Question

10) A shower pipe at the beach that is 10 cm in diameter carries water (density 1000 kg/m3 ) up a height of 2.0 m, at which point the pipe narrows to 5 cm and continues another 0.50 m up before the water squires out of the pipe 2.5 m above where it started. Ignore viscosity.

a) If the water moves at 1.5 m/s in the wider part of the pipe, what is its speed in the narrower part of the pipe?

b) What is the pressure change from the bottom of the pipe to the top?

c) What was the pressure of the water at the bottom of the pipe? (Hint: What does the pressure have to be when the water leaves the pipe and is exposed to the atmosphere?)

Answer 1

a)

Consider the point at the bottom to be 1 and top to be 2.

At the bottom,

Radius of the pipe is

r₁ = 10 cm /2

    = 5 cm

Area is

At the top,

Radius of the pipe is

r₂ = 5 cm /2

   = 2.5 cm

Area is

From the equation of continuity, the flow rate of water remains the same.

Thus,

Substitute 5 cm for r₁, 2.5 cm for r₂ and 1.5 m/s for v₁ in the above equation,

Therefore, the speed in the narrower part of the pipe is 6 m/s.

b)

The pressure changes from the bottom to the top is

P = gh

Here, density of water is , acceleration due to gravity is g and height of the water from the bottom to the top in the pipe is h.

Substitute 1000 kg/m³ for r, 9.8 m/s² for g and 2.5 m for h in the above equation,

P = rgh

= (1000 kg/m³) ( 9.8 m/s²)( 2.5 m)

= 24500 kg /m s²) (1 Pa / 1 kg /m s²)

= 24500 Pa

Therefore, the pressure changes from the bottom of the pipe to the top 24500 Pa.

c)

The magnitude of atmospheric pressure is

1 atm = 1.01 x 10⁵ Pa

The pressure at the bottom of the water is

P₁ = P + atmospheric pressure

= 24500 Pa + 1.01 x 10⁵ Pa

= 1.255 x 10⁵ Pa

Rounding off to three significant figures, the pressure of the water at the bottom of the pipe is 1.26 x 10⁵ pa.