A process requires the delivery of drops of volume (3.2 times 10^{-8} mathrm{~m}^{3}). A liquid has a

Question:

A process requires the delivery of drops of volume \(3.2 \times 10^{-8} \mathrm{~m}^{3}\). A liquid has a density of \(900 \mathrm{~kg} / \mathrm{m}^{3}\) and a surface tension of \(0.03 \mathrm{~N} / \mathrm{m}\). What size of capillary would you recommend to form these drops?

In order to find the dripping rate, a simple criterion based on the Weber number can be used (Middleman, 1988b). The Weber number is defined as the dimensionless quantity \(ho v^{2} d_{\mathrm{c}} / \sigma\), where \(v\) is the flow velocity through the capillary. The value of the Weber number is suggested to be one at the onset of dripping. From this, calculate an estimate on the upper limit on the production rate from a single capillary in the units of \(\mathrm{kg}\) of drops per hour. The drops are formed in air.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Question Posted: