Q.No. 45 A batch reactor is inoculated with 1 g/L biomass. Under these conditions, cells exhibit a lag phase of 30 min.
If the specific growth rate in the log phase is 0.00417 min^-1, the time taken for the biomass to increase to 8 g/L is __________ min
(round off to 2 decimal places).

Batch Reactor Biomass Growth: Time to Reach 8 g/L

Batch reactors follow microbial growth kinetics with distinct phases: lag, log (exponential), stationary, and death. Here, biomass starts at 1 g/L, experiences a 30-minute lag phase with no growth, then enters log phase with constant specific growth rate μ = 0.00417 min^-1 until reaching 8 g/L. The time calculation uses the exponential growth equation after lag.

Growth Kinetics Basics

Biomass concentration X follows X = X₀ e^(μt) during log phase, where X₀ = 1 g/L, X = 8 g/L, and μ = 0.00417 min^-1. No substrate limitation or death phase is mentioned, so assume pure log phase post-lag. Lag phase adds fixed 30 min with μ = 0, during which cells adapt without dividing.

Step-by-Step Calculation

Verification

• Doubling time τ = ln(2)/μ ≈ 0.6931/0.00417 ≈ 166.2 min
• From 1 to 8 g/L requires 3 doublings (2³ = 8)
• 3 × 166.2 ≈ 498.6 min growth time, confirming calculation
• Final Answer: 528.67 min (rounded to 2 decimal places)

Ultimate Guide: Time for Biomass to Reach 8 g/L

Biomass growth in batch reactors drives bioprocess engineering, especially for IIT JAM exams. This guide solves the exact query: batch reactor inoculated at 1 g/L biomass, 30 min lag phase, specific growth rate 0.00417 min^-1—time to reach 8 g/L? Master exponential kinetics, lag phase impact, and precise calculations for competitive exams.

Key Phases Explained

• Lag Phase: 30 min adaptation—no biomass increase (X remains 1 g/L)
• Log Phase: Exponential growth X = X₀ e(μt), μ constant at 0.00417 min^-1
• Total time = t_lag + t_log, ignoring stationary phase as 8 g/L occurs in log

Detailed Math Breakdown

1. Growth phase: ln(X/X₀) = μ tlog → tlog = ln(8)/0.00417
2. ln(8) = 2.07944154168
3. tlog = 2.07944154168 / 0.00417 ≈ 498.67 min
4. Total: 30 + 498.67 = 528.67 min (rounded)

Exam Tips

• Use natural log tables or calculator for ln(8)
• Verify via doublings: N = log₂(8) = 3, τ = ln(2)/μ ≈ 166.22 min
• 3τ ≈ 498.67 min
• Practice similar: vary μ or Xfinal for speed