Consider an incompressible fluid (water is almost incompressible) flowing along a pipe, as in Figure 1.

Its volume (V) is given by:

Therefore the volume passing per second (the volumetric flow rate Q) is given by:

But we can write velocity as distance moved/time, so we can replace L/t by v:

This is the FLOW EQUATION.

Now consider pipes of different areas A1 and A2 as shown in Figure 2.

The volumetric flow rate (Q) must be the same for both pipes, because we cannot gain or lose any fluid.

This is the CONTINUITY EQUATION and it is true for any number of changes in pipe diameter for a single pipe arrangement (a single flow path).

The relationship between the area inside the pipe (the pipe's internaldiameter) and the velocity of the fluid is expressed in the equation ofcontinuity, written as v1 A1 = v2 A2 . Here, v is the velocity of the fluid, and Ais the area that fluid travels through.

d1 = 0.2 m

v1= 6 m/s

d2= 0,1 m

v2 = ?

Q = ?

d1 = 0.2 m => A1 = (d12π) / 4 = (0.22π) / 4 = 0.0314 m2

d2= 0.1 m => A2= (d22π) / 4 = (0.12π) / 4 = 0.00785 m2

Q1 = A1 x v1 = 0.0314 x 6 = 0.1884 m3/ s

Q1 = Q2 = 0.1884 m3/ s

v2 = Q2 /A2 = 0.1884 / 0.00785 = 24 m/s

5.10. Determine the volume flow in over ; the volume flow out over; the mass in over ;

the mass out over!

Volume flow in over: A1=A1 x V1 x Δt

Volume flow out over: A2= A2 x V2 x Δt

Mass in over: ρ x A1 x V1 x Δt

Mass out over: ρ x A2 x V2 x Δt

ρ x A1 x V1 = ρ x A2 x V2

Consider the pipe system shown below in the Figure.

v1 = 1 m/s

v2 = 2 m/s

d1 = 100 mm

d2 = 30 mm

d3 = 50 mm

Q1 =? Q2 =? Q3 =? v3 =?

Sum of the Flows in = Sum of the Flows out

A1 x v1 = A2 x v2 + A3 x v3

A1= (0.12x π) / 4 = 0.00785 m2

A2= (0.032x π) / 4 = 0.000706 m2

A3= (0.052x π) / 4 = 0.00196 m2

Q1 = A1 x v1 = 0.00785 m3

Q2 = A2 x v2 = 0.001413 m3

Q3 = Q1 - Q2 = 0.006437 m3

v3 = Q3 / A3 = 3.284 m/s

9.3. Calculate the flow rate (Q3) and v1, v2, v3 in the pipe!

d1= 200 mm = 0.2 m

A1= (d12π) / 4 = (0.22π) / 4 = 0.0314 m2

Q1= 8.3 L/s = 8.3 dm3/s = 0.0083 m3/s

v1= Q1 / A1 = 0.0083 / 0.0314 = 0.26 m/s

d2= 300 mm = 0.3 m

A2= (d22π) / 4 = (0.32π) / 4 = 0.0706 m2

Q2= 8.8 L/s = 8.8 dm3/s = 0.0088 m3/s

v2= Q2 / A2 = 0.0088 / 0.0706 = 0.125 m/s

Q3 = Q1 + Q2 = 0.0083 m3/s + 0.0088 m3/s = 0.0171 m3/s

A3= (d32π) / 4 = (0.272π) / 4 = 0.057 m2

v3= Q3 / A3 = 0.0171 / 0.057 = 0.3 m/s

Anywhere in a perfect system (i.e. there are no frictional effects), for an incompressible fluid there are three types of energy existing:

Pressure Energy

Kinetic Energy

Potential Energy

These different forms of energy are expressed mathematically (as pressures) in the Bernoulli Equation (for a perfect system) shown below:

The terms on the left hand side of the Equation are as follows:

P1 is the pressure energy at point 1 (expressed as a pressure). [Units are N/m2 or

Pa]

ρ is the density of the fluid.[Units are Kg/m3]

v1 is the velocity of the fluid at point 1. [Units are m/s]

g is the acceleration due to earth’s gravity (9.81 m/s/s).[Units are m/s/s]

h1 is the height (from a given datum) of the fluid at point 1.[Units are m]

The terms are similar on the right hand side of the Equation, but for point 2.