Question 2 of 7: Salt dissolving from a pipe wall — effect of tripling the flow
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (diffusivities, vapour pressures, solubilities, humid-air heat capacities) is stated in the question, so each answer is self-contained.
Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, gas absorption and drying; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer and transient diffusion; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer — transient conduction/diffusion charts and the sphere in a stagnant medium; Treybal, Mass-Transfer Operations (3rd ed.) — packed-tower design; Crank, The Mathematics of Diffusion (2nd ed.) — series solutions for the sphere; supporting data from Perry's Chemical Engineers' Handbook (9th ed.).
Question 2: Salt dissolving from a pipe wall — effect of tripling the flow (Part A — equal value)
Given. Turbulent internal flow past a soluble wall; the water leaves below saturation. The outlet approach to saturation is governed by the wall mass-transfer coefficient and the residence time, both of which shift with flow rate.
Quantity
Value
Base flow $Q_1$
0.5 L/s (turbulent)
Salt solubility $c_s$
0.21 mol/L
Base outlet concentration $c_{out,1}$
0.01 mol/L
New flow $Q_2$
1.5 L/s (3×)
Find. The outlet salt concentration $c_{out,2}$ at the tripled flow.
Approach. A plug-flow mass balance gives an exponential approach to saturation; the exponent group $k_c\pi DL/Q$ scales with flow through the Linton–Sherwood correlation $Sh=0.023\,Re^{0.83}Sc^{1/3}$, so only its power of $Q$ is needed — the geometry and $Sc$ cancel between the two cases.
Plug-flow dissolution balance. Integrating $Q\,dc=k_c(c_s-c)\,\pi D\,dz$ over the pipe length gives $$\frac{c_s-c_{out}}{c_s}=\exp\!\left(-\frac{k_c\,\pi D L}{Q}\right)\equiv e^{-\Gamma},\qquad \Gamma=\frac{k_c\,\pi DL}{Q}.$$
Evaluate the base-case group. From the measured outlet at $Q_1$: $$\Gamma_1=-\ln\!\frac{c_s-c_{out,1}}{c_s}=-\ln\frac{0.21-0.01}{0.21}=0.0488.$$
Scale the group with flow. With turbulent $Sh=0.023\,Re^{0.83}Sc^{1/3}$ and $Re\propto Q$, $k_c\propto Q^{0.83}$; then $\Gamma\propto Q^{0.83}/Q=Q^{-0.17}$. Tripling the flow: $$\Gamma_2=\Gamma_1\left(\frac{Q_2}{Q_1}\right)^{-0.17}=0.0488\times3^{-0.17}=0.0488\times0.830=0.0405.$$
New outlet concentration. $$c_{out,2}=c_s\left(1-e^{-\Gamma_2}\right)=0.21\left(1-e^{-0.0405}\right)=\boxed{0.0083\ \text{mol/L}}.$$
Quantity
Result
Base transfer group $\Gamma_1$
0.0488
Tripled-flow group $\Gamma_2$
0.0405
New outlet concentration $c_{out,2}$
0.0083 mol/L (down from 0.010)
Check
The outlet concentration drops when the flow is tripled — a slightly counter-intuitive but correct result. The residence-time reduction ($\propto Q^{-1}$) outweighs the rise in the wall coefficient ($\propto Q^{+0.83}$), so the water has less opportunity to pick up salt. The answer is only mildly sensitive to the exact exponent: using $0.8$ instead of $0.83$ gives $c_{out,2}=0.0081$ mol/L (about 3% lower), and the conclusion that the outlet concentration falls is unchanged.
