2.25 Advanced Fluid Mechanics
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1 MIT Department of Mechanical Engineering.5 Advanced Fluid Mechanics Problem 6.0a This problem is from Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin Consider a steady, fully developed laminar flow in an annulus with inside radius R and outside radius. (a Find a relation between the pressure gradient dp, the volume flow rate Q, the fluid viscosity µ,, dx and R. (b Fin the limiting form of the relation for a very thin annulus by expressing it in terms of and h h, where h = R, and taking the limit 0. Compare with the formula for fully developed laminar flow between parallel flat plates separated by a distance h. (c In the opposite limit R 0, does the relation of (a reduce to the formula for Hagen-Poiseuille flow in a circular pipe of radius? Discuss your answer..5 Advanced Fluid Mechanics Copyright c 008, MIT
2 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0a Solution: From the N-S in cylindrical coordinates, the equation can be reduced to p ( v x 0= + r, (6.0aa µ x r r r where the first term is approximately a constant across the space between the cylinders (long cylinder approximation, then ( v x 0= K + r, (6.0ab r r r then, integrating, ( v x r v x r C v x Krdr = r dr, K + C = r, K + =. (6.0ac r r r r r Now, integrating again Then, applying the boundary conditions, r C v x r (K + dr = ( dr, K + C ln(r+ C = v x, (6.0ad r r v x ( =0,v x (R =0, (6.0ae the constants can be obtained. Then, R R K + C ln( + C =0, OR K + C ln(r + C =0. (6.0af Now, substracting the solutions to obtain C, then, Now, re-expressing in terms of the requested variables, K ( R + C ln =0, (6.0ag R K ( R C =. (6.0ah ln R R K K ( Φ (Φ C = R, C = R, (6.0ai ln Φ ln Φ where, Φ= R /. Now, for C, we can use any of the two equations, R R C = K C ln(, C = K C ln(r. (6.0aj.5 Advanced Fluid Mechanics Copyright c 008, MIT
3 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0a Upon substitution of C, simplifying, K R R ( R K ( R C = K + ln(, C = K + ln(r, (6.0ak ln R R ln R K ( ( R C = + ln(, (6.0al ln R R K ( ( R C = R + ln(r. (6.0am ln R R R Then the velocity is dp [ R ( r ] v x = r R ln (6.0an µ dx ln( /R R Now, to obtain the flux, let s integrate this expression, π π ( r v x rdrdθ = K + C ln(r+ C rdrdθ, (6.0ao 0 R 0 R π ( r ( r 3 K + C ln(r+ C rdrdθ = π K + C r ln(r+ C r dr (6.0ap 0 R R After integration, ( ( 3 R r Kr r r [ ] π K + C r ln(r+ C r dr = π + C + C ln(r, (6.0aq R 6 R then, finally, ( K(R R (R [ ] [ ] R R R Q = π + C + C ln( C ln(r. (6.0ar 6 Now, substituting C and C, πk ( R R R R R (R R Q = + ( R + (6.0as ln( /R Now, re-expressing in terms of the requested variables, And simplifying again, KR ( Φ ( KR C = + ln(, C = + Φ ln Φ ln(. (6.0at ln Φ ( ( πkr (Φ + (Φ Q = (6.0au ln Φ.5 Advanced Fluid Mechanics 3 Copyright c 008, MIT
4 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0a D Problem Solution by MC, Fall Advanced Fluid Mechanics Copyright c 008, MIT
5 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0b Problem 6.0b This problem is from Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin Solution: Now, factorizing taking into account that = R + H, then R = H, Q = ( R Kπ (R H +( H ln( +( H H ln( H + 8 R ln( H Now, let s substitute H using F = H/, Q = ( R Kπ (R F R +( F ln( +( F R ln( + R F F 8 R ln( F Now, taking the limit as F 0, but keeping the higher order terms, (6.0ba (6.0bb Q = KπF 3, and substituting the value of K, and the original variables, (6.0bc dp Q = πh 3, (6.0bd 6µ dx ( Q = H3 dp (π, (6.0be µ dx which corresponds to the solution of a pressure driven flow between two plates separated by a distance H, over a length equal to the average circumference of the annulus. NOTE: SEE PLOTS OF THE SOLUTIONS USING THE ATTACHED MATLAB FILES, PLAY WITH THE SOLUTIONS TILL THE LIMITS MAKE SENSE TO YOU..5 Advanced Fluid Mechanics 5 Copyright c 008, MIT
6 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0b D Problem Solution by MC, Fall Advanced Fluid Mechanics 6 Copyright c 008, MIT
7 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0c Problem 6.0c This problem is from Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin Solution: Now, taking the limit asφ 0 of part a solution, πkr ( (Φ ( (Φ lim Q = lim +, (6.0ca Φ 0 Φ 0 lnφ πkr ( lim Q = lim ( +, (6.0cb Φ 0 Φ 0 πkr lim Q =, (6.0cc Φ 0 8 R 3 ( dp lim Q = (π, (6.0cd Φ 0 6µ dx which is the solution for Poiseuille flow for a simple tube. You may have guessed that the solution did not converge to this value, i.e. the velocity profile had a hole in the center, but this is wrong. The solution converges to the simple tube flow because as the inner cylinder becomes smaller, the area that it uses to transmit vorticity decreases, and as the area decreases, its influence decreases too (Think of a small string (hot wire inside the tube for measuring flow, and think how small are the disturbances that it creates in the flow. To further verify that the solution makes physical sense, let s look at the product r τ viscous to show that the viscous force per unit length decreases as r 0. Using the velocity profile, the viscous stress can be obtained, now, let s evaluate at r = R, and multiply by R, now, taking the limit as R 0, ( dv x R µ = µk r, (6.0ce dr ln( /R r ( dv x R µr = µk R, (6.0cf dr R ln( /R dv x lim µr = lim µk(r =0, (6.0cg R 0 dr R R 0 then, the net viscous force goes to 0 as the radius approaches 0. NOTE: SEE PLOTS OF THE SOLUTIONS USING THE ATTACHED MATLAB FILES, PLAY WITH THE SOLUTIONS TILL THE LIMITS MAKE SENSE TO YOU..5 Advanced Fluid Mechanics 7 Copyright c 008, MIT
8 Viscous Flows A.H. Shapiro and A.A. Sonin 6.0c D Problem Solution by MC, Fall Advanced Fluid Mechanics 8 Copyright c 008, MIT
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