### Continuity Equation

• Recognize continuity equation as a form of conservation of mass.

### Solved Example:

#### 62-1-01

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 continuity equation, calculate the velocity in the 100 mm pipe.

Solution:

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

$v_{100} = (10 m^3/h) (1 / 3600 h/s) / (3.14 (0.1 m)^2 / 4) = 0.35 m/s$

Using equation 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 $v_{80} = (10 m^3/h) (1 / 3600 h/s) / (3.14 (0.08 m)^2 / 4) = 0.55 m/s$

### Solved Example:

#### 62-1-02

A fluid flowing through a pipe of diameter 450 mm with velocity 3 m/s is divided into two pipes of diameters 300 mm and 200 mm. The velocity of flow in 300 mm diameter pipe is 2.5 m/s, then the velocity of flow through 200 mm diameter pipe will be: (ESE Mechanical 2014)

### Solved Example:

#### 62-1-03

When 0.1 m$^3$/s water flows through a pipe of area 0.25 m$^2$, which later reduces to 0.1 m$^2$, what is the velocity of flow in the reduced pipe? (SSC JE CE Oct 2020 Morning)

### Solved Example:

#### 62-1-04

If an incompressible fluid enters a pipe with a velocity of 4 cm/s and moves out with a velocity of 2 cm/s, calculate the cross sectional area of the inlet if the diameter of the pipe at the outlet is 7 cm. (UPPCL JE CE 2016)