= Solution
In a stably stratified ambient, entrained fluid is progressively lighter with increasing height. Relative to that ambient, the plume's <buoyancy flux> satisfies $B'=-N^2Q<0$. It therefore becomes neutrally buoyant before its upward <momentum> vanishes. It overshoots, becomes negatively buoyant, and eventually turns and spreads laterally near its neutral-buoyancy level. The top-hat ascent equations cease to describe the overturning cap and lateral intrusion.
For constant $N$, the balances give a useful exact relation:
$$
(M^2)'=2BQ,\qquad (B^2/N^2)'=-2BQ,\qquad
M^2+B^2/N^2=B_0^2/N^2
$$
for an ideal zero-momentum source. The rising branch begins at $B=B_0$, attains its largest momentum at $B=0$, and reaches its terminal zero-momentum limit at $B=-B_0$. Entrainment and decreasing buoyancy make the associated rise interval finite.
An estimate follows already from the unstratified <pure plume> flux: $N^2\int_0^H Q\,dz\sim B_0$, so $N^2B_0^{1/3}H^{8/3}\sim B_0$. Hence
$$
\boxed{H=C_H(\alpha)\left(\frac{B_0}{N^3}\right)^{1/4}.}
$$
The coefficient depends on the <entrainment coefficient> and the distinction between neutral height, maximum rise and final spreading level. The scaling can also be obtained from the dimensions $[B_0]=L^4T^{-3}$ and $[N]=T^{-1}$. The sketch distinguishes the neutral level from the maximum rise.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-43-stratified-plume.png]
{title=Entrainment, loss of buoyancy, momentum overshoot and lateral intrusion of a stratified plume}
Back to article page
