What should the setting be on a chlorinator in pounds per day if the desired dosage is 2.9 mg/L and the pumping rate is 875 GPM?

Prepare for the Indiana Water Operator Certification Test. Utilize flashcards and multiple-choice questions with hints and explanations. Achieve success with confidence!

To determine the correct setting for the chlorinator in pounds per day given a desired dosage of 2.9 mg/L and a pumping rate of 875 GPM, it is essential to perform the appropriate calculations.

First, convert the flow rate from gallons per minute (GPM) to million gallons per day (MGD). There are 1,440 minutes in a day, so you can convert the rate as follows:

[

\text{Flow rate in MGD} = \frac{875 \text{ GPM} \times 1,440 \text{ minutes}}{1,000,000} = 1.26 \text{ MGD}

]

Next, to find out how many mg of chlorine are needed per day, multiply the flow rate in MGD by the desired dosage in mg/L:

[

\text{Dosage in mg per day} = 1.26 \text{ MGD} \times 2.9 \text{ mg/L} = 3.654 \text{ mg/day}

]

Since the dosage is in milligrams and we need it in pounds, convert it by knowing that there are 453,592 milligrams in a pound:

[

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy