Please find the gauge pressure of the water flowing through the smaller pipe Atmospheric pressure is 1.013 × 10^5 Pa .

Question

Water flows at speed of 5.4 m/s through a horizontal pipe of diameter 3.2 cm . Suppose

a thick pipe is connected to a thinner pipe. In the first pipe the gauge pressure P1 of the

water is 1.7 atm . The diameter of the smaller section is 2.1 cm . Please find the gauge pressure of the water flowing through the smaller pipe Atmospheric pressure is 1.013 × 10^5 Pa . The density of

water is 1000 kg/m^3. The viscosity of water is negligible. Answer in units of atm.

Answer 1

Let

p₀ = 1.013×10⁵ Pa be atmospheric pressure,

ρ = 1000 kg/m³ be the density of water,

d₁ = 3.2 cm be the diameter of the first pipe,

v₁ = 5.4 m/s be the speed of the water in the first pipe,

p₁ = 1.7 atm + p₀ = 2.736×10⁵ Pa be absolute pressure in the first pipe,

d₂ = 2.1 cm be the diameter of the second pipe,

v₂ (unknown) be the speed of the water in the second pipe,

p₂ (to be calculated) be absolute pressure in the second pipe.

In the absence of any information to the contrary, I will make the following assumptions:

Assumption 1) The pipes are concentric.

That implies that (on average) water does not need to flow higher of lower

when passing from the first pipe to the second.

Assumption 2) There is no heat transfer anywhere.

You can solve this problem by conservation of mass and energy:

The amount of water flowing each second must have the same mass and energy

in both pipes.

Conservation of mass:

The volume of water passing through the first pipe each second is

V₁ = π(d₁/2)²v₁

And through the second pipe

V₂ = π(d₂/2)²v₂

Assuming that water is non-compressible (as we were given only one density number),

we must have V₁ = V₂:

π(d₁/2)²v₁ = π(d₂/2)²v₂

From that

v₂ = (d₁/d₂)²v₁

Conservation of energy:

Let m = V₁ρ be the mass of water passing through the first pipe each second

(which is the same as the mass passing through the second pipe.)

The water in the first pipe has the following kinds of energy:

- Kinetic energy = mv₁²/2,

- Potential energy = PE -- an arbitrary constant, which by assumption 1) is the same for both pipes,

- Pressure energy = p₁V₁

- Internal energy = U -- an arbitrary constant, which by Assumptions 2) must be the same in both pipes.

The analogous energy quantities exist for the second pipe.

Conservation of energy then becomes

mv₁²/2 + PE + p₁V₁ + U = mv₂²/2 + PE + p₂V₂ + U

Thus

p₂ = ((v₁² - v₂²)m/2 + p₁V₁)/V₂

= (v₁² - v₂²)m/2V₂ + p₁V₁/V₂

= (v₁² - ((d₁/d₂)²v₁)²)ρ/2 + p₁

= v₁²(1 - (d₁/d₂)⁴)ρ/2 + p₁

Now we can substitute actual numbers, making sure that the units are consistent.

That implies that p₁ must be absolute pressure expressed in Pa.

p2 = 5.4²×(1 - (3.2/2.1)⁴)×1000/2 + 2.736×10⁵ = 2.0957×10⁵ Pa

To get gauge pressure, we need to subtract atmospheric pressure:

2.0957×10⁵ - 1.013×10⁵ = 1.0827×10⁵ Pa = 1.07 atm