This document discusses the continuity equation in fluid mechanics. It defines the continuity equation as the product of cross-sectional area and fluid speed being constant at any point along a pipe. This constant product equals the volume flow rate. The document then derives the continuity equation mathematically by considering the mass flow rate at the inlet and outlet of a pipe with varying cross-sectional areas but steady, incompressible flow. It provides an example calculation and solution for water flow rates and velocities through pipes of different diameters.

1. Presentation
Submitted To
Mr. Zahid Iqbal
Presented By
2016-BT-Mech-703
2o16-BT-Mech-726
Department
Mechanical
M.N.S U.E.T MULTAN

2. Continuity Equation in Fluid Mechanics
 The product of cross sectional area of the pipe and the
fluid speed at any point along the pipe is constant.
 This product is equal to the volume flow per second or
simply flow rate.
 Mathematically it is represented as
Av = Constant
 Continuity equation derivation
Consider a fluid flowing through a pipe of non uniform
size.The particles in the fluid move along the same lines
in a steady flow.

3. If we consider the flow for a short interval of time Δt,the
fluid at the lower end of the pipe covers a distance Δx1 with
a velocity v1 ,then:
Distance covered by the fluid = Δx1 = v1Δt
Let A1 be the area of cross section of the lower end then
volume of the fluid that flows into the pipe at the lower end
=V= A1 Δx1 = A1 v1 Δt

4. If ρ is the density of the fluid,then the mass of the fluid
contained in the shaded region of lower end of the pipe is:
Δm1=Density × volume
Δm1 = ρ1A1v1Δt ——–(1)
Now the mass flux defined as the mass of the fluid per unit
time passing through any cross section at lower end is:
Δm1/Δt =ρ1A1v1
Mass flux at lower end = ρ1A1v1 ———————(2)
If the fluid moves with velocity v2 through the upper end
of pipe having cross sectional area A2 in time Δt,then the
mass flux at the upper end is given by:
Δm2/Δt = ρ2A2v2
Mass flux at upper end =ρ2A2v2 ———————–(3)

5. Since the flow is steady,so the density of the fluid between
the lower and upper end of the pipe does not change with
time.Thus the mass flux at the lower end must be equal to the
mass flux at the upper end so:
ρ1A1v1 = ρ2A2v2 ———————-(4)
In more general form we can write :
ρ A v =constant
This relation describes the law of conservation of mass in
fluid dynamics.If the fluid is in compressible ,then density is
constant for steady flow of in compressible fluid so
ρ1 =ρ2
Now equation (4) can be written as:
A1 v1 = A2 v2
In general:
A v = constant

6. Question: Calculate the velocity if 10 m3/h of water flows through a
100 mm inside diameter pipe.If the pipe is reduced to 80 mm inside
diameter.
Solutions
(1)
 Velocity of 100 mm pipe
(10 m3/h)(1/3600 h/s)=v100(3.14(0.1 m)2/4)
Or
v100=(10 m3/h)(1/3600 h/s)(3.14(0.1)2/4)
V100 = 0.35 m/s
Answer

7. (2)
 Velocity of 80 mm pipe
According To Formula
(10 m3/h)(1/3600 h/s)=v80(3.14(0.08 m)2/4)
Or
 v80=(10 m3/h)(1/3600 h/s)(3.14(0.08 m)2/4)
V80 = 0.55 m/s
Answer:

8. Question:-Water runs through a water main of cross-sectional
area 0.4 m2 with a velocity of 6 m/s. Calculate the velocity of the
water in the pipe when the pipe tapers down to a cross-sectional
area of 0.3 m2.
Solution
using formula

9. Question:-Water enters a typical garden hose of diameter 1.6
cm with a velocity of 3 m/s. Calculate the exit velocity of water
from the garden hose when a nozzle of diameter 0.5 cm is
attached to the end of the hose.
Solutions
First, find the cross-sectional areas of the entry (A1) and
exit (A2) sides of the hose.

10. Next, apply the continuity equation for fluids to solve for
the water velocity as it exits the hose (v2).