BSL Transport Phenomena 2e Revised: Chapter 2 - Problem 2C.4 Page 1 of 9

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1 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 1 of 9 Problem C.4 Falling-cylinder viscometer see Fig. C.4. 6 A falling-cylinder viscometer consists of a long vertical cylindrical container radius R capped at both ends, with a solid cylindrical slug radius R. The slug is equipped with fins so that its axis is coincident with that of the tube. One can observe the rate of descent of the slug in the cylindrical container when the latter is filled with fluid. Find an equation that gives the viscosity of the fluid in terms of the terminal velocity of the slug and the various geometrical quantities shown in the figure. a Show that the velocity distribution in the annular slit is given by v z = 1 ξ 1 + ln1/ξ C.4-1 in which ξ = r/r is a dimensionless radial coordinate. b Make a force balance on the cylindrical slug and obtain µ = ρ 0 ρgr [ ln C.4- in which ρ and ρ 0 are the densities of the fluid and the slug, respectively. c Show that, for small slit widths, the result in b may be expanded in powers of ε = 1 to give µ = ρ 0 ρgr ε ε 13 0 ε + C.4-3 Solution See C. for information on expansions in Taylor series. 6 J. Lohrenz, G. W. Swift, and F. Kurata, AIChE Journal, 6, and 7, 6S 1961; E. Ashare, R. B. Bird, and J. A. Lescarboura, AIChE Journal, 11, F. J. Eichstadt and G. W. Swift, AIChE Journal, 1, ; M. C. S. Chen, J. A. Lescarboura, AIChE Journal, 14,

2 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page of 9 Part a For this problem we choose a cylindrical coordinate system with the origin at the bottom of the slug s center. We assume that as the slug falls, the fluid in the annular slit flows in the z-direction and varies as a function of radius r. v z = v z r As a result, only φ rz the z-momentum in the positive r-direction and φ zz the z-momentum in the positive z-direction contribute to the momentum balance. The pressure is assumed to vary with height z. p = pz Figure 1: This is the shell over which the momentum balance is made for flow through an annular slit. Rate of z-momentum into the shell at z = 0: πr rφ zz z=0 Rate of z-momentum out of the shell at z = : πr rφ zz z= Rate of z-momentum into the shell at r: πrφ rz r Rate of z-momentum out of the shell at r + r: [πr + rφ rz r+ r Component of gravitational force on the shell in z-direction: πr rρg If we assume steady flow, then the momentum balance is Rate of momentum in Rate of momentum out + Force of gravity = 0. Considering only the z-component, we have πr rφ zz z=0 πr rφ zz z= + πrφ rz r [πr + rφ rz r+ r πr rρg = 0. Factor the left side. πr rφ zz z= φ zz z=0 π[r + rφ rz r+ r rφ rz r πr rρg = 0

3 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 3 of 9 Divide both sides by π r. Take the limit as r 0. r φ zz z= φ zz z=0 + r + rφ rz r+ r rφ rz r r + ρgr = 0 r φ zz z= φ zz z=0 r + rφ rz + lim r+ r rφ rz r + ρgr = 0 r 0 r The second term is the definition of the first derivative of rφ rz. r φ zz z= φ zz z=0 Now substitute the expressions for φ rz and φ zz. + d dr rφ rz + ρgr = 0 φ rz = τ rz + ρv r v z = τ rz φ zz = pδ zz + τ zz + ρv z v z = pz + ρvz Since v z does not depend on z, the ρvz terms cancel. r p z= + ρv z z= p z=0 ρv z z=0 + d dr rτ rz + ρgr = 0 Make it so that ρgr is in the fraction. Place ρg0 in the numerator. r p z= + ρg p z=0 + d dr rτ rz = 0 r p z= + ρg p z=0 ρg0 + d dr rτ rz = 0 The point of doing this is that now we can use the modified pressure P z = pz + ρgz. r P P 0 + d dr rτ rz = 0 So we have d dr rτ rz = P P 0 r. From Newton s law of viscosity we know that τ rz = µdv z /dr, so d µr dv z = P P 0 r. dr dr The boundary conditions for this differential equation are obtained from assuming that no slip occurs between the fluid and the walls of the slug and container. That is, at r = R the fluid travels with the slug v z = and at r = R the fluid is stationary v z = 0. B.C. 1 : v z =, at r = R B.C. : v z = 0, at r = R

