UNIT V BOUNDARY LAYER INTRODUCTION

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1 UNIT V BOUNDARY LAYER INTRODUCTION The variation of velocity from zero to free-stream velocity in the direction normal to the bondary takes place in a narrow region in the vicinity of solid bondary. This narrow region of the flid is called bondary layer. The theory dealing with bondary layer flows is called bondary layer theory. 1. A very thin layer of the flid called the bondary layer in the immediate neighborhood of the solid bondary, where the variation of velocity from zero at the solid bondary to free-stream velocity in the direction normal to the bondary takes place. 2. The remaining flid which is otside the bondary layer. The velocity otside the bondary layer is constant and eqal to free-stream velocity. Laminar bondary layer: The leading edge of the srface of the plate where the thickness is small, the flow in the bondary layer is laminar thogh the main flow is trblent. This layer of the flid is said to be laminar bondary layer. Trblent bondary layer: The laminar bondary layer becomes nstable and motion of flid within, it is distrbed and irreglar which leads to a transition from laminar to trblent bondary layer. This short length over which the bondary layer flow changes from laminar to trblent is called transition zone. Frther downstream the transition zone the bondary layer is trblent and contines to grow in thickness. This layer of bondary is called trblent bondary layer. Laminar sb-layer: The region in the trblent bondary layer zone, adjacent to the solid srface of the plate. In this zone the velocity variation is inflenced only by viscos effects. Thogh the velocity distribtion wold be a parabolic crve in the laminar sb-layer zone bt in view of the very small thickness. That velocity variation is linear and so the velocity gradient can be considered constant. Therefore, the shear stress in the laminar sb-layer wold be constant and eqal to the bondary shear stress τ. Bondary layer thickness: It is defined as the distance from the bondary of the solid body measred in the y-direction to the point, where the velocity of the flid is approximately eqal to.99 times the free stream velocity of the flid. It is denoted by the symbol.

2 Displacement thickness: It is defined as the distance measred perpendiclar to the bondary of the solid body by which the bondary shold be displaced to compensate for the redction in flow rate on accont of bondary layer formation. It is denoted by *. * [ 1 (/U) ] dy It is also defined as: The distance perpendiclar to the bondary by which the free stream is displaced de to the formation of bondary layer. Momentm thickness: It is defined as the distance, measred perpendiclar to the bondary of the solid body, by which the bondary shold be displaced to compensate for the redction in momentm of the flowing flid on accont of bondary layer formation. It is denoted by θ. θ [ (/U) (/U) 2 ] dy Energy thickness: It is defined as the distance measred perpendiclar to the bondary of the solid body by which the bondary shold be displaced to compensate for the redction in kinetic energy of the flowing flid on accont of bondary layer formation. It is denoted by **. ** [ (/U) (/U) 3 ] dy Bondary condition for the velocity profiles: 1. At y, and ddy has some finite vale. 2. At y,u. 3. At y, / dy Trblent bondary layer on a flat plate: The thickness of the bondary layer, drag force on one side of the plate and co-efficient of drag de to trblent bondary layer on a smooth plate at zero pressre gradient are determined as in case of laminar bondary layer provider the velocity profile is known. Blasis on the basis of the experiment given the following velocity profile for a trblent bondary layer. U (y)n Where n1/7 for Re < 1 7 bt more than U (y)17

The above eqation is not applicable very near the bondary, where the thin laminar sb-layer of thickness ` exist. Here velocity distribtion is inflenced by viscos effects.

