Q.22 A chemostat is operated at a dilution rate of 0.6 h−1.
At steady state, the biomass concentration in the exit stream was
found to be 30 g l−1. The biomass productivity
(g l−1 h−1) after 3 h of steady state
operation will be __________.
Chemostat Biomass Productivity
D = 0.6 h-1 | X = 30 g L-1
📊 Given Parameters
Dilution Rate (D):
0.6 h-1
0.6 h-1
Biomass Conc. (X):
30 g L-1
30 g L-1
Steady State Time:
3 hours (irrelevant)
3 hours (irrelevant)
🎯 Chemostat Productivity Formula
P = D × X
🔢 Step-by-Step Calculation
0.6 × 30
= 18
18 g L-1 h-1
Why 3 hours doesn’t matter: At steady state, X remains constant regardless of operation time.
❌ Complete Options Analysis
| Option | Value | Status | Why Correct/Wrong? |
|---|---|---|---|
| A | 18 | ✅ CORRECT | Standard formula: P = D × X = 0.6 × 30 = 18 Steady state principle ✓ |
| B | 5.4 | ❌ Wrong | Batch thinking: 30/3h × 0.6 Chemostat ≠ Batch reactor |
| C | 10.8 | ❌ Wrong | Half steady state assumption 0.6 × 30/2 = incorrect logic |
| D | 54 | ❌ Wrong | Batch accumulation: 30 + 0.6×3×8 Ignores continuous flow |
🧮 Steady State Mass Balance
At steady state: dX/dt = 0
μ = D and P = D × X
μ = D and P = D × X
🎓 GATE Exam Memory Trick
- Steady state → P = D × X (memorize!)
- Ignore time** at steady state
- Sterile feed → X₀ = 0
- μ = D** prevents washout
📋 Quick Reference Values
| Parameter | Value | Units |
|---|---|---|
| Dilution Rate | 0.6 | h-1 |
| Biomass Conc. | 30 | g L-1 |
| Productivity | 18 | g L-1 h-1 |
