This example illustrates how WEAP calculates instream
water quality. The section of the river modeled is 100 km, with a water
temperature throughout of 15 °C. The river headflow of 500 units contains
initial concentrations of BOD (
= 5 mg/l), DO (
= 8 mg/l), Salt (2 mg/l) and TSS (20 mg/l). Salt is a conservative
constituent, so it will not decay. TSS follow first-order decay, with a decay rate of 0.25 (/day). The demand site withdraws 100 units, consumes 50%, and returns 50% (50 units) to the wastewater treatment plant, with the following concentrations: BOD (20 mg/l), DO (3 mg/l), Salt (10 mg/l) and TSS (5 mg/l). The wastewater treatment plant removes 90% of the BOD, 0% of salt and TSS, and the outflow concentration of DO is 4 mg/l. 10% of the water evaporates in processing, so 90% of the inflow (45 units) returns to the river.
The initial concentrations of non-conservative pollutants will decay along the reach below the headflow. If the reach is 38.8 km long, and the water velocity is 15.61 m/s, it will take 25 seconds to traverse the reach.
Salinity
Salt is conservative, so it will not decay. Therefore, the concentration at the end of the first reach will be the same as at the beginning: 2 mg/l.
TSS
TSS is simulated with first-order decay. In Eqn. 3, if c0 = 20 mg/l, k = 0.25/day, L = 38.8 km, and U = 196.36 km/day; then c = 19.036 mg/l.
DO
In Eqn. 4, if T = 15; then OS = 10.94.
It follows from Eqn. 5 that if kd=0.4/day,
ka = 0.95/day, kr
= 0.4/day, L = 38.8 km, U = 196.36 km/day, = 5 mg/l, and
DOIN = 8 mg/l; then DO = 8.157
mg/l.
BOD
Using Eqn. 7 with
= 0.25, H = 4m (i.e., 
= 0.3), and T = 15C; krBOD
= 0.288
Then, using Eqn. 6 with 
= 0.288, L = 38.8 km, U = 196.36
km/day, and = 5 mg/l; then BOD = 4.72 mg/l
Withdrawal Node
Removing water from the river does not change the concentration of the water, it just reduces the volume. Therefore, the concentrations immediately below this node will be the same as flowed into the node from the reach above.
Salinity
Salt is conservative, so it will not decay. Therefore, the concentration at the end of the reach will be the same as at the beginning: 2 mg/l.
TSS
Using Eqn. 3, with c0 = 19.036 mg/l, k = 0.25/day, L = 20.4 km, and U = 180 km/day; then c = 18.504 mg/l
DO
Using Eqn. 4 with T = 15; OS = 10.94
Then, using Eqn. 5 with kd
= 0.4/day; ka = 0.95/day, kr = 0.4/day, L = 20.4 km, U = 180
km/day, = 4.72 mg/l, DOIN
= 8.157 mg/l; then DO = 8.243 mg/l
BOD
Using Eqn. 7 with
= 0.25, H = 3.6 m (i.e., = 0.3), T = 15C; then krBOD
= 0.288.
Demand Site
Concentrations of pollution in demand site return flows does not depend on concentrations of inflows to the demand site. Therefore, the concentration of the river water supply is irrelevant.
In this example, demand site pollution generation is specified as the concentration in the return flow. The demand site withdraws 100 units, consumes 50%, and returns 50% (50 units) to the wastewater treatment plant with the following concentrations: BOD (20 mg/l), DO (3 mg/l), Salt (10 mg/l) and TSS (5 mg/l).
The mass of each pollutant is calculated as,
MonthlyPollutionGeneratedDS,p,m = DemandSiteReturnFlowDS,m x ReturnFlowConcentrationDS,m,p
Salinity
Mass = 50 * 10 = 500
TSS
Mass = 50 * 5 = 250
DO
Mass = 50 * 3 = 150
BOD
Mass = 50 * 20 = 1000
Wastewater Treatment Plant
The wastewater treatment plant removes 90% of the BOD, 0% of salt and TSS, and the outflow concentration of DO is 4 mg/l. 10% of the water evaporates in processing, so 90% of the inflow (45 units) returns to the river. Evaporation concentrates the pollutants somewhat, causing the concentrations to be higher (by 11% in this case) than they would with no evaporation. The removal rate (e.g., 90% of BOD) refers to the mass of pollutant, not the concentration.
