53:050
Natural Environmental Systems
Solutions to Homework No. 6 – Spring 2011
1.
Problem 45 from Davis and Masten.
First, draw the massbalance diagram:
Rappahannock River: Q
Rap
= 3.00 m
3
/s
and
C
Rap
= 0 mg/L
Tin Pot Run:
Q
TP
= 0.05 m
3
/s
and
C
TP
= ?
Detection limit = 1.0 mg/L
Q
TP
= 0.05 m
3
/s, C
TP
= ?
Q
Rap
= 3 m
3
/s
Q
mix
= 3 + 0.05 = 3.05 m
3
/s
C
Rap
= 0
C
mix
= 1 mg/L
Mass Balance:
Accumulation = In – OUT + Reactions
Accumulation = 0
(assume steadystate)
No Reactions
So,
0 = Q
Rap
C
Rap
+ Q
TP
C
TP
– Q
mix
C
mix
0 = 3(0) + 0.05(C
TP
) – 3.05(1)
Solving for C
TP
gives
C
TP
= 61 mg/L
Mass rate = Q
TP
C
TP
= 0.05 m
3
/s(61 mg/L)(10
3
L/m
3
)(10
6
kg/mg)(86,400 s/d) =
263.5 kg/d
2.
Problem 416 from Davis and Masten.
The general massbalance equation given in class is:
∑
±
∑
=
reactions
C
Q

C
Q
dt
d(CV)
out
out
in
in
Since this is a batch system with no flow in and now flow out, the above equation reduces to:
kCV

dt
d(CV)
=
or
kC

dt
dC
=
integrating gives:
C = C
0
e
kt
The problem states that C
0
= 2 mg/L, k = 1.0 d
1
, and t = 8/24 = 0.333 d
C
= 2e
1.0(0.333)
=
1.43 mg/L
To return to the original C = 2.0 mg/L, we need to add 2.0 – 1.43 = 0.57 mg/L.
1
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Mass required
=
mg/kg
10
)
L/m
)(1,000
m
mg/L(4,000
57
.
0
6
3
3
=
2.28 kg
3.
Problem 418 in Davis and Masten.
Actually, this problem can be worked easily by
recognizing the definition of half life – the amount of time it takes for half the material
remaining to be removed.
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 Spring '11
 Parkin
 Mass Balance, Steady State, dt, Chemical reactor, Masten, general massbalance equation

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