Which end has the higher pressure, the thick end or the fine end, give out the pressure difference between them.

Question

A water pipe has a thick and a fine end. The radius of the thick is 3cm, and the radius of the fine is 1cm. The thick end is 6 meters higher than the fine end. Ask: how about the flowing speed in fine end if the flowing speed is 9m/s at thick end. Which end has the higher pressure, the thick end or the fine end, give out the pressure difference between them. (Acceleration of gravity is 9.8m/s^2).

Answer 1

In order to solve this problem, we will use two fundamental equations. The first one is the Continuity Equation, which ensures that the same amount of water (or any other fluid) flowing into the pipe is the same as the amount flowing out of the pipe.

(1) A₁*U₁ = A₂*U₂

where A₁ and U₁ are the cross-sectional area and flow speed of the fluid going into the thick end of the pipe, respectively, and A₂ and U₂ are the cross-sectional area and flow speed of the fluid leaving the thin end of the pipe, respectively.

Since we know the radius of the ends of the pipes, we can calculate their cross-sectional area using the equation for the area of a circle:

A₂ = π(R₂)²

A₁ = π(R₁)² = π(3R₂)² = 9π(R₂)² as we know that R₁ is 3 times larger than R₂

and we know that U₁ = 9 m/s, we can rearrange the Continuity Equation (1) to calculate U₂ as such:

U₂ = U₁*(A₁/A₂) = 9U₁ = 81 m/s

Now that we know the velocity flowing out of the pipe, we can use the Bernoulli Equation to relate the energy at the beginning and end of the pipe, which due to the conservation of energy, is a constant.

(2) P + ρU²/2 + ρgz = constant

where P is the static pressure (what we are trying to find), ρ is the density of the fluid, g is gravity, and z is the height of the pipe.

We can get two separate expressions for the beginning and end of the pipe, and since they will have the same energy, we can set the equations equal to each other:

(3) P₁ + ρ(U₁)²/2 + ρgz₁ = P₂ + ρ(U₂)²/2 + ρgz₂

Thanks to the conditions provided by the question and our result from the continuity equation, we can plug in the following values:

ρ (of water) = 997 kg/m³

g = 9.81 m/s²

U₁ = 9 m/s

U₂ = 81 m/s

z₁ = z₂ + 6 m (you can just set z₂ = 0; it's all relative)

Bring both static pressures to one side of the equation and solve. You can now find the change in pressure:

ΔP = P₂ - P₁ = -3.172 MPa (MPa = 10⁶ Pa)

Since this number is negative, the pressure is higher on the thick end.