The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 9, Problem 5SP

(a)

To determine

The cross-sectional area of the wide and narrow portions of the pipe.

(a)

Expert Solution
Check Mark

Answer to Problem 5SP

The cross-sectional area of the wide and narrow portions of the pipe are 50.26 cm2 and 19.6 cm2 respectively.

Explanation of Solution

Given info: The diameter of the wider portion of the pipe is 8 cm. The diameter of the narrower portion of the pipe is 5 cm. The velocity of water in the wider portion of the pipe is 1.5 m/s.

Write the expression to find the area of the wider portion of the pipe.

Aw=πrw2=π(dw2)2

Here,

dw is the diameter of the narrow pipe

rw is the radius of the narrow pipe

Substitute 8 cm for dw in the above equation.

Aw=π(8 cm2)2=50.26 cm2

Write the expression to find the area of the narrow pipe.

An=πrn2=π(dn2)2

Here,

dn is the diameter of the narrow pipe

rn is the radius of the narrow pipe

Substitute 5 cm for dn in the above equation.

An=π(5 cm2)2=19.6 cm2

Conclusion:

Therefore, the cross-sectional area of the wide and narrow portions of the pipe are 50.26 cm2 and 19.6 cm2 respectively.

(b)

To determine

The speed of water in narrow portion of the pipe.

(b)

Expert Solution
Check Mark

Answer to Problem 5SP

The speed of water in narrow portion of the pipe is 3.85 m/s.

Explanation of Solution

Write the expression to find relation between velocity and cross-sectional area.

vnAn=vwAw

Here,

vn is the velocity of the narrower section of pipe

An is the cross-sectional area of the narrower section of pipe

vw is the velocity of the wider section of pipe

Aw is the cross-sectional area of the wider section of pipe

Substitute 1.5 m/s for vw, 50.26 cm2 for Aw, 19.6 cm2 for An and rearrange the above equation to find the speed of water in the narrowed portion of the pipe.

vn=vwAwAn=(1.5 m/s)(50.26 cm2)(19.6 cm2)=3.85 m/s

Conclusion:

Therefore, the speed of water in narrow portion of the pipe is 3.85 m/s.

(c)

To determine

Whether the pressure in the narrow portion of the pipe is greater than, less than, or equal to the pressure in the wider portion.

(c)

Expert Solution
Check Mark

Answer to Problem 5SP

The pressure in the narrow portion is less than the pressure in the wider portion of the pipe.

Explanation of Solution

The expression for the continuity equation is vnAn=vwAw. The expression for Bernoulli’s equation for a horizontal streamline is given by PwPn=12ρ(vn2vw2), here, Pn is the pressure in the narrow portion of the pipe, Pw is the pressure in the wide portion of the pipe and ρ is the density of water

The velocity of the narrow portion of the pipe is greater than the velocity of the wide portion of the pipe, vn>vw. Thus according the Bernoulli’s equation, the pressure in the narrow portion will be less than the pressure in the wider portion of the pipe.

Conclusion:

Therefore, the pressure in the narrow portion is less than the pressure in the wider portion of the pipe.

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Chapter 9 Solutions

The Physics of Everyday Phenomena

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