Fluid dynamics of lazy river ride: how do you calculate pump size?

Question

the lazy river is nearly oval in shape, 520 feet long, 3 feet dep and 10 feet wide. The experts are all over the place on required pump size ie total horespower. please provide solution and methodology.

Answer 1

You need to specify volumetric flow rate and the head loss.

Answer 2

What is important is flowrate and head loss.

You are gong to have to make an assumption on the speed.  Someplace between 1-3 ft/sec.  Head loss will go up with the square of the velocity.

Mannings' formula will give you the head loss.  http://en.wikipedia.org/wiki/Open_channe... .  You might need to take into account bends and other stuff.

This will give you the minimum horsepower.  You might need more if you have inefficiencies such as powering your lazy river with jets located through out the track.  (This would decrease the flowrate for the pump but require increased head.  It would also make for a better ride).

Answer 3

Simple approach uses water velocity and change in height of ride (from bottom of channel at the bottom of ride to bottom of channel at the top of ride):

Power = Q * pressure

So, lets calculate Q and pressure:

Q = Area * velocity

For velocity of 2 ft/sec,

Q = (3ft * 10 ft) * 2ft/sec

Also, pressure = mgh = (mg)h

Putting is all together, for v= 2ft/sec and height h is 50 feet:

Power = Q * v * Pressure

Power = (3ft * 10ft) * (2 ft/sec) * (62.4lb/ft^3 * 50ft)

This comes out in ft-lb/sec. Divide by 550 to get minimum HP.

The only problem with this approach is that if you can't measure velocity, you have to calculate it using Manning formula as indicated in the post above, which appears to be correct.

Most likely, 2 - 5x as much power is needed for efficiency, pressure drops across nozzles, etc.