Question 30.7: The dissolution of lead from lead-soldered joints in househo...
The dissolution of lead from lead-soldered joints in household piping is a possible health problem. If the water flowing through the pipe is mildly acidic, then some of the lead will dissolve from the inner surface of the lead-soldered pipe section into the water. New drinking water requirements stipulate that soluble lead (\mathrm{Pb}) concentration levels should not exceed 0.015 \mathrm{~g} / \mathrm{m}^{3}. At present, water flows through a 0.025-\mathrm{m} I.D. pipe at a bulk velocity of 0.2 \mathrm{~m} / \mathrm{s}. The water chemistry results in a soluble lead concentration of 10 \mathrm{~g} / \mathrm{m}^{3} at the surface of the soldered pipe section. Determine the soluble lead concentration in the water after it passes across three lead-soldered joints that are each 0.05 \mathrm{~m} in length. The diffusivity of soluble lead in water is 1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}, and the kinematic viscosity of water is 1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}.
It is important to realize that the concentration of soluble lead will be continuously increasing as water flows past the lead-soldered sections of pipe. A steady-state mass balance for soluble lead (species A ) on a differential volume element of pipe along axial coordinate z is
\begin{aligned} & \text { Input }(\text { moles } \mathrm{Pb} / \text { time })=\text { Output }(\text { moles } \mathrm{Pb} / \text { time }) \\ & c_{A} v_{\infty} \frac{\pi D^{2}}{4}\left|+k_{L}\left(c_{A s}-c_{A}\right) \pi D \Delta z=c_{A} v_{\infty} \frac{\pi \mathrm{D}^{2}}{4}\right| \\ & \qquad \quad z \qquad \qquad \qquad \qquad \qquad \qquad \qquad z+\Delta z \end{aligned}
where c_{A} represents the bulk concentration of soluble lead in the water stream. Upon rearrangement, we obtain
\frac{v_{\infty} D}{4}\left(\frac{\left.c_{A}\right|_{z+\Delta z}-\left.c_{A}\right|_{z}}{\Delta z}\right)=k_{L}\left(c_{A s}-c_{A}\right)
At the limit of \Delta z \rightarrow 0, the resulting differential equation is
\frac{v_{\infty} D}{4} \frac{d c_{A}}{d z}=k_{L}\left(c_{A s}-c_{A \infty}\right)
Separation of variables gives
\int_{c_{A o}}^{c_{A L}} \frac{d c_{A}}{c_{A s}-c_{A}}=\frac{4 k_{L}}{D v_{\infty}} \int_{0}^{L} d z
As c_{A s} is constant, the integral is
\ln \left(\frac{c_{A s}-c_{A o}}{c_{A s}-c_{A L}}\right)=\frac{4 L}{D} \frac{k_{L}}{v_{\infty}}
The mass-transfer coefficient \left(k_{L}\right) will be determined by both correlation and by analogy. Both approaches require the Reynolds and Schmidt numbers. For flow through a pipe, the Reynolds number is
\operatorname{Re}=\frac{v_{\infty} D}{v}=\frac{(0.2 \mathrm{~m} / \mathrm{s})(0.025 \mathrm{~m})}{1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}}=5000
The Schmidt number is
\mathrm{Sc}=\frac{v}{D_{A B}}=\frac{1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}}{1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}}=1000
If \operatorname{Re}>2000, for liquids, the Linton-Sherwood correlation is appropriate for estimation of k_{L}
\mathrm{Sh}=\frac{k_{L} D}{D_{A B}}=0.023 \operatorname{Re}^{0.83} \mathrm{Sc}^{1 / 3}=0.023(5000)^{0.83}(1000)^{1 / 3}=270
or
k_{L}=\operatorname{Sh} \frac{D_{A B}}{D}=270 \frac{1 \times 10^{-9} \mathrm{~m} / \mathrm{s}}{0.025 \mathrm{~m}}=1.08 \times 10^{-5} \frac{\mathrm{m}}{\mathrm{s}}
The Chilton-Colburn analogy can also be used to obtain k_{L}. From Figure 14.1, the friction factor is 0.0095 for flow through a smooth pipe at Re equal to 5000 . From the Chilton-Colburn analogy, we obtain
\frac{k_{L}}{v_{\infty}}=\frac{C_{f}}{2} \frac{1}{\mathrm{Sc}^{2 / 3}}=\frac{0.0095}{2} \frac{1}{(1000)^{2 / 3}}=4.75 \times 10^{-5}
From this result, k_{L}=9.5 \times 10^{-6} \mathrm{~m} / \mathrm{s} at v_{\infty} equal to 0.2 \mathrm{~m} / \mathrm{s}, which agrees with the Linton-Sherwood correlation to within 12 \%.
As there is no transfer of soluble lead into the flowing water stream between the lead-soldered pipe sections, the total transfer length (L) for the three 0.05-\mathrm{m} sections (L) is 0.15 \mathrm{~m}. The outlet concentration of soluble lead, c_{A L}, is
\begin{aligned} c_{A L} & =c_{A s}-\left(c_{A s}-c_{A o}\right) \exp \left(-\frac{4 L}{D} \frac{k_{L}}{v_{\infty}}\right) \\ & =10 \frac{\mathrm{g}}{\mathrm{m}^{3}}-(10-0) \frac{\mathrm{g}}{\mathrm{m}^{3}} \exp \left(-\frac{4 \cdot 0.15 \mathrm{~m}}{0.025 \mathrm{~m}} \frac{1.08 \times 10^{-5} \frac{\mathrm{m}}{\mathrm{s}}}{0.2 \frac{\mathrm{m}}{\mathrm{s}}}\right) \end{aligned}
or c_{A L}=0.013 \mathrm{~g} / \mathrm{m}^{3}, using k_{L} obtained by the Linton-Sherwood correlation. Based on this calculation, the soluble-lead concentration is just below drinking water standards.
