Today we will be walking through a sample calculation for sludge pumping, as a followup to yesterday’s post. We noted that we need at least four items of information to do this calculation: influent and effluent solids levels, sludge pumping rate, and sludge concentration. For our example, these will be 200 ppm, 60 ppm, 50 gpm, and 4.0 %, respectively. We actually need a fifth item — the plant flow rate — and we’ll be using 3 MGD. So here goes!
First, we need to find the pounds of sludge removed. For this we’ll use our old pal, the Pounds Formula, and recognize that our ppm entry will be the difference between the influent and effluent solids levels — that is, the ppm removed.
Pounds Removed = 8.34 x 3.0 x 140 = 3,503 lb/day
Next, we’ll use the concentration and density to convert this into a total volume of sludge to be removed each day. Absent additional lab information, we’ll assume that our sludge has the same density of water.
Sludge Volume = 3,503 lb/day ÷ 8.34 lb/gal ÷ 0.040 = 10,500 gal/day
Finally, we’ll determine how many minutes our sludge pump needs to operate each hour:
Run Time = 10,500 gal/day ÷ 24 hr/day ÷ 50 gal/min = 8.75 min/ hour
So, we’ll be setting our pump timer to run for 8.75 minutes each hour. I would initially break this in half, and run 4.37 minutes, then rest for 25.63 minutes, before repeating the cycle. Or maybe break this into three runs per hour, with a pumping time of 3 minutes, then rest for 17 minutes.
This calculation contains many assumptions, and these assumptions will definitely impact the performance of our sedimentation process. Stay tuned, as we will have a few more remarks on this topic in another upcoming blog. I know! You just can’t wait to hear more about sludge! But remember what we discussed yesterday. Sludge removal is the key to sedimentation process performance, and it is completely up to the Operator to determine the optimum sludge removal schedule. In fact, it is the MOST important Operator-controlled variable in the sedimentation process.