16.3 Flows, Loadings, Removal Efficiency & Solids Math

16.3 Flows, Loadings, Removal Efficiency & Solids Math

Quick Answer: Hydraulic loading compares flow to area (gpd/ft² or similar). Organic loading uses the pounds formula on BOD (or COD): MGD × mg/L × 8.34. Percent removal is $(C_{\text{in}} - C_{\text{out}}) / C_{\text{in}} \times 100$. Solids mass balances use the same 8.34 factor for WAS, inventory, and simple SRT/sludge age. Wet tons of cake scale dry solids by cake percent solids.

Wastewater Class C exams lean hard on loadings and solids. Water treatment operators still need removal efficiency and filter hydraulic loading. The algebra is shared.

---

1. Hydraulic Loading

Hydraulic loading rate (HLR) expresses how much flow is applied per unit area (or sometimes per unit volume for towers).

$$\text{HLR} = \frac{Q}{A}$$

Worked Example 1 — Clarifier Overflow Rate

Secondary clarifier diameter 60 ft, flow 2.0 MGD.

Area:

$$A = 0.785 \times 60^2 = 0.785 \times 3{,}600 = 2{,}826\ \text{ft}^2$$

HLR (gpd/ft²): 2.0 MGD = 2,000,000 gpd.

$$\text{HLR} = 2{,}000{,}000 / 2{,}826 \approx 708\ \text{gpd/ft}^2$$

Worked Example 2 — Filter Rate

Filter area 200 ft², flow 600 gpm.

$$\text{Rate} = 600 / 200 = 3.0\ \text{gpm/ft}^2$$

---

2. Organic Loading (lb BOD/day and Related)

$$\text{lb BOD/day} = Q\ (\text{MGD}) \times \text{BOD (mg/L)} \times 8.34$$

Same form works for TSS, COD, nitrogen, etc.:

$$\text{lb/day} = Q \times C \times 8.34$$

Volumetric organic loading (e.g., aeration tank):

$$\text{lb BOD/day per 1,000 ft}^3 = \frac{\text{lb BOD/day}}{V_{\text{ft}^3}/1{,}000}$$

Worked Example 3 — Plant BOD Load

Influent BOD 220 mg/L, flow 1.5 MGD.

$$\text{lb BOD/day} = 1.5 \times 220 \times 8.34 = 330 \times 8.34 = 2{,}752\ \text{lb/day}$$

Worked Example 4 — Aeration Tank Loading

Same 2,752 lb BOD/day into aeration volume 120,000 ft³.

$$\text{lb/day per 1,000 ft}^3 = 2{,}752 / 120 = 22.9\ \text{lb BOD/day}/1{,}000\ \text{ft}^3$$

(Because 120,000 / 1,000 = 120 “thousands.”)

---

3. Percent Removal Efficiency

$$\%\ \text{Removal} = \frac{C_{\text{in}} - C_{\text{out}}}{C_{\text{in}}} \times 100$$

Worked Example 5 — BOD Removal

Influent BOD 220 mg/L, effluent BOD 12 mg/L.

$$\%\ \text{Removal} = \frac{220 - 12}{220} \times 100 = \frac{208}{220} \times 100 = 94.5\%$$

Worked Example 6 — Required Effluent from Target Removal

Influent TSS 180 mg/L, need 90% removal. Max effluent TSS:

$$C_{\text{out}} = 180 \times (1 - 0.90) = 18\ \text{mg/L}$$

Worked Example 7 — Load Removed

Flow 2.0 MGD, influent BOD 200 mg/L, effluent 20 mg/L.

Influent load: $2 \times 200 \times 8.34 = 3{,}336$ lb/day
Effluent load: $2 \times 20 \times 8.34 = 333.6$ lb/day
Removed: $3{,}336 - 334 ≈ 3{,}002$ lb/day
% removal: $(200-20)/200 × 100 = 90\%$ (matches concentration %)

---

4. Solids Math — WAS, Inventory, Simple SRT

Mass of solids in a flow:

$$\text{lb/day solids} = Q_{\text{MGD}} \times \text{SS (mg/L)} \times 8.34$$

Mass of solids in a tank (inventory):

$$\text{lb solids} = V_{\text{MG}} \times \text{MLSS (mg/L)} \times 8.34$$

(Use MG of aeration + clarifier inventory only if the problem defines system volume that way.)

Simple sludge age / SRT (operator exam form):

$$\text{SRT (days)} \approx \frac{\text{lb solids in system}}{\text{lb solids leaving system per day}}$$

Solids leaving ≈ WAS lb/day + effluent SS lb/day (if significant). Many basic problems use WAS only when effluent SS is neglected by the stem.

Worked Example 8 — WAS Solids

Waste 0.05 MGD of sludge at 8,000 mg/L SS.

$$\text{lb/day WAS} = 0.05 \times 8{,}000 \times 8.34 = 400 \times 8.34 = 3{,}336\ \text{lb/day}$$

Worked Example 9 — Simple SRT

Aeration tank volume 0.40 MG, MLSS 2,500 mg/L. WAS solids 2,000 lb/day (ignore effluent SS).

Inventory:

$$0.40 \times 2{,}500 \times 8.34 = 1{,}000 \times 8.34 = 8{,}340\ \text{lb}$$

SRT:

$$\text{SRT} = 8{,}340 / 2{,}000 = 4.17\ \text{days}$$

Worked Example 10 — Include Effluent Solids

Same inventory 8,340 lb. WAS 2,000 lb/day. Effluent flow 1.5 MGD at 10 mg/L SS.

$$\text{lb/day effluent SS} = 1.5 \times 10 \times 8.34 = 125.1\ \text{lb/day}$$ $$\text{SRT} = 8{,}340 / (2{,}000 + 125) \approx 8{,}340 / 2{,}125 \approx 3.92\ \text{days}$$

---

5. Wet Tons of Cake (Dewatering)

Dewatered cake is reported as % total solids (wet basis).

$$\text{Dry solids lb} = \text{Wet cake lb} \times (\text{decimal \% solids})$$ $$\text{Wet tons} = \frac{\text{Dry tons}}{\text{decimal \% solids}}$$

Also: 1 ton = 2,000 lb.

Worked Example 11 — Wet Tons from Dry Solids

A belt press produces 1,200 lb/day dry solids as 20% cake.

$$\text{Wet lb/day} = 1{,}200 / 0.20 = 6{,}000\ \text{lb/day}$$ $$\text{Wet tons/day} = 6{,}000 / 2{,}000 = 3.0\ \text{wet tons/day}$$

Worked Example 12 — Hauling Check

Plant wastes 3,336 lb/day WAS solids (from Example 8). After digestion/thickening assume 80% of that mass is dewatered at 18% cake (simplified exam chain).

Dry to press ≈ $0.80 × 3{,}336 = 2{,}669$ lb/day dry.

$$\text{Wet lb/day} = 2{,}669 / 0.18 \approx 14{,}827\ \text{lb/day} \approx 7.4\ \text{wet tons/day}$$

---

6. Formula Sheet for This Section

---

7. Exam Strategy — Loadings & Solids

• If the stem gives mg/L and MGD, reach for 8.34 immediately.
• % removal almost always uses concentrations, not a mix of load and concentration.
• SRT denominators: read carefully whether effluent SS is included.
• Cake problems: % solids is wet basis unless stated otherwise—wet tons > dry tons.
• Overflow rate: convert MGD → gpd before dividing by ft².

Florida wastewater items often combine a loading question with a removal or WAS follow-on in separate stems—each still uses the same pounds engine.