5.1 5.33 The solutions for these problems are the solutions for
problems 5.1 - 5.33 inthe 2nd edition Solutions Manual.
5.34On a molar basis the C:N:P ratio for algae is 106:16:1 and
on a mass basisit is 41:7.2:1
The mass ratio of the lake water is:
1
4
13.3
03.0
12.0
40.0
P
N
C==
Therefore, carbon is limiting.
5.35 5.53 The solutions for these problems are the solutions for
problems 5.34 - 5.52 inthe 2nd edition Solutions Manual.
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5.54 a. From Table 5.11, =0.34 for a sand aquifer
( ) ( )( )
( ) dm61.2
0.0005d
m09.00.34
dd
dd
====
Lh
v'
Lh
vK
b.tcontaminan
Rv
v'= so
dm0.015
6d
m0.09tcontaminan ===
R
v'v
and the distance traveled is (0.015 m/d) (365 d) =5.48 m
c. Calculate the maximum contaminant mass dissolved based on the
aqueoussolubility and density of PCE (use Table 5.14)
( ) ( ) ( ) ( ) g2,230m5.48m4.0m2.00.34m
g1503max
=
=m
Calculate the mass leaked
( ) ( ) g59,500d365g1.63d
mL100leaked =
=
mLm
Since the mass leaked >soluble mass, the solubility limit is
exceeded and theguideline for the aqueous concentration in the
presence of NAPLs (see page258) is used: CPCE =0.10 Aq SolPCE =15
mg/L
d.
( ) ( ) ( ) ( )d
m0.490d
m0.090.34m42m223
capture === BvwQ
e.( )
( ) ( )yr22.2d8,100
mgg10
mL10
Lmg15
dm0.490
g59,500
33
33PCE
PCE ==
==QC
mt
5.55The solution for this problem is the solution for problem
5.53 in the 2nd editionSolutions Manual.
