FLUID FLOW

 

Continuity Equation

For a incompressible fluid like water the following formula is applicable:

          A1 . v1        =       A2 . v2

Where ;            

            A         =          cross sectional area (m2) A = ( p . d 2 ) /4

            v          =          velocity (m/s)

Also   Q = A . v       

Where  Q         =          Volume flow rate (m3/s)

Therefore in a section of fluid flow from point 1 to point 2 with no branches or leaks and constant volume flow:          Q1 = Q2

Bernoulli Equation:

A fluid such as water contains energy.

The three forms of energy that are relevant are:

1.     Potential energy

2.     Pressure energy

3.     Kinetic energy.

It is more convenient to use the forms of energy in values of energy head as shown below;

z 1 + ( p1 / r.g ) + ( v12 / 2 . g )       =      z 2 + ( p2 / r.g ) + ( v22 / 2 . g )

Where;

z           =          height or potential head (m)

p          =          pressure (N/m2)

r          =          density of water (1000 kg/m3)

g          =          acceleration due to gravity (9.81 m/s2)

v          =          fluid velocity (m/s)

 

In fluid flow examples, if a section of flow is examined, then the change in energy head between point 1 and 2 is as shown below.

In some examples the section of flow to be examined is a horizontal component therefore the height value (z) does not change and may be omitted from the formula, as shown below.

( p1 / r.g ) + ( v12 / 2 . g )      =      ( p2 / r.g ) + ( v22 / 2 . g )

Where;

p          =          pressure (N/m2)

r          =          density of water (1000 kg/m3)

g          =          acceleration due to gravity (9.81 m/s2)

v          =          fluid velocity (m/s)

EXAMPLE 1

A horizontal water pipe contains a uniform concentric taper where the diameter reduces from 400mm to 200mm.

The water pressure at the inlet of the taper is 1 bar or (100 kN/m2) or (100,000 N/m2)

If the inlet velocity is 2 m/s calculate the outlet velocity and the final pressure (in bar) at the outlet.

Continuity Equation:

A1 . v1        =       A2 . v2

Find A1:

Cross sectional area        =       ( p . d 2 ) /4

A1 = ( p . d 2 ) /4 = ( p . 0.4 2 ) / 4 = 0.1257 m2

Q1      = 0.1257 x 2.0 = 0.2514 m/s = Q2

A2          =       ( p . 0.2 2 ) / 4 = 0.0314 m2

v 2      =       Q2 / A2

v 2      =       0.2514 \ 0.0314      =       8.0 m/s

Use Bernoulli to find p2.

( p1 / r.g ) + ( v12 / 2 . g )         =      ( p2 / r.g ) + ( v22 / 2 . g )

There is no z value because the taper is horizontal.

(100,000 / 1000  x 9.81)   +  (22 / 2 x 9.81)    =       ( p 2 / 9810 ) +  82 / 2 x 9.81)

             10.194 + 0.204             =       ( p 2 / 9810 ) + 3.262

                             7.136           =       p 2 / 9810

                                      p 2     =       70,004 N/m2. = 0.7 bar

NOTE: 100,000 N/m2. = 1.0 bar

EXAMPLE 2

A water pipe tapers from 300mm to 150mm.

The pressure at the inlet is 500 kN/m2.

The mean flow velocity at the inlet is 3.2 m/s.

The elevation of the 300mm section is 4 metres above the 150mm section.

Using Bernoulli’s equation find:

     (a) the mean flow velocity at the 150mm section.

(b) the kinetic head at the 150mm section.

Answer (a)

Q2 = Q1

Q1 = A1 . v1.

A1 . v1 = A2 . v2

A1 = ( p . d 2 ) / 4 = ( p . 0.3 2 ) / 4 = 0.0707 m2

Q1 = 0.0707 x 3.2  =      0.226 m3/s   = Q2

v 2      =       Q2 / A2

v 2      =       0.226 / ( p . 0.15 2 ) / 4              =       12.79 m/s.

 

Answer (b)

Kinetic head     =    ( v12 / 2 . g)

                             =       12.79 2 / 2 x 9.81

                             =       8.338 m

EXAMPLE 3

A horizontal water pipe contains a uniform concentric taper where the diameter reduces from 300mm to 80mm.

The water pressure at the inlet of the taper is 5 bar.

If the inlet velocity is 2 m/s calculate the outlet velocity and the final pressure (in bar) at the outlet.

Continuity Eqn.

A1 . v1 = A2 . v2

A1 = ( p . d 2 ) / 4 = ( p . 0.30 2 ) / 4 = 0.0707 m2

A2 = ( p . d 2 ) / 4 = ( p . 0.08 2 ) / 4 = 0.00503 m2

A1 . v1 = A2 . v2

0.0707 x 2.0 =       0.00503 x v 2

v 2      =       28.11 m/s

         

          ( p1 / r.g ) + ( v12 / 2 . g )         =      ( p2 / r.g ) + ( v22 / 2 . g )

where;

z           =          height or potential head (m)

p          =          pressure (N/m2)

r          =          density of water (1000 kg/m3)

g          =          acceleration due to gravity (9.81 m/s2)

v          =          fluid velocity (m/s)

         

50.968 + 0.204                =       ( p 2 / 9810 ) + 40.274

                              p 2 / 9810             =       10.898

                                      p 2               =       106,909 N/m2.

                                                p 2               =       1.069 bar.

         

EXAMPLE 4

A horizontal water pipe contains a uniform concentric taper where the diameter reduces from 250mm to 150mm.

The water pressure at the inlet of the taper is 0.6 bar.

If the inlet velocity is 0.8 m/s calculate the outlet velocity and the final pressure (in bar) at the outlet.

z 1 + ( p1 / r.g ) + ( v12 / 2 . g )       =      z 2 + ( p2 / r.g ) + ( v22 / 2 . g )

Answer:

Velocity at outlet = 2.22 m/s

Pressure at outlet = 0.579 bar