Tutorial on DO: Environmental Engineering PDF

Tutorial on DO: Principle of Environmental Engineering and Science, Page 347 to 369 1. Derivation of water quality model: Let us now explore a simple water-quality model known as Streeter- Phelps Dissolved Oxygen Model. What are the possible sources and sinks of oxygen in a stream?...

Tutorial on DO: Principle of Environmental Engineering and Science, Page 347 to 369 1. Derivation of water quality model: Let us now explore a simple water-quality model known as Streeter- Phelps Dissolved Oxygen Model. What are the possible sources and sinks of oxygen in a stream? Sources: Photosynthesis of phytoplankton and algae, reaeration by turbulence, and other supply that available in the river. Sink: degradation of BOD, suspended matter and sludge in water and sediments; algal respiration, etc. A. How BOD affects the oxygen resources in stream: We will start with a general mass balance at the point of waste discharge to the river: Lr= BODt , T1 and DOr in river Lo= BODt , T0 and DOo in river water flow Qr before mixing water flow Q0 (Qr+Qw) after mixing C1, 3cr, 3. C2, 2cr, 4 GPA = (3x3+2x4)/(3+2) Lw= BODt, Tw and DOw in waste stream flow Qw before mixing How does BOD change as “it” moves downstream from the wastewater discharge point? Recall decay is a first-order rate expression and we are considering the river as a plug-flow system. Where Lo = BODL in the river immediately after discharge (mg/L O2) Similar to BODt = BODu e-kt t = river travel time (= distance traveled/velocity = x/u) kd = river deoxygenation rate coefficient (time-1) B. How reaeration affects oxygen resources in stream: Oxygen dissolves in water from the atmosphere. The saturation (equilibrium) concentration is determined using Henry’s Law (we have done this before). The rate at which O2 is dissolved in water is proportional to the “dissolved oxygen deficit”: C = dissolved oxygen (DO) in water, mg/L Cs = saturation DO concentration, mg/L (temperature dependent) kr = reaeration rate coefficient, time-1 D = dissolved oxygen deficit, mg/L (Cs – C) Kr depends on mixing and flow rate of the river/stream C. Derivation of Streeter-Phelps dissolved oxygen model: What are the assumptions of the Streeter-Phelps DO model? – Only CBOD (Carbonaceous BOD) point sources of pollution – Only mixing by advection, so we can treat the stream is an ideal plug flow reactor – Only oxidation of CBOD is considered – DO sources are only: reaeration and DO addition through effluents and tributaries – steady-state conditions for the flow rate and the BOD, DO consumption – The only reactions of interest for DO are reaeration from air to water across air-water interface (kr = Cs – C) and deoxygenation of UCBOD (Ultimate carbonaceous BOD) (kdL) The figure below showed the graphical representation of the Streeter-Phelps dissolved oxygen sag curves (Source: Introduction to environmental engineering and science, Masters and Gilbert) x or t x=0 x=x t=0 t=t L = Lo L = Lt DO = DO0 D=Dt=Cs-DOt D=D0=Cs-DO0 Dt Dc Cs C= Cc DO concentration (Cx) at any distance along the river (x) is given by dC x dC x  U  k d Lx  k r (C s  C x ).........(1) dt dx Deoxygenation or oxidation of UCBOD x where, Lx  Lo exp(  k r ) U At steady-state condition, dCx/dt = 0, and substitute as (Cs - Cx) = DO deficit =Dx; the eq. (1) reduced to dC x U  k d Lx  k r Dx.........(2) dx  dC x As dx/U = dt,   k d Lx  k r Dx.........(2) dt dDx dCs dC x Dx = Cs- Cx or   and dCs/dt = 0 during constant temperature, dt dt dt dDx dC x  dt dt dDx So eq. (2) reduced to  k d Lx  k r Dx.........(3) dt By integrating eq. (3) using boundary condition as (at t = 0, D = D0, L = L0 and at t = t, D = Dt, L = Lt), the DO sag/deficit equation is obtained: The DO deficit at any point downstream of a river:  This is the Streeter-Phelps dissolved oxygen-sag curve model  Note that for a constant stream cross-section, t= x/u (u=stream velocity)  If we want to plot DO versus distance downstream, we need to subtract D from Ds at each point Where, K0 = overall loss of ultimate CBOD (L) in river water (T -1) = ks + kd ks = loss of UCBOD (L) in river water due to settling of particulate UCBOD and removal of them into bottom sediments (T-1) kd = loss of