SFM_exampleBOARD

# SFM_exampleBOARD - ENU 4134 – On-Board SFM Example – D...

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Unformatted text preview: ENU 4134 – On-Board SFM Example – D. Schubring 1 Impact of slip model on the predicted void fraction – adapted from T&K Problem 11-5 For the following conditions, compute the void fraction with the indicated models: ˙ m = 0 . 15 kg s- 1 (1) A = 1 cm 2 (2) x = 0 . 125 (3) T sat = 270 ◦ C (4) The liquid and vapor densities, along with pressure, need to be obtained using steam tables as: ρ v = 28 . 1 kg m- 3 (5) ρ f = 768 kg m- 3 (6) P sat = 5 . 505 MPa (7) a. HEM HEM implies S = 1. The equation for void fraction is then { α } = 1 1 + 1- x x ρ v ρ l S (8) { α } = 1 1 + 1- . 125 . 125 28 . 1 768 1 (9) { α } = 0 . 796 (10) b. Martinelli-Nelson (Figure 11-17, page 497 – you’ll need to interpolate between lines) (2 points) One needs to visually interpolate where a fine of 55.05 bar would appear, and then figure out where (on a semilog-x plot) 0.125 is. Your instructor estimates { α } ≈ . 71, but there’s about a 5% on this estimate. c. Bankoff’s slip correlation ( i.e. , version of the drift flux model) { β } is the same as { α } HEM (0.796). This correlation then proceeds as: { α } = K { β } (11) { α } = (0 . 71 + 0 . 0001 P abs ) × . 796 (12) This pressure must be in psi for this equation to work. 5.505 MPa is equivalent to 800 psi.This pressure must be in psi for this equation to work....
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SFM_exampleBOARD - ENU 4134 – On-Board SFM Example – D...

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