4 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 4 of 9 Integrate both sides of the differential equation with respect to r. Divide both sides by µr. µr dv z dr = P P 0 r + C 1. dv z dr = P P 0 µ r C 1 µr Integrate both sides of the differential equation with respect to r once more. v z r = P P 0 r C 1 µ ln r + C Apply the boundary conditions here to determine C 1 and C. v z R = P P 0 R C 1 µ lnr + C = v z R = P P 0 R C 1 µ ln R + C = 0 Solving this system of equations, we get [ µ P P 0 R C 1 = 1 C = P P 0 R [ 1 ln R 1 ln R. So we have for the velocity distribution, v z r = P P 0 r ln r [ P P 0 R 1 + P P 0 R [ 1 ln R 1 ln R. Factor the right side. Thus, v z r = P P 0 r R + P P 0 R 1 ln R ln r + ln r ln R v z r = P P 0 R [ r R ln R r ln R r. ere we introduce the dimensionless radial coordinate ξ = r/r. v z ξ = P P 0 R [ξ ln1/ξ ln1/ξ 1 Our aim now is to eliminate the coefficient of the square brackets because it s not in the desired answer. As the slug descends in the container, it displaces a certain volume of fluid per unit time. That same volume per unit time must be what flows up the side of the slug in the annular slit. The following relation can be written from this. dv dt = dv displaced from bottom of slug dt up side of slug in annular slit

5 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 5 of 9 The volumetric flow rate is velocity times area, so on the left side it s just πr. Since the velocity varies radially in the slit, we will have to integrate the velocity over the area on the right side. ˆ π R = v z da R = ˆ R R Make the substitution, to get R R π R = ˆ R π R = π R ˆ R v z πr dr R rv z dr { P P 0 R [ r 1 r 1 + R ln R r ln R } dr r ξ = r R Rξ = r dξ = dr R R dξ = dr, { P P 0 R = Rξ [ξ ln1/ξ { = R P P 0 R [ξ 3 ξ + 1 = P P 0 R [ξ 3 ξ 1 ξ ln ξ Bring the last term on the right side over to the left. ξ 4 + ξ ln ξ ξ ln ξ dξ = P P 0 R 1 = P P 0 R ln = P P 0 R 4 1 = P P 0 R 1 = P P 0 R 1 = P P 0 R = P P 0 R = P P 0 R ξ ln1/ξ } ln1/ξ R dξ dξ + dξ ξ ln ξ dξ [ξ 3 ξ 1 ξ ln ξ dξ [ ξ 4 4 ξ 1 } ξ ln1/ξ dξ ξ 4 + ξ ln ξ 1 14 [ ln 14 4 [ [ [ [ [

6 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 6 of 9 Consequently, P P 0 R Substitute this result into equation 1. v z ξ = = [ξ ln1/ξ ln1/ξ Divide both sides by and factor the minus signs. v z = ξ ln1/ξ ln1/ξ { ξ ln1/ξ = [ ln1/ξ + } 1 ln1/ξ [ { ξ 1 + = 1 ln1/ξ ln1/ξ + } 1 ln1/ξ [ { ξ } ln1/ξ = [ [ = 1 ξ + 1 ln1/ξ [ 1 ξ + 1 ln1/ξ = Therefore, where ξ = r/r. v z = 1 ξ 1 + ln1/ξ 1 1 +, Figure : This is a plot of the velocity distribution when = 0.4 and = for ξ 1.

7 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 7 of 9 Part b To determine the viscosity, the sum of the forces in the z-direction will be considered. There are five forces that need to be taken into account in the free body diagram of the slug: 1 the gravitational force, the buoyant force, 3 the viscous force acting on the side from the flow of fluid in the annulus, 4 the weight of the water pressure acting over the top of the slug s surface, and 5 the weight of the water pressure acting over the bottom of the slug s surface. The forces due to pressure are normal to the slug s surface, and the viscous force is parallel to the slug s surface in the direction opposing the slug s motion. Figure 3: This is the free body diagram of the slug. Gravitational force = Density of slug Volume of slug Acceleration = ρ 0 πr g Buoyant force = Density of fluid Volume of slug Acceleration = ρ πr g Viscous force = Shearing stress Surface area = τ rz r=r πr Force at bottom = Pressure at bottom Surface area = P 0 πr Force at top = Pressure at top Surface area = P πr The minus sign in the shearing stress is due to the fact that the fluid in the annulus is at a higher radius acting on the slug s surface, which has a lower radius. Use Newton s law of viscosity, τ rz = µdv z /dr, to evaluate it. Viscous force = τ rz r=r πr = µ dv z πr dr r=r = dv z dr πrµ r=r = 1 R πrµ 1 = πµ

8 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 8 of 9 Because the slug is falling at terminal velocity, the acceleration is zero, so the sum of the forces in the z-direction is equal to 0. Fz = ma z = 0 Plug in the forces on the left side. The positive z-direction is chosen to be upward, so forces pointing up are positive and those pointing down are negative. ρ 0 πr g + ρπr g Divide both sides by πr. ρ ρ 0 g πµ µ R + P 0 P Solve equation for P 0 P / and substitute the expression here. ρ ρ 0 g All that s left to do is to solve for µ. Therefore, Part c + P 0 πr P πr = µ R µ R = 0 R ρ ρ 0 g = 0 µ µ = 0 R ρ ρ 0 g µ = 0 R ρ ρ 0 g1 R ρ ρ 0 g1 + µ 1 + = 0 R ρ ρ 0 g[1 1 + = µ 1 + R [ ρ ρ 0 g = µ µ = ρ 0 ρgr [ ln For small slit widths, is just barely less than one: = 1 ε, where 0 < ε 1. Substitute this into the result of part b. µ = ρ 0 ρgr [ ln = ρ 0 ρg1 ε R = ρ 0 ρgr 1 ε The Taylor series expansion for ln1 ε is [ ln 1 1 ε [ ln1 ε. 1 1 ε ε ε ε ε + ε ln1 ε = ε 1 ε 1 3 ε3 1 4 ε4 1 5 ε5.

9 BSL Transport Phenomena e Revised: Chapter - Problem C.4 Page 9 of 9 Plug this in to the formula for µ. µ = ρ 0 ρgr [ 1 ε ε 1 ε 1 3 ε3 1 4 ε4 1 5 ε5 ε ε ε + ε = ρ 0 ρgr 1 ε + ε [ε + 1 v ε + 13 ε ε ε ε5 ε + 0 ε + ε = ρ 0 ρgr [ ε + 1 ε ε ε ε5 + 1 ε + ε Use long division to obtain a series for the quotient. ε + ε ε ε + 0ε 3 ε + 1 ε 1 4 ε4 1 4 ε5 + ε ε + ε 3 ε ε 3 ε ε ε + ε 1 ε + ε ε ε ε4 1 ε4 1 ε4 + 1 ε5 1 4 ε6 1 ε ε6 aving terms up to ε 5 will suffice. µ = ρ 0 ρgr [ ε + 1 ε ε ε ε5 + 1 ε + ε ε + 1 ε 1 4 ε4 1 4 ε5 + 1 ε + ε. Multiply the series together. = ρ 0 ρgr = ρ 0 ρgr Combine like-terms. = ρ 0 ρgr [ε + ε + 1 ε ε ε ε ε ε ε ε ε ε ε5 1 4 ε4 1 4 ε ε3 1 6 ε ε5 + Factor ε 3 /3 to obtain the final result. Therefore, µ = ρ 0 ρgr ε ε 13 0 ε +. +

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