ultimate CBOD (L) in river water due to oxidation of dissolved UCBOD (T -1), sometimes using K0 instead of Kd Lx = ultimate CBOD at a distance (x) from the initial point (at x = 0) along the river water (ML-3) Lo = particulate UCBOD + dissolved UCBOD Cs = saturation oxygen concentration under the river conditions (ML-3) U = river water velocity (LT-1) a. Make list of what are the different parameters be measured by the Streeter-Phelps DO sag curves? The Streeter-Phelps equation can be used to: 1. Describe DO as f(x) or f(t) 2. Predict minimum DO at downstream 3. Predict distance (travel time, tc ) to minimum DO 4. Estimate wastewater treatment requirements b. Identify and Define Dc and tc in graphical representation of sag curve? Dc = the maximum DO deficit at the downstream point where DO level is minimum tc = the travel time at the downstream point where DO level is minimum. The mathematical solutions: b. Interpretation and explain the nature of the DO-sag curve? 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 i. At this point (called the critical point) the DO reaches a minimum ii. Downstream of the critical point, reaeration occurs faster than deoxygenation, so the DO increases Example 1: Streeter-Phelps calculations: Consider the following characteristics: parameter Wastewater River (prior to waste discharge) discharge Flow rate , Q (m3/ d) 380 7600 Initial BOD, Lo (mg/L) 500 3 Dissolved oxygen, C (mg/L) 0 8.9 T (oC) 20 20 at 20 ◦C temperature, Kd = 0.4 d-1 , kr = 0.25 d-1 , and river flow rate, u = 10 km/d, DOs = 9.17 mg/L 1. What is the DO in the river 10 km downstream? 2. What is the minimum DO at downstream? [Use the equations below] Solutions: BOD immediately after mixing waste in river: = (380 x 500 + 7600 x 3) / (380+7600) = 26.7 mg/L DO immediately after waste mixing in river: = 8.5 mg /L Do = DOs – DOo = 9.17- 8.5 = 0.67 DO deficit at 10 km downstream of the river: t = x/ u = 10/10 = 1 day Dt = ( - + = 8.25 mg /L 1. DO at 10 km downstream of river = DOs – Dt = 9.17 – 8.25 = 0.92 mg/L 2. The time at which minimum DO level reaches is: = (1/ (0.25-0.4)) ln (0.25/0.4(1-0.67(0.25-0.4)/(0.4x26.7)))= 3.2 day = 0.4/0.25 x 26.7 e- 0.4 x 3.2 = 11.88 mg/L > DOs At critical point, the river will flow at zero DO. = DOs – Dc = 9.17-11.88=-2.71 So, DOc = 0 3. What is the remaining CBOD at the 10 km downstream of the river? t=x/u= 1 d = 26.7 e-0.4x1 = 17.9 mg/L remaining BOD in the river. Qw2=500 m3/d, DOw2=4 mg/L, Lw2 = 200 mg/L 4. at 10 km, second wastewater discharge. Draw the DO deficit curve for the river. Qr =7800+380= 7980 m3/d, DOr =0.92 mg/L Lr = 17.9 mg/L DO2 = (7980x0.92 +500x4) /(7980+500) = 1.1 mg/L L0 = (7980x17.9 +500x200) /(7980+500) = 28.63 mg/L X=10 km Do= 9.17-1.1=8.07 mg/L, At the 2nd discharge at 10 km, Dt Example 2. A wastewater flow from industry mixes with a river resulting in a BOD = 10.9 mg/L, DO = 7.6 mg/L, the mixture has a temperature of 20 ◦C and saturated DO is 9.1 mg/L. Deoxygenation constant is 0.2 day-1, reoxygenation constant is 0.41 day-1. a). Find the time and distance downstream at which the oxygen deficit is a maximum? [4 marks] b). find the maximum deficit of DO? [2 marks] a). Initial deficit D0 = 9.1- 7.6 = 1.5 mg/L Calculate the time at which the maximum deficit is reached, with tc 1  k 2  Do (k 2  k1 )   tc  ln  1   k 2  k1  k1  k1 Lo   1  0.41  1.5(0.41  0.2)    ln  1  (0.41  0.2)  0.2  0.2  10.9    2.67days xc  vtc  0.3m / s  86,400s / day  2.67days  69,300m b). The maximum DO deficit is: [2marks] k1 Dc  Lo e  k1tc k2 0.2  (0.2day1 )(2.67days)  (10.9 mg/L) e 0.41  3.1 mg/L Do yourself: A city of population produces sewage at the rate of 120 gpd, and the sewage plant effluent has a BOD of 28 mg/L. The temperature of the sewage is 25.5°C, and there is 1.8 mg/L of DO in the plant effluent. The stream flow is 250 cfs at 1.2 fps. The temperature of the water is 24°C before the sewage is mixed with the stream. The stream is 90% saturated with oxygen and has a BOD of 3.6 mg/L. The kd is 0.5 d-1 at 20°C and kr =0.4 d-1. Use the given Streeter-Phelps DO Model to determine the distance downstream to the point of maximum DO deficit and the minimum DO in the stream downstream of the sewage plant. [106140 feet and 5.5 mg/L]