# Chapter 07 FE Problem Solutions

```Chapter 7 FE EXAM FORMATTED PROBLEMS
7-1.
If the BOD5 and ultimate BOD of a municipal wastewater are 200 mg/L and 457 mg/L
respectively, what is the reaction rate constant?
From Eq. 7-4
BOD5  Lo (1  e  kt )
200  457 (1  e k (5) )
200
1
 e k ( 5)
457
 200 
ln 1 
   k (5)
 457 
k  0.115 d 1
7-2.
Twenty six million gallons per day of wastewater with a DO of 1.00 mg/L is discharged
into a river with a DO of 6.00 mg/L. If the flowrate of the river is 165 x 106 gal/d and
saturation value of dissolved oxygen is 9.17 mg/L, what is the oxygen deficit after
compete mixing of the two flows?
From Eq. 7-25
(26 MGD)(1.00 mg / L)  (165 MGD)(6.00 mg / L)
 5.32mg / L
26 MGD  165 MGD
From Eq. 7 - 31
D  DOs  DO  9.17  5.32  3.85 or 3.9 mg/L
DOmix 
7-3.
What is the mass loading (in lbm/d) of a 33 MGD wastewater discharge with an ultimate
BOD of 30.0 mg/L?
Assuming 1mg/L  1ppm and a conversion factor of 8.34 lb m / gal


 30 parts 
 33  10 6 gal / day (8.34lbm / gal)  8,256 .6 or 8,300 lb m / d
10
part


7-4.
A stream survey revealed that the reaeration constant is 0.05 d-1 and the deoxygenation
constant is 0.03 d-1. If the initial deficit is 5.00 mg/L and the ultimate BOD after mixing
is 12.0 mg/L, how long will it take to reach the critical point in the DO sag?
From Eq. 7-46
 0.05  0.03  
1

  0.05 
   (50) ln (1.67)(1  0.28)  50 ln(1.20)  9.27 d
t 
 ln 
1  5.0
 0.05  0.03   0.03 
 (0.03)(12)  
```
Number theory

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