3 The above eqation is not applicable very near the bondary, where the thin laminar sb-layer of thickness ` exist. Here velocity distribtion is inflenced by viscos effects. Analysis of trblent bondary layer: (a) If Reynold nmber is more than 5 15 and less than 17the thickness of bondary layer and drag co-efficient are given as:.37(rex)1/5 and CD.72(ReL)1/5 Where xdistance from the leading edge Rex reynold nmber for length x Rel reynold nmber at the end of the plate (b) If reynold nmber is more than 1 7 bt less than 1 9, gave the empirical eqation as CD.455(log1 ReL) SEPARATION OF BOUNDARY LAYER: The loss of kinetic energy is recovered from the intermediate flid layer in contact with the layer adjacent to solid srface throgh momentm exchange process. Ths the velocity of the layer goes on decreasing. Along the length of the solid body, at a certain point a stage may come when the bondary layer may not be able to keep sticking to the solid body if it can t provide kinetic energy to overcome the resistance offer by the solid body, the bondary layer will be separated from the srface. This phenomenon is called bondary layer separation. The point on the body at which the bondary layer is on the verge of separation from the srface is called as the point of separation.

4 EFFECT OF PRESSURE GRADIENT ON THE BOUNDARY LAYER SEPARATION: Effect of pressre gradient (dpdx) on the bondary layer separation can be explained by considering the flow over a crved srface. The area of flow decreases and hence velocity increases. This means that flow gets accelerated in this region. De to increase in the velocity, the pressre decreases in the direction of the flow and hence pressre gradient (dpdx) is negative. Location of separation point: The separation point is determined from the condition, ( ) For a given velocity profile, it can be determine whether the bondary layer has separated or verge of separation or will not separate from the following condition. 1.If ( y)y is negative the flow has separated. 2.If ( y)y the flow is on the verge of separation. 3.If ( y)y is positive.the flow will not separate or flow will remain attached with the srface. Methods of preventing the separation of bondary layer: When the bondary layer separates from the srface, a certain portion adjacent to the srface has a back flow and eddies are continosly formed in this region and hence continos loss of energy takes place. Ths separation of bondary layer is ndesirable and attempts shold be made to avoid separation by varios methods. The following are the methods for preventing the separation of bondary layer: 1. Sction of the slow moving flid by a sction slot. 2. Spplying additional energy from a blower. 3. Providing a bypass in the slotted wing. 4. Rotating bondary in the direction of flow. 5. Providing small divergence in a diffser. 6. Providing gide-blades in a bend. 7. Providing a trip-wire ring in the laminar region for the flow over a sphere. Find the displacement thickness, the momentm thickness and energy thickness for the velocity distribtion in the bondary layer given by U y, where is the velocity at a distance y from the plate and U at y, where bondary layer thickness. Also calclate the vale of /θ. Given:

5 Velocity distribtion U y (i) Displacement thickness * is given by eqation, * (1 ) U [y ] 2 dy (1 y ) dy { y } U { is constant across a section} * 2 (ii) Momentm thickness, θ is given by eqation, Sbstitting the vale of U y, θ (1 ) dy U U θ y (1 y ) [ y2 y ] θ (iii) Energy thickness ** is given by eqation, as 6 dy ( y 2) dy ** 2 (1 U U 2) dy y y2 (1 2) dy { y } U [ y y3 3] dy [ y2 y ] (iv) θ ( 2 ) ( 6 ) 2 x 6 θ 3 Find the displacement thickness, the momentm thickness and energy thickness for the velocity distribtion in the bondary layer given by U 2 [ y ] - [ y ]2 Soltion: Given: Velocity distribtion (i) Displacement thickness * is given by eqation, Sbstitting the vale of * (1 ) dy U U 2 [ y ] - [ y ]2, we have U 2 [ y ] - [ y ]2

6 * {1 [2 ( y ) ( y )2 ]} dy {1 2 ( y ) ( y )2 } dy [ 2y2 + y ] * 3 (ii) Momentm thicknessθ, is given by eqation, θ (1 ) U U [ 2y [ 2y [ 2y 2] 4y2 2 5y2 2 dy ( 2y [1 2y + 2y y3 3 2 ) + y2 2] dy y2 2y3 2 3 y4 4 ] dy [1 ( 2y y4 4] dy [ 2y2 2 [ ] θ (ii) (iii) Energy thickness ** is given by eqation, ** 2 (1 U U 2) dy ( 2y ( 2y ( 2y ( 2y ( 2y ) 2 ) 2 8y3 3 (1 [ 4y2 2 (1 4y2 2 2y y4 4 2 ) + y4 4y3 4 2)] dy 5y3 + 4y4 y5 ] ([1 ( 2y 2)]2 ) dy 3 ]) dy y4 4 4y3 3 ) dy y2 + 4y4 + y6 4y5 ) y3 12y4 6y5 + y6 ) dy ( 2y y3 8y4 12y5 6y6 + y7 ) ** dy

