Page 87 - Physical Principles of Sedimentary Basin Analysis

3.19 Mass conservation 69

v(x,t) A v(x + dx,t)

x x + dx
Figure 3.18. The figure shows mass flowing through a box. The velocity of the mass entering the box
(at the left) is v(x, t) and the velocity of the mass leaving the box (at the right) is v(x + x, t).

Since mass is a conserved property we have

m in − m out = m 2 − m 1 (3.160)
which gives
∂ ∂(v )
+ = 0 (3.161)
∂t ∂x
in the limit where both x and t approach zero. This is the equation for mass conser-
vation, which is also called the continuity equation.
Exercise 3.22 Show that equation (3.161) can be rewritten as

D ∂v
+ = 0. (3.162)
dt ∂x
Exercise 3.23 Show that a constant density in 1D implies that the velocity v has to be
constant too.
Exercise 3.24 Show that equation (3.161) becomes
∂ ∂(v )
+ = S(x, t) (3.163)
∂t ∂x
when S(x, t) is the rate of mass production per unit volume and per time in the position x
at time t.

3.19 Mass conservation
The equation for mass conservation can be derived by considering a volume V (t) that
always contains the same “particles” and therefore the same mass. We then have

d
dV = 0 (3.164)
dt V (t)
where dV = dx dy dz. The time differentiation of the integral is not straightforward to
carry out because the volume V (t) is a function of time. The integral is therefore rewritten
in Lagrangian coordinates as
d

JdV = 0 (3.165)
dt V 0