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

Question

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 [latex](\mathrm{Pb})[/latex] concentration levels should not exceed [latex]0.015 \mathrm{~g} / \mathrm{m}^{3}[/latex]. At present, water flows through a [latex]0.025-\mathrm{m}[/latex] I.D. pipe at a bulk velocity of [latex]0.2 \mathrm{~m} / \mathrm{s}[/latex]. The water chemistry results in a soluble lead concentration of [latex]10 \mathrm{~g} / \mathrm{m}^{3}[/latex] 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 [latex]0.05 \mathrm{~m}[/latex] in length. The diffusivity of soluble lead in water is [latex]1 imes 10^{-9} \mathrm{~m}^{2} / \mathrm{s}[/latex], and the kinematic viscosity of water is [latex]1 imes 10^{-6} \mathrm{~m}^{2} / \mathrm{s}[/latex].

Accepted Answer

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 [latex]A[/latex] ) on a differential volume element of pipe along axial coordinate [latex]z[/latex] is

[latex] \begin{aligned} & ext { Input }( ext { moles } \mathrm{Pb} / ext { time })= ext { Output }( ext { moles } \mathrm{Pb} / ext { time }) \ & c_{A} v_{\infty} \frac{\pi D^{2}}{4}\left|+k_{L}\left(c_{A s}-c_{A} ight) \pi D \Delta z=c_{A} v_{\infty} \frac{\pi \mathrm{D}^{2}}{4} ight| \ & \qquad \quad z \qquad \qquad \qquad \qquad \qquad \qquad \qquad z+\Delta z \end{aligned} [/latex]

where [latex]c_{A}[/latex] represents the bulk concentration of soluble lead in the water stream. Upon rearrangement, we obtain

[latex]\frac{v_{\infty} D}{4}\left(\frac{\left.c_{A} ight|_{z+\Delta z}-\left.c_{A} ight|_{z}}{\Delta z} ight)=k_{L}\left(c_{A s}-c_{A} ight)[/latex]

At the limit of [latex]\Delta z ightarrow 0[/latex], the resulting differential equation is

[latex]\frac{v_{\infty} D}{4} \frac{d c_{A}}{d z}=k_{L}\left(c_{A s}-c_{A \infty} ight)[/latex]

Separation of variables gives

[latex]\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[/latex]

As [latex]c_{A s}[/latex] is constant, the integral is

[latex]\ln \left(\frac{c_{A s}-c_{A o}}{c_{A s}-c_{A L}} ight)=\frac{4 L}{D} \frac{k_{L}}{v_{\infty}}[/latex]

The mass-transfer coefficient [latex]\left(k_{L} ight)[/latex] 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

[latex]\operatorname{Re}=\frac{v_{\infty} D}{v}=\frac{(0.2 \mathrm{~m} / \mathrm{s})(0.025 \mathrm{~m})}{1 imes 10^{-6} \mathrm{~m}^{2} / \mathrm{s}}=5000[/latex]

The Schmidt number is

[latex]\mathrm{Sc}=\frac{v}{D_{A B}}=\frac{1 imes 10^{-6} \mathrm{~m}^{2} / \mathrm{s}}{1 imes 10^{-9} \mathrm{~m}^{2} / \mathrm{s}}=1000[/latex]

If [latex]\operatorname{Re}>2000[/latex], for liquids, the Linton-Sherwood correlation is appropriate for estimation of [latex]k_{L}[/latex]

[latex]\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[/latex]

or

[latex]k_{L}=\operatorname{Sh} \frac{D_{A B}}{D}=270 \frac{1 imes 10^{-9} \mathrm{~m} / \mathrm{s}}{0.025 \mathrm{~m}}=1.08 imes 10^{-5} \frac{\mathrm{m}}{\mathrm{s}}[/latex]

The Chilton-Colburn analogy can also be used to obtain [latex]k_{L}[/latex]. 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

[latex]\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 imes 10^{-5}[/latex]

From this result, [latex]k_{L}=9.5 imes 10^{-6} \mathrm{~m} / \mathrm{s}[/latex] at [latex]v_{\infty}[/latex] equal to [latex]0.2 \mathrm{~m} / \mathrm{s}[/latex], which agrees with the Linton-Sherwood correlation to within [latex]12 \%[/latex].

As there is no transfer of soluble lead into the flowing water stream between the lead-soldered pipe sections, the total transfer length [latex](L)[/latex] for the three [latex]0.05-\mathrm{m}[/latex] sections [latex](L)[/latex] is [latex]0.15 \mathrm{~m}[/latex]. The outlet concentration of soluble lead, [latex]c_{A L}[/latex], is

[latex] \begin{aligned} c_{A L} & =c_{A s}-\left(c_{A s}-c_{A o} ight) \exp \left(-\frac{4 L}{D} \frac{k_{L}}{v_{\infty}} ight) \ & =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 imes 10^{-5} \frac{\mathrm{m}}{\mathrm{s}}}{0.2 \frac{\mathrm{m}}{\mathrm{s}}} ight) \end{aligned} [/latex]

or [latex]c_{A L}=0.013 \mathrm{~g} / \mathrm{m}^{3}[/latex], using [latex]k_{L}[/latex] obtained by the Linton-Sherwood correlation. Based on this calculation, the soluble-lead concentration is just below drinking water standards.

Question 30.7: The dissolution of lead from lead-soldered joints in househo...

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