- #1

ag123

- 32

- 5

i've this question that i can't seem to work out an answer.

supposing that a (centrifuge) pump transfer an amount of energy (power) p to a column of water height h so that it maintains that as a static head. i.e. the water level in that tube is at a fixed height h due to the power transferred by the pump.

how do i calculate this height (head) h based on the power transferred into the water. note that the column of water is non moving.

if i assume the column of water has potential energy = mass * g * h/2 = density * area * h^2 * g / 2

where area is the cross-section area of the tube and the input power being constant. it would seem that the pump can support a much higher column (head) of liquid by simply reducing the diameter since:

h = sqrt(2 x power / ( density x area x g ))

h = sqrt(2 x power / ( density x pi x (d/2)^2 x g))

this calc would look absurd as if d diameter -> 0 , h -> infinity

so if i use a very small diameter, i can support a huge column of water with very little power

assuming no other forces act other than gravity on the water itself

i.e. how do i work the formula for how much power from the pump is required to maintain a static head a column of water height h?

(for that matter, you can assume all that power (energy) is transferred into the water, but it is a fixed amount of energy per sec watts)

supposing that a (centrifuge) pump transfer an amount of energy (power) p to a column of water height h so that it maintains that as a static head. i.e. the water level in that tube is at a fixed height h due to the power transferred by the pump.

how do i calculate this height (head) h based on the power transferred into the water. note that the column of water is non moving.

if i assume the column of water has potential energy = mass * g * h/2 = density * area * h^2 * g / 2

where area is the cross-section area of the tube and the input power being constant. it would seem that the pump can support a much higher column (head) of liquid by simply reducing the diameter since:

h = sqrt(2 x power / ( density x area x g ))

h = sqrt(2 x power / ( density x pi x (d/2)^2 x g))

this calc would look absurd as if d diameter -> 0 , h -> infinity

so if i use a very small diameter, i can support a huge column of water with very little power

assuming no other forces act other than gravity on the water itself

i.e. how do i work the formula for how much power from the pump is required to maintain a static head a column of water height h?

(for that matter, you can assume all that power (energy) is transferred into the water, but it is a fixed amount of energy per sec watts)

Last edited: