Continuity Equation

Continuity Equation

The continuity equation describes the transport of some quantities like fluid or gas. The equation explains how a fluid conserves mass in its motion. Many physical phenomena like energy, mass, momentum, natural quantities and electric charge are conserved using the continuity equations.

This equation provides very useful information about the flow of fluids and its behaviour during its flow in a pipe or hose. The hose, a flexible tube, whose diameter decreases along its length has a direct consequence. The volume water flowing through the hose must be equal to the flow rate on the other end. The flow rate formula.

The Equation of Continuity and can be expressed as:

\(m = \rho _{i 1} \ v _{i 1} \ A _{i 1} + \rho _{i 2} \ v _{i 2} \ A _{i 2} + ….. + \rho _{i n} \ v _{i n} \ A _{i m}\)

\(m = \rho _{o 1} \ v _{o 1} \ A _{o 1} + \rho _{o 2} \ v _{o 2} \ A _{o 2} + ….. + \rho _{o n} \ v _{o n} \ A _{o m}……….. (1)\)

Where,

\( m \) = Mass flow rate

\( \rho \) = Density

\( v \) = Speed

\( A \) = Area

With uniform density equation (1) it can be modified to:

\(q = v _{i 1} \ A _{i1} + v _{i2} \ A _{i2} + …. + v _{i n} \ A _{i m}\)

\(q = v _{o 1} \ A _{o1} + v _{o2} \ A _{o2} + …. + v _{o n} \ A _{o m}………..(2)\)

Where,

\(q\) = Flow rate

\(\rho _{i 1} = \rho _{i 2} .. = \rho _{i n} = \rho _{o 1} = \rho _{o 2} = …. = \rho _{o m}\)

Fluid Dynamics

The continuity equation in fluid dynamics describes that in any steady state process, the rate at which mass leaves the system is equal to the rate at which mass enters a system.

The differential form of the continuity equation is:

\(\frac{\partial \rho}{\partial t } + \bigtriangledown \cdot \left ( \rho u \right ) = 0\)

Where,

\( t \) = Time

\( \rho \) = Fluid density

\( u \) = Flow velocity vector field.

Continuity Equation Example

Question: Calculate the velocity if \( \small 10 \ m^{3}/h\) of water flows through a 100 mm inside diameter pipe. If the pipe is reduced to 80 mm inside diameter.

Solution

Velocity of 100 mm pipe

Using the equation (2), to calculate the velocity of 100 mm pipe

\(\left ( 10 \ m^{3}/h \right )\left ( 1 / 3600 \ h/s \right ) = v_{100} \left ( 3.14\left ( 0.1 \ m \right )^{2} / 4 \right )\)

Or

\(v_{100} = \frac{\left ( 10 \ m^{3} / h \right )\left ( 1/3600 \ h/s \right )}{\left ( 3.14 \left ( 0.1 \right )^{2} / 4 \right )}\)

\(= 0.35 \ m/s\)

Velocity of 80 mm pipe

Using the equation (2), to calculate the velocity of 80 mm pipe

\(\left ( 10 \ m^{3} / h \right )\left ( 1 / 3600 \ h/s \right ) = v_{80} \left ( 3.14 \left ( 0.08 \ m \right )^{2} / 4 \right )\)

Or

\(v_{80} = \frac{\left ( 10 \ m^{3} / h \right )\left ( 1 / 3600 \ h/s \right )}{\left ( 3.14 \left ( 0.08 \ m \right )^{2} / 4 \right )}\)

\(= 0.55 \ m/s\)


Practise This Question

An external electric field is applied on a metal plate, high enough to force the electrons to eject out of the surface. Is this an example of Photo electric effect?