7 DRAG FORCE ON A FLAT PLATE DUE TO BOUNDARY LAYER Consider the flow of a flid having free-stream velocity eqal to U, over a thin plate a shown in Fig. The drag force on the plate can be determined if the velocity profile near the plate is known. Consider a small length x of the plate at a distance of x from the leading edge as shown in Fig.(a). The enlarged view of the small length of the plate is shown in Fig.(b) Fig. Drag force on a plate de to bondary layer The shear stress τ is given by τ ( d ), where ( d ) is the velocity distribtion near the plate at dy y dy y y. The drag force or shear force on a small distance x is given by F D shear stress x area τ x x x b (1) [Taking width of plate b] Where F D drag force on distance x The drag force F D mst also be eqal to the rate of change of momentm over the distance x. Consider the flow over the small distance x. Let ABCD is the control volme of the flid over the distance x as shown in Fig. (b). The edge DC represents the oter edge of the bondary layer. Let velocity at any point within the bondary layer b width of plate Then mass rate of flow entering throgh the side AD ρ x velocity x area of strip of thickness dy ρ x x b x dy ρb. dy Mass rate of flow leaving the side BC mass throgh AD + x (mass throgh AD) x x ρbdy x [ (ρbdy)] x x From continity eqation for a steady incompressible flid flow, we have Mass rate of flow entering AD + mass rate of flow entering DC mass rate of flow leaving BC [ Area of strip bxdy] Mass rate of flow entering DC mass rate of flow throgh BC-mass rate of flow throgh AD

8 ρbdy x + x [ (ρbdy)] x x [ (ρbdy)] x x - ρbdy The flid is entering throgh side DC with a niform velocity U. Now let s calclate momentm flx throgh control volme. Momentm flx entering throgh AD momentm flx throgh strip of thickness by mass throgh strip x velocity Momentm flx leaving the side BC ρ 2 bdy (ρbdy) + [ x ρ2 bdy] x x Momentm flx entering the side DC mass rate throgh DC x velocity x x x (ρ2 bdy) [ ρbdy] x x x U ( Velocity U) [ ρbdy] x x As U is constant and so it can be taken inside the differential and integral Rate of change of momentm of the control volme Momentm flx throgh BC Momentm flx throgh AD momentm flx throgh DC ρ 2 bdy + [ x ρ2 bdy] x x - ρ 2 bdy - x [ ρubdy] x x [ x ρ2 bdy] x x - [ ρubdy] x x x [ x ρ2 bdy] - [ ρubdy] x x [ x (ρ2 b ρub)dy] x x [ρb x (2 U)dy] x x {For incompressible flid ρ is constant} ρb [ x (2 U)dy] x x.(2) Now the rate of change of momentm on the control volme ABCD mst be eqal to the total force on the control volme in the same direction according to the momentm principle. Bt for a flat plate p x. Which means there is no external pressre force on the control volme. Also the force on the side DC is negligible as the velocity is constant and velocity gradient is zero approximately. The only external force acting on the control volme is the shear force acting on the side AB in the direction from B to A as shown in Fig.(b). The vale of this force is given by eqation (1) as, F D τ x x x b The external force in the direction of rate of change of momentm - τ x x x b.(3)

