Chapter 5: Isothermal Reactor Design: Conversion
Effect of Reducing Particle Size on Conversion in a PBR
Currently the gas phase reaction
\(\text{A} + \text{B} \rightarrow \text{C} + \text{D}\)
is carried out in a packed bed reactor. For an entering pressure of 20 atm, a pressure drop of 4 atm is realized for 100 kg of catalyst. What pressure drop would be realized if the particle size is reduced by a factor of 4 and the catalyst weight charged to 15 kg. The flow is laminar.
Hint 1: What is the pressure drop parameter alpha?
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Hint 1
For Laminar flow:\(\alpha_2 = \alpha_1 \left( \frac{D_{P1}}{D_{P2}} \right)^2 \left( \frac{A_{c1}}{A_{c2}} \right)^2\)
Back to Problem
Start with the differential equation and realize that epsilon is zero because the same number of moles
are formed as are reacted.
\(\frac{dy}{dW} = -\frac{\alpha (1 + \varepsilon X)}{2y}\)
\(\varepsilon = 0\)
\(\frac{dy}{dW} = -\frac{\alpha}{2y}\)
\(y \, dy = -\frac{\alpha}{2} \, dW\)
Now integrate both sides and evaluate the constant. Remember that at W = 0, P = P0.
\(y^2 = -\alpha W + 1\)
\(\alpha_2 = \alpha_1 \left(\frac{D_{P1}}{D_{P2}}\right)^2 \left(\frac{A_{c1}}{A_{c2}}\right)^2\)
\(\alpha_2 = \alpha_1 \left(\frac{D_{P1}}{D_{P2}}\right)^2\)
\(\alpha_2 = \alpha_1 \left(\frac{1}{\frac{1}{4}}\right)^2 = \alpha_1 (4)^2\)
\(\alpha_2 = 16\alpha_1\)
Now we can evaluate \( \alpha_1 \) and \( \alpha_2 \).
\( \alpha_1 = \frac{1 - y^2}{W} = \frac{1 - 0.8^2}{100} \)
\( \alpha_1 = 0.0036 \)
\( \alpha_2 = 0.0576 \)
\( y^2 = -(0.0576)(15) + 1 \)
\( y = 0.369 \)
\( \Delta P = 0.369 \times 20 \)
\( \Delta P = 7.38 \)