Integrating Bulk Modulus to Find Pressure at Variable Depth

[reddawg]

Homework Statement

See image.

Homework Equations

pressure = density*gravity*depth

The Attempt at a Solution

The pressure at 5000 ft according to the book is 322,000 psf. This makes sense because density*gravity*depth = 2*32.2*5000 = 322,000 psf. How do I apply the bulk modulus equation to find the pressure at 5000 ft factoring a variable density with depth (The book says its 323,200 psf which makes sense because density increases with depth, although very slightly)?

 

[Chestermiller]

What is the hydrostatic equation, expressed in terms of the derivative of pressure with respect to depth?

 

[reddawg]

That would be just [density*gravity]. So how do I apply that?

 

[Chestermiller]

That would be just [density*gravity]. So how do I apply that?

Right. $$\frac{dp}{dz}=\rho g$$
Now, from the bulk modulus equation, if the density approaches $\rho_0$ at low pressures, what is the density at pressure p?

 

[reddawg]

That's where I have trouble. I always end up with just [density*gravity*depth]. Is it just (1/g)*(dp/dz) ?

 

[Chestermiller]

That's where I have trouble. I always end up with just [density*gravity*depth]. Is it just (1/g)*(dp/dz) ?

The relationship between density and pressure does not involve z. It's strictly a physical property relationship (sort of like the ideal gas law, except for a liquid).

Chet

 

[reddawg]

So, solving for density using the Bulk Modulus equation:

ρ = B*(Δρ/p)

 

[Chestermiller]

So, solving for density using the Bulk Modulus equation:

ρ = B*(Δρ/p)

Actually, the equation is $$\frac{1}{\rho}\frac{d\rho}{dp}=\frac{d(\ln \rho)}{dp}=\frac{1}{B}$$ What do you get if you integrate that, subject to the initial condition $\rho=\rho_0$ at p --> 0?

 

[reddawg]

I get:

(1/B)*p = ln(ρ/ρ₀) when factoring in the initial conditions.

 

[Chestermiller]

I get:

(1/B)*p = ln(ρ/ρ₀) when factoring in the initial conditions.

Good. Now solve for $\rho$ in terms of p. What do you get?

 

[reddawg]

ρ = ρ₀e^p/B

 

[Chestermiller]

ρ = ρ₀e^p/B

OK. Now substitute that into the hydrostatic equation in post #4. What do you get? Can you integrate that from z =0?

 

[reddawg]

(dp/dz) = ρ₀ge^p/B

How do I rearrange that to integrate from z=0 to h?

 

[Chestermiller]

(dp/dz) = ρ₀ge^p/B

How do I rearrange that to integrate from z=0 to h?

Cmon man.

$$e^{-\frac{p}{B}}dp=\rho_0 g dz$$