Advanced Fluid Mechanics Problems And Solutions __link__ <HOT - METHOD>

vr=𝜕ϕ𝜕r=1r𝜕ψ𝜕θ=m2πrv sub r equals partial phi over partial r end-fraction equals 1 over r end-fraction partial psi over partial theta end-fraction equals the fraction with numerator m and denominator 2 pi r end-fraction

u(y,t)=U0[1−erf(y2νt)]=U0erfc(y2νt)u open paren y comma t close paren equals cap U sub 0 open bracket 1 minus erf open paren the fraction with numerator y and denominator 2 the square root of nu t end-root end-fraction close paren close bracket equals cap U sub 0 space erfc open paren the fraction with numerator y and denominator 2 the square root of nu t end-root end-fraction close paren is the complementary error function. advanced fluid mechanics problems and solutions

ρ(𝜕u𝜕t+u𝜕u𝜕x+v𝜕u𝜕y)=−𝜕p𝜕x+μ(𝜕2u𝜕x2+𝜕2u𝜕y2)rho open paren partial u over partial t end-fraction plus u partial u over partial x end-fraction plus v partial u over partial y end-fraction close paren equals negative partial p over partial x end-fraction plus mu open paren partial squared u over partial x squared end-fraction plus partial squared u over partial y squared end-fraction close paren Step 1: Establish Boundary Layer Equations

M22=2+0.4(4)2.8(4)−0.4=2+1.611.2−0.4=3.610.8=13cap M sub 2 squared equals the fraction with numerator 2 plus 0.4 open paren 4 close paren and denominator 2.8 open paren 4 close paren minus 0.4 end-fraction equals the fraction with numerator 2 plus 1.6 and denominator 11.2 minus 0.4 end-fraction equals 3.6 over 10.8 end-fraction equals one-third advanced fluid mechanics problems and solutions

η=yU∞νxeta equals y the square root of the fraction with numerator cap U sub infinity end-sub and denominator nu x end-fraction end-root

over a thin flat plate aligned with the flow. Derive the using the boundary layer scaling parameters and state the relevant boundary conditions. Step 1: Establish Boundary Layer Equations