Equipment Design — Decanter/Settler: 23-Chem-A5 Chemical Plant Design and Economics, May 2015, Q5

Question 5: Equipment Design — Decanter/Settler (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A liquid–liquid gravity separation:

Property	Light oil (dispersed)	Water (continuous)
Mass flow	1000 kg/hr	5000 kg/hr
Density, $\rho$	900 kg/m³	1000 kg/m³
Viscosity, $\mu$	1 mN·s/m² ($10^{-3}$ Pa·s)	$10^{-3}$ Pa·s
Droplet diameter, $d$	150 µm ($150\times10^{-6}$ m)

Design constraints: dispersion band = 10 % of decanter height; oil–water interface at 50 % of height; light-oil take-off at 90 % of height.

Find. A decanter (horizontal gravity settler) sized so the oil droplets separate, together with the piping arrangement (feed, oil take-off, and water draw-off leg heights).

Approach. Follow Sinnott's decanter method: compute the Stokes settling (rise) velocity of the dispersed oil droplet, confirm laminar (Stokes) validity, size the interfacial area so the continuous-phase velocity stays below that settling velocity, choose vessel dimensions from that area, then fix the water draw-off leg height by a manometric balance across the interface.

Settling (rise) velocity of the oil droplet — Stokes' law. The light oil droplet rises through the continuous water phase; for a small droplet the terminal velocity is $$u_d = \frac{d^2(\rho_c-\rho_d)\,g}{18\,\mu_c}= \frac{(150\times10^{-6})^2(1000-900)(9.81)}{18(10^{-3})}= 1.23\times10^{-3}\ \text{m/s}$$ i.e. $u_d \approx 1.23$ mm/s.
Check Stokes' law is valid. The droplet Reynolds number is $$Re = \frac{\rho_c\,u_d\,d}{\mu_c}= \frac{(1000)(1.23\times10^{-3})(150\times10^{-6})}{10^{-3}}= 0.18 \;(<1)$$ so creeping (Stokes) flow holds. The value is also well below Sinnott's recommended design cap of 4 mm/s, so $u_d$ is used directly.
Volumetric flows of each phase. Dividing mass flow by density (and by 3600 s/hr): $$Q_{oil}=\frac{1000}{900\cdot3600}=3.09\times10^{-4}\ \text{m}^3/\text{s},\qquad Q_{water}=\frac{5000}{1000\cdot3600}=1.39\times10^{-3}\ \text{m}^3/\text{s}$$ total $Q_t = 1.70\times10^{-3}$ m³/s ($\approx 6.1$ m³/hr).
Interfacial (settling) area. For clean separation the upward/through velocity of the continuous phase must not exceed the droplet settling velocity, $u_c = Q_{water}/A_i \le u_d$, giving the minimum interfacial area $$A_i = \frac{Q_{water}}{u_d}= \frac{1.39\times10^{-3}}{1.23\times10^{-3}}= \boxed{1.13\ \text{m}^2}$$
Vessel dimensions. Use a horizontal cylindrical decanter with the interface at the mid-diameter, so the interfacial area is $A_i = L\times D$. Taking a length-to-diameter ratio $L/D = 4$ gives $4D^2 = 1.13 \Rightarrow D = 0.53$ m, $L = 2.13$ m. Rounding up to a standard vessel, select $$\boxed{D = 0.6\ \text{m}, \quad L = 2.4\ \text{m}\ \;(A_i = 1.44\ \text{m}^2)}$$ which gives an actual continuous-phase velocity $u_c = Q_{water}/A_i = 0.96$ mm/s $< u_d$ — a comfortable margin.
Piping arrangement — water draw-off leg height. The interface position is fixed by balancing the two liquid legs about the interface, $\rho_d(z_{oil}-z_{if})=\rho_c(z_{w}-z_{if})$, so the heavy-liquid (water) take-off height is $$z_w = z_{if}+\frac{\rho_d}{\rho_c}\,(z_{oil}-z_{if}) = 0.50H + \frac{900}{1000}(0.90H-0.50H)= 0.86H$$ With $H=D=0.6$ m: interface at 0.30 m, dispersion band $0.10H=0.06$ m straddling it, light-oil take-off at $0.90H=0.54$ m (near the top), and the water draw-off leg set at $0.86H=0.52$ m so the interface sits at mid-height as specified.

Figure 5.1 — Horizontal gravity decanter. Feed enters at mid-height; oil droplets rise at $u_d=1.23$ mm/s through the water and coalesce in the dispersion band ($0.10H$) at the interface ($0.50H$). Light oil overflows at $0.90H$; the water draw-off leg, set at $0.86H$, holds the interface at mid-height by the density balance. The interfacial area $L\times D = 1.44$ m² keeps the continuous-phase velocity ($0.96$ mm/s) below $u_d$.

Quantity	Value
Droplet settling velocity, $u_d$ (Stokes)	$1.23\times10^{-3}$ m/s (1.23 mm/s)
Droplet Reynolds number	0.18 (Stokes valid)
Water / oil / total volumetric flow	$1.39\times10^{-3}$ / $3.09\times10^{-4}$ / $1.70\times10^{-3}$ m³/s
Minimum interfacial area, $A_i$	1.13 m²
Selected vessel (horizontal)	$D=0.6$ m, $L=2.4$ m ($A_i=1.44$ m²)
Water draw-off leg height	$0.86H = 0.52$ m

Check: the design assumes the oil droplets are 150 µm and monodisperse and that Stokes' law applies ($Re=0.18<1$, confirmed). A hydrocyclone or coalescer would be needed if the true droplet size were much smaller; the 10 % dispersion-band and mid-height-interface assumptions are given design targets, met by keeping $u_c

Concept & Theory

Topic. Gravity liquid–liquid separation (decanter/settler) sizing by Stokes settling and interfacial area.

Key relations. Stokes terminal velocity $u_d = d^2(\rho_c-\rho_d)g/18\mu_c$; validity $Re=\rho_c u_d d/\mu_c < 1$; sizing criterion $u_c = Q_c/A_i \le u_d$; interface balance $\rho_d(z_{oil}-z_{if})=\rho_c(z_w-z_{if})$ setting the heavy-phase leg height.

Why this works. A decanter simply gives the dispersed droplets enough residence time to rise (or fall) to the interface before the continuous phase carries them out. Sizing the interfacial area so the continuous-phase velocity is below the droplet settling velocity guarantees the droplets reach the interface; the density-balanced draw-off leg then holds the interface at the chosen height without active control.

Common pitfall. Sizing on the dispersed-phase flow instead of the continuous-phase flow; forgetting to check $Re<1$ (Stokes only); or mis-placing the water leg — if it is too high the vessel floods with water, too low it drains the water layer and entrains oil. Using $\Delta\rho$ with the wrong sign (dispersed heavier than continuous) is another frequent slip.

Source. Sinnott & Towler, Chemical Engineering Design §10.6 (decanters, gravity separators); Coulson & Richardson Vol. 2 (particle/droplet settling, Stokes' law).

Question 5 of 6: Equipment Design — Decanter/Settler

Question 5: Equipment Design — Decanter/Settler (20 marks)