The no slip effect causes the fluid near the surface of the plate to be

The no slip effect causes the fluid near the surface

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The no-slip effect causes the fluid near the surface of the plate to be subjected to largeshear forces. These shear forces lead to a local rotation of the fluid particles, known asvorticity of the fluid field. It turns out that this vorticity is advected by the upward flowaway from the plate. At this point in the course, we do not yet have the mathematicaltools needed to understand and verify this effect, so we will just accept it as an experi-mental fact. The part of the flow field above the plate where there are strong shear stressin the fluid is known asthe boundary layer.
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fluid mechanics chapter 4 lecture notesCopyright©Wayne Hacker 2015. All rights reserved.47The fluid far from the plate will barely be affected by the vorticity shedding off of theplate. The shear forces outside the boundary layer in the flow field are typically non-existent, or at least small enough that they can be ignored. We refer to this region asthe free-flow field. This vorticity-free region surrounding the plate is typically classifiedas a region of irrotational flow (no vorticity).The boundary between the boundary layer and the free-stream is known as the edgeof boundary layer, and is mathematically denoted byy=δ(x), wherexrepresents thedistance downstream of the leading edge (see figure below).
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fluid mechanics chapter 4 lecture notesCopyright©Wayne Hacker 2015. All rights reserved.48
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fluid mechanics chapter 4 lecture notesCopyright©Wayne Hacker 2015. All rights reserved.49
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fluid mechanics chapter 4 lecture notesCopyright©Wayne Hacker 2015. All rights reserved.50
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fluid mechanics chapter 4 lecture notesCopyright©Wayne Hacker 2015. All rights reserved.51Example 4(A time-dependent problem).A rigid metal air tank of volumeV-= 1 m3,shown in figure 9 below, contains high-pressure air. The initial pressure, temperature, anddensity of the air in the tank areP0= 8 kPa (absolute),T0= 15C, andρ0= 6 kg/m3.At timet= 0, a valve is partially opened and air begins escaping out of the exit pipe withcross-sectional flow area of 1 cm2with a speed of 200 m/s. Determine the instantaneousrate of change of density in the tank att= 0.Figure 9: Diagram of air escaping out the end of the exit pipe of a rigid tank with openvalve. The exit velocity~Veis assumed to be uniform.Solution:Let’s write down what we’re given, what we want, which equations we’llapply, and any assumptions.Given:V-= 1 m3(volume of tank fixed and the control volume)P0= 8kPa (absolute)(initial pressure)T0= 15C(initial temperature)ρ0= 6 kg/m3(initial density)Ae= 1 cm2= 10-4m2(exit velocity)Ve= 200 m/s(exit velocity)Want:dtAssumptions:1. The compressed air behaves like an ideal gas.2. The air exiting the pipe has a uniform velocity profile.We could just computethe average of the velocity profile, and replace the actual velocity profile by theaverage, uniform, velocity profile.
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fluid mechanics chapter 4 lecture notesCopyright©Wayne Hacker 2015. All rights reserved.
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