00:01
So let's start by drawing our needle.
00:03
So this is the needle, and this needle is attached to a pipe, and that pipe ultimately leads to the transfusion bottle.
00:15
And normally the transfusion bottle has a hole at the top, so that the pressure at the top would be equal to the atmospheric.
00:23
So what we're talking about here is this would be, so the pressure at the end of the needle, p1, is equal to the atmospheric pressure, and the pressure of the fluid just before it enters the needle, p2, is equal to p atmospheric plus rhogh, where roh gh is the hydrostatic pressure of the blood in the transfusionary bottle and the pipe leading into the needle.
00:53
So first thing is we can figure out that p2 minus p1 would actually be equal to p atmospheric plus rog phd minus p atmospheric so that's equal to row gh.
01:13
Next, flow rate is given by pi times r to the power four times p2 minus p one over 8 eta l.
01:26
So the question says that the flow rate is 4 .5 times 10 to the negative 8 meter cube per second, that equals pi, times the radius of the needle is 0 .25 millimeters...