9 According to momentm principle, the two vales given by eqations (3) and (2) shold be the same. - τ x x x b ρb [ x (2 U)dy] x x Cancelling x x b, to both sides, we have - τ ρ [ x (2 U)dy] or τ ρ [ x (2 U)dy] ρ [ (U x 2 )dy] ρ [ x U2 ( 2 U 2) dy] ρu 2 [ [1 ] dy ] x U U U τ ρu 2 [ [1 ] dy ] (4) x U U In eqation (4), the expression [1 ] U U also written as τ θ ρu 2 x dy is eqal to momentm thickness θ. Hence eqation (4) is..(5) Eqation (5) is known as Von Karman momentm integral eqation for bondary layer flows. This is applied to: 1. Laminar bondary layers 2. Transition bondary layers and 3. Trblent bondary layer flows. For a given velocity profile in laminar zone, transition zone or trblent zone of a bondary layer, the shear stress τ is obtained from eqation (4) or (5). Then drag force on a small distance x of the plate is obtained from eqation (1) as F D - τ x x x b Then total drag on the plate of length L on one side is L F D F D τ x b x dx {change x dx}..(6) Local Co-efficient of Drag [C D ]. It is defined as the ratio of the shear stress τ to the qantity 1 2 ρu2. It is denoted by C D Where Hence C D τ 1 2 ρu2 A Area of the srface (or plate) U Free-stream velocity ρ Mass density of flid (7) Bondary Conditions for the Velocity Profiles. The followings are the bondary conditions which mst be satisfied by any velocity profile, whether it is in laminar bondary layer zone, or in trblent bondary layer zone:

10 1. At y, and d dy 2. At y, U 3. At y, d dy has some finite vale For the velocity profile given in, find the thickness of bondary layer at the end of the plate and the drag force on one side of a plate 1m long and.8m wide when placed in water flowing with a velocity of 15 mm per second. Calclate the vale of co-efficient of drag also. Take μ for water.1 poise. Given: Length of plate, L 1m Width of plate, b.8m Velocity of flid (water) U 15mm/s.15 m/s μ for water μ.1 poise.1 1 Soltion: Ns Ns m2.1 m 2 Reynold nmber at the end of the plate i.e., at a distance of 1m from leading edge is given by R el ρul μ.15 x 1. 1 x.1 1 x.15 x ( ρ 1) (i) As laminar bondary layer exists pto Reynold nmber 2 x 1 5. Hence this is the case of laminar bondary layer. Thickness of bondary layer at x 1.m is given by eqation as, 5.48 x 5.48 x m 14.15mm. R 15 el (ii) Drag force on one side of the plate is given by eqation, F D.73 bμu ρul μ F D.73 x.8 x.1 x.15 x 15 { ρul μ R e L }.338N. (iii) Co-efficient of drag. C D is given by eqation as, C D 1.46 C D

11 S.No Velocity Distribtion C D U 2(y ) (y )2 U 3 2 (y ) 1 2 (y )3 U 2 (y ) 2 (y )3 + ( y )4 U sin (π 2 y ) 5.48 x R ex x R ex x R ex x R ex Blasis s Soltion 4.91 x R ex For the velocity profile in laminar bondary layer as, U 3 2 (y ) 1 2 (y )3 find the thickness of the bondary layer and the shear stress 1.5m from the leading edge of a plate. The plate is 2m long and 1.4 m wide and is placed in water which is moving with a velocity of 2 mm per second. Find the total drag force on the plate if μ for water.1 poise. Given: Velocity profile is U 3 2 (y ) 1 2 (y )3 Distance of x from leading edge, x 1.5m Length of plate, L 2m Width of plate, b 1.4m Velocity of plate U 2 mm/s.2 m/s Viscosity of water, μ.1 poise.1 1 Soltion:.1 Ns/m2 For the given velocity profile, thickness of bondary layer is given by eqation as 4.46 x R e x 4.46 x m mm. Shear stress (τ ) is given by τ.323 Ux x [Here, R ex ρux μ R ex.323 x.1 x τ.235 N/m 2..2x1.5 1 x 3].1

