Capillary tensile force of a slender thread ($T=3\pi \mu a^2{w}_z+\pi \gamma a$)

For an axisymmetric slender viscous thread, the Young–Laplace equation and radial extensional viscous stress give $p=p_e+\gamma /a-\mu w_z$. The axial liquid stress relative to ambient pressure is $3\mu w_z-\gamma /a$. Its cross-sectional integral plus the circumferential surface tension force $2\pi \gamma a$ gives the displayed tensile force. Thus the net capillary contribution is $\pi \gamma a$, and axial force balance with downward gravity is $T_z+\pi \rho g{a}^2=0$.

For a hollow slender viscous thread, $A=R_e^2-R_i^2$ and both surfaces contribute axial surface tension. The liquid cut force relative to exterior pressure, including these surface terms, is $\pi [3\mu A{w}_z+(p_i-p_e)R_i^2+\gamma (R_e+R_i)]$. The axial traction on the sloping inner gas surface contributes $-\pi (p_i-p_e)(R_i^2{)}_z$ to force balance. For constant gas pressures the extra derivative cancels, giving $3\mu (A{w}_z{)}_z+\rho gA+\gamma (R_e+R_i{)}_z=0$. The displayed $T$ is this effective tension after accounting for gas-pressure work, not the unadjusted liquid cut force.

For steady equal-pressure flow, mass conservation gives $Aw=A_0{w}_0$ and the kinematic boundary condition gives $(D\sqrt{w}{)}_z=\gamma /(2\mu \sqrt{w})$, where $D=R_e-R_i$. Therefore finite-height closure occurs exactly whenFor positive velocity the integral is increasing. If it diverges, closure is finite; if its limiting value lies below the right-hand threshold, there is no finite closure. Quadratic freely falling profiles have logarithmically divergent integrals, while sufficiently strongly drawn exponentially accelerating profiles can have finite integrals. Thus nozzle speed alone is insufficient to guarantee closure: the axial tension or downstream condition matters.