Question

A horizontal pipe contains water at a pressure of 140 kPa flowing with a speed of 1.2 m/s . The pipe narrows to one-half its original diameter.

What is the speed of the water?

Express your answer using two significant figures.

What is the pressure of the water?

Express your answer using two significant figures.

Answer 1

This horizontal pipe contains water at a pressure of 140 kPa.

Force = Pressure x Area................................................... (i)

also,
Force = Mass x Acceleration
where,
Acceleration = dv/dt, ie rate of change of Velocity.[v=velocity]
So,
Force = M x dv/dt = P x A
Now for a particular fluid- the Mass of the fluid will remain constant.
Hence,
P x A ~ dv/dt [mass is constant]
or (P x A) x dv/dt = constant.
for Constant flowrate-
Area x Velocity = Constant flowrate................................... (ii)

here if the diameter of the pipe become half,

Let previous diameter was 'd' and new diameter is 'd/2'.

so previous area were A₁=.

and new area is A₂=.

According to equation no. (ii)

1.2⁼ . V [V=present velocity]

V=4.8 m/s

Here as force is constant due to decrease of the length of the diameter,area will decrease.

Now from equation (i),we can write,

P₁A₁=P₂A_{2 [}where P₁ and P₂ are previous pressure and new pressure respectively]

or, 14010³=P_²

or, P₂=56010³

So, speed of the water is 4.8 m/s and pressure of the water is 560 kPa.