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 100RQ
To determine

The error involved in each case and the accuracy of Stokes law.

Expert Solution & Answer
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Explanation of Solution

Given:

The density of aluminum balls is, ρs=2600 kg/m3.

Temperature of glycerin is, T=22°C.

The density of glycerin is, ρf=1274 kg/m3.

The viscosity of glycerin is, μ=1 kg/ms.

The diameter of the aluminum balls are, D=2,4,10 mm.

The terminal velocities are, Vexp=3.2,12.8,60.4mm/s.

Calculation:

From the force balance,

  FD=WFB3πμVD+9π16ρsV2D2=ρsgVρfgV3πμVD+9π16ρsV2D2=(ρsρf)gπD36V=3πμD+(3πμD)24(9π16ρsD2)[(ρsρf)gπ6D3]2(9π16ρsD2)

Case 1: D=2 mm and Vexp=3.2mm/s.

The terminal velocity is,

  V=3πμD+(3πμD)24(9π16ρsD2)[(ρsρf)gπ6D3]2(9π16ρsD2)=3π(1 kg/ms)(0.002 m)+[3π(1 kg/ms)(0.002 m)]2{4[9π16(2600 kg/m3)(0.002 m)2][(2600 kg/m31274 kg/m3)×(9.81 m/s2)π6(0.002 m)3]}2[9π16(2600 kg/m3)(0.002 m)2]=0.00310 m/s=3.10m/s

The error in terminal velocity is,

  Error=VexpVVexp×100%=3.23.103.2×100%=3.1%

Thus, the error in the terminal velocity is 3.1%.

The Reynolds number is,

  Re=ρfVDμ=(1274 kg/m3)(0.0032m/s)(0.002m)1 kg/ms=0.0081

Case 2: D=4 mm and Vexp=12.8mm/s.

The terminal velocity is,

  V=3πμD+(3πμD)24(9π16ρsD2)[(ρsρf)gπ6D3]2(9π16ρsD2)=3π(1 kg/ms)(0.004 m)+[3π(1 kg/ms)(0.004 m)]2{4[9π16(2600 kg/m3)(0.004 m)2][(2600 kg/m31274 kg/m3)×(9.81 m/s2)π6(0.004 m)3]}2[9π16(2600 kg/m3)(0.004 m)2]=0.0121 m/s=12.1m/s

The error in terminal velocity is,

  Error=VexpVVexp×100%=12.812.112.8×100%=5.5%

Thus, the error in the terminal velocity is 5.5%.

The Reynolds number is,

  Re=ρfVDμ=(1274 kg/m3)(0.0128m/s)(0.004m)1 kg/ms=0.0651

Case 3: D=10 mm and Vexp=60.4mm/s.

The terminal velocity is,

  V=3πμD+(3πμD)24(9π16ρsD2)[(ρsρf)gπ6D3]2(9π16ρsD2)=3π(1 kg/ms)(0.010 m)+[3π(1 kg/ms)(0.010 m)]2{4[9π16(2600 kg/m3)(0.010 m)2][(2600 kg/m31274 kg/m3)×(9.81 m/s2)π6(0.010 m)3]}2[9π16(2600 kg/m3)(0.010 m)2]=0.0597 m/s=59.7m/s

The error in terminal velocity is,

  Error=VexpVVexp×100%=60.459.760.4×100%=1.2%

Thus, the error in the terminal velocity is 1.2%.

The Reynolds number is,

  Re=ρfVDμ=(1274 kg/m3)(0.0604m/s)(0.010m)1 kg/ms=0.770<1

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Fundamentals of Thermal-Fluid Sciences

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