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Fluid Mechanics: Compressible Fluids, density function

  • #1

Homework Statement


"Suppose that a liquid has an appreciable compressibility. Its density therefore varies with depth and pressure. The density at the surface is ρ0
(a) Show that the density varies with pressure according to ρ=ρ0ekp where P is gauge pressure at any depth and k is the compressibility, a constant.
(b) Find P as a function of depth y.

Homework Equations


ρ = m/V (Density definition)
k = (1/V)(ΔV/ΔP) (Compressiblity definition)
dP/dy = ρg (Pascal's Law)
My professor also hinted that I should replace any deltas with differentials, that I need to eliminate V, and to find dρ/dV (which is -m/V2

The Attempt at a Solution


From the second equation, I multiplied both sides by dP (after changing ΔV to dV and ΔP to dP) to obtain the following:
k dP = (1/V) dV
I then integrated both sides:
∫k dP (from P0 to P) = ∫(1/V) dV (from 0 to V)
kP(gauge)=ln(V)
ekP(gauge) = V
I get the feeling that I'm going in the correct direction, but haven't found a viable solution from this point. I fooled around with the density definition and Pascal's Law to no avail. Any assistance would be much appreciated.
 

Answers and Replies

  • #2
20,235
4,265
Good start!!! Your integration limits on V are incorrect. They should be from V0 to V. Also, for a given amount of mass the product of density and volume is constant.
 
  • #3
Good start!!! Your integration limits on V are incorrect. They should be from V0 to V. Also, for a given amount of mass the product of density and volume is constant.
When I intergrate on those limits I get kP(gauge)=ln(V)-ln(V0), which seems to be worse than before.
Also, if m = ρV is constant, how does that work with ρ and V both being varying functions?
 
  • #4
20,235
4,265
When I intergrate on those limits I get kP(gauge)=ln(V)-ln(V0), which seems to be worse than before.
Also, if m = ρV is constant, how does that work with ρ and V both being varying functions?
The compressibility condition should have a minus sign (the volume decreases with increasing pressure):

k dP = -(1/V) dV

[tex]\frac{V}{V_0}=\frac{\rho_0}{\rho}=e^{-kP}[/tex]
 
  • #5
I understand where the v/v0 came from (the ln) but how did you derive p0/p? Did it come from another integral somewhere?
EDIT: nvm, I understand now. Thanks a bunch!
 

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