completely mixed lake or cstr
play

Completelymixed lake or CSTR Often useful to assume perfect mixing - PDF document

CEE 572 Lecture #4 9/21/2017 Updated: 21 September 2017 Print version Lecture #4 (mass balance, loadings & steady state solutions) Chapra L3 David A. Reckhow CEE 577 #4 1 Completelymixed lake or CSTR Often useful to assume


  1. CEE 572 Lecture #4 9/21/2017 Updated: 21 September 2017 Print version Lecture #4 (mass balance, loadings & steady state solutions) Chapra L3 David A. Reckhow CEE 577 #4 1 Completely‐mixed lake or CSTR  Often useful to assume perfect mixing  same concentration throughout system     Accumulation loading outflow reaction settling Outflow C C Loading Q C in reaction Q V settling David A. Reckhow CEE 577 #4 2 1

  2. CEE 572 Lecture #4 9/21/2017 Accumulation   C M Accumulation V  t So: M  and c V   Vc  M Vc Accumulation  t And if volume is constant: Equals zero  c V dc   at steady Accumulation V  t dt state David A. Reckhow CEE 577 #4 3 Loading   ( ) ( ) Loading W t Qc t in  Point Sources  Municipal Wastewater Well defined origin  Industrial Wastewater easily measured more constant  Tributaries  Non ‐ point sources  agricultural  silvicultural Diffuse origin  atmospheric more transient  urban & suburban runoff often dependent on precipitation  groundwater David A. Reckhow CEE 577 #4 4 2

  3. CEE 572 Lecture #4 9/21/2017 Reported Values Of Selected Waste Input Parameters In The United States (Table 1.3 from Thomann & Mueller) Units a CSO c Variable Municipal Urban Agriculture Forest Atmosphere Influent b Runoff d (lb/mi 2 -d) e (lb/mi 2 -d) e (lb/mi 2 -day) f Average gcd 125 daily flow Total mg/L 300 410 610 2500 400 suspended solids g CBOD5 mg/L 180 170 27 40 8 g CBODU mg/L 220 240 g NBOD mg/L 220 290 Total mg-N/L 50 9 2.3 15 4 8.9-18.9 nitrogen Total mg-P/L 10 3 0.5 1.0 0.3 0.13-1.3 phosphorus 6 /100 Total 10 30 6 0.3 coliforms mL Cadmium  g/L 1.2 10 13 0.015 Lead  g/L 22 190 280 1.3 Chromium  g/L 42 190 22 0.088 Copper  g/L 159 460 110 Zinc  g/L 241 660 500 1.8 Total PCB  g/L 0.9 0.3 - 0.002-0.02 David A. Reckhow CEE 577 #4 5 Footnotes for T&M Table 1.3 a Units apply to municipal, CSO (combined sewer overflow), and urban runoff sources; gcd = gallons per capita per day. b Thomann (1972); heavy metals and PCB, HydroQual (1982). c Thomann (1972); total coli, Tetra Tech, (1977); heavy metals Di Toro et al. (1978): PCB. Hydroscience (1978). d Tetra Tech (1977): heavy metals, Di Toro et al. (1978). e Hydroscience (1976a). f Nitrogen and phosphorus, Tetra Tech (1982): heavy metals and PC13, HydroQual (1982). g CBOD5 = 5 day carbonaceous biochemical oxygen demand (CBOD); CBODU = ultimate CBOD; NBOD = nitrogenous BOD. David A. Reckhow CEE 577 #4 6 3

  4. CEE 572 Lecture #4 9/21/2017 Loading: Flow as a function of precipitation  Non point sources are difficult to characterize  Empirical approach: export coefficients (see Table 3.1 in T&M)  Mechanistic approach: relate to meteorology, topology, etc.  Flow: use the rational formula: Q R = cIA Drainage Area [L 2 ] Runoff flow [L 3 /T] Rainfall Intensity [L/T] Runoff coefficient 0.1-0.3 for rural areas (1 person/acre) 0.7-0.9 for heavy commercial areas Note: 1 acre-in/hr  1 cfs David A. Reckhow CEE 577 #4 7 Runoff: Contrasting approaches  Lumped model  Empirical  Built on a single rainfall intensity from rain gage data  Distributed model  Mechanistic  Built on radar data for rainfall  Spatial & temporal resolution  Combine with overland flow models  Many computer codes  CASC2D, CUHP, CUHP/SWMM, DR3M, HEC ‐ 1, HSPF, PSRM, SWMM, TR20 David A. Reckhow CEE 577 #4 8 4

