In a section of horizontal pipe with a diameter of 3.00 cm the pressure is 5.21 kPa and water is flowing with a speed of 1.50 m/s. The pipe narrows to 2.50 cm. What is the pressure in the narrower region if water behaves like an ideal fluid of density 1000 kg/m3?

To solve this problem, we can use the principle of conservation of energy for fluids, known as the Bernoulli's equation. Bernoulli's equation states that for an incompressible, non-viscous fluid, the total mechanical energy remains constant along any streamline. The equation is given by:

P + 0.5 * ρ * v^2 + ρ * g * h = constant

where: - P is the fluid pressure, - ρ is the fluid density, - v is the fluid velocity, - g is the acceleration due to gravity, and - h is the fluid's elevation.

Since the pipe is horizontal, the elevation term (ρ * g * h) does not change along the pipe, and we can exclude it. So the Bernoulli's equation simplifies to:

P + 0.5 * ρ * v^2 = constant

Now, let's apply Bernoulli's equation to the two sections of the pipe (the wider section and the narrower section).

For the wider section (section 1) with diameter d1 = 3.00 cm and pressure P1 = 5.21 kPa: v1 = 1.50 m/s P1 = 5.21 kPa

For the narrower section (section 2): d2 = 2.50 cm We need to find pressure P2.

First, convert the pressures from kPa to Pa since the standard SI unit for pressure is Pascal (1 kPa = 1000 Pa): P1 = 5.21 kPa = 5210 Pa

Next, we need to find the velocity v2 in the narrower section of the pipe. We can use the equation of continuity, which states that the volumetric flow rate (Q) is constant through a pipeline:

Q = A1 * v1 = A2 * v2

Where A is the cross-sectional area of the pipe. For a circular pipe, A = π * (d/2)^2.

Now let's find A1 and A2 from the diameters:

A1 = π * (d1/2)^2 = π * (0.030 m / 2)^2 A2 = π * (d2/2)^2 = π * (0.025 m / 2)^2

Then, to find v2:

A1 * v1 = A2 * v2 v2 = (A1 * v1) / A2

We calculate v2 so we can find the pressure P2 in the narrower section using Bernoulli's principle.

Now we'll substitute the known values for ρ (1000 kg/m^3), A1, v1, and A2 into the equation to find v2:

v2 = (π * (0.030/2)^2 * 1.50) / (π * (0.025/2)^2) v2 = ((π * (0.015)^2 * 1.50) / (π * (0.0125)^2)) v2 = ((π * 0.000225 * 1.50) / (π * 0.00015625)) v2 = (0.0003375 / 0.00015625) v2 ≈ 2.16 m/s

Finally, we apply Bernoulli's equation with conservation of total mechanical energy from section 1 to section 2:

P1 + 0.5 * ρ * v1^2 = P2 + 0.5 * ρ * v2^2

Now, solve for P2:

P2 = P1 + 0.5 * ρ * v1^2 - 0.5 * ρ * v2^2 P2 = 5210 Pa + 0.5 * 1000 kg/m^3 * (1.50 m/s)^2 - 0.5 * 1000 kg/m^3 * (2.16 m/s)^2 P2 = 5210 Pa + 0.5 * 1000 * 2.25 - 0.5 * 1000 * 4.6656 P2 = 5210 Pa + 1125 - 2332.8 P2 = 5210 + 1125 - 2332.8 P2 = 6335 - 2332.8 P2 = 4002.2 Pa

So, the pressure in the narrower region is approximately 4002.2 Pa or 4.002 kPa.