The drag force.

Question

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Answer 1

Question

Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

(b)

To determine

The drag force.

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

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Answer 2

Question

Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

(b)

To determine

The drag force.

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

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Answer 3

Question

Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

(b)

To determine

The drag force.

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

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Answer 4

Question

Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

(b)

To determine

The drag force.

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

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Answer 5

Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

(b)

To determine

The drag force.

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

Answer 6

Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

Answer 7

Chapter 15, Problem 45P

(a)

To determine

The drag force.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

Pressure is 83.4 kPa.

Temperature is 25°C.

Wind velocity is 9m/s.

Dimension of the plate is 2.5 m×5 m.

Calculation:

Obtain the dynamic viscosity of air as 1.849×10−5kg/m⋅s.

Calculate the density of air.

ρ=PRT=83.4 kPa(0.287kPa⋅m3/kg⋅K)(298 K)=0.9751kg/m3

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×10−5kg/m⋅s)=2.373×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(2.373×106)0.2−1742(2.373×106)=0.003194

The drag force is,

FD=CfAρV22=(0.003194)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.58 N

Thus, the drag force is 1.58 N.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The drag force.

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

Answer 13

(b)

To determine

The drag force.

Answer 14

(b)

Expert Solution

Explanation of Solution

Calculation:

The Reynolds number is,

Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×10−5kg/m⋅s)=1.187×106

Thus the flow is turbulent. Hence the friction coefficient is given by,

Cf=0.074Re0.2−1742Re=0.074(1.187×106)0.2−1742(1.187×106)=0.003044

The drag force is,

FD=CfAρV22=(0.003044)(2.5 m×5 m)(0.9751kg/m3)(9m/s)22=1.50 N

Thus, the drag force is 1.50 N.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 15 - Prob. 1P Ch. 15 - Define the frontal area of a body subjected to...Ch. 15 - Prob. 3P Ch. 15 - Prob. 4P Ch. 15 - Prob. 5P Ch. 15 - Prob. 6P Ch. 15 - Name some applications in which a large drag is...Ch. 15 - Prob. 8P Ch. 15 - Prob. 9P Ch. 15 - Prob. 10P

The drag force.

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The drag force.

Explanation of Solution

The drag force.

Explanation of Solution

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