Download presentation

Presentation is loading. Please wait.

Published byJames Aduddell Modified over 2 years ago

1
Sphere Drag Force Jace Benoit Pamela Christian Amy Stephens February 11, 2007

2
Problem Statement SAE 10 oil at 20°C flows past an 8-cm- diameter sphere. At flow velocities 1, 2, and 3 m/s, the measured sphere drag forces are 1.5, 5.3, and 11.2 N, respectively. Estimate the drag force if the same sphere is tested at a velocity of 15 m/s in glycerin at 20°C.

3
Sketch 1 m/s 1.5 N 2 m/s 5.3 N 3 m/s 11.2 N 15 m/s ? in Glycerin

4
Assumptions Incompressible Flow Incompressible Flow Steady Flow Steady Flow Constant Temperature Constant Temperature Scaling Law is Valid Scaling Law is Valid

5
Scaling Law Scaling Laws “convert data from a cheap, small model to design information for an expensive, large prototype” (White 294). In this problem, oil can be used to model information for glycerin. When scaling laws apply, the Reynolds number and force coefficient for the model and prototype (oil and glycerin) are the same.

6
Solution The densities of the SAE 10 oil and glycerin are found in Table A.3 to be as follows: ρ oil =870 kg/m 3 ρ glycerin =1260 kg/m 3

7
Solution Knowing the density (ρ), length (L or diameter), velocity (V), and force (F) acting on the sphere, White (5.2) can be modified to determine the force coefficients (C F ) for each corresponding velocity and force in the oil. C F = F / ρV 2 L 2 (White 5.2)

8
Solution Substituting the known values into the previous equation yields the following force coefficients for the oil: C F1 = 1.5 N / (870 kg/m 3 )*(1 m/s) 2 *(0.08 m) 2 C F1 = 0.2694 C F2 = 5.3 N / (870 kg/m 3 )*(2 m/s) 2 *(0.08 m) 2 C F2 = 0.2380 C F3 = 11.2 N / (870 kg/m 3 )*(3 m/s) 2 *(0.08 m) 2 C F3 = 0.2235

9
Solution Averaging these three force coefficients gives a single value to compare to the unknown drag force in the glycerin. Since the scaling law applies, this averaged force coefficient can be substituted back into the same equation with some rearranging to yield the drag force for glycerin. F D = C F *ρ glycerin *V glycerin 2 *L 2 F D = (0.2436)*(1260 kg/m 3 )*(15 m/s) 2 *(0.08 m) 2 F D = 442.0 N

10
Biomedical Application Instead of finding a particular force for a given velocity through experimentation, the force on an object is found through scaling. Instead of finding a particular force for a given velocity through experimentation, the force on an object is found through scaling. So instead of testing 20 different velocities to find a force, we can test 3-4 velocities and develop a curve of the forces to determine any forces for that fluid. So instead of testing 20 different velocities to find a force, we can test 3-4 velocities and develop a curve of the forces to determine any forces for that fluid. This can be useful when designing implants such as knee replacement or stints. By knowing the velocity and force an implant can withstand, we can determine the size of the implant for maximum effectiveness. This can be useful when designing implants such as knee replacement or stints. By knowing the velocity and force an implant can withstand, we can determine the size of the implant for maximum effectiveness. Also using scaling, if an implant is found to have certain characteristics in a particular fluid, the characteristics of the implant in another fluid can be predicted. Also using scaling, if an implant is found to have certain characteristics in a particular fluid, the characteristics of the implant in another fluid can be predicted.

11
Reference White, Frank M. Fluid Mechanics. 5 th Ed. McGraw-Hill Companies, Inc.: New York. 2003.

Similar presentations

OK

Chapter 10 Approximate Solutions of the Navier-Stokes Equation

Chapter 10 Approximate Solutions of the Navier-Stokes Equation

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google