Parabolic flow
\[\begin{align}
\frac{c_A - c_{A,s}}{c_{A,i} - c_{A,s}}
&= 5.5
\left[
\frac{W}{D_{AB}\,\rho\,L}
\right]^{-\tfrac{2}{3}}
\end{align}\]
\(c_A\): exit concentration
\(c_{A,i}, c_{A,s}\): inlet & surface concentration
\(W\): flow rate in (kg/s)
\(k_c'\) can be calculated by \(j_D\)
Turbulent flow
\[\begin{align}
N_{Sh}
&= k_c'\left(\frac{D}{D_{AB}}\right) \\
&= \frac{k_c\,p_{BM}}{P}
\left(\frac{D}{D_{AB}}\right) \\
&= 0.023
\left(\frac{\rho D v}{\mu}\right)^{0.83}
\left(\frac{\mu}{\rho D_{AB}}\right)^{0.33} \\
&= 0.023\,
N_{Re}^{0.83}\,
N_{Sc}^{0.33}
\end{align}\]
- Similar to the \(j_D\) analog
- Just need \(N_{Re}\) and \(N_{Sc}\) to determine \(k_c'\)
- Characteristic length \(D\) is pipe diameter!
Case 2: Flow Past Parallel Plates
- Can be used for gases or evaporation of liquid
- Distinguished between laminar & turbulent flow
- \(N_{Re}\) regime cutoff different in gas & liquid!
- Characteristic length \(L\): length of plate in flow direction
Flow Past Parallel Plates: Results
Laminar flow (\(N_{Re} < 15,000\))
\[\begin{align}
j_D &= 0.664 N_{Re, L}^{-0.5} \\
\frac{k_c' L}{D_{AB}} &= 0.664 N_{Re, L}^{0.5}
N_{Sc}^{1/3}
\end{align}\]
- This follows our derivation of boundary layer theory
Turbulent flow
- Gas: $15,000 < N_{Re}< 300,000 $
\[\begin{align}
j_D = 0.036 N_{Re, L}^{-0.2}
\end{align}\]
- Liquid: \(600 < N_{Re}< 50,000\)
\[\begin{align}
j_D = 0.99 N_{Re, L}^{-0.5}
\end{align}\]
Case 3: Flow Past Single Sphere
- Frequent geometry in particle solutions
- Low Reynolds regime 👉 solution for stagnant diffusion on spherical surface
- High Reynolds regime 👉 correct \(N_{Sh}\) and back calculate \(k_c'\)
Flow Past Single Sphere: Results
Low Reynolds (\(N_{Re} < 2\))
\[\begin{align}
N_A &= \boxed{\frac{2 D_{AB}}{D_p}} (c_{A1} - c_{A2}) \\
&= k_c (c_{A1} - c_{A2}) \\
&= \frac{k_c'}{x_{Bm}}(c_{A1} - c_{A2}) \\
\end{align}\]
- For \(x_{Bm} \approx 1\), we have:
\[
k_c' = \frac{2D_{AB}}{D_p}
\]
- Sherwood number: \(N_{Sh} = 2\)
High Reynolds (\(N_{Re} > 2\))
\[\begin{align}
&N_{Sh} = 2 + 0.552 N_{Re}^{0.53} N_{Sc}^{1/3} \\
&0.6 < N_{Sc} < 2.7\;\ N_{Re} < 48000
\end{align}\]
\[\begin{align}
N_{Sh} &= 2 + 0.95 N_{Re}^{0.5} N_{Sc}^{1/3};\ N_{Re} < 2000\\
N_{Sh} &= 0.347 N_{Re}^{0.62} N_{Sc}^{1/3};\ 2000 < N_{Re} < 17000
\end{align}\]
- Back calculate \(k_c' = N_{Sh} \frac{D_{AB}}{D_p}\)
Case 4: Mass Transfer for Packed Beds
- Very common geometry for chemical engineering
- Adsorption and desorption through solid particles (gases and liquids)
- Catalytic processes with very large surface area
- Geometry characteristics: void fraction \(\varepsilon\): \[
\varepsilon
= \frac{\text{void space}}{\text{total space}}
= \frac{\text{void space}}{\text{void space} + \text{solid space}}
