Question:
A volatile organic compound (VOC) is to be adsorbed from air onto a bed of activated carbon. The equilibrium capacity of activated carbon at the feed conditions is 0.4 grams VOC per gram of activated carbon. The column contains 4 grams of activated carbon per cm2 of cross-section. The feed rate into the adsorber column is 0.2 grams VOC cm-2 h-1. Breakthrough time is defined as the time at which the concentration at the exit of the bed (c) reaches a value of \(0.05c_0\), where \(c_0\) is the concentration of VOC in the feed. The breakthrough time for the bed is 2.1 h. The area under \(c/c_0\) curve between the initial and breakthrough times is 0.1 h. Which one of the following is the fraction of unused bed at breakthrough?
A volatile organic compound (VOC) is to be adsorbed from air onto a bed of activated carbon. The equilibrium capacity of activated carbon at the feed conditions is 0.4 grams VOC per gram of activated carbon. The column contains 4 grams of activated carbon per cm2 of cross-section. The feed rate into the adsorber column is 0.2 grams VOC cm-2 h-1. Breakthrough time is defined as the time at which the concentration at the exit of the bed (c) reaches a value of \(0.05c_0\), where \(c_0\) is the concentration of VOC in the feed. The breakthrough time for the bed is 2.1 h. The area under \(c/c_0\) curve between the initial and breakthrough times is 0.1 h. Which one of the following is the fraction of unused bed at breakthrough?
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Compare the VOC mass actually adsorbed by breakthrough (breakthrough time minus the area under the c/c0 curve, times feed rate) with the total equilibrium capacity of the whole bed; the shortfall is the unused fraction.
Updated On: Aug 10, 2026
- 0
- 0.25
- 0.50
- 0.75
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The Correct Option is D
Solution and Explanation
Step 1: Recall the length/fraction-of-unused-bed (LUB) concept.
The fraction of the bed capacity still unused at breakthrough is found by comparing the capacity actually used up to breakthrough with the total capacity the whole bed could hold.
Step 2: Compute total bed capacity per unit area.
\[ W = 4 \times 0.4 = 1.6\ \text{g VOC/cm}^2 \]
Step 3: Convert to an equivalent ideal saturation time.
\[ t_{total} = 1.6/0.2 = 8\ \text{h} \]
Step 4: Compute usable capacity as equivalent time.
\[ t_u = t_b - \text{area} = 2.1 - 0.1 = 2.0\ \text{h} \]
Step 5: Compute fraction used and unused.
\[ f_{used} = 2.0/8.0 = 0.25 \]\[ f_{unused} = 1 - 0.25 = 0.75 \]
\[ \boxed{f_{unused} = 0.75} \]
The fraction of the bed capacity still unused at breakthrough is found by comparing the capacity actually used up to breakthrough with the total capacity the whole bed could hold.
Step 2: Compute total bed capacity per unit area.
\[ W = 4 \times 0.4 = 1.6\ \text{g VOC/cm}^2 \]
Step 3: Convert to an equivalent ideal saturation time.
\[ t_{total} = 1.6/0.2 = 8\ \text{h} \]
Step 4: Compute usable capacity as equivalent time.
\[ t_u = t_b - \text{area} = 2.1 - 0.1 = 2.0\ \text{h} \]
Step 5: Compute fraction used and unused.
\[ f_{used} = 2.0/8.0 = 0.25 \]\[ f_{unused} = 1 - 0.25 = 0.75 \]
\[ \boxed{f_{unused} = 0.75} \]
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