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ECH 141Problem Set #61.A solid piston with radius D0is pushed by a horizontal force F into a cylindrical cavity with diameterD1as shown below.Use the macroscopic momentum balance to derive an expression for the relationbetween the force F and the piston velocity u0if u0is constant.Except for the space occupied by thepiston, the entire cylindrical cavity as well as the region outside the cavity contains an incompressibleliquid.
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Start with a mass balance (see also Deen Example 11.2-1).The macroscopic mass balance isddtρdVVa(t)∫=−ρnAa(t)∫⋅v−w()dA.Use the control volume as shown above by the dashed line, with all boundaries fixed except for the partalong the right side of the piston.The outer edges, on the top, bottom and far right, can be arbitrarily farfrom the piston.The density is constant, and the rate of change of the volume equals the rate at whichthe piston enters the cylinder, i.e.ddtρdVVa(t)∫=π4ρD02u0.The only place where the right side of the mass balance is non-zero is at the thin annulus where the fluidexits the cylinder.The velocity of the piston is known, so substituting gives (remember the normalvector points out of the control volume)π4D02u0=vπ4D02⎛⎝⎜⎞⎠⎟D12D02−1⎛⎝⎜⎞⎠⎟.We see that the average fluid velocity exiting the annulus isv=u0D12D02−1⎛⎝⎜⎞⎠⎟.In the text, the same result is derived using a different control volume.Themacroscopic momentum balanceis given byddtρvdVVa(t)∫+ρvv−w()⋅ndAAa(t)∫=ρgdVVa(t)∫+n⋅TdAAa(t)∫.Note that, if the diameter of the cylinder is similar in magnitude to the diameter of the piston, then thecontrol volume used in Fig. 11.2 of Deen makes it difficult to extract the applied force, because thestresses on the cylinder wall are not likely to be negligible relative to the stresses on the piston.Thecontrol volume shown here, which is outside the cylinder, appears to be a better choice for the forcecalculation on the piston.We assume quasi-static conditions, so the time derivative term is negligible.Since we are interested inthe x-component of the force, dot with ex.Note also that v=weverywhere except at the opening of theannulus, then
ρvxv⋅ndAAannulus∫=ex⋅n⋅TdAAa(t)∫The surface integral on the right side is over the entire surface of the control volume.However, on thesurfaces other than the hydraulic ram, and the fluid exiting the annulus, it is reasonable to assume thatthere is little flow and hence negligible stress.In the fluid exiting the annulus, we furthermore neglectviscous stresses and assume that the exit pressure is the ambient pressure outside the cylinder.Then thebalance becomesρvxv⋅ndAAannulus∫=ex⋅n⋅TdAAram∫Since the normal vector here points out of the control volume, at the ram surface it points into the ram.The force being exerted on the ram is therefore, in terms of the normal nto the control volume,Fx=−ex⋅n⋅TdAAram∫