Question

A completely mixed activated sludge plant is not performing particularly well. You calculate a F:M of 1.5 mg/mg·day^-1. What would you recommend to solve the problem?

	a.	Raise the pH
	b.	Increase the dissolved oxygen concentration in the aeration tank
	c.	Lower the pH
	d.	Reduce the amount of sludge wasted

Answer 1

F/M ratio means Food to micro organisms ratio. In other words mass of food applied per unit mass of micro organism.  

Range of F/M ratio = 0.2 to 0.6 for conventional activated sludge. In this case 1.5 indicates high F/M ratio.

High F/M ratio means high quantity of food available to less number of micro organism. In this situation micro organism will be very active. This is will create highly turbid effluent. Because as a result of high microbial activity, sludge produced is not dense. This sludge will not settle easily. That's why high F/M ratio result in turbid effluent.

Problem: sludge is not settling properly. Highly turbid effluent.

Solution : Produce dense sludge. So it will settle essily.

Now we will see how to produce dense floc.

1,3) Lowering or raising ph will adversely affect microbial activity. Activated sludge process works good in neutral ph (6-8).

2) Increasing DO concentration will increase the microbial activity. Which is not desirable in this situation. Reason already explained above.

3) ANSWER- Reducing the sludge wasted. Means incresing the sludge age. That means waste water stays in aeration basin for long time. Increased sludge age will produce dense floc. This will settle very easily. This sludge is darker and dense ( Pin floc). Hence is produce less turbid effluent.