Averaging_Example_onboard

Averaging_Example_onboard - s-1 . Consider time-averaging...

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An Example of Averaging Parameters 1 Direction of Flow 6 cm (center-to- center) 3 cm (center-to- center) 3 cm A B C Spherical vapor bubbles, D = 1 cm Rectangular duct, 3 cm side length. Assume ρ l is 1000 kg m - 3 and ρ v is 1 kg m - 3 . When one bubble is shown at a fixed z , one is present. Two shown means four are present (two forward, two back). Compute: 1. { α } along area C. There is no vapor here { α } = 0. 2. { α } along area A. { α } = A k /A A = (3 cm ) 2 = 9 cm 2 (1) A k = 0 . 25 πD 2 bubble (2) A k = 0 . 25 π (1 cm ) 2 (3) A k = 0 . 785 cm 2 (4) { α } = 0 . 785 / 9 = 0 . 0873 (5) 3. { α } along area B. { α } = A k /A A = (3 cm ) 2 = 9 cm 2 (6) A k = 4 × 0 . 25 πD 2 bubble (7) A k = π (1 cm ) 2 (8) A k = 3 . 142 cm 2 (9) { α } = 3 . 142 / 9 = 0 . 349 (10)
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An Example of Averaging Parameters 2 Direction of Flow 6 cm (center-to- center) 3 cm (center-to- center) 3 cm Spherical vapor bubbles, D = 1 cm A (area) d (point) 4. < α > in the volume. < α > = V k /V (11) V = 6 cm × A = 54 cm 3 (12) V k = 5 πD 3 bubble 6 ! (13) V k = 2 . 618 cm 3 (14) < α > = 2 . 618 / 54 = 0 . 0485 (15) 5. < ρ > in the volume. < ρ > = < α > ρ v + < 1 - α > ρ l (16) < ρ > = 0 . 0485 × 1 kg m - 3 + (1 - 0 . 0485) 1000 kg m - 3 (17) < ρ > = 951 . 5 kg m - 3 (18) Continue with the earlier assumption + bubbles are moving upwards at a velocity of 10 cm
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Unformatted text preview: s-1 . Consider time-averaging at area A. 6. Appropriate t ? – the time between passages of group of 4 bubbles (or of single bubble) t ? = 6 cm 10 cm s-1 = 0 . 6 s (19) An Example of Averaging Parameters 3 7. ˜ α at point d. In 0.6 s, 1 bubble passes through point d. It requires 0.1 s to pass (= 1 cm/ 10 cm/s ), so ˜ α is 0.167. 8. g { α } = { ˜ α } for the area A. By computing { α } at all z and the transforming z to t by considering the velocity, this could be solved analytically. However, there is only one type of coherent structure moving. As a result, a fixed axial location will be exposed to the entire volume in time, so < α > = g { α } . That is, g { α } = 0 . 0485....
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This note was uploaded on 10/22/2011 for the course ENU 4134 taught by Professor Schubring during the Fall '11 term at University of Florida.

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Averaging_Example_onboard - s-1 . Consider time-averaging...

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