 Solution

For a neutrally buoyant horizontal turbulent round jet, a segment's horizontal momentum flux changes only if an external horizontal force or ambient momentum input exists. Neither is present, soPositive buoyancy adds a vertical force, not a horizontal one. Therefore $d{M}_x/ds=0$ remains true for an inclined forced plume even when $F_0\ne 0$; in general $M_x=M_0\cos {\theta }_0$.

The net upward force on a segment is $g({\rho }_0-\rho )\pi b^{2}ds$. Since $F=g({\rho }_0-\rho )\pi b^{2}V$ and $V=|\mathbf{M}|/Q$, vertical momentum balance givesIn kinematic plume fluxes, the equations are $m_x^{\prime }=0$, $m_z^{\prime }=fq/m$, where $m=(m_x^2+m_z^2{)}^{1/2}$. The centreline geometry satisfiesFor $F_0>0$, $Q>0$ beyond the source, and an initially nonvertical turbulent round jet pointing right, $M_z$ increases and the centreline turns toward the vertical. The magnitude of momentum flux is not conserved once the tangent has a vertical component.