# Equation of Continuity

## The Equation of Continuity is a statement of mass conservation

Sponsored Links

The Law of Conservation of Mass states that mass can be neither created or destroyed. Using the Mass Conservation Law on a **steady flow** process - flow where the flow rate do not change over time - through a control volume where the stored mass in the control volume do not change - implements that

- inflow equals outflow

This statement is called **the Equation of Continuity. **Common application where **the Equation of Continuity** are used are pipes, tubes and ducts with flowing fluids or gases, rivers, overall processes as power plants, diaries, logistics in general, roads, computer networks and semiconductor technology and more.

**The Equation of Continuity** and can be expressed as:

m = ρ_{i1 }v_{i1 }A_{i1}+ρ_{i2 }v_{i2 }A_{i2}+..+ρ_{in }v_{in }A_{im}

=ρ_{o1 }v_{o1 }A_{o1}+ρ_{o2 }v_{o2 }A_{o2}+..+ρ_{om }v_{om }A_{om}(1)

where

m= mass flow rate (kg/s)

ρ= density (kg/m^{3})

v= speed (m/s)

A= area (m^{2})

With uniform density equation (1) can be modified to

q=v_{i1 }A_{i1}+ v_{i2 }A_{i2}+..+ v_{in }A_{im}

= v_{o1 }A_{o1}+ v_{o2 }A_{o2}+..+ v_{om }A_{om}(2)

where

q= flow rate (m^{3}/s)

ρ_{i1}= ρ_{i2}= . . = ρ_{in}= ρ_{o1}= ρ_{o2}= . .= ρ_{om}

### Example - Equation of Continuity

*10 m ^{3}/h* of water flows through a pipe with

*100 mm*inside diameter. The pipe is reduced to an inside dimension of

*80 mm*.

Using equation (2) the velocity in the 100 mm pipe can be calculated as

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

or

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

= 0.35 m/s

Using equation (2) the velocity in the 80 mm pipe can be calculated

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

or

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

= 0.55 m/s

Sponsored Links