Bulk Modulus of Elasticity and Compressibility - Fluid Mechanics

Bulk Modulus of Elasticity and Compressibility - Fluid Mechanics - Hydraulics 1
Following the last lecture where we mentioned bulk modulus, here we look at the theory in more detail, deriving the definitions for it and then working through an example to test our understanding.
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If you do have any questions following this lecture, or would like to request any specific topics, please put them in the comment section and I will respond as soon as possible.

[@Ahelot]

Thank you ♥️ your videos should be used on facultys, great job 👍

[@Learn_Civil_Engineering]

Thank you 😊 your support is much appreciated!!

[@sumitkumarsahoo7601]

you cleared my concept, thanks a lot, love from India

[@Learn_Civil_Engineering]

I'm very glad to hear that! And thanks for the comment 🙂

[@sumitkumarsahoo7601]

@@Learn_Civil_Engineering you are creating a great change in students' life, have a great go for this great work.

[@suyashsharnagat2912]

Keep posting videos
It helped a lot
Thanks man !!!

[@Learn_Civil_Engineering]

I'm glad to hear they are helping! Thank you for the support 😀

[@xavierhenriques1116]

Very well done
Thank you

[@Learn_Civil_Engineering]

Thanks for the support Xavier!

[@Learn_Civil_Engineering]

Subscribe to the channel to stay up to date with a new video released every Wednesday!

[@Kurosaka]

can you make a vid about gases? for bulk modulus?

[@tuyishimealexis4306]

Thanks for your video ,but try to use
Standard camera

[@Derimmerjung]

If I use the water density = 1000 kg/m3. The change in pressure would be 2.1x10 exp 3. It´s logical?

[@orkungoksel5346]

Thanks

[@manuelseque8055]

nice video. good work

[@Learn_Civil_Engineering]

Thank you for your comment Manuel Seque

[@manuelseque8055]

is it possible to prepare video with shear modulus too

[@Learn_Civil_Engineering]

Hi there, yes I will try to post a video on shear modulus as soon as possible.

[@isharauditha4257]

How did you assume this ^v/v=^raw/raw

[@ken-cf7tf]

its actually -dV/V = dρ/ρ , you can derive this by differentiation of ρ=m/v
ρV=m
ρdV+Vdρ = 0
Vdρ=-ρdV
dρ/ρ = -dV/V

[@Learn_Civil_Engineering]

It's not actually an assumption, delta Rho is equal to (delta V / V) times Rho due to the fluid's mass being the same before an after the compression.
Therefore, as there is a decrease in volume, there must be a corresponding increase in fluid density. So the the above is an expression representing this.

[@isharauditha4257]

@@ken-cf7tf thanks

[@isharauditha4257]

@@Learn_Civil_Engineering thanks

[@arashmokhtarbaf5826]

Why v or change in v? Why not change in V over V?

[@malumognekusangeloc.7374]

How did the change of Rho (0.1) turned to 0.001?

[@wildankoso]

By converting the percentage into a number; multiplying it by 100