Minimum and Maximum Oxygen Transfer to a Wastewater: 18-Env-B5 Industrial & Hazardous Waste Management, December 2014, Q20

Question 20: Minimum and Maximum Oxygen Transfer to a Wastewater (6 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the source prints C_s in "g/L" (11.27, 9.02, 7.44). These are the standard textbook dissolved-oxygen saturation values at 10/20/30°C in mg/L (= g/m³); read literally as g/L they would be physically impossible — oxygen's solubility limit in water is only about 14 mg/L even at 0°C, roughly 1000× smaller than 11.27 g/L. Solved below using mg/L, consistent with the standard DO saturation table this data is drawn from. Also, since no tank volume is given, the result is reported as a specific (per-unit-volume) transfer rate; the operating dissolved-oxygen concentration C_L in the aeration tank is not given and is assumed to be 0 (the maximum available driving force / theoretical transfer capacity, consistent with "amount of oxygen that could be transferred").

Given. A clean-water mass-transfer coefficient K_La = 1.1 hr⁻¹ measured at 20°C, saturation DO concentrations C_s at 10/20/30°C, a temperature-correction factor θ = 1.0241, an alpha factor α = 0.55 (rate correction, wastewater vs. clean water) and a beta factor β = 0.95 (saturation-concentration correction, wastewater vs. clean water).

Given data
Quantity	Symbol	Value
Saturation DO at 10°C	C_s,10	11.27 mg/L
Saturation DO at 20°C	C_s,20	9.02 mg/L
Saturation DO at 30°C	C_s,30	7.44 mg/L
Mass-transfer coefficient at 20°C	K_La,20	1.1 hr⁻¹
Temperature correction factor	θ	1.0241
Alpha factor	α	0.55
Beta factor	β	0.95

Find. The minimum and maximum field oxygen-transfer rate (mg O₂ per litre of wastewater per hour) across the three given temperatures.

Approach. Correct K_La from 20°C to each target temperature via $K_{La}(T) = K_{La,20}\,\theta^{(T-20)}$, then compute the field oxygen-transfer rate at each temperature via $\text{OTR}(T) = \alpha\,K_{La}(T)\,\big(\beta\,C_{s,T} - C_L\big)$ with C_L = 0, and compare all three to identify the minimum and maximum.

Correct K_La to 10°C and 30°C. $$K_{La}(10) = 1.1 \times 1.0241^{(10-20)} = 1.1 \times 0.7881 = 0.8669\ \text{hr}^{-1}$$ $$K_{La}(30) = 1.1 \times 1.0241^{(30-20)} = 1.1 \times 1.2689 = 1.3958\ \text{hr}^{-1}$$
Compute the field oxygen-transfer rate at each temperature. With $\alpha\beta = 0.55\times0.95 = 0.5225$: $$\text{OTR}(10) = 0.5225 \times 0.8669 \times 11.27 = 5.10\ \text{mg/L}\cdot\text{hr}^{-1}$$ $$\text{OTR}(20) = 0.5225 \times 1.1 \times 9.02 = 5.19\ \text{mg/L}\cdot\text{hr}^{-1}$$ $$\text{OTR}(30) = 0.5225 \times 1.3958 \times 7.44 = 5.43\ \text{mg/L}\cdot\text{hr}^{-1}$$
Identify the minimum and maximum. Although C_s falls as temperature rises (less oxygen can dissolve), K_La rises faster with temperature than C_s falls, so the transfer rate is highest — not lowest — at the highest temperature: $$\boxed{\text{OTR}_{min} \approx 5.11\ \text{mg/L}\cdot\text{hr}^{-1}\ \text{(at }10^{\circ}\text{C)}, \quad \text{OTR}_{max} \approx 5.43\ \text{mg/L}\cdot\text{hr}^{-1}\ \text{(at }30^{\circ}\text{C)}}$$

Final results
Quantity	Value
K_La at 10°C	0.867 hr⁻¹
K_La at 30°C	1.396 hr⁻¹
Oxygen transfer rate at 10°C	5.10 mg/L·hr⁻¹
Oxygen transfer rate at 20°C	5.19 mg/L·hr⁻¹
Oxygen transfer rate at 30°C	5.43 mg/L·hr⁻¹
Minimum oxygen transfer rate	≈ 5.11 mg/L·hr⁻¹, at 10°C
Maximum oxygen transfer rate	≈ 5.43 mg/L·hr⁻¹, at 30°C

The result is a useful, slightly counter-intuitive design check: a naive read of the C_s table alone would suggest cold weather (higher C_s) is when the most oxygen can be transferred, but because K_La itself is temperature-dependent (bubble/liquid-film mass transfer accelerates with temperature per θ), the field transfer rate actually increases with temperature over this range — so an aeration system's capacity should not be assumed to be tightest in summer without checking both effects together, not C_s alone.

Topic: Field (actual) oxygen transfer rate correction for temperature, alpha and beta factors

Key relations: $K_{La}(T) = K_{La,20}\,\theta^{(T-20)}$; $\text{OTR}(T) = \alpha\,K_{La}(T)(\beta C_{s,T}-C_L)$; standard field aeration performance equation $\text{AOTR}=\text{SOTR}\times[(\beta C_{\infty,T,H}-C_L)/C_{\infty,20}]\times\alpha\times1.024^{(T-20)}$

Why this works: a clean-water K_La measured at 20°C must be corrected on three independent axes before it represents the actual field transfer rate — temperature (via θ, because both bubble dynamics and diffusivity are temperature-dependent), and the wastewater's own chemistry (via α for the rate and β for the saturation ceiling) — and all three act multiplicatively, so none can be applied in isolation.

Common pitfall: assuming the transfer rate must be highest wherever C_s is highest (i.e. at the lowest temperature), which ignores that K_La itself rises with temperature and can dominate the product.

Source: Metcalf & Eddy, Wastewater Engineering, 5th ed. (aeration system design, α/β/θ correction factors).

Question 20 of 20: Minimum and Maximum Oxygen Transfer to a Wastewater

Question 20: Minimum and Maximum Oxygen Transfer to a Wastewater (6 marks)