Respuesta :
This question is incomplete, the complete question is;
A manufacturing facility with a wastewater flow of 0.011 m³/sec and a BOD₅ of 590 mg/L discharges into the Cattaraugus Creek. The creek has a 10-year, 7-day low flow of 1.7 m³/sec. Upstream of the facility, the BOD₅ of the creek is 0.6 mg/L. The BOD rate constants k are 0.115 d⁻¹ for the wastewater and 3.7 d⁻¹ for the creek. The temperature of both the creek and tannery of wastewater is 20°C. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing
Answer:
a) Ultimate BOD of wastewater is 1349.188 mg/L
b) Ultimate BOD of creek is 0.6 mg/L
c) the initial ultimate BOD after mixing is 9.27 mg/L
Explanation:
Given the data in the question;
Q[tex]_{wastewater[/tex] = 0.011 m³/s
BOD[tex]_{wastewater[/tex] = 590 mg/L
Q[tex]_{creek[/tex] = 1.7 m³/sec
BOD[tex]_{creek[/tex] = 0.6 mg/L
time t = 5
rate constants k for wastewater = 0.115 d⁻¹
rate constants k for creek = 3.7 d⁻¹
a) UBOD of wastewater.
The Ultimate BOD of wastewater is;
BOD[tex]_{wastewater[/tex] = L₀[tex]_{wastewater[/tex]( 1 - [tex]e^{-kt[/tex] )
where BOD[tex]_{wastewater[/tex] is the BOD of wastewater after 5 days, L₀[tex]_{wastewater[/tex] is the ultimate BOD of wastewater, k is the rate constant of wastewater and t is the time( days ).
we make L₀[tex]_{wastewater[/tex] the subject of formula
BOD[tex]_{wastewater[/tex] = L₀[tex]_{wastewater[/tex]( 1 - [tex]e^{-kt[/tex] )
L₀[tex]_{wastewater[/tex] = BOD[tex]_{wastewater[/tex] / ( 1 - [tex]e^{-kt[/tex] )
so we substitute
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - [tex]e^{(-0.115*5)[/tex] )
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - [tex]e^{(-0.575)[/tex] )
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - 0.5627 )
L₀[tex]_{wastewater[/tex] = 590 / 0.4373
L₀[tex]_{wastewater[/tex] = 1349.188 mg/L
Therefore, Ultimate BOD of wastewater is 1349.188 mg/L
b) UBOD of creek
The Ultimate BOD of creek is;
BOD[tex]_{creek[/tex] = L₀[tex]_{creek[/tex]( 1 - [tex]e^{-kt[/tex] )
we make L₀[tex]_{creek[/tex] the subject of formula
L₀[tex]_{creek[/tex] = BOD[tex]_{creek[/tex] / (1 - [tex]e^{-kt[/tex] )
we substitute
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - [tex]e^{(-3.7*5)[/tex] )
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - [tex]e^{(-18.5)[/tex] )
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - (9.2374 × 10⁻⁹) )
L₀[tex]_{creek[/tex] = 0.6 / 0.99999
L₀[tex]_{creek[/tex] = 0.6 mg/L
Therefore, Ultimate BOD of creek is 0.6 mg/L
c) the initial ultimate BOD after mixing;
Lₐ = [( Q[tex]_{wastewater[/tex] × L₀[tex]_{wastewater[/tex] ) + ( Q[tex]_{creek[/tex] × L₀[tex]_{creek[/tex] )] / [ Q[tex]_{wastewater[/tex] + Q[tex]_{creek[/tex] ]
we substitute
Lₐ = [( 0.011 × 1349.188 ) + ( 1.7 × 0.6 )] / [ 0.011 + 1.7 ]
Lₐ = [ 14.841068 + 1.02 ] / 1.711
Lₐ = 15.861068 / 1.711
La = 9.27 mg/L
Therefore, the initial ultimate BOD after mixing is 9.27 mg/L