Salinity
TreatmentPlantPollOutflowTP,p = (1 - RemovalRateTP,p)x TreatmentPlantPollInflowTP,p
Mass outflow = (1 - 0) * 500 = 500
TSS
Mass outflow = (1 - 0) * 250 = 250
BOD
Mass outflow = (1 - 0.9) * 1000 = 100
DO
DO is specified as a concentration, rather than a removal rate. Therefore, the inflow of DO is not used.
TreatmentPlantPollOutflowTP,p = OutflowConcentrationTP,p x TreatmentPlantReturnFlowTP,p
Mass outflow = 4 * 45 = 180
Reach below Withdrawal Node
Salinity
Salt is conservative, so it will not decay. Therefore, the concentration at the end of the reach will be the same as at the beginning: 2 mg/l.
TSS
Using Eqn. 3, with c0 = 19.036 mg/l, k = 0.25/day, L = 20.4 km, U = 180 km/day; then c = 18.504 mg/l.
DO
Using Eqn. 4, with T = 15; then OS = 10.94.
Using Eqn. 5 with kd = 0.4/day,
ka = 0.95/day, kr
= 0.4/day, L = 20.4 km, U = 180 km/day, = 4.72 mg/l,
and DOIN = 8.157 mg/l; then
DO = 8.243 mg/l
BOD
Using Eqn. 7 with
= 0.25, H = 3.6 m (i.e., = 0.3), T = 15C; then krBOD
= 0.288.
Using Eqn. 6 with = 0.288, L = 20.4 km, U = 180 km/day,
= 4.72 mg/l; then BOD = 4.57 mg/l
Return Flow Node
Treated effluent from the wastewater treatment plant mixes with the river water, using the following weighted average:
Eqn. 8
c is the new concentration (mg/l)
Qw is the inflow of wastewater = 45
Qr is the flow from upstream = 400
Cw is the concentration of pollutant in the wastewater
Cr is the concentration of pollutant in the flow from upstream
Mw = Qw Cw, the mass of pollutant in wastewater
Salinity
With Mw = 500 and Cr = 2 mg/l; then c = 2.92 mg/l in Eqn. 8.
TSS
With Mw = 250 and Cr = 18.5 mg/l; then c = 17.2 mg/l
DO
With Mw = 180 and Cr = 8.24 mg/l; then c = 7.81 mg/l
BOD
With Mw = 100 and Cr = 4.57 mg/l; then c = 4.33 mg/l
Reach below Return Flow Node
Salinity
Salt is conservative, so it will not decay. Therefore, the concentration at the end of the reach will be the same as at the beginning: 2.92 mg/l.
TSS
TSS uses first-order decay. Using Eqn. 3 with c0 = 17.2 mg/l, k = 0.25/day, L = 40.8 km, and U = 187.92 km/day; then c = 16.29 mg/l
DO
Using Eqn. 4 with T = 15; then OS = 10.94
Using Eqn. 5 with kd = 0.4/day,
ka = 0.95/day, kr
= 0.4/day, L = 40.8 km, U = 187.92 km/day, = 4.33 mg/l,
and DOIN = 7.81 mg/l; then DO
= 8.07 mg/l
BOD
Using Eqn. 7 with
= 0.25, H = 3.78m (i.e., = 0.3), and T = 15C; then krBOD
= 0.291
Using Eqn. 6 with krBOD
= 0.288, L = 40.8 km, U = 187.92 km/day, and = 4.33 mg/l;
then BOD = 4.066 mg/l
Here is a summary of surface water quality:
