final_sol

# final_sol - 2.25 Final Exam 2005 SOLUTION Question 1 U A...

This preview shows pages 1–4. Sign up to view the full content.

2.25 Final Exam 2005 SOLUTION Question 1 U A R d U Toroidal rim Cross Section Two identical jets of water impact on one another forming a sheet of liquid as shown in the ﬁgure (and as shown in the surface tension movie). The sheet grows to a radius R ending in a toroidal rim that ejects droplets of ﬂuid. Consider incoming jet velocities, U = 1m/s, and cross-sectional jet areas, A = π/ 4cm 2 . 1. Calculate the Reynolds number, Capillary number, and Froude number for this ﬂow. Use these numbers to argue which eﬀects (e.g. inertia, viscosity, gravity, etc.) are important in this system. Solution: U = 100 cm/s, A = 2 D =1 cm , µ =10 2 dyne-s/cm 2 , ρ = 1 gm/cm 3 , σ = 70 dynes/cm. Hence: Re = ρUD 10 4 inertia ± viscosity (1) µ Ca = µU σ 10 2 surface tension ± viscosity (2) Fr = U 2 gD 10 inertia gravity (3) If the sheet is ﬂat, we can neglect gravity (if the sheet is curved, as in part 4, we must include gravity). 2. Derive an expression for the maximum radius R of the liquid sheet and calculate R using the values given above. Solution: The maximum radius occurs when the sheet is ﬂat so we neglect gravity. Apply conservation of momentum on the following control volume: 1 Photo removed for copyright reasons. Source: Figure 17l in Clanet, C. "Dynamics and stability of water bells." J. Fluid Mech 430 (2001): 111-147.

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

View Full Document
σ σ 2 1 By Bernoulli, u 1 = u 2 = U . Conservation of momentum yields (note: p = p a ): d ρ v dV + ρ v ( v v c ) · n dA = p n dA 2 σd± (4) dt CV CS CL ρU 2 2 πrd = 2 σ 2 πr (5) where d = d ( r ) is the thickness of the sheet. Thus inertia balances surface tension at steady state when 2 σ d = . (6) ρU 2 By conservation of mass (2 on the LHS from two incoming jets): A 2 UA =2 πrdU r = . (7) πd Combining this with (6) we ﬁnd ρU 2 A r = R rim = 18 cm (for water) (8) 2 πσ 3. Using energy arguments, show that the toroidal rim is unstable and will break into droplets. Derive a lower bound for the size of the ejected droplets. Solution: E surf = σA surf (9) Thus, if the torus has a larger surface area that the droplets (where the volume of ﬂuid is identical in both conﬁgurations), surface tension will break the torus into drops. 4 0 (2 πR rim )= V drop = N 3 (10) V torus = 2 drop 3 R 2 N = 0 (2 rim ) 3 (11) 3 4 r drop The torus is unstable if S torus 2 0 (2 rim ) >S drop = N 4 2 (12) drop 3 r drop > R 0 (13) 2 3 Thus the torus will break into droplets with radii > 2 R 0 to lower the surface energy. (Note, tiny droplets are no good since they create extra surface area and hence cost more energy.) To calculate the actual drop radius rather than a lower bound, we need to consider how to move ﬂuid from the torus into the droplets. This calculation is a bit more involved and results in a critical wavelength of 9 . 02 R 0 (see Rayleigh-Plateau instability). 2
± ² ³ ´ µ µ µ µ µ µ µ · ¸¹ º ± ² ± ² ³ ´ 4. If gravity becomes important in the problem, the sheet will sag and become a “water bell” (instead of a ﬂat sheet) as sketched below. Show that the shape of the “bell” is given by the solution to the following

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

final_sol - 2.25 Final Exam 2005 SOLUTION Question 1 U A...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online