Question

A circular copper tube with an inner diameter of \(2 \mathrm{~cm}\) and a length of \(100 \mathrm{~m}\) is used to transport drinking water. Water flows in the tube at an average velocity of \(0.11 \mathrm{~m} / \mathrm{s}\) at $20^{\circ} \mathrm{C}$. At the inner tube surface, the mass concentration of copper in water is \(50 \mathrm{~g} / \mathrm{m}^{3}\). The Environmental Protection Agency (EPA) sets the standards for the National Primary Drinking Water Regulations (NPDWR) that apply to public water systems. The drinking water regulations limit the levels of contaminants in drinking water to protect public health. The maximum contaminant level for copper in drinking water, set by the NPDWR, is \(1.3 \mathrm{mg} / \mathrm{L}\). Above that, additional steps are required to treat the water before it is considered safe for the public. Determine whether the water from the tube has a safe level of copper as per the NPDWR. The diffusion coefficient for copper in water is \(1.5 \times\) \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Short answer

Answer: No, the water in the copper tube does not meet the safe level of copper, as the calculated mass concentration of copper at the output of the tube (8.79 mg/L) is higher than the NPDWR limit (1.3 mg/L).

Step-by-step solution

Calculate the cross-sectional area of the tube

Firstly, we need to calculate the cross-sectional area of the copper tube for further calculations. The given inner diameter of the tube is \(2 \mathrm{~cm}\), and thus the radius is \(1 \mathrm{~cm}.\) Using the formula for the area of a circle, we can calculate the cross-sectional area of the tube. The formula for the area of a circle is: \(A = \pi r^2\). Substituting the given values: \(A = \pi (1 \mathrm{~cm})^2 \Rightarrow A = \pi \mathrm{~cm}^{2}.\)

Calculate the water flow rate in the tube

The given average velocity of water in the tube is \(0.11 \mathrm{~m} / \mathrm{s}\) or \(11 \mathrm{~cm} / \mathrm{s}\). Using the cross-sectional area calculated in step 1, we can determine the water flow rate (Q) in the tube with the following formula: \(Q = A \times v\). \(Q = (\pi \mathrm{~cm}^{2}) (11 \mathrm{~cm} / \mathrm{s}) \Rightarrow Q = 11\pi \mathrm{~cm}^3 / \mathrm{s}.\)

Calculate the mass concentration of copper at the output

Now, we will calculate the concentration profile of copper ions in the water using the diffusion coefficient for copper in water, \(1.5 \times10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) or \(1.5\times10^{-5} \mathrm{~cm}^{2} / \mathrm{s}\). The mass concentration of copper at the surface is \(50 \mathrm{~g} / \mathrm{m}^{3}\) or \(50 \mathrm{~mg} / \mathrm{L}\). To determine the concentration profile C(x) of copper ions in the water, we can use Fick's law of diffusion with the following equation: \(C(x) = C_s\left[1 - erf\left(\dfrac{d}{2\sqrt{Dt}}\right)\right]\). Here, \(C_s = 50 \mathrm{~mg} / \mathrm{L}\) is the surface concentration of copper, \(D = 1.5 \times10^{-5} \mathrm{cm}^{2} / \mathrm{s}\) is the diffusion coefficient, and \(t = \dfrac{L}{v} = \dfrac{100 \times 10^2\mathrm{~cm}}{11 \mathrm{~cm} / \mathrm{s}}\). The quantity \(d\) represents the characteristic length, which is equal to the radius of the tube, i.e., \(1 \mathrm{~cm}\). Substituting the values in the equation: \(t = \dfrac{100 \times 10^2 \mathrm{~cm}}{11 \mathrm{~cm} / \mathrm{s}} \approx 909.09 \mathrm{~s}.\) \(C(1) = 50 \mathrm{~mg} / \mathrm{L} \left[1 - erf\left(\dfrac{1}{2\sqrt{(1.5 \times 10^{-5}) (909.09)}}\right)\right].\) Upon solving the equation using a calculator: \(C(1) \approx 8.79 \mathrm{~mg} / \mathrm{L}\).

Compare the calculated concentration with the NPDWR limit

Now that we have obtained the mass concentration of copper at the output, we need to compare this value with the maximum contaminant limit set by NPDWR, which is \(1.3 \mathrm{mg} / \mathrm{L}\). Since the calculated mass concentration of copper at the output of the tube is \(8.79 \mathrm{~mg} / \mathrm{L}\), which is higher than the NPDWR limit of \(1.3 \mathrm{mg} / \mathrm{L}\), the water from this tube does not meet the safe level of copper as required by the National Primary Drinking Water Regulations. Therefore, additional steps are necessary to treat the water before it is considered safe for the public.