Question

Indoor waterfalls are installed in commercial centers and private houses to offer a refreshing view, humidity and a pleasant sound effect. You are asked to choose water pumps for a new indoor waterfall project. The submersible pumps are to be installed in a water reservoir and connected by flexible hoses to the top of the 3 m-high waterfall. The interior designer estimates that a water flow rate of 1500 liters per minute is required for an optimum effect and 4 identical water pumps are needed, installed at equal separations along the wall. Standard pumps have the same inlet and outlet crosssections.a) Determine the power rating of each of the 4 identical pumps, assuming 95% efficiency.b) What is the change in flow power between the inlet and outlet of the pump?c) If the inlet pressure is 1 atm, what is the pressure directly at the outlet of each pump?d) Suppose the pump outlet has a diameter of 5.0 cm, what would the flow velocity be?

Answer 1

a)0.1936KW

b)9.68W

c)P=130.755kPa

d)3.18m/s[/tex]

Step-by-step explanation:

A pump is a device capable of raising the pressure of a fluid, and thus making it flow through a pipe.

For one of the 4 pumps you have the following data

h=3m( height)

Q=water flow=1500lpm/4=375lpm=6.25x10^-3 m^3/S

γ= especific weight of water=9.81KN/m^3

η=efficiency of the pump=95%

A)

To calculate the power of the pump we use the following equation

P=hγQ/η

solving

P=(3)(9.81)(6.25x10^-3)/0.95)=0.1936KW

B)

Efficiency is defined as the percentage that the pump converts electrical energy into fluid power, so the power at the input is that found at point a, and for the calculation of the output power, we must only multiply it by the efficiency

input power=0.1936KW

output power=0.1936*0.95=0.1839KW

change in flow power

P=0.1936KW-0.1839KW=9.68W

C)

we must take into account that the pump's output will have the maximum pressure, then all we have to do is add the pressure of 1atm = 101.325Kpa plus the maximum pump pressure obtained from the height.

P=101.325kPa+ 3m*9.81KN/m^3

P=130.755kPa

D)For this part we only use the equation that relates the flow and velocity diameter

Q=VA

V=Q/A

A=Area of circle=

`A=circle-  area =(\pi D^(2) )/(4) =(\pi (0.05)^(2) )/(4) =</p><p>1.9635X10^-3M^2</p><p></p><p>[tex]V=(6.25x10^-3)/(1.9635X10^-3) =3.18m/s`