ENV 2101 Principles of Environmental Engineering

Sag Curves

1. DO content is one of the most widely used indicators of overall ecological health of a body of water
• fish need 4 to 5 mg/L to survive
• under anaerobic conditions, undesirable (smelly) microbes can take over
• many factors affect the DO level
2. If a river was healthy before we began discharging wastewater, a significant factor in its continued health or illness is the BOD added to it by wastewater
1. At the outfall, BOD of the river/wastewater mixture (L₀) is given by:


• Note: We found this formula earlier when we used a mass balance model for mixing
2. This BOD is comparable to what we have in our stoppered bottle at the beginning of our BOD test
• As time passes (ie, the water moves downstream) the oxygen content of the river water is consumed in just the same way oxygen is consumed in the test
• Remember that BOD (L_t) in a test bottle at time t is given by:

L_t = L₀e^-kDt

• This formula holds in the river too (k_D is the deoxygenation constant that we previously just called k; it can be adjusted for temperature using k_T = k_20q ^T-20)
• If we know an average velocity of flow, we can calculate the BOD for a given distance downstream
3. We are probably more interested in how much DO remains, which depends both on the rate of deoxygenation (as in our bottle) and on the rate of reoxygenation or reaeration (which doesn’t occur in our bottle)
• The rate of reaeration, r_R, is given by:

r_R = -k_R*D with

k_R = reaeration time constant

D = DO deficit = DO_s-DO

• The reaeration time constant can be estimated from Table 3-2, or calculated by:

k_R,20°C = 3.9u^1/2/H^3/2

u = average stream velocity

H = average stream depth

Table 3-2 Reaeration constants
Water body	Ranges of kR at 20°C, base e
Small ponds and backwaters	0.1-0:23
Sluggish streams and large lakes	0.23-0.35
Large streams of low velocity	0.35-0.46
Large streams of normal velocity	0.46-0.69
Swift streams	0.69-1.15
Rapids and waterfalls	Greater than 1.15
Source: Peavy, Rowe and Tchobanoglous, 1985

• To start with, the waste has some oxygen deficit which causes an initial DO deficit in the stream
• Water can only hold so much oxygen (DO_sat), depending on the the water temperature
• Calculate the initial dissolved oxygen (DO₀) using the same formula we used for L₀ above
• Subtracting that from the initial DO_sat:
D₀ = DO_sat - DO₀
4. The DO at any point downstream depends on these competing processes:

rate of deficit increase = rate of deoxygenation - rate of reaeration

• This gives us a differential equation with the solution:



• This is the Streeter-Phelps oxygen-sag curve formula
• Note that for a constant stream cross-section, t=x/u (with u=stream velocity); therefore:



• If we want to plot DO versus distance downstream (that’s what I’m used to seeing) we need to subtract D from D_s at each point
5. To start with, DO is being depleted faster than it can be replenished
• As long as this occurs, the DO of the stream will continue to drop
• Since the BOD is decreasing as time goes on, at some point, the rate of deoxygenation decreases to just the rate of reaearation
• At this point (called the critical point) the DO reaches a minimum
• Downstream of the critical point, reaeration occurs faster than deoxygenation, so the DO increases
• Using calculus and the Streeter-Phelps equation we get:

Typo corrected 6/4/2010, otherwise this is unchanged since 2001. There's a more detailed treatment of oxygen depletion in rivers at rivers.pdf.

ANTHONY G BENOIT abenoit@trcc.commnet.edu
(860) 885-2386

Revised 6/4/10