Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 38B (Fluid Dynamics II)

The outer fluid is at rest, so its pressure is independent of $x$ and $dP/dx=0$. Using incompressibility, the longitudinal boundary-layer equation becomes the conservative momentum equationIntegrating across the jet and using $u,u_y\to 0$ and $uv\to 0$ at both transverse infinities givesPut $\nu =\mu /\rho$. The momentum-flux scaling gives $U^2\delta \sim F/\rho$, while balancing longitudinal convection with transverse viscous diffusion gives $U^2/x\sim \nu U/{\delta }^2$. Eliminating $U$ or $\delta$ yields

Choose the precise width convention $U{\delta }^2=\nu x$ and take $U$ to be the centre-line velocity. For the self-similar solution with streamfunction $\psi =U\delta f(\eta )$, $\eta =y/\delta$,The $\eta f^{\prime }{f}^{\prime \prime }$ terms cancel in $u{u}_x+v{u}_y$, leaving $-U^2[(f^{\prime }{)}^2+f{f}^{\prime \prime }]/(3x)$. Since $u_{yy}=U{f}^{\prime \prime \prime }/{\delta }^2$, the dimensionless profile equation isFor a symmetric jet choose its streamfunction zero at the axis. The boundary and amplitude conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(0)=1$, and $f^{\prime }(\eta )\to 0$ as $\eta \to \pm \infty$, with $f$ bounded and $f^{\prime \prime }\to 0$. The momentum normalization isIt determines the dimensional constants in $U,\delta$; one should not independently impose $J=1$ after fixing both the centre-line and width conventions.

Indeed integrating the profile equation once and using its far-field conditions gives $f^{\prime \prime }+f{f}^{\prime }/3=0$. With the axis conditions, $f^{\prime }=1-f^2/6$, and thereforeConsequently $\delta =(\rho J{\nu }^2/F{)}^{1/3}{x}^{2/3}$ and $U=[F^2/({\rho }^2{J}^2\nu ){]}^{1/3}{x}^{-1/3}$. The transverse velocity describes entrainment; the profile need not have $f(\infty )=0$ when its axial velocity vanishes there.