Engineering Fundamentals
Engineering Fundamentals
6th Edition
ISBN: 9780357112144
Author: Saeed Moaveni
Publisher: MISC PUBS
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Chapter 8, Problem 27P
To determine

Express the volume flow rate of the duct in ft3s, ft3min, m3s and find the average speed of air in the duct.

Expert Solution & Answer
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Answer to Problem 27P

The volume flow rate of the duct in ft3s, ft3min and m3s are 16.7ft3s2004ft3min,0.95m3s_ respectively and the average speed of air is 19.1fts_.

Explanation of Solution

Given data:

Refer to the Figure problem 8.27 in the textbook for the duct system.

Formula used:

Write the expression for the volume flow rate.

Q=VaverageAc (1)

Here,

Ac is cross-sectional area, and

Vaverage is average velocity.

Calculation:

Substitute 30fts for Vaverage, 8in.×10in. for Ac in equation (1) to find the volume flow rate in each of the 8in.×10in. duct.

Q=(30fts)(8in.×10in.)=(30fts)(8×0.083333ft×10×0.083333ft){1in=0.083333ft}=16.7ft3s

Give the volume flow rate in 18in.×14in. duct as below.

Qtotal=16.7ft3s+16.7ft3s

Qtotal=33.4ft3s (2)

Convert the unit volume flow rate in equation (2) from ft3s to ft3min.

Qtotal=33.4ft3s(60s1min){1min=60s}=2004ft3min

Convert the unit volume flow rate in equation (2) from ft3s to m3s.

Qtotal=33.4ft3s(1m3.28ft)3{1m=3.28ft}=0.95m3s

Find the average speed of air in the 18in.×14in. duct.

Substitute 33.4ft3s for Qtotal, 18in.×14in. for Ac in equation (1).

33.4ft3s=Vaverage(18in.×14in.)Vaverage=33.4ft3s(18in.×14in.)Vaverage=33.4ft3s(18×0.083333ft×14×0.083333ft){1in=0.083333ft}Vaverage=19.1fts

Conclusion:

Hence, the volume flow rate of the duct in ft3s, ft3min and m3s are 16.7ft3s2004ft3min,and0.95m3s_ respectively and the average speed of air is 19.1fts_.

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