Conservation of Mass Equation - Di↵erential Form

As seen previously in Ch.V, the equation of conservation of mass can be written

in integral form as:

@

@t

Z⇢dV +

Z⇢(

~

V · n)dA = 0

If we apply this equation to a rectangular prism with sides �x, �y, �z as shown:

�y �x

�z

x y

z

We can obtain the conservation of mass equation in di↵erential form. For sim-

plicity we apply our analysis to a 2-D rectangle and then we extend our conclu-

sion to 3-D.

Let’s represent the x�, y� and z� components of the velocity by u, v and w,

respectively.

Now, before proceeding any further, let’s remember what a Taylor series ex-

pansion is. Assume that we know the value of a function f(x) at x = x1, and

additionally f(x) is di↵erentiable n-times. Then the value of f(x) at x = x1+�x

is given by:

f(x1 + dx) = f(x1) +�x

d

dx

f(x) +

�x

2

2

d

2

dx

f(x) +

�x

3

3!

d

3

dx

f(x) + · · ·

1

Now that we have revised Taylor series expansion, we can proceed with our

derivation so let’s take into consideration the following element:

�x

�y

m

out�y

m

out�x

⇢v�x�z

⇢u�y�z

As shown in the figure, the mass flow rate entering the left side of the element is

⇢u�y�z where as the mass flow rate entering through the bottom of the element

is ⇢v�y�z. ⇢ is the density of the fluid, u and v are the x� and y�components

of the velocity and �x, �y, �z are the dimensions of the sides of the rectangular

prism.

Now, we know how much fluid is entering the control volume, but we do not

know how much of it is leaving the control volume. Then we use Taylor series

expansion to find the amount of fluid leaving the c.v.

Then if ⇢u�y�z is entering through the left side of the di↵erential element, then:

m

out�x

=

⇢u+

@(⇢u)

@x

�x+

�x

2

2

@

2(⇢u)

@x

2+

�x

3

3!

@x

3(⇢u)

@x

3+ · · ·

��y�z

If we assume that our di↵erential element is very small, we can neglect the

higher order terms such that:

m

out�x

=

⇢u+

@(⇢u)

@x

�x

��y�z

In a similar way, we have that:

m

out�y

=

⇢v +

@(⇢v)

@y

�y

��x�z

2

Therefore, our element can be shown as:

�y

�x

✓⇢u+

@

@y

(⇢u)�x

◆�y�z

✓⇢v +

@

@y

(⇢v)�y

◆�x�z

⇢u�y�z

⇢v�x�z

Now, we go back to the equation of conservation of mass:

@

@t

Z⇢dV +

Z⇢(

~

V · n)dA = 0

Beginning with the first term, applied to our control volume we have:

@

@t

Z⇢dV =

@

@t

(⇢ �x�y�z)

Now, lets analyze the second term:

Z⇢(

~

V · n)dA = �⇢u�y�z+

⇢u+

@(⇢u)

@x

�x

��y�z�⇢v�x�z+

⇢v +

@(⇢v)

@y

�y

��x�z

Simplifying:

Z⇢(

~

V · n) dA =

@(⇢u)

@x

+

@(⇢v)

@y

��x�y�z

Substituting these terms into the conservation of mass equation:

@

@t

(⇢�x�y�z) +

@(⇢u)

@x

+

@(⇢v)

@y

��x�y�z = 0

or

@⇢

@t

+

@(⇢u)

@x

+

@(⇢v)

@y

= 0

3

We can extend our analysis to 3-D:

@⇢

@t

+

@(⇢u)

@x

+

@(⇢v)

@y

+

@(⇢w)

@z

= 0

This is the equation of conservation of mass in di↵erential form. This equation

is also known as the continuity equation. Quite often you will see this equation

presented as:

@⇢

@t

+r · (~V ⇢) = 0

Where r =

@

@x

ı+

@

@y

|+

@

@z

ˆ

k

For the case when the fluid is incompressible (⇢ = const), the continuity equa-

tion is written as:

r · ~V = 0

@u

@x

+

@v

@y

+

@w

@z

= 0

If flow is in addition 2-D then:

@u

@x

+

@v

@y

= 0

4