Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 15, Problem 45P

(a)

To determine

The drag force.

(a)

Expert Solution
Check Mark

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 1.849×105kg/ms.

Calculate the density of air.

  ρ=PRT=83.4kPa(0.287kPam3/kgK)(298 K)=0.9751kg/m3

The Reynolds number is,

  Re=ρVLμ=(0.9751kg/m3)(9m/s)(5 m)(1.849×105kg/ms)=2.373×106

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

  Cf=0.074Re0.21742Re=0.074(2.373×106)0.21742(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
Check Mark

Explanation of Solution

Calculation:

The Reynolds number is,

  Re=ρVLμ=(0.9751kg/m3)(9m/s)(2.5 m)(1.849×105kg/ms)=1.187×106

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

  Cf=0.074Re0.21742Re=0.074(1.187×106)0.21742(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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Chapter 15 Solutions

Fundamentals of Thermal-Fluid Sciences

Ch. 15 - Prob. 11PCh. 15 - What is the difference between skin friction drag...Ch. 15 - Prob. 13PCh. 15 - What is the effect of streamlining on (a) friction...Ch. 15 - What is flow separation? What causes it? What is...Ch. 15 - Prob. 16PCh. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - The resultant of the pressure and wall shear...Ch. 15 - Prob. 20PCh. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 24PCh. 15 - Prob. 25PCh. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - A submarine can be treated as an ellipsoid with a...Ch. 15 - Prob. 29PCh. 15 - During major windstorms, high vehicles such as RVs...Ch. 15 - Prob. 31PCh. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - Prob. 34PCh. 15 - Prob. 35PCh. 15 - Prob. 36PCh. 15 - Prob. 37PCh. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Prob. 40PCh. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - Prob. 45PCh. 15 - Prob. 46PCh. 15 - Prob. 48PCh. 15 - Prob. 49PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Prob. 54PCh. 15 - Prob. 55PCh. 15 - Prob. 56PCh. 15 - Prob. 57PCh. 15 - Prob. 58PCh. 15 - Prob. 59PCh. 15 - Prob. 60PCh. 15 - Prob. 61PCh. 15 - Prob. 62PCh. 15 - A 2-m-long, 0.2-m-diameter cylindrical pine log...Ch. 15 - Prob. 64PCh. 15 - Prob. 65PCh. 15 - Prob. 66PCh. 15 - Prob. 67PCh. 15 - Prob. 68PCh. 15 - Prob. 69PCh. 15 - Prob. 70PCh. 15 - Prob. 71PCh. 15 - Prob. 72PCh. 15 - What is induced drag on wings? Can induced drag be...Ch. 15 - Prob. 74PCh. 15 - Prob. 75PCh. 15 - Prob. 76PCh. 15 - Prob. 77PCh. 15 - Consider an airplane whose takeoff speed is 220...Ch. 15 - Prob. 79PCh. 15 - Prob. 80PCh. 15 - Prob. 82PCh. 15 - Prob. 83PCh. 15 - Prob. 84PCh. 15 - Prob. 85PCh. 15 - Prob. 86RQCh. 15 - A 1.2-m-external-diameter spherical tank is...Ch. 15 - Prob. 88RQCh. 15 - Prob. 89RQCh. 15 - Prob. 91RQCh. 15 - Prob. 92RQCh. 15 - Prob. 93RQCh. 15 - Prob. 94RQCh. 15 - Prob. 95RQCh. 15 - Prob. 96RQCh. 15 - Prob. 97RQCh. 15 - Prob. 98RQCh. 15 - Prob. 99RQCh. 15 - Prob. 100RQCh. 15 - Prob. 102RQ
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