Tapered tubing

Level pending

What is the velocity of the water when it exits the thicker end with area 5 5 square centimeters ( A 2 ) (A_2) after flowing into the thinner end with area 4 4 square centimeters ( A 1 ) (A_1) at a velocity of 2.5 2.5 m/s ( v 1 ) (v_1) ?

4 m/s 2.5 m/s 5 m/s 2 m/s

Lakshan Dumpa by Lakshan Dumpa , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Craig Haydock
Feb 3, 2014

It's a matter of volumetric flow rate. What goes in must come out (volume/time@1=volume/time@2). Volumetric flow rate = Velocity * Area. Simply set the two equal to each other and then rearrange to solve for the unknown. Keep in mind the units of measure for area. If they are not the same, you'll need to convert them to a common unit first. Otherwise, it's a simple ratio and the units cancel out leaving you with a ratio times velocity like so... V1 (A1/A2)=V2. In our case, (2.5m/s) (4/5)=2m/s

0 Helpful 0 Interesting 0 Brilliant 0 Confused

is it not necessary to mention about the ideal flow of water?

sagnik lahiri - 7 years, 3 months ago

Log in to reply

I suppose one could go into greater constraint stating that it is an ideal fluid with no viscosity, no flow work, no thermal heat gain or loss along the path, no phase transformations etc.. I just didn’t feel that it was necessary to go into that much detail for this problem. You do however bring up a good point that a real world system can be much more complex and many additional factors would need to be accounted for.

Craig Haydock - 7 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...