Time to Dissolve a Pipe Deposit: 23-Chem-A3 Heat and Mass Transfer, May 2014, Q3

Question 3: Time to Dissolve a Pipe Deposit (equal value)

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. Turbulent internal flow dissolving a soluble solid from the pipe wall; the Linton–Sherwood correlation gives the wall coefficient, and the deposit inventory sets the time.

Quantity	Value
Bore / deposit	$D=20$ mm, $\delta=3$ mm ⇒ open $d=14$ mm
Water velocity $u_\infty$	1.75 m/s
Solubility $c_s$	1.2 g/L $=1.2$ kg/m³
Solid density $\rho_A$	1400 kg/m³
Diffusivity $D_{AB}$	$1.35\times10^{-5}$ cm²/s $=1.35\times10^{-9}$ m²/s
$\rho_{H_2O}$, $\mu_{H_2O}$	1000 kg/m³, $1.0\times10^{-3}$ Pa·s

Find. The time to dissolve the deposit from a 1 m length.

Figure 3 — Cross-section of the fouled pipe: a 3 mm-thick deposit lines the 20 mm bore, leaving a 14 mm open channel. Flowing water dissolves the deposit off the bore wall until the full 20 mm bore is restored.

Approach. Compute $Re$ and $Sc$ on the open bore, get the wall $k_c$ from the Linton–Sherwood correlation, then divide the deposit mass by the dissolution rate evaluated at the initial (smallest, slowest) bore for a conservative estimate.

Flow numbers on the open bore. $Re = \dfrac{\rho u d}{\mu} = \dfrac{(1000)(1.75)(0.014)}{1.0\times10^{-3}} = 24{,}500$ (turbulent); $\ Sc = \dfrac{\mu}{\rho D_{AB}} = \dfrac{1.0\times10^{-3}}{(1000)(1.35\times10^{-9})} = 741.$
Wall mass-transfer coefficient. Linton–Sherwood, $Sh = 0.023\,Re^{0.83}Sc^{1/3} = 0.023(24{,}500)^{0.83}(741)^{1/3} = 915$, so $$k_c = \frac{Sh\,D_{AB}}{d} = \frac{915(1.35\times10^{-9})}{0.014} = \boxed{8.8\times10^{-5}\ \text{m/s}}.$$
Mass of deposit. The deposit is an annulus between $R_o=10$ mm and $R_i=7$ mm over $L=1$ m: $m = \rho_A\pi(R_o^2-R_i^2)L = 1400\,\pi(0.010^2-0.007^2)(1) = 0.224\ \text{kg}.$
Dissolution rate and time. At the initial bore the rate is $\dot m = k_c\,c_s\,(\pi d L) = (8.8\times10^{-5})(1.2)(\pi\cdot0.014\cdot1) = 4.66\times10^{-6}\ \text{kg/s}$, so $$t = \frac{m}{\dot m} = \frac{0.224}{4.66\times10^{-6}} = 4.82\times10^{4}\ \text{s} = \boxed{13.4\ \text{h}}.$$

Quantity	Result
Bore $Re$ / $Sc$	24,500 / 741
Wall coefficient $k_c$	$8.8\times10^{-5}$ m/s
Deposit mass (1 m)	0.224 kg
Dissolution time	≈ 4.8×10⁴ s (≈ 13.4 h)

Check

The rate is evaluated at the initial 14 mm bore, where the wetted area is smallest and the rate slowest, giving a conservative (upper-bound) time; as the bore widens to 20 mm the rate rises, so the true clean-up is somewhat faster. Velocity is taken constant at 1.75 m/s.

Concept & Theory

Topic. Dissolution (mass transfer) from a pipe wall in turbulent internal flow.

Key relations. Linton–Sherwood $Sh=0.023\,Re^{0.83}Sc^{1/3}$; $\dot m = k_c c_s A$; deposit $m=\rho_A\pi(R_o^2-R_i^2)L$.

Why this works. The dissolving wall is analogous to the heated wall of the Dittus–Boelter problem, with $Sc$ replacing $Pr$; the water leaves the wall saturated ($c_s$) and dilutes into the bulk ($\approx0$), so the driving force is just the solubility. Sizing the rate at the initial bore is the slowest instant, giving a safe time estimate.

Common pitfall. Using the full 20 mm bore for $Re$ and area instead of the open 14 mm channel; also, the deposit is an annulus, so subtract the inner from the outer radius (not the full disc).

Source. Geankoplis §7.6 (mass transfer in pipes); Treybal §3.3 (wall coefficients).

Question 3 of 7: Time to Dissolve a Pipe Deposit

Question 3: Time to Dissolve a Pipe Deposit (equal value)