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Elementary Mechanics of FluidsSome Simple FlowsSlide 3Fluid DeformationSlide 5Shear in Different FluidsViscosityFlow between 2 platesSlide 9Slide 10Example: Journal BearingExample: cont.Example: Rotating DiskElementary Mechanics of FluidsCE 319 FDaene McKinneyViscositySome Simple Flows•Flow between a fixed and a moving plateFluid in contact with the plate has the same velocity as the plateu = x-direction component of velocityu=VMoving plateFixed plateyxVu=0ByBVyu )(FluidSome Simple Flows•Flow through a long, straight pipeFluid in contact with the pipe wall has the same velocity as the wallu = x-direction component of velocityrxR21)(RrVruVFluidFluid Deformation•Flow between a fixed and a moving plate•Force causes plate to move with velocity V and the fluid deforms continuously. u=VMoving plateFixed plateyxu=0Fluid t0t1t2u=V+VMoving plateFixed plateyxu=VFluid Fluid Deformationtt+txyLtShear stress on the plate is proportional to deformation rate of the fluidVLtyLyVtyVShear in Different Fluids•Shear-stress relations for different typesof fluids•Newtonian fluids: linear relationship•Slope of line (coefficient of proportionality) is “viscosity”dydVdydVViscosity•Newton’s Law of Viscosity•Viscosity•Units•Water (@ 20oC)–  = 1x10-3 N-s/m2•Air (@ 20oC)–  = 1.8x10-5 N-s/m2•Kinematic viscositydydV /22///msNmsmmN V V+dv dydVFlow between 2 platesu=VMoving plateFixed plateyxVu=0ByBVyu )(Fluid Force acting ON the plate2121222111AAFAAF2211dydudyduThus, slope of velocity profile is constant and velocity profile is a st. lineForce is same on topand bottomFlow between 2 platesu=VMoving plateFixed plateyxVu=0ByBVyu )(BVdyduShear stress anywherebetween platesShearon fluidmBsmVCSAEmsNo02.0/3)38@30(/1.0222/15)02.0/3)(/1.0(mNmsmmsNFlow between 2 plates•2 different coordinate systemsrxB21)(BrVruVyx  yByCyu )(Example: Journal Bearing•Given–Rotation rate,  = 1500 rpm–d = 6 cm–l = 40 cm–D = 6.02 cm–SGoil = 0.88– oil = 0.003 m2/s•Find: Torque and Power required to turn the bearing at the indicated speed.Example: cont.•Assume: Linear velocity profile in oil film2/1242/)0002.0()2/06.0(1500*602)003.0*998*88.0(2/)()2/( StressShearmkNdDddydVmNdldM281206.0)4.0*206.0*000,124*2(2)22(TorquekWsmNMP 1.44/100,441.157*281Power Example: Rotating Disk•Assume linear velocity profile: dV/dy=V/y=r/y•Find shear


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UT CE 319F - Viscosity

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