Question

A liquid ( = 1.65 g/cm^3) flows through a horizontal pipe of varying cross section as in the figure below. In the first section, the cross-sectional area is 10.0 cm^2, the flow speed is 293 cm/s, and the pressure is 1.20 * 10^5 Pa. In the second section, the cross-sectional area is 3.00 cm^2. Answer parts a-b.

Answer 1

a) What is the flow speed in the second section? Assume the pressure remains constant.

Given:

Section 1 area = 10.0 cm^2

Section 1 flow speed = 293 cm/s

Section 2 area = 3.00 cm^2

Pressure = 1.20 x 105 Pa (constant)

Since pressure remains constant, by Bernoulli's equation:

P1/ρ + 1/2*v12 = P2/ρ+ 1/2*v22

Since P1 = P2 and ρ (density) is constant:

v22 = 2*(v12 - v22)

v22 = 2*(293^2 - v22)

Solving for v2:

v2 = √(2*293^2) = 846 cm/s

b) What is the kinetic energy loss per second between the two sections?

Kinetic energy = 1/2 * mass * velocity^2

Mass flow rate = density * area * velocity

= 1.65 g/cm^3 * 10.0 cm^2 * 293 cm/s = 485.5 g/s

Kinetic energy loss = 1/2 * (485.5 g/s) * (293^2 - 846^2) cm^2/s^2

= 1/2 * (485.5)*(86124 - 716256)

= 20617 J/s

So the kinetic energy loss per second between the two pipe sections is 20617 J/s.

Step-by-step explanation: