Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.70 ✕ 105 Pa and the pipe radius is 2.50 cm. At the higher point located at

y = 2.50 m, the pressure is 1.21 ✕ 105 Pa and the pipe radius is 1.70 cm.
(a) Find the speed of flow in the lower section.
(b) Find the speed of flow in the upper section.
(c) Find the volume flow rate through the pipe.

In order to answer these questions, we need to apply Bernoulli's equation and the principle of conservation of mass.

(a) To find the speed of flow in the lower section, we can use Bernoulli's equation, which relates the pressure, height, and velocity of a fluid. Bernoulli's equation states that the total mechanical energy per unit volume of a fluid remains constant in steady, ideal flow.

At the lower point (section 1), the given pressure is 1.70 × 10^5 Pa and the radius is 2.50 cm. We can convert the radius to meters by dividing by 100, so the radius becomes 0.025 m.

The equation can be written as: P₁ + ½ρv₁^2 + ρgy₁ = P₂ + ½ρv₂^2 + ρgy₂

Since we are dealing with a vertical pipe, the gravitational potential energy term ρgy cancels out. Also, the fluid density ρ is the same throughout the pipe. Thus, the equation simplifies to: P₁ + ½ρv₁^2 = P₂ + ½ρv₂^2

Plugging in the given values, we get: 1.70 × 10^5 + ½ρv₁^2 = 1.21 × 10^5 + ½ρv₂^2

Since we are interested in the speed of flow in the lower section (v₁), we can rearrange the equation to solve for v₁: v₁ = √((2(P₂ - P₁))/ρ)

(b) Similarly, to find the speed of flow in the upper section (v₂), we use the same equation, but with the pressure and radius at the higher point.

At the higher point (section 2), the given pressure is 1.21 × 10^5 Pa and the radius is 1.70 cm. Converting the radius to meters, we have a radius of 0.017 m.

Again, using the equation: v₂ = √((2(P₁ - P₂))/ρ)

(c) The volume flow rate through the pipe is given by the equation: Q = Av, where A is the cross-sectional area perpendicular to the flow direction, and v is the velocity of flow.

Substituting the values for the lower section (section 1), the area is given by A₁ = πr₁^2, and the velocity is v₁ (from part a).

For the upper section (section 2), the area is A₂ = πr₂^2, and the velocity is v₂ (from part b).

Therefore, the volume flow rate through the pipe is given by: Q = A₁v₁ = A₂v₂

We can plug in the values for the respective areas and velocities to find the volume flow rate.

Remember to use the correct units and keep track of unit conversions throughout the calculations.