\]
- Typically \(0.3 < \varepsilon < 0.5\)
- Void fraction is difficult to measure experimentally
Correlation Equations In Packed Bed
Correlation 1, applicable to:
- gase with \(10 < N_{Re} <10,000\)
- liquid with \(10 < N_{Re} < 1500\)
\[\begin{align}
j_D &= j_H = \frac{0.4548}{\varepsilon} N_{Re}^(-0.4069) \\
N_{Re} &= \frac{D_p v' \rho}{\mu}
\end{align}\]
- \(D_p\): (average) particle diameter
- \(v’\): superficial velocity in the tube without packing
Correlation Equations In Packed Bed (II)
Correlation 2, applicable to:
- liquid with \(0.0016 < N_{Re} < 55\), \(165 < N_{Sc} < 70000\)
\[\begin{align}
j_D = \frac{1.09}{\varepsilon}
N_{Re}^(-2/3)
\end{align}\]
- liquid with \(55 < N_{Re} < 1500\), \(165 < N_{Sc} < 10690\)
\[\begin{align}
j_D = \frac{0.250}{\varepsilon}
N_{Re}^(-0.31)
\end{align}\]
Correlation Equations In Packed Bed (III)
Correlation 3, applicable to fluidized beds
- \(10 < N_{Re} < 4000\) (gas & liquid)
\[\begin{align}
j_D = \frac{0.4548}{\varepsilon}
N_{Re}^(-0.4069)
\end{align}\]
- \(1 < N_{Re} < 10\) (liquid only)
\[\begin{align}
j_D = \frac{1.1068}{\varepsilon}
N_{Re}^(-0.72)
\end{align}\]
Packed Bed Calculation Steps
- Known value from operational column: \(\varepsilon\), \(V_b\) (total volume), \(D_p\), \(D_{AB}\), \(\mu\), \(\rho\), etc.
- Depend on the operational range, calculate \(N_{Re}\), \(N_Sc\) 👉 choose the equation for \(j_D\)
- Obtain \(k_c\) from \(j_D\) value
- Calculate flux \(N_A\)
- Estimate effective area \(A_{eff}\) inside the columne 👉 \(\overline{N}_A = A_{eff} N_A\)
Caveats In Packed Bed Problems (1)
- Estimate the effective area?
- First calculate the effective surface area per volume \(a\) then \(A_{eff}\)
\[\begin{align}
a &= \frac{6 (1 - \varepsilon)}{D_p} \\
A_{eff} &= a V_b
\end{align}\]
Caveats In Packed Bed Problems (2)
- Use log-mean driving force correction
\[\begin{align}
\overline{N}_A &= A_{eff} N_A \\
&= A_{eff} k_c \frac{
(c_{A, i} - c_{A1}) - (c_{A, i} - c_{A2})}
{
\ln\!\frac{(c_{A, i} - c_{A1})}{(c_{A, i} - c_{A2})}
}
\end{align}\]
where
- \(c_{A, i}\): surface concentration
- \(c_{A1}, c_{A2}\): in- and outlet concentrations
Caveats In Packed Bed Problems (3)
\[\begin{align}
\overline{N}_A &= A_{eff} N_A \\
&= V (c_{A2} - c_{A1})
\end{align}\]
where \(V\) is the volumetric flow rate.
These equations will give rise to solving the flow in packed bed problem.
Summary
- Dimensionless numbers can be used to correlate mass transfer problems in different flow rate, dimension etc
- Typically, start with a known geometry (pipe? parallel plate? sphere? packed bed?)
- Find the correlation with dimensionless numbers \(N_{Re}\), \(N_{Sc}\)
- Calculate the final mass transfer rate