5.9 Equation of Continuity

5.9 Equation of Continuity

Statement:

The product of cross-sectional area of the pipe and the fluid speed (i.e., $Av$) at any point along the pipe is a constant. This constant is equal to the volume flow per second of the fluid or simply the flow rate.

Thus $Av = \text{Constant} = \text{Volume}/\text{Time}$

Consider a fluid flowing through a pipe of non-uniform size.

The particles in the fluid move along the same lines in a steady state flow as shown in Fig 5.15.

If we consider the flow for a short interval of time $\Delta t$, the fluid at the lower end of the tube covers a distance $\Delta x_{1}$ with a velocity $v_{1}$, then distance covered by the fluid is:

Let $A_{1}$ be the area of cross-section of the lower end, then volume of the fluid that flows into the tube at $A_{1}$ is:

If $\rho_{1}$ is the density of the fluid, then the mass of the fluid contained in the shaded region (through $A_{1}$) is:

Similarly, the mass of the fluid that moves with velocity $v_{2}$ through the upper end of the pipe having cross-sectional area $A_{2}$ in the same time $\Delta t$ is given by:

where $\rho_{2}$ is the density of the fluid flowing out through $A_{2}$ and $\Delta m_{2}$ indicates small mass.

If the fluid is incompressible and the flow is steady, the mass of the fluid is conserved. That is the mass flowing into the bottom of the pipe through $A_{1}$ in a time $\Delta t$ must be equal to the fluid flowing out through $A_{2}$ in the same time. Therefore:

So $A_{1}v_{1}\Delta t \times \rho_{1} = A_{2}v_{2}\Delta t \times \rho_{2}$

or

Equation (5.15) is called the equation of continuity. Since density is constant for the steady flow of incompressible fluid, therefore, the equation of continuity becomes:

Equation (5.16) states that in steady flow, the rate of flow of the fluid inward is equal to the rate of flow of the fluid outward.

This equation justifies the conservation of mass of the fluid which is flowing through a pipe.

Example 5.4

A water hose with an internal diameter of $20\,\mathrm{mm}$ at the outlet discharge $30\,\mathrm{kg}$ of water in $60\,\mathrm{s}$. Calculate the water speed at the outlet. Assume the density of water is $1000\,\mathrm{kg/m^{3}}$ and its flow is steady.

Solution

Internal diameter of water hose $D = 20\,\mathrm{mm} = 0.02\,\mathrm{m}$

Radius $r = D/2 = 0.02/2 = 0.01\,\mathrm{m}$

Mass of water $m = 30\,\mathrm{kg}$

Time taken $t = 60\,\mathrm{s}$

Density of water $\rho = 1000\,\mathrm{kg/m^{3}}$

Speed of water $v = ?$

Mass flow per second $= m/t = 30\,\mathrm{kg}/60\,\mathrm{s} = 0.5\,\mathrm{kg\,s^{-1}}$

Cross-sectional area $A = \pi r^{2} = 3.14 \times (0.01\,\mathrm{m})^{2} = 3.14 \times 10^{-4}\,\mathrm{m^{2}}$

From equation of continuity, the mass of water discharging per second through area $A$ is: