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A tube of length l and radius R carries a steady flow of fluid whose density is rho and viscosity eta. The fluid flow velocity depends on the distance r from the axis of the tube as v=v_0(1-r^2//R^2). Find: (a) the volume of the fluid flowing across the section of the tube per unit time: (b) the kinetic energy of the fluid within the tube's volume, (c) the friction force exerted on the tube by the fluid, (d) the pressure difference at the ends of the tube. |
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Answer» Solution :(a) LET `dV` be the volume flowing per second through the cylindrical SHELL of thickness `dr` then, `dV=-(2pirdr)v_0(1-(r^2)/(R^2))=2piv_0(r-(r^3)/(R^2))dr` and the total volume, `V=2piv_0underset(0)overset(R)int(r-r^3/R^2)dr=2piv_0(R^2)/(4)=pi/2R^2v_0`  (b) Let, `dE` be the KINETIC energy, within the above cylindrical shell. Then `dT=1/2(dm)v^2=1/2(2pirldrrho)v^2` `=1/2(2pilrho)rdrv_0^2(1-(r^2)/(R^2))=pilrhov_0[r-(2r^3)/(R^2)+(r^5)/(R^4)]dr` Hence, total energy of the fluid, `T=pilrhov_0^2underset(0)overset(R)int(r-(2r^3)/(R^2)+(r^5)/(R^4))dr=(piR^2rholv_0^2)/(6)` (c) Here frictional force is the shearing force on the tube, exerted by the fluid, which equals `-etaS(dv)/(dt)`. Given, `v=v_0(1-r^2/R^2)` So, `(dv)/(dt)=-2v_0(r)/(R^2)` And at `r=R`, `(dv)/(dt)=-(2v_0)/(R)` Then, viscous force is given by, `F=-eta(2piRl)((dv)/(dr))_(r=R)` `=-2piRetal(-(2v_0)/(R))=4pietav_0l` (d) Taking a cylindrical shell of thickness `dr` and radius r viscous force, `F=-eta(2pirl)(dv)/(dr)`, Let `Deltap` be the pressure difference, then net force on the element `=Deltappir^2+2pietalr(dv)/(dr)` But, since the flow is steady, `F_(n et)=0` or, `Deltap=(-2pil etar(dv)/(dr))/(pir^2)=(-2pil etar(-2v_0(r)/(R^2)))/(pir^2)=4etav_0l//R^2`
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