Question

Using Bernoulli Equation :

1) A water pump forces water up a vertical pipe with diameter 12 cm at a flow rate of 300 L/s. The pressure at the pump is 2.8 x 106 N/m2 . Two meters above the pump, the pipe reduces to a diameter of 8 cm. What is the pressure at that point?

Answer 1

let point below is point 1 and point above is point 2.

at 1, let pressure be P1, speed be v1 and height be h1.

at 2, let pressure be P2, speed be v2, height h2.

let h1=0

then h2=2 m

given P1=2.8*10^6 N/m^2

v1=flow rate/cross sectional area

=300 L/s/(pi*radius^2)

=0.3 m^3/(pi*(0.12/2)^2)

=26.526 m/s

v2=flow rate/cross sectional area

=300 L/s/(pi*radius^2)

=0.3 m^3/(pi*(0.08/2)^2)

=59.683 m/s

using Bernoulli principle:

P1+rho*g*h1+0.5*rho*v1^2=P2+rho*g*h2+0.5*rho*v2^2

where rho=density of water=1000 kg/m^3

using the given values,

2.8*10^6+1000*9.8*0+0.5*1000*26.526^2=P2+1000*9.8*2+0.5*1000*59.683^2

=>P2=2.8*10^6+1000*9.8*0+0.5*1000*26.526^2-(1000*9.8*2+0.5*1000*59.683^2)=1.3512*10^6 N/m^2

pressure at that point is 1.3512*10^6 N/m^2.