Energy Equation with a Pump - Example Problem

[@austinswrld]

Excellent video!

[@BrianBernardEngineering]

Thanks so much, have a great semester!

[@versed77]

i applied bernoullis from the entrant at the bottom of the tank to the exit at atmospheric pressure and got a pump head of 17.022m, so the pump power would be 1335kw. i used the free surface as the datum as calculated the hydrostatic pressure at the entrant (z=-20m). why do we get different pump power? i used g=9.81, pressure at entrant = 196.2kpa

[@BrianBernardEngineering]

"calculated the hydrostatic pressure at the entrant (z=-20m)" - This is the problem. From the surface of the tank, to the entrance to the pipe at the bottom of the tank is not a hydrostatic problem. You can't treat this as a manometer. The fluid at the entrance of the tube is moving, there is velocity that you are not accounting for. You would need to use Bernoulli Eqn from top of tank to entrance to pipe to account for velocity. This will lead to lower pressure at pipe entrance than you calculated, since instead of all the height difference becoming pressure difference, some of the height difference becomes pressure difference but some of it becomes velocity.

[@versed77]

@@BrianBernardEngineering thanks. i figured. i will retry it from the free surface. is there a way i can solve using bernoullis from the entrant? how would i do it just or curiousity. what additional info, if any, would i need thanks!

[@BrianBernardEngineering]

@@versed77 You'll need to solve for velocity first, before you can solve for pressure at the pipe entrance. Energy Equation from Top of tank to pipe exit will give you velocity. Then Bernoulli Eqn from top of tank to pipe entrance would give you pressure at pipe entrance. This assumes pipe diameter is constant, so velocity is same based on continuity condition. If diameter changed, you could solve for the different velocity using continuity if you know the diameters.

[@crazygood1716]

thanks pogi

[@BrianBernardEngineering]

anytime