InterviewSolution
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InterviewSolution
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A tube of length 1 and radius R carries a steady flow of fluid whose density is `rho` and viscosity `eta`. The velocity v of flow is given by `v=v_(0)(1-r^(2)//R^(2))` Where r is the distance of flowing fluid from the axis.A. the volume of fluid flowing across the section. Of the tube, in unit time is `2piv_(0)((R^(2))/(4))`B. the kinetic energy of the fluid within the volume of the tube is `K.E=pirholv_(0)^(2)((R^(2))/(6))`C. the frictional force exerted on the tube by the fluid is `F=4pietakv_(0)`D. the pressure difference at the ends of tube is `P=(4etalv_(0))/(R^(2))` |
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Answer» Correct Answer - A::B::C::D The volume of fluid flowing through this section per second `dv=(2pirdr)v_(0)((1-r^(2))/(R^(2)))` Total volume `V=int_(0)^(R)(2pirdr)v_(0)((1-r^(2))/(R^(2)))` `=2piv_(0)((R^(2))/(4))` (ii). The kinetic energy of the fluid within the volme element of thickness dr K.E. of fluid within the tube is `=(1)/(2)(2pil)rhov_(0)^(2)int_(0)^(R)((1-r^(2))/(R^(2)))rdr` we get `K.E pirholv_(0)^(2)((R^(2))/(6))` (iii). The viscous drag exerts a force on the tube `F=-etaA((dv)/(dr))_(e=R)` here `((dv)/(dr))_(r=R)=v_(0)((-2r)/(R))_(r-R)=-2v_(0)//R` `:. F=4pietalv_(0)` (iv). `DeltaP=P_(2)-P_(1)=P` where `P_(1)=0` and `P_(2)=p` `P=("force"(F))/("area"(piR^(2)))=(4etalv_(0))/(R^(2))` |
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