Foot of head.
Foot of head.
(OP)
I have heard a "Foot of head" described as the amount of pressure that a one foot high column of water exerts. This defenition however fails to mention the diameter of this column. Is there something that I am missing here? Shouldn't the column of water have a specific volume?
RE: Foot of head.
"Head" as you have termed it is a measure of pressure and will therefore be independent of the volume. Remember using a U-tube manometer in the school physics lessons? If you raise the level of the reservoir by a foot, the water travels a foot up the tube, regardless of the diameter of the tube (and provided your reservoir holds sufficient water).
Andy Machon
RE: Foot of head.
RE: Foot of head.
I think the thing that you failed to see was the difference between PRESSURE and FORCE. A big column will have a bigger FORCE at the bottom, but it is also distributed over a bigger area. Hence, the pressure will be the same no matter what the diameter is--for the same fluid, anyway.
RE: Foot of head.
Area of cross section x height of the liquid column x specific weight of the liquid.
Therefore pressure is given by
Force/area = height of the liquid column x specific weight of the liquid.
Since, the specific weight of the liquid is a constant (of course at a given temperature), the pressure is expressed as so many'feet' of the liquid column. Hence, It is also important that the 'head' should always be associated with the fluid of reference, such as, 10 m. of water or 76 cm. of mercury etc. The pressure is obtained by multiplying the head by the specific weight of the fluid.
Trilinga