McKay, Donald. "Front matter" Multimedia Environmental Models ...

Answer
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5.1 kg, 0.255 g/m 3 , 0.102 mg/g
In all these examples, chemical is flowing or reacting, but observed conditions
in the envelope are not changing with time, thus the steady-state condition applies.
In Example 2.7, the concentration will change in a “sawtooth” manner but, over the
long term, it is constant.
2.4.3 Unsteady-State Equations
Whereas the first two types of mass balances lead to simple algebraic equations,
unsteady-state conditions give differential equations in time. The simplest method
of setting up the equation is to write
d(contents)/dt = total input rate – total output rate
The input and output rates should be in units of amount/time, e.g., mol/h or g/h.
The “contents” must be in consistent units, e.g., in mol or g, and dt, the time
increment, in units consistent with the time unit in the input and output terms, (e.g.,
h). The differential equation can then be solved along with an appropriate initial or
boundary condition to give an algebraic expression for concentration as a function
of time. The simplest example is the first-order decay equation.
Worked Example 2.8
A lake of 10 6 m 3 with no inflow or outflow is treated with 10 mol of piscicide
(a chemical that kills fish), which has a first-order reaction (degradation or decay)
rate constant k of 10 –2 h –1 . What will the concentration be after 1 and 10 days,
assuming no further input, and when will half the chemical have been degraded?
The contents are VC or 10 6 C mol. The output is only by reaction at a rate of
VCk or 10 6 ¥ 10 –2 C or 10 4 C mol/h. There is zero input, thus,
Thus,
d (10 6 C)/dt = 10 6 dC/dt = 0 – 10 4 C mol/h
dC/dt = –10 –2 C mol/h
This differential equation is easily solved by separating the variables C and t to give
Integrating gives
dC/C = –10 –2 dt
lnC = –10 –2 t + IC