Outer ligament-mediated spray formation of annular liquid sheet by an inner round air stream

Abstract

The interfacial jetting phenomena of coaxial air-assisted water jets are studied using high-speed digital camera. Here, an inner round air jet is injected into annular water sheet spray. The experimental photographs show that the outside interface of liquid sheet shoots out large numbers of violent ligaments at high air velocity. The ligament velocity, ligament angle, ligament diameter, fragment size, and distribution are measured and analyzed. There are two kinds of ligament evolution that are breakup and contraction. An empirical model is also proposed for the ligament evolution process. At last, we obtain the criterion of critical Weber number on the ligament breakup based on the experimental results. This suggestion agrees well with the experimental data.

Appendix: An empirical model for the ligament evolution process

Here, we suggest an empirical model for the ligament evolution process. When the ligament is long, the shape of ligament can be considered as a cylinder approximately. The surface tension plays an essential role in the ligament evolution process. So the acceleration generated by surface tension is

$$ a_{\sigma } = \frac{{F_{\sigma } }}{{m_{L} }} \approx \frac{{\pi D_{L0} \sigma }}{{\frac{\pi }{4}D_{L0}^{2} L\rho_{l} }} = \frac{4\sigma }{{D_{L0}^{{}} L\rho_{l} }} $$

(7)

In this test, it is the inviscid flow; here, the Euler equations now can be written as

$$ a_{\sigma } = - \left( {\frac{{\partial u_{L} }}{\partial t} + u_{L} \frac{{\partial u_{L} }}{\partial L}} \right) $$

(8)

Due to the ligament diameter is constant approximately, so the convective acceleration in Eq. (8) can be neglected. And the expression of acceleration can be deduced as $\frac{{du_{L} }}{dt} = \frac{{du_{L} }}{dL}\frac{dL}{dt} = u_{L} \frac{{du_{L} }}{dL}$. Following the method suggested by Miskin and Jaeger (2012) recently, there is

$$ u_{L} \frac{{du_{L} }}{dL} \approx - \frac{4\sigma }{{D_{L0}^{{}} L\rho_{l} }} $$

(9)

The integral equation will be

$$ \int_{{u_{L0} }}^{{u_{L} }} {u_{L} du_{L} } = - \frac{4\sigma }{{D_{L0}^{{}} \rho_{l} }}\int_{{L_{0} }}^{L} \frac{dL}{L} $$

(10)

where L ₀ is initial length of ligament. Here, we suggest that D _L0 ≈ L ₀, so the equation can be written as

$$ \frac{1}{2}u_{L}^{2} - \frac{1}{2}u_{L0}^{2} = - \frac{4\sigma }{{D_{L0}^{{}} \rho_{l} }}\left[ {\ln \left( L \right) - \ln \left( {L_{0} } \right)} \right] = - \frac{4\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right) $$

(11)

So the equation can be written as

$$ \frac{1}{2}u_{L}^{2} = \frac{1}{2}u_{L0}^{2} - \frac{4\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right) $$

(12)

When the length of ligament increases, there is

$$ u_{L}^{{}} = \left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right]^{0.5} $$

(13)

And when the length of ligament decreases, there is

$$ u_{L}^{{}} = - \left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right]^{0.5} $$

(14)

So we suggest that the processes of ligament length increase and decrease are symmetrical approximately. Based on Eq. (13) and $u_{L} = \frac{dL}{dt}$, there is

$$ \frac{dL}{dt} = \left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right]^{0.5} $$

(15)

The integral equation will be

$$ \int_{{L_{0} }}^{L} {\left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right]^{ - 0.5} dL} = \int_{{t_{0} }}^{t} {dt} $$

(16)

So the time–length relationship can be written as

$$ t = \int_{{L_{0} }}^{L} {\left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right]^{ - 0.5} dL} $$

(17)

We set $x = \ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)$, so there is dL = D _L0 e ^x dx. Then, Eq. (17) will be

$$ t = \int_{0}^{x} {\left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}x} \right]^{ - 0.5} D_{L0}^{{}} e^{x} dx} $$

(18)

So when the length of ligament increases, the final evolution relationship between L and t will be

$$ t = \sqrt {\frac{{\pi D_{L0}^{3} \rho_{l} }}{8\sigma }} \exp \left( {\frac{{We_{{}} }}{8}} \right)\left\{ {erf\left[ {\left( {\frac{{We_{{}} }}{8}} \right)^{0.5} } \right] - erf\left[ {\left( {\frac{{We_{{}} }}{8} - \ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right)^{0.5} } \right]} \right\} $$

(19)

where the expression of error function is

$$ erf\left( y \right) = \frac{2}{\sqrt \pi }\int_{0}^{y} {e^{{ - t^{2} }} } dt $$

(20)

When $\frac{{We_{{}} }}{8} = \ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)$, the length of ligament is the maximum. L _m is the maximum length of ligament. t _m is the time when the length of ligament is the maximum. The expressions are

$$ L_{m} = D_{L0}^{{}} \exp \left( {\frac{{We_{{}} }}{8}} \right) $$

(21)

and

$$ \begin{gathered} t_{m} = \int_{{L_{0} }}^{{L_{m} }} {\left[ {u_{L0}^{2} - \frac{8\sigma }{{D_{L0}^{{}} \rho_{l} }}\ln \left( {\frac{L}{{D_{L0}^{{}} }}} \right)} \right]^{ - 0.5} dL} \hfill \\ = \sqrt {\frac{{\pi D_{L0}^{3} \rho_{l} }}{8\sigma }} \exp \left( {\frac{{We_{{}} }}{8}} \right)erf\left[ {\left( {\frac{{We_{{}} }}{8}} \right)^{0.5} } \right] \hfill \\ \end{gathered} $$

(22)

So when the length of ligament decreases; finally, the relationship between L and t will be

$$ t = 2t_{m} - \sqrt {\frac{{\pi D_{L0}^{3} \rho_{l} }}{8\sigma }} \exp \left( {\frac{{We_{{}} }}{8}} \right)\left\{ {erf\left[ {\left( {\frac{{We_{{}} }}{8}} \right)^{0.5} } \right] - erf\left[ {\left( {\frac{{We_{{}} }}{8} - \ln \left( {\frac{L}{{D_{L0} }}} \right)} \right)^{0.5} } \right]} \right\} $$

(23)