Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 19, Problem 19.1P

The wing on a Piper Cherokee general aviation aircraft is rectangular, with a span of 9.75 m and a chord of 1.6 m. The aircraft is flying at cruising speed ( 141 mi/h ) at sea level. Assume that the skin-friction drag on the wing can be approximated by the drag on a flat plate of the same dimensions. Calculate the skin-friction drag:

a. If the flow were completely laminar (which is not the case in real life)

b. If the flow were completely turbulent (which is more realistic) Compare the two results.

(a)

Expert Solution
Check Mark
To determine

The value of skin friction drags for laminar flow.

Answer to Problem 19.1P

The value of skin friction drag for laminar flow is 38.38N .

Explanation of Solution

Given:

The length of the wing is l=9.75m .

The chord of the wing is C=1.6m .

The cruising speed of the aircraft is V=141mi/h .

Formula used:

The expression for the area of wing is given as,

  A=l×C

The expression for the Reynolds number is given as,

  Re=ρ0VCμ0

Here, Re is the Reynolds number, μ0 is the dynamic viscosity of the air, ρ0 is the density of air.

The expression for the coefficient of drag is given as,

  Cd=1.328Re

The expression for the drag is given as,

  D=Cdρ0V2A2

Calculation:

The area of the wing can be calculated as,

  A=l×CA=9.75m×1.6mA=15.6m2

It is known that the density of the air is 1.225kg/m3 and the dynamic viscosity of the air is 1.78×105Pas .

The Reynolds number can be calculated as,

  Re=ρ0VCμ0Re=1.225kg/ m 3×( 141 mi/ h× 0.4470m/s 1 mi/h )×1.6m1.78× 10 5PasRe=6.90×106

The coefficient of the drag can be calculated as,

  Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×104

The drag can be calculated as,

  D=Cdρ0V2A2D=5.05×104×1.225kg/ m 3× ( 141 mi/ h× 0.4470m/s 1 mi/h ) 2×15.6 m 22D=38.38N

Conclusion:

Therefore, the value of skin friction drag for laminar flow is 38.38N .

(b)

Expert Solution
Check Mark
To determine

The value of skin friction drags for turbulent flow.

Answer to Problem 19.1P

The value of skin friction drag for turbulent flow is 240.93N .

Explanation of Solution

The length of the wing is l=9.75m .

The chord of the wing is C=1.6m .

The cruising speed of the aircraft is V=141mi/h .

Formula used:

The expression for the area of wing is given as,

  A=l×C

The expression for the Reynolds number is given as,

  Re=ρ0VCμ0

Here, Re is the Reynolds number, μ0 is the dynamic viscosity of the air, ρ0 is the density of air,

The expression for the coefficient of drag is given as,

  Cd=0.074( Re)15

The expression for the drag is given as,

  D=Cdρ0V2A2

Calculation:

The area of the wing can be calculated as,

  A=l×CA=9.75m×1.6mA=15.6m2

It is known that the density of the air is 1.225kg/m3 and the dynamic viscosity of the air is 1.78×105Pas .

The Reynolds number can be calculated as,

  Re=ρ0VCμ0Re=1.225kg/ m 3×( 141 mi/ h× 0.4470m/s 1 mi/h )×1.6m1.78× 10 5PasRe=6.90×106

The coefficient of the drag can be calculated as,

  Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×103

The drag can be calculated as,

  D=Cdρ0V2A2D=3.17×103×1.225kg/ m 3× ( 141 mi/ h× 0.4470m/s 1 mi/h ) 2×15.6 m 22D=240.93N

The value of skin friction drag for turbulent flow is 240.93N and The value of skin friction drag for laminar flow is 38.38N .

By comparing these two results, it is showing that the skin drag force in turbulent flow is 6.27 times greater than the skin drag force in laminar flow.

Conclusion:

Therefore, the value of skin friction drag for turbulent flow is 240.93N .

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