As we learn about conservation of mass in chemistry and other branches , such a way equation of continuity is a mathematical expression of the law of conservative of mass in fluid dynamics .
This equation defines the steady flow of a fluid in a tube. It states that in steady flow mass of the liquid enter in one tube will be same as the mass of liquid leaving from other end .
Derivation of equation of continuity.
Fig. 1 shows a section of a tube in which at the ends , the cross sectional areas are A1 and A2 and the velocity of the fluid are v1 and v2 respectively .
According to the equation of continuity we can say that , if there is study flow then , mass of fluid entering at end A1 per second will be equal to mass of fluid leaving the end A2 per second .
In time dt the fluid enters a distance v1dt at end A , so the volume entered in time dt is dV = A1v1dt and the volume entered per second is
dV/dt = A1V1
Hence mass entering per second at A1 is = A1V1ρ
Similarly mass leaving per second at A2 is = A2V2ρ
According to the definition of steady flow A1V1ρ = A2V2ρ
A1V1 = A2V2
Therefore, the velocity of the liquid is smaller in the wider parts of the tube and larger in the narrower parts.
or V2>V1 as A2<A1
Note – The product Av is the volume flow rate dV/dt , the rate at which volume crosses a section of the tube.
dV/dt = volume flow rate = Av
The mass flow rate is the mass flow per unit time through a cross-section . This is equal to density (ρ) times the volume flow rate dV/dt .
We can generalize the continuity equation for the case in which the fluid is not incompressible. If ρ1 and ρ2 are the densities at sections 1 and 2 then ,
ρ1A1V1 = ρ2A2V2
So, this is the continuity equation for a compressible fluid .
- Example 1. In a horizontal tube of non-uniform cross section area , water is flowing . At a place, the radius of the tube is 5.0 cm and the velocity of water is 9 m/s. What will be the velocity of water where the radius of the pipe is 3.0 cm ?
Solution – Using equation of continuity, A1V1 = A2V2
or V2 = ( πR12/πR22 ) V1 = ( R1/R2 )2 V1
Substituting the values , we get
V2 = ( 5.0×10-2/3.0×10-2)2(9)
V2 = 25 m/s
- Example 2 . A broad pipe having a radius 10 cm branches into two pipes of radii 5 cm and 3 cm. If the velocity of flowing water in the pipe of radius 3 cm is 5 cm/s, determine the velocities of water in the remaining two pipes. Given that the rate of discharge through the main branch is 600π cm3/s.
Solution – Consider any three sections (1), (2) and (3) in the three pipes of different radii as shown in fig. 2
If v1 , and v2 , be the velocities of water at sections (1) and (2), respectively; then ,
a1 = π(10)2 cm2 , a2 = π(5)2 cm2 and a3 = π(3)2 cm2
v1 = ? , v2 = ? , v3 = 5 cm/s
Discharge rate through the three pipes are
a1v1 = 100πv1 cm3/s
a2v2 = 25πv2 cm3/ s
a3v3 = 9π×5 cm3 /s
Now , Q = a1v1 = a2v2 + a3v3
600π= 100πv1 = 25πv2 + 45π
Solving we get , v1 = 6 cm/s and v2 = 22.2 cm/s
Example 3 . Water coming from a tap , as shown in the fig. 3 If it’s speed near the mouth of the tap is Vo= 10m/s , and Area of cross section of tap is 1cm2 . Then calculate the area of the stream water at point M , that is below h= 5 m from tap .
Solution – Velocity when it fall 5m
v2 = u2 + 2as
v2 = (10)2 + 2(10)(5)
v = 10√2 m/s
Apply equation of continuity, A1V1 = A2V2
1×10 = A2 × 10√2
A2 = 1/√2 cm2