   Chapter 9, Problem 36P

Chapter
Section
Textbook Problem

A liquid (ρ = 1.65 g/ cm3) flows through a horizontal pipe of varying cross section as in Figure P9.36. In the first section, the cross- sectional area is 10.0 cm2, the flow speed is 275 cm/s, and the pressure is 1.20 × 105. In the second section, the cross-sectional area is 2.50 cm2. Calculate the smaller section’s (a) flow speed and (b) pressure. Figure P9.36

(a)

To determine
The flow speed of a liquid in the smaller section of the horizontal pipe.

Explanation
The equation of continuity is used for the speed of the liquid at the smaller end of the pipe as shown here. A2v2=A1v1v2=(A1/A2)v1 .

Given info: The cross sectional area of the first section is 10.0cm2 , the cross sectional area of second section is 2.50cm2 , and the flow speed of the liquid in the first section of the pipe is 275cm/s .

The formula for the speed of a liquid in the smaller section of the pipe is,

v2=(A1A2)v1

• A1 is cross sectional area of the first section of the pipe.
• A2 is cross sectional area of the second section of the pipe.
• v1 is speed of the liquid in the first section of the pipe

(b)

To determine
The pressure of a liquid in the smaller section of the horizontal pipe.

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