PressureDrop_Examples_SteamWaterBOARD

PressureDrop_Examples_SteamWaterBOARD - steamwater examps...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
steamwater examps Equations Parameters of Problem T = 290 [C] (1) G m = 2000 kg / m 2 · s (2) D = 0 . 01 [m] (3) Property lookups – note different values than T&K ρ g = ρ ( water, T = T, x = 1) (4) ρ f = ρ ( water, T = T, x = 0) (5) μ g = μ ( water, T = T, x = 1) (6) μ f = μ ( water, T = T, x = 0) (7) P = P ( water, T = T, x = 0) (8) Quality info x in = 0 (9) x out = 0 . 3 (10) x here = 0 . 15 Here is halfway up (11) L = 4 [m] (12) dxdz = x out - x in L (13) Martinelli-Nelson – Integral Approach for entire channel Gravity Δ P grav = L · ρ f · g grav · r 4 (14) g grav = 9 . 81 m / s 2 (15) r 4 = 0 . 42 From Figure 11-20 (16) Acceleration α out = 0 . 78 From Figure 11-17 (17) Δ P accel = G 2 m ρ f · (1 - x out ) 2 1 - α out + x 2 out · ρ f α out · ρ g - 1 ! (18) 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Friction Re lo = G m · D/μ f (19) f lo = 0 . 184 · Re - 0 . 2 lo (20) Δ P fric = f lo · G 2 m · L · r 3 2 · D · ρ f (21) r 3 = 6 From Figure 11-16 (22) Total Δ P MN = Δ P fric + Δ P accel + Δ P grav (23) Armand-Treschev (at the midplane) β = 1 1 + 1 - x here x here · ρ g f (24) α AT = β · 0 . 833 + 0 . 05 · ln 10 · P (10 6 ) [Pa] (25) Friction dpdz AT,fric = φ lo · f lo · G 2 m 2 · ρ f · D (26)
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern