please show full explanation and steps for answer A city of 100,000 people disch

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please show full explanation and steps for answer

A city of 100,000 people discharges 51, 840 m^3/d of treated wastewater into the Bergelmir River. The treated wastewater has a BOD_3 of 15 mg/L and a BOD decay constant (i.e., k) of 0.20 day^-1 at 20 degree C. The DO of the wastewater is 1.0 mg/L. Bergelmir River has a flow rate of 2.50 m Vs and a velocity of 0.50 m/s. The ultimate BOD and DO in the river are 4.0 and 7.0 mg/L, respectively. Assume complete mixing and that the velocity in the river is the same upstream (i.e., before mixing) and downstream (i.e., after mixing) of the outfall. Given that the river and wastewater temperature is 20 degree C. Saturation values of dissolved oxygen can be found in Table A-2 in Appendix of the textbook. Calculate the initial oxygen deficit and initial ultimate BOD after mixing. The average depth (i.e., H) of the river after mixing is 3.0 m and the bed activity coefficient (i.e., H) is 0.50, calculate the deoxygenation rate constant and reaeration rate constant. Based on the calculated rate constants, what is the DO concentration 16 km downstream? Determine the critical time (i.e., t_c) and the minimum DO.

Explanation / Answer

a) First convert the wastewater flow rate(Qw) to compatible units (m3/s):

Qw = 51,840m3/d / 86,400s/d = 0.6 m3/s

DO after mixing:

DO = Qw.DOw + Qr.DOr / Qw + Qr

where, Qw = flow rate of wastewater, m3/s

Qr = flow rate of river, m3/s

DOw = DO concentration in wastewater , mg/L

DOr = DO concentration in river, mg/L

So, DO = 0.6m3/s*1.0mg/L + 2.50m3/s*7.0mg/L / (0.6+2.50)m3/s = 5.84mg/L

Saturation value of DO at 200C = DOsat = 9.1 mg/L

So, Initial oxygen deficit, D0 = DOsat - DO = 9.1mg/L - 5.84mg/L = 3.26mg/L

Now determine the ultimate BOD of wastewater:

BODt = Lw (1-e-kt)

where, Lw is ultimate BOD of wastewater, mg/L

k is BOD rate constant,day-1

and t is time in days

So, Lw = BODt / (1-e-kt) = 15mg/L / (1-e-0.20*3) = 33.25 mg/L

Now , we can determine the ultimate BOD after mixing:

La = QwLw + QrLr / Qw+Qr

where, La is ultimate BOD after mixing

Lr is ultimate BOD of river and Lw is ultimate BOD of wastewater

La = 0.6m3/s*33.25mg/L + 2.50m3/s*4.0mg/L / (0.6+2.50)m3/s = 9.66mg/L

b) Deoxygenation rate constant, kd = k + v/H*n

where, kd - deoxygenation rate constant, d-1

v - velocity,m/s

k - BOD rate constant, d-1

H - average depth of river, m and

n - bed activity coefficient

kd = 0.20d-1 + 0.50m/s*0.50 / 3.0m = 0.20 d-1

Reaeration constant , kr = 3.9*v0.5/H1.5 = 3.9 * 0.500.5m/s / 3.01.5 = 0.53 d-1

Now calculate the DO concentration 16 km downstream

first calculate the time in days

t = 16,000m/0.50m/s * 1hr/3600s * day/24 hr = 0.37 days

D = kd.La/(kr-kd) (e-kd*t - e-kr*t) + D0*e-kr*t

D- oxygen deficit

La - initial ultimate BOD after mixing

t= time of travel of wastewater discharge downstream,d

D = 0.20d-1*9.66mg/L/(0.53-0.20)d-1*(e-0.20*0.37 - e-0.53*0.37) + 3.26mg/L * e-0.53*0.37)

D = 3.31 mg/L

So, DO = DOsat - D = 9.1 - 3.30 = 5.8 mg/L

c) Critical time, tc = 1/kr-kd*ln[kr/kd (1-D0*kr-kd/kd*La)]

substitute values and we get tc = 0.49 days

now calculate minimum DO:

Dmax = kd*La/kr-kd (e-kd*t - e-kr*t ) + D0 e-kr*t

here t= tc = 0.49 days

substitute values and get Dmax

Dmax = 3.31mg/L

So, DOmin = DOsat - Dmax = 9.1 - 3.31 = 5.79 mg/L

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