Question

A liquid

(ρ = 1.65 g/cm³)

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 cm², the flow speed is 281 cm/s, and the pressure is 1.20  10⁵ Pa. In the second section, the cross-sectional area is 4.50 cm².

The flow within a horizontal tube is depicted by five lines. The tube extends from left to right, with the left end wider than the right end. The five lines start at the left end, go horizontally to the right, curve slightly toward the center of the tube such that all five lines come closer together, and again go horizontally to the right to exit at the right end. Arrows on the lines point to the right to represent the direction of flow.

(a) Calculate the smaller section's flow speed. (Enter your answer to at least two decimal places.)
m/s

(b) Calculate the smaller section's pressure.

Answer 1

(a) The cross sectional area of the first section = A1 = 10 cm2

The cross-sectional area of the second section = A2 = 4.5 cm2

velocity (flow speed) of first section = v1 = 281 cm/sec = 2.81 m/sec

let v2 be the velocity (flow speed) of second section

then conservation of mass will give A1v1 =A2v2

v2 = A1v1/A2 = 10*281/4.5 = 624.44 cm/sec = 6.244 m/sec

(b) Applying Bernoulli's law to friction less flow,

the total mechanical energy will remain constant.

1.65 gm/cm3 = 1650 kg/m3

or

Since the tube lies horizontal, hence h in both the sections will be constant.

P2 = 1.2*105 Pa+25650.43 Pa

=12*104 + 2.56*104 = 14.56*104 Pa = 1.456*105 Pa  ( the smaller section's pressure)

Please check the calculations.