  5. CEE 572 Lecture #4 9/21/2017 Loading: conc. as a function of flow  It is common for pollutant concentrations from uncontrolled sources (e.g. tributaries) to be correlated with flow  establish a 1000 log-log Concentration (mg/L) relationship 100  c=aQ b 10 Log(C) = log(a) + b*log(Q) 1 1 10 100 1000 Flow (cfs) David A. Reckhow CEE 577 #4 9 Loading Example: #3.1 from T&M  Data: Runoff from 100 mi 2 of agricultural lands drains to a point in a river where a city of 100,000 people is located. The city has a land area of 10 mi 2 and its sanitary sewers are separated from its storm drains. A sewage treatment plant discharges to the river immediately downstream of the city. The area receives an annual rainfall of 30 in. of which 30% runs off the agricultural lands and 50% drains off the more impervious city area.  Problem: Using the loading data from Table 1.3 and the residual fractions cited in the table below, compare the contributions of the atmospheric, agricultural and urban sources to annual average values of flow, CBOD5, total coliform bacteria, and lead in the river. Neglect any decay mechanisms for all parameters. (at) (ag) (ur) Wastewater Treatment Plant Item Atmospheric Agricultural Urban Runoff Influent Resid. Fract. Fow 30% precip. 50% precip. 125 gcd 1.00 40 lb/mi 2 -d CBOD5 27 mg/L 180 mg/L 0.15 3x10 5 /100mL 3x10 6 /100mL Total coliform 100/100 mL 0.0001 1.3 lb/mi 2 -d Lead 280  g/L 22  g/L 0.05 David A. Reckhow CEE 577 #4 10 5

  6. CEE 572 Lecture #4 9/21/2017 Solution to loading problem  Flow contributions     2  5280 1 1 2 ( ) 100 30 / 0 . 3 ft ft yr 1 Q ag mi in yr d 12 365 86 , 400 mi in d s  66 . 3 cfs     2  2 5280 1 1 ( ) 10 30 / 0 . 5 ft ft yr 1 d Q ur mi in yr 12 365 86 , 400 mi in d s  11 . 1 cfs 125 gal  ( ) 100 , 000 1 MG Q wwtp cap  6 10 gal cap d    1 . 548 12 . 5 cfs MGD MGD  19 . 4 cfs David A. Reckhow CEE 577 #4 11 Solution to loading problem (cont.)  CBOD5 loading   lb  2 ( ) 100  40  W ag mi 2   mi d lb  4000 d /   lb d  ( ) 11 . 1 27 / 5 . 4 W ur cfs mg L  / cfs mg L lb  1620 d      ( ) 12 . 5 180 / 0 . 15 8 . 34 / lb d W wwtp MGD mg L * / MGD mg L lb  2810 d David A. Reckhow CEE 577 #4 12 6

  7. CEE 572 Lecture #4 9/21/2017 Solution to loading problem (cont.)  Lead loading   lb  ( ) 100 2  1 . 3  0 . 1 W atm mi 2   mi d lb  13 d    3 / 10   lb d mg     ( ) 11 . 1 280 / 5 . 4 W ur cfs g L     /   cfs mg L g lb  16 . 8 d      3   10 ( ) 12 . 5 22 / 0 . 05 8 . 34 / mg lb d W wwtp MGD g L  * / MGD mg L g lb  0 . 11 d David A. Reckhow CEE 577 #4 13 Other Terms in the Mass Balance  Outflow  Outflow Qc  Reaction   Re action kM kVc  Settling J=vc  Settling vA c s A s  k Vc s Sediment- water interface  s  v H V A H Since: k s David A. Reckhow CEE 577 #4 14 7

  8. CEE 572 Lecture #4 9/21/2017 Combining all terms: V dc     ( ) W t Qc kVc vA c s dt  Dependent variable: c  Independent variable: t  Forcing function: W(t), the way in which the external world “forces” the system  Parameters: V, Q, k, v, A s David A. Reckhow CEE 577 #4 15 Steady State Case V dc       Mass Balance 0 ( ) W t Qc kVc vA c s dt W  Solution W W  c  or c   Q kV vA s a  Assimilation factor    a Q kV vA s  Where  The assimilation or “cleansing” factor David A. Reckhow CEE 577 #4 16 8

  9. CEE 572 Lecture #4 9/21/2017 Steady State Example #3.1 from Chapra (pg.52) A lake has the following characteristics:  50 000 , 3 Volume m Mean Depth = 2 m  3 1 Inflow = Outflow = 7500 m d o Temperature = 25 C The lake receives the input of a pollutant from three sources: a factory discharge of 50 kg d -1 , a flux from the atmosphere of 0.6 g m -2 d -1 , and the inflow stream that has a concentration of 10 mg/L. If the pollutant decays at the rate of 0.25/d at 20 o C (note: Ɵ=1.05). a. compute the assimilation factor b. steady state concentration c. show breakdown for each term David A. Reckhow CEE 577 #4 17 Example 3.1: Solution First correct the decay rate for temperature     25 20  25 20 0 25 . 0 25 105 . ( . ) k   1 0 319 . d Now the assimilation factor   a Q kV   7500 0 . 319 ( 50 , 000 )   3 1 23 , 454 m d David A. Reckhow CEE 577 #4 18 9

  10. CEE 572 Lecture #4 9/21/2017 Example 3.1: Solution (cont.) The surface area of the lake is: 50 000 , V s    2 25 000 , A m 2 H The atmospheric and inflow load is then:    0 6 25 000 . ( , ) 15 000 , / W JA g d atmosphere s   7500 10 ( ) 75 000 , / W g d inf low Combining all loads: W    W W W inf factory atmosphere low    50 000 , 15 000 , 75 000 ,  140 000 , / g d David A. Reckhow CEE 577 #4 19 Example 3.1: Solution (cont.) And finally, the concentration: W  c a 140 , 000 / g d  3 23 , 454 / m d  5 . 97 / mg L David A. Reckhow CEE 577 #4 20 10

Recommend


More recommend