Help Needed: Concentrations from Flow

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BradyD

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Hi All -

I'm in the midst of studying for the PE, but there's one problem on the NCEES practice exam that I can't quite wrap my head around.  I'm going to truncate it for fear of reprisal from NCEES.  You are given 20,000 gpd of sludge flow, with 2% solids and 75% volatile.  They somehow get 20,000 mg/L solids out of the flow rate and solids percentage.  Could anybody explain this leap in logic to me?  It must be simple and I'm overthinking it, but I'm having trouble getting there.  

Thanks!

 
A  liter of water weighs 1,000,000 miligrams. 

1,000,000 x 2% = 20,0000 mg

 
Last edited by a moderator:
Hi Brady, I used to get very confused about this as well, to the point I wrote a blog post to help myself remember it! Ignore the other information such as volatile solids percentage. 2% solids  is equivalent to 20,000 mg/L. Here's why:

2% solids = 2 lb solids per 100 lbs of water. (unless stated otherwise). So:

(2 lb solids/100 lb water) x (62.4 lb water/ft3 water) x (1 ft3 water/28.3 L water) x (1 kg solids/ 2.2 lb solids) x (1E6 mg solids/kg solids) = 20,000 mg/L

OR, more simply:

(2 kg solids/100 kg water) x (1 kg water/L water) x (1,000,000 mg solids/kg solids) = 20,000 mg/L

Just clever manipulation of the units. Write it down and cross out the units to see for yourself. Whenever you see % solids, multiply by 10,000 to get mg/L.

 
Last edited by a moderator:
Thank, this is a huge help!  I think the link I was missing was assuming that the % mass of the flow is assumed to be the weight of the sludge.  This kept me up last night thinking about it, and I realized generally the SG of a sludge solution is assumed to be 1.0 per the CERM, so the leap started to make sense.  

Are either of you hydrograph experts?  I'm going to be posting a question on that momentarily...

Thanks!!!

 
I think the closest I got to being an expert was studying for the exam lol. I'll take a look though

 

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