Pressure Drop through a Bed of Ion-Exchange Resin: 23-Chem-A2 Unit Operations and Separation Processes, Undated paper, Q3

Question A3: Pressure Drop through a Bed of Ion-Exchange Resin (25 marks: a 7.5, b 10, c 7.5)

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.

Quantity	Value
Sieve openings (Table D.1): 20-mesh / 50-mesh	0.841 mm / 0.297 mm
Arithmetic-mean particle size $d_p$	0.569 mm
Superficial flux (10 gpm/ft²) $\Rightarrow u_s$	$6.79\times10^{-3}$ m/s
Published gradient (0.8 psi/ft) $\Rightarrow \Delta P/L$	18 096 Pa/m
Water $\mu$ / $\rho$	$1\times10^{-3}$ Pa·s / 1000 kg/m³

Find. (a) predicted $\Delta P/L$; (b) $d_p$ for agreement; (c) $\varepsilon$ for agreement.

Low Reynolds bed ($Re_p=\rho u_sd_p/\mu\approx3.9$): the viscous (Blake–Kozeny) term of the Ergun equation dominates, so $\Delta P/L\propto (1-\varepsilon)^2/(\varepsilon^3 d_p^2)$.

Approach. Use the Ergun equation; at $Re_p\approx4$ the viscous Blake–Kozeny term controls, so $\Delta P/L$ scales as $1/d_p^2$ and as $(1-\varepsilon)^2/\varepsilon^3$. Predict (a), then invert those scalings for (b) and (c).

Superficial velocity. $10\ \mathrm{gpm/ft^2}=\dfrac{10(3.785\times10^{-3})/60}{0.0929}=6.79\times10^{-3}\ \mathrm{m/s}$; check $Re_p=\rho u_s d_p/\mu=(1000)(6.79\times10^{-3})(5.69\times10^{-4})/10^{-3}=3.9$ — deep in the viscous regime.
(a) Predicted gradient (Blake–Kozeny viscous term of the Ergun equation): $$\frac{\Delta P}{L}=\frac{150\,\mu u_s(1-\varepsilon)^2}{\varepsilon^3 d_p^2}=\frac{150(10^{-3})(6.79\times10^{-3})(0.65)^2}{(0.35)^3(5.69\times10^{-4})^2}=3.1\times10^{4}\ \mathrm{Pa/m}.$$ In field units, $$\frac{\Delta P}{L}=3.1\times10^{4}\times\frac{0.3048}{6895}=\boxed{1.37\ \mathrm{psi/ft}}.$$ The model over-predicts the published 0.8 psi/ft — the real resin is either coarser or more porous than assumed.
(b) Particle size for agreement ($\varepsilon=0.35$ fixed). Since $\Delta P/L\propto 1/d_p^2$ in this regime, $$d_p'=d_p\sqrt{\frac{1.37}{0.8}}=0.569\sqrt{1.714}=0.745\ \mathrm{mm}.$$ Required $d_p\approx0.75$ mm (about a 20–24 mesh equivalent — the bed behaves as if made of larger grains).
(c) Void fraction for agreement ($d_p=0.569$ mm fixed). The porosity group scales the gradient by $(1-\varepsilon)^2/\varepsilon^3$; setting $$\frac{(1-\varepsilon)^2/\varepsilon^3}{(1-0.35)^2/0.35^3}=\frac{0.8}{1.37}$$ and solving gives $\varepsilon\approx0.398$. A modest increase in voidage (0.35 → 0.40) is enough to bring the prediction onto the published value.

Part	Result
(a) Predicted $\Delta P/L$ ($d_p=0.569$ mm, $\varepsilon=0.35$)	$3.1\times10^{4}$ Pa/m = 1.37 psi/ft
(b) $d_p$ for agreement with 0.8 psi/ft	0.745 mm
(c) $\varepsilon$ for agreement with 0.8 psi/ft	0.398

Concept & Theory

Topic. Single-phase flow through a packed bed — the Ergun equation and its viscous (Blake–Kozeny) limit.

Key relations. Ergun $\dfrac{\Delta P}{L}=\dfrac{150\mu u_s(1-\varepsilon)^2}{\varepsilon^3 d_p^2}+\dfrac{1.75\rho u_s^2(1-\varepsilon)}{\varepsilon^3 d_p}$; particle Reynolds $Re_p=\rho u_s d_p/\mu$; viscous limit valid for $Re_p\lesssim10$.

Why this works. The bed pressure gradient is set by the hydraulic radius of the void channels, which shrinks as particles get finer or as the bed packs tighter (lower $\varepsilon$). At $Re_p\approx4$ inertial losses are negligible, so the gradient is linear in velocity (hence “proportional to flow rate,” as the problem states) and inversely proportional to $d_p^2$. That single scaling lets you back out either the effective grain size or the porosity from one measured point.

Common pitfall. Reading sieve numbers as sizes (a “20-mesh” screen has a 0.841 mm opening, not 20 mm), and forgetting to check $Re_p$ before dropping the inertial term. Including the turbulent term here (a rigorous Ergun solve) raises (a) to $\approx1.47$ psi/ft and shifts (b),(c) by only a few percent.

Source. Geankoplis §3.1C (Flow through packed beds); McCabe, Smith & Harriott Ch. 7.

Question 3 of 6: Pressure Drop through a Bed of Ion-Exchange Resin

Question A3: Pressure Drop through a Bed of Ion-Exchange Resin (25 marks: a 7.5, b 10, c 7.5)