Equation of Continuity

The Equation of Continuity is a statement of mass conservation

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/m³)

v = speed (m/s)

A = area (m²)

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 (m³/s)

ρ_i1 = ρ_i2 = . . = ρ_in = ρ_o1 = ρ_o2 = . .= ρ_om

Example - Equation of Continuity

10 m³/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 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 m)² / 4)

    = 0.35 m/s

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

(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)

= 0.55 m/s

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