Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 17D (Fluid Dynamics)

For a steady inviscid incompressible fluid of constant density in uniform gravity $\mathbf{g}=-g{\mathbf{e}}_z$, the Euler equation for fluid motion and continuity equation areTaking the scalar product with $\mathbf{u}$ givesThus Bernoulli's quantity is constant along each streamline. For an irrotational connected flow it is constant throughout the flow; constancy across different streamlines is not assumed in general.

Since incompressibility makes $\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=\rho (\mathbf{u}\cdot \nabla )\mathbf{u}$, integrating over a fixed control volume and applying the divergence theorem yields the integral momentum balanceHere $\mathbf{n}$ is outward and the left side is the net outgoing momentum flux. Uniform atmospheric pressure has zero resultant on a closed control surface, so it may be subtracted from all pressures in the jet calculations.

The free jet is at atmospheric pressure. Bernoulli's equation from the nozzle to height $z$ gives $U(z{)}^2/2+gz=U_0^2/2$. Conservation of volume flux gives $A(z)U(z)=A_0{U}_0$. HenceFor the prescribed well-below-apex regime, $U(z)=U_0[1-gz/U_0^2+O((gz/U_0^2{)}^2)]$ and $A(z)=A_0[1+gz/U_0^2+O((gz/U_0^2{)}^2)]$. The slowing jet therefore thickens. These expressions use the slender-jet, nearly uniform cross-sectional speed approximation underlying the question; the formal vanishing-speed apex is outside that approximation.

Choose a control volume containing the small impingement region, with an inlet section across the incident jet and an outlet annulus across the conical sheet. The remaining boundary follows free surfaces and the wetted cone. Subtract atmospheric pressure and neglect gravity. Bernoulli's equation gives the same speed $U$ in the outgoing sheet as in the incoming jet, and its total volume flux is $AU$.

The incoming axial momentum rate is $\rho A{U}^2$. The outgoing velocity makes angle $\alpha$ with the cone axis, so its axial momentum rate is $\rho A{U}^2\cos \alpha$; transverse components cancel by axisymmetry. The integral momentum equation says that the cone's axial force on the water is their difference, $-\rho A{U}^2(1-\cos \alpha )$. By equal and opposite reaction, the jet's force on the cone is along the incident jet, with magnitudeThe outlet control surface is across the sheet, not along it, so its entire outgoing volume flux is counted.

At the cone's vertex height $h$, the first part gives $U(h)=U_0\sqrt{1-2gh/U_0^2}$ and $A(h)U(h)=A_0{U}_0$. Insert these into the assumed local impingement force from part (ii). Static support requires upward force equal to $Mg$, givingEquivalently,For a nonnegative support height the ratio in parentheses must not exceed one, and the prescribed $h\ll U_0^2/(2g)$ requires that ratio to be close to one. If the weight exceeds the lift available at the nozzle, this model has no supported height above the nozzle.