Basic Equation Of Fluid Flow

[Chemist]

Water is being pumped from an open water reservoir at a rate of 2.0 kg/s at 10°C to an open storage tank 1500 m away. The pipe is 3 ½ - in. schedule 40 pipe and the fictional losses in the system are 625 J/kg. The surface of the water reservoir is 20 m above the level of the storage tank. The pump has an efficiency of 75%. What is the kW power required for the pump?

[Zacs-Master]

What answer did you get?

[gingejbee]

Chemist: pity you're not "Physicist", isn't it?

[Peter Pedant]

as a retiree with phys A level here goes:

2 kg/s water with fictional losses of 625 J kg-1 is 1250 j s-1 ( fictional losses to be overcome by potential energy + power unit)

P E , per sec is mgh 2. 20. 9.81around 360 j s-1 (or watts)

so in terms of j s-1 .. 1250 = 360 + power
so power needed is c 890 j s -1

so teh power reqd for the pump at 75% efficiency is 100/75 times 890

which is around 1200 w and since they want kW 1.2 kw

data which is irrel is 10'C, 1500 m (because the frictional losses are not given per metre) and the 3 1/2 in
also not given are the density and viscosity of water
so I dont think Poiseuille is needed.

well that's my answer - you can see why I didnt do engineering at uni !

[Bert]

If the losses are fictional, does that mean that they don't exist? :-)

[Peter Pedant]

yeah and what about non-fictional flow ?