 64%
Tank volume (\(V_{tank}\)) = \(\pi/4 \times D^2 \times H = \pi/4 \times 1^2 \times 2 = 1.5708\) m³
Working volume = \(.64 \times 1.5708 = 1.0053\) m³
- Air flow per min = \(1.2 \times 1.0053 = 1.206\) m³/min
Per second: \(Q = \frac{1.206}{60} = .0201\) m³/s
#### **Calculate \(Q / (N d^3)\):**
\[
N d^3 = 5 \times (.5)^3 = 5 \times .125 = .625
\]
\[
\frac{Q}{N d^3} = \frac{.0201}{.625} = .0322
\]
\[
(.0322)^{.56} = .156
\]
\[
P_g = 20.46 \times [1 - .15 \times .156] = 20.46 \times [1 - .0234]
\]
\[
P_g = 20.46 \times .9766 = 19.98 \ \text{kW}
\]
---
## **Summary Table for (3)**
| Power Type | Value (kW) |
|-----------------|-----------|
| Without aeration (\(P\)) | 20.46 |
| With aeration (\(P_g\)) | 19.98 |
---
# Step 2: **Scale-Up Calculation**
### **Annual Production Target:**
- Lipase required = 30,000 kg/year
- Working days/year = 300
- Daily production target = \( \frac{30,000}{300} = 100 \) kg/day
### **Per-batch lipase produced (pilot):**
- Final concentration = 18.18 g/L = .01818 kg/L
- Working volume = 1.0053 m³ = 1005.3 L
\[
\text{Lipase produced per batch} = 18.18 \ \frac{g}{L} \times 1005.3 \ L = 18,287 \ g = 18.29 \ kg
\]
- **Fermentation time**: 180 h = 7.5 days + 60 h (operation) = 10 days/cycle
\[
\text{Batches per year (pilot)} = \frac{300}{10} = 30
\]
\[
\text{Total annual lipase (pilot)} = 18.29 \times 30 = 548.7 \ \text{kg/year}
\]
### **Scale-up Factor:**
\[
\text{Scale-up factor} = \frac{30,000}{548.7} = 54.66
\]
### **Required Production-scale Bioreactor Volume:**
\[
\text{Production scale volume} = 1.0053 \times 54.66 = 54.94 \ \text{m}^3
\]
---
## **Summary Table for (4)**
| Parameter | Value |
|-----------------------------------|-----------|
| Production-scale volume | 54.94 m³ |
| Scale-up factor | 54.66 |
---
# Step 3: **Production Tank Main Dimensions, Linear Velocity of Air, Rotational Speed, and Impeller Power**
### **A. Tank Dimensions**
Assume same proportions (height/diameter = 2):
- Let \(D_p\) = production tank diameter, \(H_p = 2D_p\):
\[
V = \frac{\pi}{4} D_p^2 H_p \text{ (ignore headspace for design)}
\]
\[
54.94 = \frac{\pi}{4} D_p^2 (2 D_p) = \frac{\pi}{2} D_p^3
\]
\[
D_p^3 = \frac{54.94}{\pi/2} = \frac{54.94 \times 2}{\pi} = \frac{109.88}{3.1416} = 34.97
\]
\[
D_p = (34.97)^{1/3} = 3.28 \ \text{m}
\]
\[
H_p = 2 \times 3.28 = 6.56 \ \text{m}
\]
### **B. Impeller Diameter**
Keep same ratio as pilot:
\[
\frac{d}{D} = \frac{.5}{1} = .5 \implies d_p = .5 \times 3.28 = 1.64 \ \text{m}
\]
### **C. Rotational Speed (Constant Tip Speed)**
Tip speed \(u = \pi d N\)
Keep same tip speed as pilot:
\[
u_{pilot} = \pi \times .5 \times 5 = 7.85 \ \text{m/s}
\]
\[
N_p = \frac{u_{pilot}}{\pi d_p} = \frac{7.85}{\pi \times 1.64} = \frac{7.85}{5.153} = 1.52 \ \text{s}^{-1}
\]
\[
\text{rpm} = 1.52 \times 60 = 91.2 \ \text{rpm}
\]
### **D. Linear Air Velocity**
Use same VVM:
\[
Q_p = V_p \times 1.2 = 54.94 \times 1.2 = 65.93 \ \text{m}^3/\text{min}
\]
\[
Q_p = \frac{65.93}{60} = 1.099 \ \text{m}^3/\text{s}
\]
Superficial velocity (\(u_s\)):
\[
u_s = \frac{Q_p}{\frac{\pi}{4} D_p^2} = \frac{1.099}{\frac{\pi}{4} \times (3.28)^2} = \frac{1.099}{8.454} = .13 \ \text{m/s}
\]
### **E. Impeller Power (Production Scale)**
\[
P_p = N_P \rho N_p^3 d_p^5
\]
\[
N_p = 1.52 \ \text{s}^{-1} \quad d_p = 1.64 \ \text{m}
\]
\[
(1.64)^5 = 11.51
\]
\[
(1.52)^3 = 3.51
\]
\[
P_p = 5 \times 105 \times 3.51 \times 11.51 = 5 \times 105 \times 40.41
\]
\[
= 5 \times 42,430.5 = 212,152.5 \ \text{W} \approx 212.2 \ \text{kW}
\]
With aeration (using same Michel’s factor as pilot, approx .9766),
\[
P_{g,p} = 212.2 \times .9766 = 207.2 \ \text{kW}
\]
---
## **Final Summary Table: Production Tank**
| Parameter | Value |
|---------------------------------|-------------|
| Volume (\(V_p\)) | 54.94 m³ |
| Diameter (\(D_p\)) | 3.28 m |
| Height (\(H_p\)) | 6.56 m |
| Impeller diameter (\(d_p\)) | 1.64 m |
| Rotational speed (\(N_p\)) | 91.2 rpm |
| Linear air velocity (\(u_s\)) | .13 m/s |
| Impeller power (\(P_p\)) | 212.2 kW |
| With aeration (\(P_{g,p}\)) | 207.2 kW |
---
# **Final Answers**
### **(3) Agitator Power (Pilot Tank)**
- **Without aeration (\(P\)):** **20.46 kW**
- **With aeration (\(P_g\)):** **19.98 kW**
### **(4) Scale-up Calculation**
- **Production-scale reactor volume:** **54.94 m³**
- **Scale-up factor:** **54.66**
- **Production tank main dimensions:**
- **Diameter:** 3.28 m
- **Height:** 6.56 m
- **Impeller diameter:** 1.64 m
- **Rotational speed:** 91.2 rpm
- **Linear air velocity:** .13 m/s
- **Impeller power:** 212.2 kW (without aeration), **207.2 kW** (with aeration)
---
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