An incompressible fluid is flowing through a vertical pipe with a constriction. The wide section is 2.00 cm in diameter and is at the top of the pipe. The pressure of the fluid in the wide section at the top is 200 kPa. The velocity of the fluid in the wide section is 4.00 m/s. The narrow section is located 4.00 m below the wide section. What is the diameter of the narrow section , if the pressure of the fluid in the narrow section is equal to the pressure in the wide section? (density of the fluid is 680 kg/m3)

To solve this problem, we need to use the Bernoulli's equation which states that the sum of the pressure energy, kinetic energy and potential energy per unit volume is constant for an incompressible, non-viscous fluid in steady flow. The equation is given by:

P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2

where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height.

Given that the pressure in the wide section (P1) is 200 kPa, the velocity in the wide section (v1) is 4.00 m/s, the height difference between the wide and narrow sections (h1 - h2) is 4.00 m, the density of the fluid (ρ) is 680 kg/m^3, and the pressure in the narrow section (P2) is equal to P1, we can substitute these values into the Bernoulli's equation to find the velocity in the narrow section (v2).

200 kPa + 1/2 * 680 kg/m^3 * (4.00 m/s)^2 + 680 kg/m^3 * 9.81 m/s^2 * 4.00 m = 200 kPa + 1/2 * 680 kg/m^3 * v2^2 + 680 kg/m^3 * 9.81 m/s^2 * 0

Solving for v2, we get v2 = sqrt[(2 * (200 kPa - 680 kg/m^3 * 9.81 m/s^2 * 4.00 m) / 680 kg/m^3) + (4.00 m/s)^2] = 14.28 m/s

Next, we use the continuity equation which states that the product of the cross-sectional area and the velocity is constant for an incompressible fluid in steady flow. The equation is given by:

A1 * v1 = A2 * v2

where A is the cross-sectional area.

Given that the diameter of the wide section (d1) is 2.00 cm, we can find the cross-sectional area of the wide section (A1) using the formula for the area of a circle:

A1 = π * (d1 / 2)^2 = π * (2.00 cm / 2)^2 = 3.14 cm^2

Substituting the values of A1, v1, and v2 into the continuity equation, we can find the cross-sectional area of the narrow section (A2):

A2 = A1 * v1 / v2 = 3.14 cm^2 * 4.00 m/s / 14.28 m/s = 0.88 cm^2

Finally, we can find the diameter of the narrow section (d2) using the formula for the diameter of a circle:

d2 = 2 * sqrt(A2 / π) = 2 * sqrt(0.88 cm^2 / π) = 1.06 cm

Therefore, the diameter of the narrow section is approximately 1.06 cm.