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by: Jonah Leary

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# Handouts ME 3350 - 01

Jonah Leary
WSU
GPA 3.284

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The main information from some of the handouts compiled into one 3-page document. I did not have time to include information from all of the handouts, so I chose the handouts that relate most to t...
COURSE
Fluid Dynamics
PROF.
Philippe Sucosky
TYPE
Study Guide
PAGES
6
WORDS
KARMA
50 ?

## Popular in Mechanical and Materials Engineering

This 6 page Study Guide was uploaded by Jonah Leary on Thursday October 6, 2016. The Study Guide belongs to ME 3350 - 01 at Wright State University taught by Philippe Sucosky in Fall 2016. Since its upload, it has received 81 views. For similar materials see Fluid Dynamics in Mechanical and Materials Engineering at Wright State University.

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Date Created: 10/06/16
Handouts 1.2 Measures of Fluid Mass and Weight m kg 3 ρ= [ 3] v= = V [ ] Density: V m Specific Volume: ρ m kg Specific N Weight: γ=ρg[ 3] m SG= ρ Specific Gravity: ρwater @4°Cunitless] 1.3 Measures of Fluid Compressibility −dP dP E v = [Pa] Bulk Modulus: dV dρ Isothermal Compressibility: V ρ 1 1 α= [ ] Ev Pa ∆V − ∆ρ V ρ 1 Coefficient of Volume Expansion: β= = [ ] ∆T ∆T K 1.4 Viscosity and Fluid Deformation τ= F Shear stress acting on fluid layer: A Boundary Conditions: y=0→u=0 y=l→u=V V Velocity Profile :(y)= y l ' dβ=tan dβ = BB =Vδt Shear Strain: l l dβ V du Shearing Rate: dt=γ= l= dy τ∝γ→τ=μ du Relation between shearing rate and shear stress: (dy µ=dynamic viscosity 2.3 Hydrostatic Force on Plane Surface Procedure to calculate resultant force: 1. dp 1. Determine pressure distribution in fluid =±ρg dz 2. Find resultant force by integration over surface dF=pdA F R ∫ p∗dA surface Procedure to calculate coordinates of center of pressure: 1. Choose point to use as origin 2. Calculate moment exerted by resultant pressure force about origin M 1F ×R c 3. Calculate moment resulting from all moments produced by infinitesimal forces along surface of submerged body d M 2dF ×l=pdAl M = pdAl therefore: 2 ∫ surface 4. Solve for c by setting equality M 1M 2 2.5 Hydrostatic Force on Curved Suface dF=−pdA Pressure acting on infinitesimal surface element dA is Resultant force calculated by integrating over gate area: FR=− p∫A A Decompose into x- and z- components: F RF i+Rxk Rz To find component, take dot product of force with unit vector: FR=F ∙iR−^ ∫ pdA∙i=− pd∫A x x A A F =− pdz If gate has unit width perpendicular to page, then R x ∫ A Now, find y-component: FR=− ∫ pd Az=− ∫dx z A A dp By definition: dz=−ρg , therefore: gatesurface p=− ∫ ρgdz freesurface Therefore: gatesurface x22 FR=− ∫ ∫ ρgdz dx=− ∫∫ ρgdzdx z A (freesurface) x11 where x1, 2 ,1z , and2z are limits of fluid domain above curved surface. When curved surface is below liquid, vertical resultant force is equal to vertical hydrostatic force plus weight of fluid block. When curved surface is above liquid, vertical resultant force is equal to vertical hydrostatic force minus weight of fluid block. 3.1 CV Analysis – Conversion of Mass Determine ρ2 as a function of ρ1 , device geometry, and velocities at inlet and outlet. 1. Choose CV to be fluid contained within device δ ∫ ρdV+ ∮ ρv∙dA=0 2. General expression of governing equation: δt CV CS δ 3. Assumptions: Steady flow ( =0 ), uniform flow and fluid properties across δt inlet/outlet, two surfaces crossed by flow (inlet 1 and outlet 2) 4. Reduced governing equation: ∫ ρ1v1∙d A1+ ∫ v2∙d2A =02 1 2 dA=n ^ dA 5. Study of fluxes across CS: Only across inlet and outlet. Convention: velocity directed into CV represents negative contribution to efflux, dA outward the CV Flux across inlet: v1=v 1^ d A 1d A ^1n=−d A i1 ^ v ∙d A =−v d A 1 1 1 1 Flux across exit: v2=v 2^ d A 1d A n2^=d A i2^ v ∙d A =v d A 2 2 2 2 6. Solution to governing equation: −ρ 1 1∫ dA 1ρ v2 2∫ dA 20→ρ v A2=2 v2A 1 1 1 1 2 3.3 Linear Momentum Balance Determine force exerted by fluid on elbow. 1. Choose CV to be fluid within elbow 2. Free-body diagram 3. General expression for governing ρvdV + ρv∙dA=¿ F ∮ ∑ CV CS equation: δ ∫ ¿ δt CV 4. P 2 Assumptions: Steady W flow ( P1 δ =0 ), δt incompressible: ρ =constant 5. Reduced governing equation: ∫ ρ v v ∙d A + ρ∫v v ∙d A = ∑ F 1 1 1 1 1 2 2 2 2 2 CV R y 6. Study of fluxes across CS: Rx Inlet: v1 1∙d A 1−v iv1d 1 1 ^ ^ Exit: v2v2∙d A2=−v ^2nv 2 A 2v (2osθi+sinθ j)v d A2 2 7. Solution to governing equations: ∫ −ρv d A + ρv ∫osθd A =P A −P A cosθ+R x: 1 1 2 2 1 1 2 2 x 1 2 2 y: ∫ ρv 2osθd A =R 2W−P y sinθ 2 2 2 2 2 2 −ρv A1+ρ1 cosθ2A =P A −2 A c1sθ1R 2 2 x ρv c2sθA =R −2−P Aysinθ 2 2 Apply conservation of mass: ∫ ρ1v 1d A 1− ρ ∫ ∙d2A2 2 1 2 A 1 A 1 P A −P A cosθ+R =ρv A 2 −1 R −W−P A sinθ=ρv A 2 sinθ 1 1 2 2 x 1 1(A)2 y 2 2 1 1(A)2 Solving for R axd R givey force exerted by elbow on fluid. Force on elbow by flow is negative of this.

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