Uh≤h2μdpdx⟹dpdx≥2μUh2the fraction with numerator cap U and denominator h end-fraction is less than or equal to the fraction with numerator h and denominator 2 mu end-fraction d p over d x end-fraction ⟹ d p over d x end-fraction is greater than or equal to the fraction with numerator 2 mu cap U and denominator h squared end-fraction
Integrate (assuming $\delta=0$ at $x=0$): $$ \frac\delta^22 = \frac15 \nu xU_\infty $$ $$ \delta(x) = \sqrt\frac30 \nu xU_\infty = \frac5.48 x\sqrtRe_x $$
Below is a guide to solving some of the most critical advanced problems in the field, including the rigorous procedure for tackling the Navier-Stokes equations and turbulent flow . 1. The Exact Solution Procedure for Navier-Stokes
Determine the condition for instability at the interface of two parallel, inviscid, incompressible fluids moving at different velocities ( ) with densities ( advanced fluid mechanics problems and solutions
Differentiating the velocities to find the acceleration terms:
While there are only about 80 known exact analytical solutions to the Navier-Stokes equations (NSE), mastering the procedure to derive them is essential for any advanced student.
2μUh2the fraction with numerator 2 mu cap U and denominator h squared end-fraction 2. Boundary Layer Theory and Similarity Solutions Problem: Blasius Boundary Layer Flow Over a Flat Plate 2μUh2the fraction with numerator 2 mu cap U
ψ(r,θ)=f(r)sin2θpsi open paren r comma theta close paren equals f of r sine squared theta Substituting this into
This is solved numerically to find the wall shear stress ( \tau_w = \mu r f''(0) ). The value ( f''(0) \approx 1.312 ) is a universal constant.
Applying Newton's Second Law to a fluid control volume: $$ \rho \fracD\mathbfVDt = \sum \mathbfF $$ Where $\fracDDt$ is the material derivative. The forces are surface forces (pressure and viscous stresses) and body forces (gravity). Applying Newton's Second Law to a fluid control
M2=M2nsin(β−θ)=0.700sin(36.95∘−15∘)=0.700sin(21.95∘)=0.7000.3738≈1.873cap M sub 2 equals the fraction with numerator cap M sub 2 n end-sub and denominator sine open paren beta minus theta close paren end-fraction equals the fraction with numerator 0.700 and denominator sine open paren 36.95 raised to the composed with power minus 15 raised to the composed with power close paren end-fraction equals the fraction with numerator 0.700 and denominator sine open paren 21.95 raised to the composed with power close paren end-fraction equals 0.700 over 0.3738 end-fraction is approximately equal to 1.873 Step 5: Calculate Post-Shock Static Pressure ( Using the normal shock jump equation for pressure:
β≈36.95∘beta is approximately equal to 36.95 raised to the composed with power
A viscous jet impinges normally on an infinite flat plate. The external potential flow is ( u_e = a x ), ( w_e = -2a z ) (axisymmetric). Determine the exact velocity profile.
The flow is turned by the wall back to horizontal. The effective deflection for the reflected shock is ( \delta = 15^\circ ) again. The pre-shock Mach is ( M_2=2.26 ). Solve ( \theta-\beta-M ) again for ( M_2, \delta=15^\circ ): ( \beta_2 \approx 40.5^\circ ).