Now, by construction, the volume of the differential control volume is

while the mass of fluid in the control volume is

The rate of change of mass or accumulation in the control volume is then

For the net flow through the control volume, we deal with it one face at

a time. Starting with the r faces, the net inflow is

r,in = vr rd dz

while the outflow in the r direction is

r,out=(vr +

)(r+dr)ddz

So that the net flow in the r direction is Being O(dr^2), the last term in this equation can be dropped so that the net flow on the r faces is

Net rate of mass out flow in r-direction = out -in

= ( vr +

)(r+dr)ddz vrrddz

=( vr +

r)drddz

The net flow in the direction is slightly easier to compute since the

areas of the inflow and outflow faces are the same. At the outset, the

net flow in the theta direction is

,in = v dr dz

,out=(v +

d) dr dz

Net rate of mass out flow in -direction = out -in

= ( v +

d)drdz vdrdz

=

ddrdz

We now turn our attention to the z direction. The face area is that of

a sector of angle

Az =

( r + dr )

2

r

2 d

= r dr d +

d

2r d

= r dr d

then, the inflow at the lower z face is

z,in = vz r dr d

while the outflow at the upper z face is

z,out=(vz +

dz) r dr d

Finally, the net flow in the z direction is

Net rate of mass out flow in z-direction = out -in

= (vz +

dz) r dr d vz r dr d

=

r dr ddz

the rate of change of mass in the control volume =

r dr ddz

Now we can put things together to obtain the continuity equation

Net rate of mass outflow

=

r dr ddz + ( vr +

r)drddz +

ddrdz +

r dr ddz

By dividing ( r dr ddz )

=

+

v

r +

+

+

=

+

+

+

=

+

+

+