Suppose we want to increase the entering total pressure. How does the pressure drop parameter, \(\alpha\), vary with the entering total pressure, \(P_0\)? [see Preface Table P1, Q5 -- Question that probes implications and consequences. I.e., what would be the consequence of changing the inlet pressure, \(P_0\)?]

a)   a increases with increasing P_o.

b)   a decreases with increasing P_o.

c)   a does not change with P_o.

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Solution

\(\alpha = \frac{2 \beta_0}{A_c (1 - \phi_c) \rho_c} \left(\frac{1}{P_0}\right)\)

\(\beta_0 = \frac{G}{\rho_0 g D_p} \left[\frac{(1 - \phi)}{\phi^3}\right] \left[ \frac{150 (1 - \phi)}{D_p} \mu + 1.75 G \right]\)

\(\alpha \sim \frac{\beta_0}{P_0}\)

\(\beta_0 \sim \frac{1}{\rho_0}, \, \rho_0 = \left(\text{MW}\right) \frac{P_0}{RT}\)

\(\beta_0 \sim \frac{1}{P_0}\)

\(\alpha \sim \frac{1}{P_0^2}\)

\(\alpha_2 = \alpha_1 \left(\frac{P_{01}}{P_{02}}\right)^2\)

a decreases with increasing P_o

We note that increasing the total pressure increases the concentration

\(C_{A0} = y_{A0} \frac{P_0}{RT_0}\)

for a fixed y_ao. For Pure A, y_ao=1.0.  Therefore, the rate of reaction (A --> B) would increase.

\(-r_A = kC_A = kC_{A0}(1 - X) \frac{P}{P_0} = k \left(\frac{P_0}{RT_0}\right)(1 - X) \left(\frac{P}{P_0}\right)\)

\(-r_A = \frac{kP_0}{RT_0}(1 - X) y \quad \text{where} \quad y = \frac{P}{P_0}\)

we note we have competing effects

1)   Increasing P_o increases the rate initially

2)   Increasing P_o increases the rate at which y decreases and therefore the rate decreases

To determine which of the above effects will dominate and cause the conversion to increase or decrease will depend on a number of things such as reaction order, value of a, and other parameter values. Consequently, we would need to run the simulation. However, you will find that as P_o increases, so will X, as a loose "rule of thumb".