Fluids Reviewver. 1ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20091Areas of Interest:Areas of Interest:Mechanical Engineering• Mechanics•ThermodynamicsThermodynamics• Heat TransferFl id•FluidsME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20092Fluid Flow• Reynold’s number• Bernoulli’s equationq• ContinuityOsborne ReynoldsDaniel BernoulliME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20093Osborne Reynolds1842-1912Daniel Bernoulli1700-1782Reynold’s number(Re)Reynold s number (Re)ratio of momentum (inertia) to viscosityratio of momentum (inertia) to viscosityρVd()()()Re =μρVd()()()()viscositydiametervelocitydensity∗∗=()ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20094Critical Reynold’s NumberCritical Reynold s Number•Re~2 000Re 2,000– laminar to turbulent transitioneddies begin to form–eddies begin to form• Re > 20,000–very turbulentME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20095Reynold’s Number-Ex 1-1Reynold s Number -Ex. 1-1How fast would a stream of honey 1 inHow fast would a stream of honey 1 in. in diameter need to be to be turbulent?turbulent?• Density (ρ) = 1.43 g/cm3(at 20oC)• Viscosity (μ) = 189 poise (at 20.6oC)y(μ)p()ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20096Reynold’s Number - Ex. 1-2Re =Vdρ025401430∗∗Vμ9180254.01430∗∗=V9.18ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20097Reynold’s Number-Ex. 1-3Reynold s Number Ex. 13So, for turbulent flow transition,Re = 2 000 = 1 430*V*0 0254/18 9Re = 2,000 = 1,430 V 0.0254/18.9V = 1,040 m/s (Mach 3.1)(This ignores shear thinning.)ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20098Bernoulli’s EquationBernoulli s EquationUsed to calculate flow velocities:fghvPghvPo+++=++212ρρρρP = pressureg = gravityfghPghPoo+++=++1122ρρP = pressureg = gravityρ = density h = heightv=velocityf=losses due to frictionv velocityf losses due to frictionME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20099ContinuityContinuityAv=AvA0v0= A1v1where: A = areav = velocityME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200910Manufacturing - Ex. 1-1• You are pouring liquid iron into a mold.•The mold has a sprue height of 2 inches.The mold has a sprue height of 2 inches. • The bottom of the sprue has a diameter of 0.2 in.SprueRisersGateCopeDParting LineGateDragCastingME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200911Manufacturing - Ex. 1-2g• You wish to pour the metal so that you do not entrain airnot entrain air.•What should the diameter of the gate•What should the diameter of the gate (dgate) be?ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200912ManufacturingEx 13Manufacturing -Ex. 1-3•Here we need to use:•Here we need to use:– Reynold’s number values below 20 000 are OK in casting•values below 20,000 are OK in casting– Bernoulli’s equationContin it–ContinuityME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200913Manufacturing - Ex. 1-4• Iron data:density (ρ)= 7860 kg/m3–density (ρ)= 7860 kg/m3– viscosity at pouring temp (μ) = 2.25 cp =225x10-3N*s/m2= 2.25 x 103Ns/m2•h0= 2 in. = 0.051 m•h=0m•h1= 0 m• g = 9.8 m/s2A23 14 * 0 002542•A1 = πr2= 3.14 * 0.002542 =2.03 x 10-5m2ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200914Manufacturing - Ex. 1-5• Now we need to determine the velocity at the bottom of the sprue yp(v1) using Bernoulli’s equation.ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200915Manufacturing - Ex. 1-6• We can assume that the velocity at the top of the mold (v)iszeroat the top of the mold (vo) is zero, if there is a pouring basin, which is typicaltypical.ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200916ManufacturingEx 17Manufacturing -Ex. 1-7Ifitifft(f0)•Ignore friction effects (f=0).• Assume the mold is open to atmospheric pressure (P0=P1=Patm).ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200917Manufacturing - Ex. 1-8g• Substituting into Bernoulli’s equation:fghvPghvPooo+++=++1211222ρρρρ220510897860078602∗∗∗P008978607860051.08.97860221∗=∗∗++vPPatm008.97860278601+∗∗++vPatmME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200918ManufacturingEx 19Manufacturing -Ex. 1-9• And solving: v1= 1 m/s• Checking Reynolds numberRe = 7860*1*0.00508/2.25x10-3=17,746 < 20,000ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200919Manufacturing-Ex 1-10Manufacturing Ex. 110• Now using continuity:gyAv=Av=A1v1= Agatevgate=203 105*1 A2.03 x 10-5* 1 = AgatevgatedA2and Agate= πrgate2ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200920Manufacturing - Ex. 1-11g• Now, Reynold’s number < 20,0007860∗∗dvVdρ000,20 10 2.257860Re3-=∗∗∗==gategatedvVdμρ• Solving gives:vgate*dgate= 5.72 x 10-3m2/s ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200921Manufacturing - Ex. 1-12g• Combining the following equations:• 2.03 x 10-5* 1 = Agatevgategategate•Agate= πrgate2gategate•vgate*dgate= 5.72 x 10-3m2/sgategateME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200922MftiE113Manufacturing -Ex. 1-13•And solving gives:•And solving gives:dgate= 4.5 mm = 0.18 in.A not unreasonable answer, given gthe sprue is 5 mm in diameter. ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200923Summaryy• We have reviewed some basic fl idfluids.• The combination of the three equations discussed can be used to determine processing conditions.ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200924ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT