Question

A packed bed bioreactor with length 1 m and inside diameter 10 cm has liquid flowing at an interstitial velocity of 1 cm s\(^{-1}\). Given a volumetric flow rate of 0.025 L s\(^{-1}\), the void fraction of the packed bed is . (rounded off to two decimal places)

Hint:

The interstitial velocity is higher than the superficial velocity because the liquid only flows through the void space. Compare the two velocities to get the void fraction.

Answer 1

Correct Answer: 0.315

Step 1: Understand the two kinds of velocity in a packed bed.
In a packed bed, the solid packing takes up part of the cross-section, and liquid can only flow through the empty gaps between the particles, called the voids.
The superficial velocity \(v_s\) is a fictitious velocity found by pretending the whole cross-section is open to flow, with no packing at all.
The interstitial velocity \(v_i\) is the real average velocity of the liquid moving through the void spaces only.
Since the liquid actually squeezes through a smaller open area (only the fraction \(\varepsilon\) of the total cross-section), it must speed up compared to the superficial value, so
\[ v_i = \frac{v_s}{\varepsilon} \]
where \(\varepsilon\) is the void fraction (the fraction of the bed volume that is empty space).

Step 2: Find the cross-sectional area of the bed.
The inside diameter is \(d = 10\) cm, so the radius is \(r = 5\) cm.
The total (empty-bed) cross-sectional area is
\[ A = \pi r^2 = \pi (5)^2 = 25\pi \approx 78.54 \text{ cm}^2 \]
The length of the bed, 1 m, is not needed here, since velocity and area calculations only use the cross-section.

Step 3: Find the superficial velocity.
The volumetric flow rate is given as \(Q = 0.025\) L s\(^{-1}\). Since \(1\) L \(= 1000\) cm\(^3\),
\[ Q = 0.025 \times 1000 = 25 \text{ cm}^3\text{ s}^{-1} \]
The superficial velocity is the flow rate divided by the total cross-section:
\[ v_s = \frac{Q}{A} = \frac{25}{78.54} \approx 0.318 \text{ cm s}^{-1} \]

Step 4: Solve for the void fraction.
Rearrange \(v_i = v_s/\varepsilon\) to get \(\varepsilon = v_s/v_i\). Using the given interstitial velocity \(v_i = 1\) cm s\(^{-1}\):
\[ \varepsilon = \frac{v_s}{v_i} = \frac{0.318}{1} = 0.318 \]

Final Answer:
Rounded off to two decimal places, the void fraction is \(0.32\), which lies inside the accepted range of 0.31 to 0.32 for this question.
\[ \boxed{\varepsilon \approx 0.32} \]