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. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing

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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

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 = 0.011 m³/s

BOD = 590 mg/L

Q = 1.7 m³/sec

BOD = 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 = L₀( 1 - )

where BOD is the BOD of wastewater after 5 days,  L₀ is the ultimate BOD of wastewater, k is the rate constant of wastewater and t is the time( days ).

we make L₀ the subject of formula

BOD = L₀( 1 - )

L₀ = BOD / ( 1 - )

so we substitute

L₀ = 590 / ( 1 - )

L₀ = 590 / ( 1 - )

L₀ = 590 / ( 1 - 0.5627 )

L₀ = 590 / 0.4373

L₀ = 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 = L₀( 1 - )

we make L₀ the subject of formula

L₀ = BOD / (1 - )

we substitute

L₀ = 0.6 / ( 1 - )

L₀ = 0.6 / ( 1 - )

L₀ = 0.6 / ( 1 - (9.2374 × 10⁻⁹) )

L₀ = 0.6 / 0.99999

L₀ = 0.6 mg/L

Therefore, Ultimate BOD of creek is 0.6 mg/L

c) the initial ultimate BOD after mixing;

Lₐ = [( Q × L₀ ) + ( Q × L₀ )] / [ Q + Q ]

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

do you have any options?

answer:

yes

explanation: