The friction drag force for the given condition and when the velocity is doubled.

Question

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Answer 1

Question

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.

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Answer 2

Question

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.

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Answer 3

Question

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.

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Answer 4

Question

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.

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Answer 5

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.

Answer 6

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.

Answer 7

Chapter 15, Problem 53P

To determine

The friction drag force for the given condition and when the velocity is doubled.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Pressure is 1 atm.

Temperature is 5°C.

Wind velocity is 55km/h.

Dimension of the wall is 10 m×4 m.

Calculation:

Refer Table A-22 “Properties of air at 1 atm pressure” from Appendix 1 and obtain the following properties of air:

Density=1.269kg/m3Viscosity=1.382×10−5m2/s

The Reynolds number is,

Re=VLυ=(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=11.05×106

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

Cf=0.074Re0.2−1742Re=0.074(11.05×106)0.2−1742(11.05×106)=0.002730

The drag force is,

FD=CfAρV22=(0.002730)(10 m×4 m)(1.269kg/m3)(55km/h×5m/s18km/h)22=16 N

Thus, the drag force is 16 N.

When the velocity is doubled:

The Reynolds number is,

Re=VLυ=2(55km/hkm/h×5m/s18km/h)(10 m)(1.382×10−5m2/s)=22.10×106

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

Cf=0.074Re0.2−1742Re=0.074(22.10×106)0.2−1742(22.10×106)=0.002435

The drag force is,

FD=CfAρV22=(0.002435)(10 m×4 m)(1.269kg/m3)(2×55km/h×5m/s18km/h)22=58 N

Thus, the drag force when the velocity is doubled is 58 N.

Treating the flow over the side wall as a flow over a flat plate is unrealistic because the separation bubble exists over most of the side panels of the house and the boundary layer equations does not hold good.