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multiplication of % gives the mass, since mass is volume times density.) The flow out
of ˝ during the time t is %q.b/t . The net mass flow is hence %t .q.a/
q.b//.
The total mass inside ˝ is obtained by integrating the density, but in a small space
interval dx. Not all of the space is occupied by the substance, only a fraction c.We
must hence multiply % by c and dx to get the mass in dx. The change in total mass
in ˝ is therefore
R
b
a
%cdx,wherec is the change in concentration.
The balance of the net inflow of mass and the increase in mass, in an interval
˝
D
Œa; b, can now be summarized as
%q.a/t
%q.b/t
D
Z
b
a
%c
dx
:
(7.11)
Sometimes we can inject or extract mass by some means, and this will influence
the mass balance. Suppose f.x;t/ reflects the injected volume of the substance, per
unit volume and unit time. In a small time interval t and in a small space interval
x, the injected mass then becomes approximately f.x;t/%tx,wherex is an
arbitrary x value inside the interval x,andt is an arbitrary t value inside the
interval t . The total injection of mass in the domain Œa; b in the time interval t
is found by summing up all the f.x;t/%tx contributions, from x
D
a to x
D
b,
as x
!
0. This sum is just the integral of %f t ,or
Z
b
%f t
dx
:
a
When f>0we have mass injection, whereas f<0means mass extraction.
We need to add the f -integral to the balance in (7.11), i.e., the sum of the mass
injection and the net flow into ˝ must equal the net increase of the mass of the
substance inside ˝:
%q.a/t
%q.b/t
C
Z
b
a
%f t
dx
D
Z
b
a
%c
dx
:
(7.12)
An increase c in the concentration in the time interval t can be expressed by
the time-rate-of-change of c, multiplied by t :
c
@c
@t
t :
(7.13)
Alternatively, one can arrive at the same result using a Taylor series,
c.x; t
C
t/
D
c.x; t/
C
@c
.t
2
/;
@t
t
C
O
which, when neglecting higher-order terms in t ,gives(7.13), since c.x; t
C
t/
c.x; t/
D
c.
