Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 11, Problem 48P

The local atmospheric pressure in Denver, Colorado (elevation 1610 m) is 83.4 kPa. Air at this pressure and at 25’C flows with a velocity of 9 m/s over a 2.5-rn 5-rn flat plate. Determine the drag force acting on the top surface of the plate if the air flows parallel to the (a) 5-rn-long side and (b) the 2.5-m-long side.

Expert Solution
Check Mark
To determine

(a)

The drag force acting on the top surface of the plate when the air flows parallel to the 5 m long side.

Answer to Problem 48P

FD=1.58N

Explanation of Solution

Given information:

Size of plate=2.5m×5m

Atmospheric pressure=83.4kPa

Temperature=25C=298K

Velocity=9ms

Concept used:

ρ=PRT

Re=ρVLμ

Cf=0.074ReL1/51742ReL

FD=CfAρV22

ReReynolds numberFDDrag forceCfAverage friction coefficientρDensityμDynamic viscosityAAreaVVelocity

Calculation:

The gas constant is, R=0.287kPam3kgK.

ρ=PRT=83.40.287×298=0.975kgm3

For the given temperature, μ=1.849×105kgm-s.

ReL=ρVLμ=0.975×9×51.849× 10 5=2.373×106>Recritical(5× 105)

Cf=0.074 ReL 1/51742 ReL=0.074 ( 2.373× 10 6 ) 1/517422.373× 106=0.0032

FD=CfAρV22=0.0032×(5×2.5)×0.975×922×( 1N 1kg m/s 2 )=1.58N

Conclusion:

The drag force acting on the top surface of the plate when the air flows parallel to the 5 m long side is 1.58N.

Expert Solution
Check Mark
To determine

(b)

The drag force acting on the top surface of the plate when the air flows parallel to the 2.5 m long side.

Answer to Problem 48P

FD=1.48N

Explanation of Solution

Given information:

Size of plate=2.5m×5m

Atmospheric pressure=83.4kPa

Temperature=25C=298K

Velocity=9ms

Concept used:

ρ=PRT

Re=ρVLμ

Cf=0.074ReL1/51742ReL

FD=CfAρV22

ReReynolds numberFDDrag forceCfAverage friction coefficientρDensityμDynamic viscosityAAreaVVelocity

Calculation:

The gas constant is, R=0.287kPam3kgK.

ρ=PRT=83.40.287×298=0.975kgm3

For the given temperature, μ=1.849×105kgm-s.

ReL=ρVLμ=0.975×9×2.51.849× 10 5=1.186×106>Recritical(5× 105)

Cf=0.074 ReL 1/51742 ReL=0.074 ( 1.186× 10 6 ) 1/517421.186× 106=0.003

FD=CfAρV22=0.003×(5×2.5)×0.975×922×( 1N 1kg m/s 2 )=1.48N

Conclusion:

The drag force acting on the top surface of the plate when the air flows parallel to the 2.5 m long side is 1.48N.

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ASAP

Chapter 11 Solutions

Fluid Mechanics: Fundamentals and Applications

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