tag:blogger.com,1999:blog-9122514974659083342.post4117594408531005187..comments2024-02-02T22:30:20.736-05:00Comments on Chemical Facility Security News: Reader Comment – 01-13-14 – ContainmentPJCoylehttp://www.blogger.com/profile/03390039682578324978noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-9122514974659083342.post-65118726660711133472014-01-13T09:37:52.391-05:002014-01-13T09:37:52.391-05:00Assuming the hole is big enough that the liquid is...Assuming the hole is big enough that the liquid is flowing freely, the theoretical distance can be calculated as follows. Note that things get complicated unless we assume that the loss rate is small relative to the volume of the tank. Otherwise we have to use calculus to allow for the change in the height of the liquid as the liquid escapes. This is based on Torricelli's Theorem.<br /><br />Procedure :- <br /><br />Let, <br /><br />ha = height of the hole above tank's bottom; <br />hb = height of water in the tank; <br />h = height of water column above the hole-level; <br />v = speed at which the stream gushes out of the tank's hole horizontally;<br />t = time taken by the stream gushing out of the hole to fall to the ground; <br />g = acceleration due to gravity; <br />s = distance traveled by the gushing horizontal stream before it falls to the ground; <br /><br />The tank is assumed to be placed on the ground; <br /><br />Given, <br /><br />ha=15ft<br />hb=30ft<br />g=32ft/sec**2<br /><br />We have to find out the value of "s";<br /><br />h = hb-ha = 15ft <br /><br />v = sqrt(2*g*h) = 30.98ft/sec<br /><br />t = sqrt(2*ha/g) = 0.97 sec; <br /><br />s = v*t = 30ft<br />Jimhttps://www.blogger.com/profile/15581989619120654184noreply@blogger.com