SME
Environmental Engineering
Wastewater treatment
Use this for “Wastewater has an ultimate BOD of 250 mg/L and k = 0.23 per day. What is its 5-day BOD?”
Chart
BOD(5) ≈ 170.8
Level L₀ = 250 never meets the curve
This is an example. Type your own values to start a new problem.
BODₜ (mg/L)
=
L₀ (mg/L)
×
(1
−
e
−
day⁻¹
×
days
)
Formulas
BODₜ
=
L₀
×
(1
−
e
−
k
×
t
)
170.8
=
250
×
(1
−
e
−0.23 × 5
)
Assumptions
•
BOD (biochemical oxygen demand) is the oxygen microbes use to break down the organic matter.
•
First-order decay: the demand left, L₀e−kt, falls at a rate in step with what remains.
•
k is base e at 20 °C; a base-10 k is 2.303 times smaller, so multiply it by 2.303.
Step-by-step
WE KNOW
L₀ = 250 mg/L, k = 0.23 day⁻¹, t = 5 days
FIND
BOD exerted (BODₜ)
Step 1 · Find BOD exerted (BODₜ)
Use BODₜ = L₀ × (1 − e−k × t)
What
is
left
decays
as
e
−
kt
,
so
the
share
used
so
far
is
1
−
e
−
kt
.
BODₜ
=
250
×
(1
−
e
−0.23 × 5
)
BODₜ
=
250
×
(1
−
e
−1.15
)
BODₜ = 250 × (1 − 0.3166)
BODₜ = 250 × 0.6834
BODₜ = 170.8 mg/L
Check
Put the numbers back in. Both sides should match.
170.8
=
250
×
(1
−
e
−0.23 × 5
)
✓
Topic check
See what you know: 5 questions on this topic
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