SolutionHW_C2 - Homework Set #2: SOLUTIONS Due date:...

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Homework Set #2: SOLUTIONS Due date: 09/22/06, by 11:00 am. Problem #1: In the falling film problem, it is assumed that flow is laminar. This flow regime can only obtained if the Reynolds number, Re, defined by 4 Re z u ρ δ µ = is smaller or equal 10, where z u is the average fluid velocity 1. If the film liquid is water at 20 0 C (kinematic viscosity = 0.010037 cm 2 /s), what is the maximum volume flow rate per unit wall width W to ensure that Re 10. 2. For Re = 10, and the wall is vertical, calculate the film thickness. 3. If one wants to choose a different coordinate system as follows: the y and z coordinates are the same, but the x coordinate is: x = 0 at the solid wall, and δ at the gas liquid interface. Show that the velocity is then given by 2 2 cos 1 2 z gx u ρδ β µδ x ⎛⎞ ⎛⎞ =− ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ Give your comment on this choice of coordinate Answer: 1. Maximum volume flow rate per unit wall width Volume flow rate, Q, is given as ( ) z Qu W = . For the given expression of the Reynolds number, one can solve for ( ) z u as: () Re Re 44 z v u == , where kinematic viscosity, v = Therefore, the volume flow rate per unit wall width is: Re 4 Qv W = To obtain the laminar flow: 4 Re 10 Q W ν =≤ Thus the maximum volume flow rate per unit width for the laminar flow of the falling film is at Re = 10. 2 2 max Re 10*0.010037 / 0.0251 / Q vc m s cm s W =
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2. Film thickness Using the relation Re 4 = w W ν ρ , the film thickness can be calculated from the equation of mass flow rate: where 0 = β (Fig 2.2-3 of BSL) δ βρ = ⎟ = == = 33 4 1 9 80665 0 009167 13 2 g w Wg gm s cm cos cos Re cos , . / . / / 3. Velocity profile in a different coordinate system From the z-momentum balance cos (1) xz d g dx τ ρβ = Integrating equation (1) gives 1 cos xz gx C = + Applying the boundary conditions 1 ,0 , c o s cos 1 cos 1 cos 1( xz xz z z at x C g x g du x g dx du dx δτ ρδ τρδ µρδ µδ === ⎛⎞ =− ⎜⎟ ⎝⎠ −= 2 ) Integrating equation (2) gives 2 2 cos 1 2 z u x 0, 0 z Note that the constant C2 resulting from the above integration is zero due to the boundary condition at xu
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Problem #2: Solve problem 2A.2 at the end of Chapter 2 of BSL. Answer: Equation 2.3-21 is expressed as such: ( ) 4 PR w 8L π ∆ ρ = µ The capillary radius can the be solved for: 4 8Lw R P µ = πρ∆ Inserting the data renders: () ( ) 53 13 4 2 4 4 35 8 4.03x10 0.5002 2.997x10 R 3.186x10 7.51x10 m 7.51x10 cm 0.9552x10 4.829x10 −− == = π =
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This note was uploaded on 12/14/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.

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SolutionHW_C2 - Homework Set #2: SOLUTIONS Due date:...

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