need help understanding "bulk modulus" in fluids
\[B=-V \frac{dp}{dV}\] thats the formula for bulk modulus, and it has a specific value for a specific substance, but "V" in the formula suggests that it depends on amount of substance under consideration... can someone explain
and how do I interpret the value for B for example.. water has B= 2 GPa.. so 2 Gpa is the pressure required to ___?
pressure required to double the original volume
what about "V" ?
here's the general concept. An increase in pressure, reduces the volume.
i got that, but bulk modulus is a particular value for a particular substance... (like density) so how does the V in the formula fit in?
we are focusing on the ratio of pressure stress and volumetric strain
well we have volume and density is defined as?
mass/vol = density?
yes, that density since you are the one asking about density. now let us look how that factors into the concept density = mass/volume you have probably learned about the volume and pressure relationship, as well the PV=nRT and thermal expansion.
ok...
you used B so I will use it as well \(\sf B = -\dfrac{dP}{dV} V \rightarrow -\dfrac{dP}{d\rho}\rho \) the density changes proportionally
baru, remember how we discussed that for a given material, application of a given stress results in the same percentage change in length? it's just the same.
@nincompoop yea, i purposely did not post the formula in that form xD...
this means that the density can be used to calculate the bulk modulus since we have already learned the relationship of mass and density
you r saying that why is bulk modulus a constant when V can take diff values okay \(B=-\large V\frac{p}{\Delta V}\) here p is constant and now we only gotta prove that \(\frac{V}{\Delta V}\) is constant well :/ a specific pressure will always bring about the same compression alwys so \(\frac{V}{\Delta V}\) must be constant so B also becomes const..
|dw:1451904234163:dw|
So, I guess what you were discussing with beauregard is about the elasticity of material objects |dw:1451904423365:dw|
think about it, if the stress of the pressure is going the opposite |dw:1451904640754:dw|
Join our real-time social learning platform and learn together with your friends!