Metal plating facilities dip plated parts in a rinse bath to remove \"drag out\"
ID: 988507 • Letter: M
Question
Metal plating facilities dip plated parts in a rinse bath to remove "drag out", or the excess plating solution that clings to the part after plating. This rinse water is contaminated with metal and must be treated before discharge. Jefferson Plating Company's plating solution contains 85 g/L nickel, and the parts drag out 0.05 L/min of plating solution to the rinse tank. Assume that the rinse tank(s) are completely mixed and that no reactions take place. When a single rinse tank is employed, 150 L/min of rinse water is required to keep the concentration of the rinse bath drag out low enough for further finishing. What is the steady-state concentration of nickel in the rinse water? To reduce the rinse water flow rate, an engineer at Jefferson Plating Company proposes using the countercurrent rinse system shown below. Assuming that the rinse bath drag out concentration must remain the same (28 mg/L, the answer to part a), estimate the new rinse water flow rate.Explanation / Answer
a.
Total flow to rinse bath = 150 + 0.05 L/min = 150.05 L/min
Nickel flow to rinse bath = 0.05 * 85 = 4.25 g/min
Since the rinse bath is completely mixed
Concentration of nickel in rinse water = 4.25/150.05 g/L = 0.028 g/L
= 28 mg/L
b.
Total flow to rinse bath 1 = QR + 0.05 L/min
Nickel flow to rinse bath 1 = 0.05 * 85 = 4.25 g/min
Concentration of nickel in rinse water, C1 = 4.25 / (QR + 0.05) g/L
Total flow to rinse bath 2 = QR + 0.05 L/min
Nickel flow to rinse bath 2 = 0.05 * C1 = 0.05 * 4.25 / (QR + 0.05) g/min
Concentration of nickel in rinse water, C2 = 0.05 * 4.25 / (QR + 0.05)2 g/L = 0.028
(QR + 0.05)2 = 0.05 * 4.25 / 0.028 = 7.6
QR + 0.05 = 2.75
QR = 2.7 L/min