The flow rate per unit width is $Q = \int_0^B u(y) dy$. $$ Q = \int_0^B \left[ \fracU yB + \frac12\mu \fracdPdx (By - y^2) \right] dy $$ $$ Q = \fracU B2 + \frac12\mu \fracdPdx \left[ \fracB y^22 - \fracy^33 \right]_0^B $$ $$ Q = \fracUB2 + \frac12\mu \fracdPdx \left( \fracB^32 - \fracB^33 \right) $$ $$ Q = \fracUB2 + \fracB^312\mu \fracdPdx $$ Alterotic200713karmarxblackmailedtosubm Extra Quality Apr 2026
Substitute $\theta$ and $\tau_w$ into the momentum equation: $$ \fracddx \left( \frac215 \delta \right) = \frac2 \mu U_\infty\delta \rho U_\infty^2 $$ $$ \frac215 \fracd\deltadx = \frac2 \nu\delta U_\infty $$ $$ \delta , d\delta = \frac15 \nuU_\infty dx $$ Jack Reacher Never Go Back Hindi Dubbed Filmyzilla Verified
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: $$ \tau_w(x) \approx 0.365 \rho U_\infty^2 \sqrt\frac\nuU_\infty x = 0.365 \rho U_\infty^2 Re_x^-1/2 $$