How many pounds of alum are needed to achieve 30 ppm if the plant treats 4 MGD?

To determine how many pounds of alum are needed to achieve a concentration of 30 parts per million (ppm) when treating 4 million gallons per day (MGD), it's essential to understand the relationship between ppm and the volume of water treated.

First, recognize that "parts per million" translates to milligrams of substance per liter of water. Given that 1 ppm equals 1 mg/L, the calculation involves converting the daily flow from MGD to liters and subsequently calculating the required alum in pounds.

1. Convert the flow rate from MGD to liters:

• There are approximately 3.78541 liters in one gallon. Therefore, for 4 MGD:

[

4,000,000 \text{ gallons/day} \times 3.78541 \text{ L/gallon} = 15,141,640 \text{ L/day}

]

2. Determine the total milligrams of alum needed for 30 ppm:

• Since 30 ppm means 30 mg of alum per liter, the total amount needed for the entire daily flow is:

[

30 \text{ mg/L} \times 15,141,640 \text{ L} =