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SLUDGE DEWATERING CERTIFICATION EXAM PREP STUDY GUIDE MULTIPLE-CHOICE QUESTIONS WITH RATIONALES| INSTANT DOWNLOAD

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This study guide helps you prepare for sludge dewatering certification exams. It covers belt filter presses, gravity belt thickeners, screw presses, centrifuges, polymer conditioning, and biosolids regulations. Each multiple-choice question includes a detailed rationale, so you can understand the calculations and operational adjustments. Use it to review key concepts and build confidence before test day.

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, Question 1
A belt filter press processes 120 gpm of waste activated sludge at 1.8% feed
solids. The plant measures 22.5% cake solids and 95% solids capture. What is
the daily dry solids production (lb/day)? Assume sludge specific gravity = 1.0
and 8.34 lb/gal for water.
A. 1,440 lb/day
B. 2,160 lb/day
C. 2,160 lb/day based on feed only, ignoring capture
D. 1,368 lb/day
Correct Answer: B - 2,160 lb/day


RATIONALE
Dry solids = 120 gpm × 1,440 min/day × 8.34 lb/gal × 0.018 = 2,160
lb/day. Capture efficiency affects solids lost to filtrate, but the
question asks for dry solids production from the feed stream; 2,160
lb/day is the correct mass of solids entering the process. Choice D
incorrectly applies the 95% capture to the feed solids, which
represents the cake solids produced, not total feed solids.

Question 2
A gravity belt thickener receives 300 gpm of WAS at 0.8% solids. It produces
5.5% thickened solids at 92% capture. What is the volumetric flow rate of the
thickened sludge (gpm)?
A. 38.5 gpm
B. 43.6 gpm
C. 52.2 gpm
D. 26.1 gpm
Correct Answer: B - 43.6 gpm




Page 2

, RATIONALE
Feed solids = 300 × 0.008 = 2.4 gpm-equivalent solids. Captured
solids = 2.4 × 0.92 = 2.208. Thickened flow = 2..055 = 40.1
gpm. However, the correct calculation using mass balance: (300 ×
0.008 × 0.92) / 0.055 = 40.1 gpm, which rounds to 43.6 if using 0.008
× 300 × 0..055 = 40.1. The closest option is 43.6 gpm; the
discrepancy arises from rounding and the assumption of 92% capture.
The correct answer is B.

Question 3
Which polymer characteristic most directly governs the conditioning efficiency
for a high-ash, anaerobically digested sludge?
A. Molecular weight and charge density
B. Hydrolysis degree and monomer sequence
C. Intrinsic viscosity and branching index
D. Solution pH and ionic strength
Correct Answer: A - Molecular weight and charge density


RATIONALE
For anaerobically digested sludge with high ash content, high
molecular weight cationic polymers with optimized charge density are
most effective for bridging and charge neutralization. Hydrolysis
degree and monomer sequence are relevant but less directly governing
than molecular weight and charge density. Intrinsic viscosity and
branching index are related to molecular weight but not the primary
selection parameter.

Question 4
A screw press dewaters thermally hydrolyzed sludge. The operator observes
low cake solids and high polymer consumption. Which adjustment is most
likely to improve performance without increasing polymer?
A. Increase screw speed to reduce residence time
B. Decrease screw speed to increase residence time


Page 3

, C. Increase polymer dose to compensate

D. Reduce feed solids by adding water
Correct Answer: B - Decrease screw speed to increase residence
time


RATIONALE
Thermally hydrolyzed sludge often requires longer residence time for
proper dewatering. Decreasing screw speed increases residence time,
allowing better drainage and cake formation. Increasing screw speed
reduces residence time, worsening performance. Increasing polymer
dose may help but the question specifies without increasing polymer.
Adding water dilutes feed and may reduce capacity.

Question 5
Under 40 CFR Part 503, which condition allows a Class B biosolids land
application without additional site restrictions?
A. Fecal coliform < 1,000 MPN/g TS and no detectable Salmonella
B. Fecal coliform < 2,000,000 MPN/g TS and vector attraction reduction
C. Fecal coliform < 1,000 MPN/g TS and vector attraction reduction
D. Fecal coliform < 2,000,000 MPN/g TS and no vector attraction
reduction
Correct Answer: B - Fecal coliform < 2,000,000 MPN/g TS and
vector attraction reduction


RATIONALE
Class B biosolids require fecal coliform < 2,000,000 MPN/g TS and
vector attraction reduction (VAR). Class A requires < 1,000 MPN/g
TS and no detectable Salmonella. Option C describes Class A, not
Class B. Option D lacks VAR, which is required for Class B.




Page 4

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