Total drag force $F_D = \int_0^L \tau_w W , dx$. First, find $\tau_w(x)$ using our new $\delta(x)$: $$ \tau_w(x) = \frac2 \mu U_\infty\sqrt\frac30 \nu xU_\infty = \frac2 \mu U_\infty^3/2\sqrt30 \nu x \sqrt\fracU_\inftyU_\infty = \frac2 \rho \nu U_\infty\sqrt30 \nu x / U_\infty $$ Simplifying constants: $$ \tau_w(x) \approx 0.365 \rho U_\infty^2 \sqrt\frac\nuU_\infty x = 0.365 \rho U_\infty^2 Re_x^-1/2 $$
The core challenge in advanced fluid mechanics is the , which describe the motion of viscous fluids. While a general solution is one of the unsolved Millennium Prize Problems , exact solutions exist for specific "reduced" scenarios where non-linear terms cancel out. Problem: Combined Couette-Poiseuille Flow advanced fluid mechanics problems and solutions
Non-Newtonian and complex fluids
To find the relationship between average velocity $V$ and $u_max$, we integrate over the pipe area $A = \pi R^2$: $$ V = \frac1\pi R^2 \int_0^R u_max \left(1 - \fracrR\right)^1/7 (2 \pi r) dr $$ Let $y = 1 - r/R$, so $r = R(1-y)$ and $dr = -R dy$. $$ V = \frac2 \pi R^2 u_max\pi R^2 \int_0^1 y^1/7 (1-y) dy $$ $$ V = 2 u_max \left[ \fracy^8/78/7 - \fracy^15/715/7 \right] 0^1 $$ $$ V = 2 u max \left( \frac78 - \frac715 \right) = 2 u_max \left( \frac105 - 56120 \right) $$ $$ V = 2 u_max \left( \frac49120 \right) = u_max \left( \frac4960 \right) \approx 0.817 u_max $$ Total drag force $F_D = \int_0^L \tau_w W , dx$
Treat the thin annular clearance as flow between parallel plates (Plane Poiseuille Flow). The Result: The leakage rate is proportional to Find the velocity profile and total flow rate
For a power-law fluid: ( \tau_rz = K \left| \fracdudr \right|^n-1 \fracdudr ) (( n>0 )), laminar steady flow in a circular pipe of radius ( R ) driven by pressure gradient ( -\fracdpdz = G > 0 ). Find the velocity profile and total flow rate.