Static Pressure and Pressure Head in Fluids

[Esam]

The pressure indicates the normal force per unit area at a given point acting on a given plane. Since there is no shearing stresses present in a fluid at rest - the pressure in a fluid is independent of direction.
For fluids - liquids or gases - at rest the pressure gradient in the vertical direction depends only on the specific weight of the fluid.
How pressure changes with elevation can be expressed as

dp = - γ dz (1)
where
dp = change in pressure
dz = change in height
γ = specific weight

The pressure gradient in vertical direction is negative - the pressure decrease upwards.
[h=Specific Weight]3[/h] Specific Weight can be expressed as:

γ = ρ g (2)
where
γ = specific weight
g = acceleration of gravity

In general the specific weight - γ - is constant for fluids. For gases the specific weight - γ - varies with the elevation.

[h=Static Pressure in a Fluid]3[/h] For a incompressible fluid - as a liquid - the pressure difference between two elevations can be expressed as:

p₂ - p₁ = - γ (z₂ - z₁) (3)
where
p₂ = pressure at level 2
p₁ = pressure at level 1
z₂ = level 2
z₁ = level 1

(3) can be transformed to:

p₁ - p₂ = γ (z₂ - z₁) (4)
or
p₁ - p₂ = γ h (5)
where
h = z₂ - z₁ difference in elevation - the dept down from location z₂.
or
p₁ = γ h + p₂ (6)

[h=Example - Pressure in a Fluid]4[/h] The absoute pressure at water depth of 10 m can be calulated as:

p₁ = γ h + p₂
= (1000 kg/m³) (9.81 m/s²) (10 m) + (101.3 kPa)
= (98100 kg/ms² or Pa) + (101300 Pa)
= 199.4 kPa

where
ρ = 1000 kg/m³
g = 9.81 m/s²
p₂ = pressure at surface level = atmospheric pressure = 101.3 kPa
The gauge pressure can be calulated setting p₂ = 0

p₁ = γ h + p₂
= (1000 kg/m³) (9.81 m/s²) (10 m)

= 98.1 kPa
[h=The Pressure Head]3[/h] (6) can be transformed to:

h = (p₂ - p₁) / γ (6)

h express the pressure head - the height of a column of fluid of specific weight - γ - required to give a pressure difference of (p₂ - p₁).
[h=Example - Pressure Head]4[/h] A pressure difference of 5 psi (lbf/in²) is equivalent to

(5 lb_f/in²) (12 in/ft) (12 in/ft) / (62.4 lb/ft³)
= 11.6 ft of water
(5 lb_f/in²) (12 in/ft) (12 in/ft) / (847 lb/ft³)
= 0.85 ft of mercury

when specific weight of water is 62.4 (lb/ft³) and specific weight of mercury is 847 (lb/ft³).

Heads at different velocities are indicated in the table below:

Velocity (ft/sec)	Head Water (ft)
0.5	0.004
1.0	0.016
1.5	0035
2.0	0.062
2.5	0.097
3.0	0.140
3.5	0.190
4.0	0.248
4.5	0.314
5.0	0.389
5.5	0.470
6.0	0.560
6.5	0.657
7.0	0.762
7.5	0.875
8.0	0.995
8.5	1.123
9.0	1.259
9.5	1.403
10.0	1.555
11.0	1.881
12.0	2.239
13.0	2.627
14.0	3.047
15.0	3.498
16.0	3.980
17.0	4.493
18.0	5.037
19.0	5.613
20.0	6.219
21.0	6.856
22.0	7.525

• 1 ft (foot) = 0.3048 m = 12 in = 0.3333 yd

[Esam]

[h=Recommended water flow velocity on suction side of pump]2[/h]
Capacity problem, cavitation and high power consumption in a pump, is often the result of the conditions on the suction side. In general - a rule of thumb - is to keep the suction fluid flow speed below the following values:

Pipe bore		Water
inches	mm	m/s	ft/s
1	25	0.5	1.5
2	50	0.5	1.6
3	75	0.5	1.7
4	100	0.55	1.8
6	150	0.6	2
8	200	0.75	2.5
10	250	0.9	3
12	300	1.4	4.5

More about NPSH suction problems can be read here:

[Esam]

[h=Pump suction head as affected by altitude]2[/h]
Suction head affected by altitude as:

Elevation above Sea Level		Pressure			Practical Suction Lift
(ft)	(m)	(psig)	(ft H2O)	(bar)	(ft H2O)	(m H2O)
0*	0*	14.71	33.95	1	22	6.7
1320	402	14.02	32.38	0.97	21	6.4
2640	805	13.33	30.79	0.92	20	6.1
3960	1207	12.66	29.24	0.87	18	5.5
5280	1609	12.02	27.76	0.83	17	5.2
6600	2012	11.42	26.38	0.79	16	4.9
7920	2414	10.88	24.13	0.75	15	4.6
10560	3219	9.88	22.82	0.68	14	4.3

^* Sea Level

[aseptman]

Thanks for these several posts. Are they fro single source like:

[link Point to another website Only the registered members can access] or can you pl direct all users to the source sites. That will be more helpful because interests of others can be more than what has been put up by you.
However thanks again

[Esam]

Thanks for these several posts. Are they fro single source like:

[link Point to another website Only the registered members can access] or can you pl direct all users to the source sites. That will be more helpful because interests of others can be more than what has been put up by you.
However thanks again

very thanks aseptman , yes most of this articles from engineeringtoolbox web site but i select the best on it also i will share any interested articles i found while i search internet ans i will referee to it's source site if any one can want to read more

i think that will be good idea to all of us , to share any interested articles or web site we found while we search for any information on the internet