{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problem 3.38

# Problem 3.38 - Problem 3.38[Difficulty:2 Fluid 1 Fluid 2...

This preview shows page 1. Sign up to view the full content.

Problem 3.38 [Difficulty :2] Fluid 1 Fluid 2 σπ D cos θ ρ 1 g Δ h π D 2 /4 Given: Two fluids inside and outside a tube Find: (a) An expression for height Δ h (b) Height difference when D =0.040 in for water/mercury Assumptions: (1) Static, incompressible fluids (2) Neglect meniscus curvature for column height and volume calculations Solution: A free-body vertical force analysis for the section of fluid 1 height Δ h in the tube below the "free surface" of fluid 2 leads to F 0 = Δ p π D 2 4 ρ 1 g Δ h π D 2 4 π D σ cos θ ( ) + = where Δ p is the pressure difference generated by fluid 2 over height Δ h , Δ p ρ 2 g Δ h = Hence Δ p π D 2 4 ρ 1 g Δ h π D 2 4 ρ 2 g Δ h π D 2 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern