ADVANCED TRANSPORT PHENOMENA Assignment 1 Shinta Leonita (0906635772) Problem 1 A viscous fluid is in laminar flow in a slit formed by two parallel walls a distance 2B apart. Make a differential momentum balance and obtain the expression for the distributions of momentum flux and velocity. What is the ratio of average to maximum velocity in the slit? Obtain the analog of the Hagen-Poiseuille law of the slit. Answer: Laminar flow in a narrow slit: Figure 1. Flow Through a Slit, with B << W << L From figure above, thickness of a system is Ξx, length L, and listing the various distributions to the momentum balance in the z-direction: ο· Rate of momentum in across slit surface at x (ππΏππ₯π§ )|π₯ ο· Rate of momentum out across slit surface at x + Ξx (ππΏππ₯π§ )|π₯+Ξx ο· (1) (2) Rate of momentum in across beam surface at z = 0 (Ξxππ£π§ )(ππ£π§ )|π§=0 (3) ο· ο· ο· Rate of momentum out across beam surface at z = L (Ξxππ£π§ )(ππ£π§ )|π§=πΏ (4) (ΞxππΏ)ππ (5) Gravity force acting on slit shell Pressure force acting on beam surface at z = 0 (Ξxπ)π0 ο· (6) Pressure force acting on beam surface at z = L β(Ξxπ)ππΏ (7) Momentum balance (assuming fluid is incompressible, vz is the same at z = 0 and z = L): (ππΏππ₯π§ )|π₯ β (ππΏππ₯π§ )|π₯+Ξx + (Ξxππ£π§ )(ππ£π§ )|π§=0 β (Ξxππ£π§ )(ππ£π§ )|π§=πΏ + (ΞxππΏ)ππ + Ξxπ(π0 β ππΏ ) (8) Dividing eq. (8) by WL Ξx and take the limit as Ξx goes to zero; this gives: ππ₯π§ |π₯+Ξx βππ₯π§ |π₯ lim ( Ξx Ξxβ0 )= (π0 βππΏ ) πΏ + ππ (9) The expression on the left side is the definition of the first derivative, and π = π + ππβ = π β πππ§. Hence eq. (9) may be written as: π ππ₯ ππ₯π§ = (π0 βππΏ ) (10) πΏ Integrating eq. (10) gives: ππ₯π§ = (π0 βππΏ ) πΏ π₯ + πΆ1 (11) The constant C1 must be zero if the momentum flux is not to be infinite at x = 0. Hence the momentum flux distribution is: πππ = (π·π βπ·π³ ) π³ Substituting Newtonβs law of viscosity ππ₯π§ = βπ ππ£π§ ππ₯ =β ππ£π§ ππ₯ π (12) to eq. (12) gives: (π0 βππΏ ) ππΏ π₯ (13) Integrating eq. (13) gives: (π0 βππΏ ) π£π§ = β 2ππΏ π₯ 2 + πΆ2 (14) Because of the boundary condition that vz be zero at x = ±B, the constant C2 has the value (π0 βππΏ ) 2ππΏ π΅ 2. Hence the velocity distribution is: ππ = (π·π βπ·π³ ) πππ³ π π π©π [π β ( ) ] π© (15) The maximum velocity vz,max at the middle of the slit occurs at x = 0 and has the value: (π0 βππΏ ) π£π§,πππ₯ = 2ππΏ π΅2 (16) Hence the ratio of the average to the maximum velocity is then: π£π§ π£π§,πππ₯ π₯ 2 =1β( ) π΅ πΎ π© β©ππ βͺ ππ,πππ = β«π β«βπ©[πβ(π/π©)π ]π ππ π πΎ π© β«π β«βπ© π ππ π π = β«π (πβππ )π π π β«π π π (17) π π π π = (π β ) = (18) The analog of the Hagen-Poiseuille law: 2 π = (2π΅π)β©π£π§ βͺ = (2π΅π) (3) π£π§,πππ₯ πΈ= π (π·π βπ·π³ )π©π πΎ π ππ³ (19) (20) Problem 2 A droplet of substance A is suspended in a stream of gas B. The droplet radius is r1. We postulate that there is a spherical stagnant gas film of radius r2. the concentration of A in the gas phase is xA1 at r = r1 and xA2 at r = r2. a) By a shell balance, show that for steady state diffusion r2NAr is a constant and set the constant equal to r12NAr1 at the droplet surface. b) Show that: r12 N Ar1 ο½ ο cD AB 2 dx A r ο¨1 ο x A ο© dr c) Integrate that equation between the limits r1 and r2 to get: N Ar1 ο½ ο cDAB ο¦ r2 οΆ xB 2 ο§ ο· ln ο¨r2 ο r1 ο© ο§ο¨ r1 ο·οΈ xB1 d) If a mass transfer coefficient kv is defined by NAr1 = kv (pA1 β pA2), show that: kv ο½ 2cD AB / D ο¨ p B ο©ln Answer: Figure 2. Diffusion through a Spherical Film (a) Showing that for steady state diffusion r2NAr is a constant and set the constant equal to r12NAr1 at the droplet surface: A mass balance on A over a spherical shell of thickness Ξr is (in molar units) at a steady state: 4ππ 2 . ππ΄π |π=π β 4π(π + βπ)2 . ππ΄π |π=π+βπ = 0 (1) (4ππ 2 . ππ΄π )|π=π β (4ππ 2 . ππ΄π )|π=π+βπ = 0 (2) Or, equivalently: Dividing eq. (2) by 4ΟΞr and take the limit as Ξr goes to zero gives: π π π (ππ . π΅π¨π ) = π (3) Eq. (3) may be integrated to give π 2 . ππ΄π = πΆ1 and we may use the boundary condition that ππ΄π = ππ΄π1 at r = r1 (the gas liquid interface) to evaluate the constant and obtain that r2NAr = r12NAr1 (b) Equation for the radial component in spherical coordinates is: ππ΄π = βππ·π΄π΅ ππ₯π΄ ππ + π₯π΄ (ππ΄π + ππ΅π ) (4) If gas B is not moving, then ππ΅π may be set equal to zero and the equation may be solved for the molar flux of A: ππ· ππ΄π = β 1βπ₯π΄π΅ π΄ ππ₯π΄ (5) ππ Multiplying by r2 and using the result obtained in (a), then we get: ππ« πππ π΅π¨ππ = β πβππ¨π© ππ π¨ π ππ¨ π π (6) π2 π₯ π12 ππ΄π1 β«π1 1/π 2 ππ = βππ·π΄π΅ β«π₯ π΄2 (c) ππ₯π΄ π΄1 (1βπ₯π΄ ) (7) Integrated that equation gives: 1 1 (1βπ₯ 1 2 ) π΄2 π12 ππ΄π1 ( β ) = ππ·π΄π΅ ππ π π (1βπ₯ ) π2 βπ1 π12 ππ΄π1 ( π΅π¨ππ = 1 1 1 2 π2 π1 ππ«π¨π© (π₯ ) ) = ππ·π΄π΅ ππ (π₯π΅2) (9) π΅1 π ππ βππ (8) π΄1 (π ) (ππ) ππ (ππ©π ) π (π₯ (10) π©π ) (d) At π β β π12 ππ΄π1 (π β π ) = ππ·π΄π΅ ππ (π₯π΅2 ) π΅1 ππ΄π1 = (π₯ ππ·π΄π΅ π1 ππ (π₯π΅2 ) (π₯π΅1 ) (11) βπ₯π΅1 ) π΅2 /π₯π΅1 ) But, (π₯π΅ )ππ = ππ(π₯π΅2 Therefore, ππ΄π1 = ππ΄π1 = ππ΄π1 = ππ·π΄π΅ 1 π1 (π₯π΅ )ππ ππ·π΄π΅ 1 π1 (π₯π΅ )ππ ππ·π΄π΅ 1 π1 (ππ΅ )ππ (π₯π΅2 β π₯π΅1 ) (12) (π₯π΄1 β π₯π΄2 ) (13) (ππ΄1 β ππ΄2 ) (14) But, ππ΄π1 = ππ£ (ππ΄1 β ππ΄2 ) So, ππ = ππ«π¨π© ππ(π ) π© ππ = πππ«π¨π© π«(ππ© )ππ (15)
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