I need your help in this problem .. A pulp and paper factory produces wastewater
ID: 874037 • Letter: I
Question
I need your help in this problem ..
A pulp and paper factory produces wastewater with an ultimate BOD of 450 mg/L. The pulp and paper wastewater is being discharged at a rate of 0.5 m^3/s into a river with a flow of 12 m^3/s. Upstream of the discharge point, the river has an ultimate BOD of 2.8 mg/L. The deoxygentation constant kd is estimated at 0.21/d a) What is the ultimate BOD just downstream of the pulp and paper discharge point? Assume complete and instantaneous mixture. b) What will be the ultimate BOD at 20000 m downstream of the discharge point? Assume a constant cross section area for the stream of 50 m^2.Explanation / Answer
a- Lo = QwLw +Qr Lr /Qw+Qr = .5m^3/sx 450mg/L + 12m^3/sx2.8mg/L / 0.5 + 12m^3/s=225+33.6 m^3 mg /s L/12.5m^3/s
Lo = 258.6/12.5 mg/l=20.688mg/L
b- cross section =50m^2, distance = 20000m
time = distance /speed = 20000/.5/50=2000000/3600x24 days=23.15days
Kd=0.21/d
L = Lo e^-Kdxt= 20.688 e^- 0.21x23.15= 20.688e^-4.8615
L=0.16025mg/l
the way we solve the equation
ln L = ln Lo - kdxt
lnL- lnLo = -kd x t
2.303 log (L/Lo) = -4.8615=>log (L/Lo) = -2.1109
taking anti log
L/Lo=0.0077464
L= 0.007746x20.688=0.16025mg/l