# Static Pressure or Head in Fluids

## Static pressure and pressure head

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 (Pa, psi)

dz = change in height (m, in)

γ = specific weight (N/m^{3}, lb/ft^{3})

The pressure gradient in vertical direction is negative - the pressure decrease upwards.

### Specific Weight

Specific Weight can be expressed as:

γ = ρ g(2)

where

ρ= density (kg/m^{3}, slugs/ft^{3})

g= acceleration of gravity(9.81 m/s^{2},32.174 ft/s^{2})

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

The pressure exerted by a static fluid depends only upon

- the depth of the fluid
- the density of the fluid
- the acceleration of gravity

### Static Pressure in a Fluid

For a incompressible fluid - as a liquid - the pressure difference between two elevations can be expressed as:

p_{2}- p_{1}= - γ (z_{2}- z_{1})(3)

where

p_{2}= pressure at level 2 (Pa, psi)

p_{1}= pressure at level 1(Pa, psi)

z_{2}= level 2 (m, ft)

z_{1}= level 1(m, ft)

(3) can be transformed to:

p_{1}- p_{2}= γ (z_{2}- z_{1})(4)

or

p_{1}- p_{2}= γ h(5)

where

h=z_{2}- z_{1}difference in elevation - the dept down from locationz_{2}(m, ft)

or

p_{1}= γ h + p_{2}(6)

#### Example - Pressure in a Fluid

The absolute pressure at water depth of *10 m* can be calculated as:

*p _{1} = γ h + p_{2}*

* = (1000 kg/m ^{3}) (9.81 m/s^{2}) (10 m) + (101.3 kPa)*

* = (98100 kg/ms ^{2} or Pa) + (101300 Pa)*

* = 199.4 kPa*

*where *

*ρ = 1000 kg/m ^{3}*

*g = 9.81 m/s ^{2}*

*p _{2} = pressure at surface level = atmospheric pressure = *

*101.3 kPa*

The gauge pressure can be calculated by setting *p _{2} = 0*

*p _{1} = γ h + p_{2}*

* = (1000 kg/m ^{3}) (9.81 m/s^{2}) (10 m) *

* = 98.1 kPa*

### The Pressure Head

(6) can be transformed to:

h = (p_{2}- p_{1}) / γ(7)

*h* express **the pressure head** - the height of a column of fluid of specific weight - *γ* - required to give a pressure difference of (*p _{2} - p_{1}).*

#### Example - Pressure Head

A pressure difference of *5 psi (lb _{f}/in^{2})* is equivalent to

(5 lb_{f}/in^{2}) (12 in/ft) (12 in/ft) / (62.4 lb/ft^{3})

= 11.6 ft of water

(5 lb_{f}/in^{2}) (12 in/ft) (12 in/ft) / (847 lb/ft^{3})

= 0.85 ft of mercury

when specific weight of water is* 62.4 (lb/ft ^{3})* and specific weight of mercury is

*847 (lb/ft*.

^{3})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*

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