Wastewater Collection and Treatment Study Guide

PE Water Resources Depth (PM) Exam Preparation

6 Exam Questions Water Resources Depth (PM)

Overview

PE depth exam study guide for wastewater: sewer design, BOD loading, activated sludge, clarifier design, SVI, and disinfection.

This topic accounts for 6 out of 40 questions on the PE Civil Water Resources Depth (PM) exam.

Key Concepts

Sewer Design

Gravity sewers designed using Manning's equation. Minimum velocity: 2 ft/s (self-cleansing). Maximum velocity: 10 ft/s. Minimum slopes vary by pipe size (8-inch: 0.40%, 12-inch: 0.22%). Design flow = peak wet weather flow. Pipe sizes: minimum 8 inches for sanitary, 12 inches for storm. Invert elevations determine if pump station is needed.

BOD Loading and Removal

Mass loading: lb/day = Q(MGD) x C(mg/L) x 8.34. First-order BOD decay: BODt = L0(1 - e^(-kt)). Ultimate BOD (L0) = BOD5 / (1 - e^(-5k)). Typical k = 0.1-0.23/day at 20C. BOD removal efficiency: E = (C_in - C_out)/C_in x 100%. Primary clarifiers remove 25-40% BOD. Activated sludge removes 85-95% BOD.

Activated Sludge Process

Key parameters: SRT (solids retention time) = mass of solids in reactor / mass wasted per day. F/M ratio = Q x S0 / (V x X). Typical F/M: 0.2-0.5 for conventional. MLSS: 1500-3500 mg/L. Return sludge ratio: R = X/(Xr - X). Aeration tank sizing: V = Q x SRT x Y x (S0-S) / X. Oxygen demand = 1.5 x BOD removed + 4.6 x NH3 oxidized.

Clarifier Design

Surface overflow rate (SOR) = Q/A. Primary: 800-1200 gpd/ft^2. Secondary: 400-800 gpd/ft^2. Detention time: primary 1.5-2.5 hr, secondary 2-4 hr. Weir loading rate: 10,000-20,000 gpd/ft. Solids loading rate (SLR) for secondary: 20-30 lb/day/ft^2. Depth: 10-16 ft.

Disinfection

CT concept: C x T = required disinfection. C = disinfectant residual (mg/L), T = contact time (min) at T10 (10th percentile). T10 = BF x V/Q. Baffling factors: 0.1 (unbaffled), 0.3 (poor), 0.5 (average), 0.7 (superior). Chlorine demand = dose - residual. Breakpoint chlorination occurs when all ammonia is oxidized.

Common Exam Question Types

Exam Tips & Strategies

lb/day = MGD x mg/L x 8.34. This is the most important conversion in wastewater problems
F/M = Q*S0/(V*X). Lower F/M = longer SRT = better treatment but more sludge
CT = C x T10 (not theoretical detention time). T10 = BF x V/Q
SOR for secondary clarifiers is about half that of primary clarifiers

Practice This Topic

Test your knowledge with free practice problems

Practice Problems Browse Equations

Related Study Guides