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Mivida ## Sample Equations

No.NameEquations
1 Discharge = area x velocity Q = a x v
2 Loss of head due to sudden enlargement ( V1 – V 2 )2
= ----------------
2 g

V1= velocity of liquid at section 1-1.
V2 = velocity of liquid at section 2-2
(enlarged section)
g(acceleration due to gravity 9.81 m/s2)

3

Loss of head due to sudden contraction 0.375( V2)2
= --------------
2g

V2 = velocity of liquid at section 2-2
(contracted section)
g(acceleration due to gravity 9.81 m/s2)

4  Loss of head at entrance     0.5 v2
= -------
2 g
5  Loss of head at exit       v2
= -------
2 g
6

Time of emptying a horizontal
tank through an orifice at its bottom 0.550x LR
T = -------------
Cd X a
7  Time of empty for tank      2 ×A (  H1 -  H 2 )2
= -------------------------
Cd a×  2 g
8  Water hammer ***           WLV
P = --------------
G T

Where: ***

P = Rise of pressure.
W = Density of water ( 1000 kg / m3 ).
L = length of the pipe
V= velocity of water in the pipe.
T = time in seconds in which the water is brought to rest by the valve closure

Pressure losses in straight pipes:
For the purpose of calculating the pressure loss in seamless extruded thermoplastic pipes the roughness factor may be taken to be K= 0.007mm

Pressure losses in fittings:
The pressure losses depend upon the type of fitting as well as on the flow in the fitting. Table below contains the ζ- values for common fittings

Out side pipe diameter mm203250≥63
Fitting type Coefficient of resistance ζ 0.5 0.6 1.0

1.5 0.3 1.5 Inlet 0.5 Outlet 1

To calculate the total pressure loss in all fittings in a pipeline take the sum of the individual losses. the pressure loss can then be calculated according to the following formula.
V2
Δ P = Σ ζ · --------- · ρ
2 · g

Where:
Δ P = pressure loss in all fittings in mm
Σ ζ = sum of the individual losses
V = flow velocity in m / s
g = acceleration due to gravity 9.81 m/s2
ρ = density of the transported medium in g/cm3

Hydraulic coefficient of discharge

The following three coefficients are known as hydraulic coefficients or orifice coefficients :
1= coefficients of contraction.
2= coefficients of velocity , and
3= coefficients of discharge.

Coefficients of discharge:
The ratio of actual discharge through an orifice to the theoretical discharge is known as coefficient of discharge.Mathematically coefficient of discharge,

Actual discharge
Cd= ----------------------------
Theoretical discharge

Actual velocity          Actual area
= -------------------------x ------------------------
Theoretical velocity   Theoretical area

= Cv x Ca

The value of coefficient of discharge varies with the values of Cv and Ca. An average is about 0.62

## Water Hammer

The water, while flowing in pipe possesses some momentum on account of it,s motion . It has been experienced that if the flowing water is suddenly brought to rest. By closing the valve, it’s momentum will be destroyed, which shall cause a very high pressure on the valve.
This high pressure will be followed by a series of pressure vibrations . The pressure vibration may set up noises in the pipe , known as knocking . Such a knocking is often heard , in water pipes, if a top is turned off quickly . The sudden rise of pressure in a pipe known as hammer blow or water hammer . Sometimes the hammer blow is so high that it may even burst the pipe .It is thus, obvious, that the valves of the pipe lines or pen stocks should always be closed gradually .Consider a uniform pipe line through which the water is flowing with a uniform velocity.

Let:
W = density of water ( 1000 kg / m3 )
L = length of the pipe.
a = area of the pipe.
V = velocity of water in the pipe.
T = time, in seconds, in which the water is brought to rest by the closure of valve.
Mass of water in the pipe

Wal
= -----------                                                        ( i )
g

rate of retardation of the water

velocity
=  -------------
time

v
= -----------                                                        ( i i )
t

we know that the force  = mass X acceleration.

Wal             v
= ------------- X ------------                                      ( i i i )
g                t

intensity of pressure rise in the valve
wal          v
force     ----------  x  --------
g            t
P = ----------- = -----------------------
area                  a

wal
P = -----------
gt

## Example : -

Water is flowing through a pipe 800 m metres long with a velocity of 1.2 m / sec. Find the rise in pressure, if the valve near the end of the pipe is closed in 30 seconds.

Solution : -

Given :

Length of the pipe
L = 800 m
Velocity of water
V = 1.2 m / sec.
Time of valve clouser .
T = 30 seconds

Let:
P = rise of pressure
Using the relation ,
Wlv
P = ---------         with usual notations
Gt

1000 x 800 x 1.2
= -----------------------------        kg/m2 P
9.81 x 30

= 3260 kg / m2

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