ChE150B PSET2

# Either c1 is 0 or r r we choose the former as

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Unformatted text preview: dcA −RA 1 3 = [ r ] + C1 dr DAB 3 CHEMICAL ENGINEERING 150B PROBLEM SET #2 7 Where C1 is a constant. Separating variables yields: dcA = C1 −RA r + 2 dr ⇒ 3DAB r cA (r) = We know by continuity that: dcA dr =0 r=0 dcA = −RA C1 r + 2 dr 3DAB r −RA 2 C1 r− + C2 6DAB r and assume that the concentration on the exterior skin of the bubble is simply the initial value stipulated by broth concentration: cA (R) = cA,0 1 With the ﬁrst boundary condition, cA (r) = α(r = 0) + C2 + 0. Either C1 is 0 or r → ∞; r we choose the former as physically reasonable. For the second condition: cA,0 = Which means: −RA R2 RA R2 + C2 ⇒ C2 = cA,0 + 6DAB 6DAB cA = RA R2 r2 1 − 2 + cA,0 6DAB R Since we would like to know R when cA → 0 at r = 0, we substitute: 0= RA R2 (1 − 0/R2 ) + cA,0 ⇒ 6DAB −6cA,0 DAB =R RA 6(1.8 × 10−9 )(0.005)) = 0.0005 m 0.0066 From the problem statement, O2 is being consumed, which means RA = Roxygen ρ = (−3.3 × 10−7 )(20, 000) ⇒ R = Problem 4 (a) 5 Reasonable Assumptions. Steady-state. No reaction. Constant c, DAB , P, T . Convective forces and diﬀusive forces never occur simultaneously along a singular path. No penetration: no ﬂux across solid boundaries. 8 ARJAN SINGH PUNIANI (b). We know that in the z -direction, convection dominates diﬀusion phenomenon. From the general equation ∂cA − RA = 0 ·N+ ∂t We are left with d dcA −DAB + yA (NA + NB dz dz by assumptions (1), (2), (3), and (4). Due to the dominance of convection only the yA (NA + NB ) term remains, which means: V= cA vA + cB vB cA + cB cB vB = vB cB So if V = vB = vz , then NA = cA V = cA vz . For consideration of the x-direction, we know that ∂cA ·N+ − RA = 0 ∂t Will reduce to dyA d −cDAB + yA (NA + NB ) = 0 dx dx By the same assumptions used to reduce the z axis consideration, Examining ﬂux more carefully dcA NA,x = −DAB dx By (3), but since cA is clearly aﬀected in the z-direction, change the total diﬀerentials to partials: NA,x = −DAB ∂cA ∂x...
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