−U∞22xff′′=U∞2xf′′′negative the fraction with numerator cap U sub infinity end-sub squared and denominator 2 x end-fraction f f double prime equals the fraction with numerator cap U sub infinity end-sub squared and denominator x end-fraction f triple prime Divide both sides by
By definition, the velocity components using the stream function
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( u_\tau = \kappa y \fracdudy ).
In the 18th century, Jean le Rond d'Alembert used "ideal" fluid math to prove that an object moving through a fluid experiences . The Problem
Using Bernoulli's equation between the pipe (1) and the nozzle exit (2), assuming horizontal flow and negligible losses:
No stagnation points exist on the cylinder surface. The stagnation point lifts off the surface and moves into the fluid core ( Summary Reference Table Flow Domain Primary Governing Equation Core Solution Technique Key Physical Metrics Navier-Stokes Equations Exact Integration via Boundary Conditions Shear stress, Volumetric flow rate, Drag profiles Boundary Layer Prandtl Boundary Layer Equations Similarity Transformations (Blasius Method) Boundary layer thickness ( ), Skin friction coefficient Potential Flow Laplace Equation ( Linear Superposition of Flow Singularities Lift generation via Kutta-Joukowski, Pressure distributions advanced fluid mechanics problems and solutions
, rearrange the equation into a solvable ordinary differential equation (ODE):
u𝜕u𝜕x+v𝜕u𝜕y=ν𝜕2u𝜕y2(Momentum)u partial u over partial x end-fraction plus v partial u over partial y end-fraction equals nu partial squared u over partial y squared end-fraction space (Momentum)
u(y)=UyB+12μ(dPdx)(y2−By)u open paren y close paren equals the fraction with numerator cap U y and denominator cap B end-fraction plus the fraction with numerator 1 and denominator 2 mu end-fraction open paren the fraction with numerator d cap P and denominator d x end-fraction close paren open paren y squared minus cap B y close paren AI responses may include mistakes
An ideal, irrotational, incompressible fluid with uniform velocity U∞cap U sub infinity end-sub flows past a circular cylinder of radius . The cylinder experiences an added clockwise circulation Γcap gamma Construct the total velocity potential and stream function in polar coordinates.
(U∞f′)(−U∞2xηf′′)+[−12νU∞x(f−ηf′)](U∞U∞νxf′′)=ν(U∞2νxf′′′)open paren cap U sub infinity end-sub f prime close paren open paren negative the fraction with numerator cap U sub infinity end-sub and denominator 2 x end-fraction eta f double prime close paren plus open bracket negative one-half the square root of the fraction with numerator nu cap U sub infinity end-sub and denominator x end-fraction end-root open paren f minus eta f prime close paren close bracket open paren cap U sub infinity end-sub the square root of the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction end-root f double prime close paren equals nu open paren the fraction with numerator cap U sub infinity end-sub squared and denominator nu x end-fraction f triple prime close paren Step 4: Simplify to the Blasius Equation Expand the terms and cancel out the common factors:
𝜕2u𝜕y2=𝜕𝜕y(U∞f′′U∞νx)=U∞f′′′U∞νx=U∞2νxf′′′partial squared u over partial y squared end-fraction equals the fraction with numerator partial and denominator partial y end-fraction open paren cap U sub infinity end-sub f double prime the square root of the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction end-root close paren equals cap U sub infinity end-sub f triple prime the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction equals the fraction with numerator cap U sub infinity end-sub squared and denominator nu x end-fraction f triple prime Step 4: Substitute into Momentum Equation The Problem Using Bernoulli's equation between the pipe
u(y)=12μ(dPdx)y2+C1y+C2u open paren y close paren equals the fraction with numerator 1 and denominator 2 mu end-fraction open paren the fraction with numerator d cap P and denominator d x end-fraction close paren y squared plus cap C sub 1 y plus cap C sub 2 Applying boundary conditions yields:
1 over r end-fraction d over d r end-fraction open paren r d u over d r end-fraction close paren equals the fraction with numerator 1 and denominator mu end-fraction the fraction with numerator d cap P and denominator d x end-fraction Since the pressure gradient is constant, we write: