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.

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Answer 1

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

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 $38.38 N$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

(b)

Expert Solution

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 $240.93 N$ .

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

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Answer 2

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

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 $38.38 N$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

(b)

Expert Solution

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 $240.93 N$ .

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

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Answer 3

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

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 $38.38 N$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

(b)

Expert Solution

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 $240.93 N$ .

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

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Answer 4

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

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 $38.38 N$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

(b)

Expert Solution

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 $240.93 N$ .

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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 $38.38 N$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

(b)

Expert Solution

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 $240.93 N$ .

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

Answer 8

(a)

Expert Solution

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 $38.38 N$ .

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The value of skin friction drags for laminar flow.

Answer 13

Answer to Problem 19.1P

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

Answer 14

Explanation of Solution

Given:

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=1.328 ReCd=1.328 6.90× 10 6 Cd=5.05×10−4$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=5.05×10−4×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=38.38 N$

Conclusion:

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

Answer 15

(b)

Expert Solution

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 $240.93 N$ .

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The value of skin friction drags for turbulent flow.

Answer 20

Answer to Problem 19.1P

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

Answer 21

Explanation of Solution

The length of the wing is $l=9.75 m$ .

The chord of the wing is $C=1.6 m$ .

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

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,

$D=Cd⋅ρ0V2A2$

Calculation:

The area of the wing can be calculated as,

$A=l×CA=9.75 m×1.6 mA=15.6 m2$

It is known that the density of the air is $1.225 kg/m3$ and the dynamic viscosity of the air is $1.78×10−5 Pa⋅s$ .

The Reynolds number can be calculated as,

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

The coefficient of the drag can be calculated as,

$Cd=0.074 ( Re ) 1 5 Cd=0.074 ( 6.90× 10 6 ) 1 5 Cd=3.17×10−3$

The drag can be calculated as,

$D=Cd⋅ρ0V2A2D=3.17×10−3×1.225 kg/ m 3× ( 141 mi/ h× 0.4470 m/s 1 mi/h ) 2×15.6 m 22D=240.93 N$

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

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 $240.93 N$ .

Answer 22

Ch. 19 - The wing on a Piper Cherokee general aviation...Ch. 19 - For the case in Problem 19.1, calculate the...Ch. 19 - For the case in Problem 19.1, calculate the...Ch. 19 - Consider Mach 4 flow at standard sea level...Ch. 19 - Repeat Problem 19.4 for the case of all turbulent...Ch. 19 - Prob. 19.6P Ch. 19 - Prob. 19.7P

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