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3. Pressure Flow, Flat Plates. Adverse pressure gradients in flow over bodies lead to flow reversal & separation, wakes, and increased drag. To examine such behavior, consider fully developed, laminar flow between two flat pates a distance 8 apart. The bottom plate is fixed, and the top plate moves with speed U (representing the outer flow over a boundary layer of thickness 8). (a) Derive the critical adverse pressure gradient, (dp/dx)*, above which there will be flow reversal, by solving the N-S equation in the coordinates shown, to get a relation for u(y), du/dy, and the critical (dp/dx)* in terms of 8, U, and u above which there μ will be flow reversal at the lower plate. (b) If 8 = 5 mm and U = 10 m/s, evaluate (dp/dx)* if the fluid is air. Moving plate Fixed plate 2/ 1.0 0.8 0.6 0.4 0.2 Back- flow P=-3 0 -0.4 -0.2 0 0.2 0.4 u U (b) U 0.6 0.8 1.0 1.2 b 1.4 (a) FIGURE 6.30 Viscous flow between parallel plates with the bottom plate fixed and the upper plate moving (Couette flow): (a) coordinate system and notation used in analysis and (b) velocity distribution as a function of parameter, P, where P = -(b²/2μU)dp/dx. (From Ref. 8, used by permission.) Assume Patm = 10^5, Pa = 14.7 psi; pwater ~ 1000 kg/m3; pair ~ 1.2 kg/m3; μwater ~ 10^-3 N•s/m2; uair ~ 2 x 10^-5 N.s/m2; Vwater ~ 10^-6 m2 /s; Vair ~ 1.67 x 10^-5 m2 /s; g = 9.8 m/s^2 .; 1 m/s = 2.24 mph; 1 lbf = 4.45 N; 1 m^3 = 264 gallons Note : when du/ dy < 0 at y = 0 Ans OM: (a) u(y) ~ A(y/8) + B(y/8)² ; (dp/dx)* ~ C(µU/8²); (b) (dp/dx)*: 10¹Pa/m

3. Pressure Flow, Flat Plates. Adverse pressure gradients in flow over bodies lead to flow reversal & separation, wakes, and increased drag. To examine such behavior, consider fully developed, laminar flow between two flat pates a distance 8 apart. The bottom plate is fixed, and the top plate moves with speed U (representing the outer flow over a boundary layer of thickness 8). (a) Derive the critical adverse pressure gradient, (dp/dx)*, above which there will be flow reversal, by solving the N-S equation in the coordinates shown, to get a relation for u(y), du/dy, and the critical (dp/dx)* in terms of 8, U, and u above which there μ will be flow reversal at the lower plate. (b) If 8 = 5 mm and U = 10 m/s, evaluate (dp/dx)* if the fluid is air. Moving plate Fixed plate 2/ 1.0 0.8 0.6 0.4 0.2 Back- flow P=-3 0 -0.4 -0.2 0 0.2 0.4 u U (b) U 0.6 0.8 1.0 1.2 b 1.4 (a) FIGURE 6.30 Viscous flow between parallel plates with the bottom plate fixed and the upper plate moving (Couette flow): (a) coordinate system and notation used in analysis and (b) velocity distribution as a function of parameter, P, where P = -(b²/2μU)dp/dx. (From Ref. 8, used by permission.) Assume Patm = 10^5, Pa = 14.7 psi; pwater ~ 1000 kg/m3; pair ~ 1.2 kg/m3; μwater ~ 10^-3 N•s/m2; uair ~ 2 x 10^-5 N.s/m2; Vwater ~ 10^-6 m2 /s; Vair ~ 1.67 x 10^-5 m2 /s; g = 9.8 m/s^2 .; 1 m/s = 2.24 mph; 1 lbf = 4.45 N; 1 m^3 = 264 gallons Note : when du/ dy < 0 at y = 0 Ans OM: (a) u(y) ~ A(y/8) + B(y/8)² ; (dp/dx)* ~ C(µU/8²); (b) (dp/dx)*: 10¹Pa/m

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter5: Analysis Of Convection Heat Transfer
Section: Chapter Questions
Problem 5.49P
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3. Pressure Flow, Flat Plates. Adverse pressure gradients in flow over bodies lead to flow reversal & separation, wakes, and increased drag. To examine such behavior, consider fully developed, laminar flow between two flat pates a distance 8 apart. The bottom plate is fixed, and the top plate moves with speed U (representing the outer flow over a boundary layer of thickness 8). (a) Derive the critical adverse pressure gradient, (dp/dx)*, above which there will be flow reversal, by solving the N-S equation in the coordinates shown, to get a relation for u(y), du/dy, and the critical (dp/dx)* in terms of 8, U, and u above which there μ will be flow reversal at the lower plate. (b) If 8 = 5 mm and U = 10 m/s, evaluate (dp/dx)* if the fluid is air. Moving plate Fixed plate 2/ 1.0 0.8 0.6 0.4 0.2 Back- flow P=-3 0 -0.4 -0.2 0 0.2 0.4 u U (b) U 0.6 0.8 1.0 1.2 b 1.4 (a) FIGURE 6.30 Viscous flow between parallel plates with the bottom plate fixed and the upper plate moving (Couette flow): (a) coordinate system and notation used in analysis and (b) velocity distribution as a function of parameter, P, where P = -(b²/2μU)dp/dx. (From Ref. 8, used by permission.) Assume Patm = 10^5, Pa = 14.7 psi; pwater ~ 1000 kg/m3; pair ~ 1.2 kg/m3; μwater ~ 10^-3 N•s/m2; uair ~ 2 x 10^-5 N.s/m2; Vwater ~ 10^-6 m2 /s; Vair ~ 1.67 x 10^-5 m2 /s; g = 9.8 m/s^2 .; 1 m/s = 2.24 mph; 1 lbf = 4.45 N; 1 m^3 = 264 gallons Note : when du/ dy < 0 at y = 0 Ans OM: (a) u(y) ~ A(y/8) + B(y/8)² ; (dp/dx)* ~ C(µU/8²); (b) (dp/dx)*: 10¹Pa/m
Transcribed Image Text:3. Pressure Flow, Flat Plates. Adverse pressure gradients in flow over bodies lead to flow reversal & separation, wakes, and increased drag. To examine such behavior, consider fully developed, laminar flow between two flat pates a distance 8 apart. The bottom plate is fixed, and the top plate moves with speed U (representing the outer flow over a boundary layer of thickness 8). (a) Derive the critical adverse pressure gradient, (dp/dx)*, above which there will be flow reversal, by solving the N-S equation in the coordinates shown, to get a relation for u(y), du/dy, and the critical (dp/dx)* in terms of 8, U, and u above which there μ will be flow reversal at the lower plate. (b) If 8 = 5 mm and U = 10 m/s, evaluate (dp/dx)* if the fluid is air. Moving plate Fixed plate 2/ 1.0 0.8 0.6 0.4 0.2 Back- flow P=-3 0 -0.4 -0.2 0 0.2 0.4 u U (b) U 0.6 0.8 1.0 1.2 b 1.4 (a) FIGURE 6.30 Viscous flow between parallel plates with the bottom plate fixed and the upper plate moving (Couette flow): (a) coordinate system and notation used in analysis and (b) velocity distribution as a function of parameter, P, where P = -(b²/2μU)dp/dx. (From Ref. 8, used by permission.) Assume Patm = 10^5, Pa = 14.7 psi; pwater ~ 1000 kg/m3; pair ~ 1.2 kg/m3; μwater ~ 10^-3 N•s/m2; uair ~ 2 x 10^-5 N.s/m2; Vwater ~ 10^-6 m2 /s; Vair ~ 1.67 x 10^-5 m2 /s; g = 9.8 m/s^2 .; 1 m/s = 2.24 mph; 1 lbf = 4.45 N; 1 m^3 = 264 gallons Note : when du/ dy < 0 at y = 0 Ans OM: (a) u(y) ~ A(y/8) + B(y/8)² ; (dp/dx)* ~ C(µU/8²); (b) (dp/dx)*: 10¹Pa/m
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