Chapter 8
Solutions
83
Rate constant
Given:
BOD
5
= 220.0; L = 320.0 mg/L.
Solution:
a. Set up Eqn 86
BOD
t
= L
o
(1 – e
kt
)
220.0 = 320.0 (1  e
k(5)
)
0.6875 = 1  e
k(5)
0.3125 = e
k(5)
b. Divide through by 1 and take the natural log of both sides
ln (0.3125) = ln (e
k(5)
)
1.163 = k(5)
k = 0.2326 d
1
86
Ultimate BOD
5
Given:
BOD rate constant = 0.433 d
1
, BOD
5
= 272 mg/L.
Solution:
a.
From Eqn 86
BOD
t
= L
o
(1 – e
kt
)
272
= L
o
(1 – e
(0.433)(5)
)
272
= L
o
(0.885)
L
o
= 307.26
or
307 mg/L
89
BOD
5
123
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentGiven:
Oxygen consumption = 2.00 mg/L, 1.00% sample.
Solution:
a. This an application of Eqn 88 with the dilution factor defined by Eqn 89
100%
BOD
5
= (2.00)() = 200. mg/L
1.00%
813
NBOD of glutamic acid
Given: Problem 81, 150 mg/L of glutamic acid.
Solution:
a. Calculate the amount of nitrogen that is oxidized.
14 g N
(150 mg/L)() = 14.29 mg N/L
147 g C
5
H
9
O
4
N
b. Using the relationship from Eqn 811:
Theo. NBOD = (14.29 mg N/L)(4.57 mg O
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 VIGIL
 Environmental Engineering, Trigraph, Reaction rate constant, mg/L, glutamic acid, Eqn

Click to edit the document details