12 Drag Force (F D ) on the side of the plate is given by as F D.646 μu ρul μ x b.646 x.1 x.2 x 1x.2x2..1 x x.1 x.2 x 4 x N F D.1138N Total drag force Drag force on both sides of the plate 2x N Total drag force.2276 N Air is flowing over a smooth plate with a velocity of 1 m/s. The length of the plate is 1.2 m and width.8 m. If laminator bondary layer exists p to a vale of R e 2 x 1 5, find the maximm distance from the leading edge pto which laminar bondary layer exists. Find the maximm thickness of laminar bondary layer if the velocity profile is given by 2 U (y) (y )2. Take kinematic viscosity for air.15 strokes. Given: Velocity of air, Length of plate, Width of plate, U 1 m/s L 1.2 m b.8 m Reynold nmber pto which laminar bondary exists 2 x 1 5 υ for air.15 stokes.5x1-4 m 2 /s ρ 1.24 kg/m 3 Soltion: If R ex 2x1 5, then x denotes the distance from leading edge pto which laminar bondary layer exists 1 x x 2x1 5.15x1 4 x 2x15 x.15x1 4 1 x 3mm..3 m Maximm thickness of the laminar bondary for the velocity profile, U 2 (y ) (y )2 is given by eqation as 5.48x x R e x 5.48x.3 2x m 3.67mm

13 Air is flowing over a flat plate 5 mm long and 6 mm wide with a velocity of 4 m/s. The kinematic viscosity of air is given as.15x1-4 m 2 /s 2. Find (i) the bondary layer thickness at the end of the plate, (ii) Shear stress at 2mm from the leading edge and (iii) drag force on one side of the plate. Take the velocity profile over the plate as U sin(π 2, y ) and density of air 1.24 kg/m3. Given: Length of plate, L 5 mm.5m Width of plate, Velocity of air, b 6 mm.6 m U 4 m/s Kinematic viscosity, υ.15 x 1-4 m 2 /s Mass density, ρ 1.24 kg/m 3 For the velocity profile, U sin(π, y ), we have 2 Soltion: (i) Bondary layer thickness at the end of the plate means vale of at x.5m. First find Reynold nmber. R ex ρux μ Ux υ 4x.5.15x x15. Hence bondary layer is laminar over the entire length of the plate as Reynold nmber at the end of the plate is 1.33x1 5. at x.5 m for the given velocity profile is given by eqation as x R e x 4.795x m 6.56 mm. 1.33x15 (ii) Shear stress at any distance from leading edge is given by τ.327 μu x R ex At x 2 mm.2 m, R ex U x x υ 4x.2.15x x τ.327 x μ x 4 x Bt μ υ x ρ { υ ρ.15 x 1-4 x x1-4, μ υxρ} τ.327 x.186 x 1 4 x 4 x N/m 2.2 (iii) Drag force on one side of the plate is given by eqation F D.655 x μu x b x ρul υ

14 .655 x.186 x 1-4 x 4. x.6 x UL υ { υ ρ }.29234x1-4 x 4 x.5.15 x 1 4 F D.186 N A thin plate is moving in still atmospheric air at a velocity of 5 m/s. The length of the plate is.6m and width.5m. Calclate (i) the thickness of the bondary layer at the end of the plate, and (ii) drag force on one side of the plate. Take density of air as 1.24 kg/m 3 and kinematic viscosity.15 stokes. Given: Velocity of air, Length of plate, Width of plate, U 5 m/s L.6 m b.5 m Density of air, ρ 1.24 kg/m 3 Kinematic viscosity, υ.15 stokes.15x1-4 m 2 /s Soltion: Reynold nmber, R e UL υ 5x x1 4 As R e is less than 5x1 5, hence bondary layer is laminar over the entire length of the plate. (i) Thickness of bondary layer at the end of the plate by Blasis s soltion is 4.91 x 4.91 L 4.91 x.6 2 R e x.658 m 6.58 mm R e x (iii) Drag force on one side of the plate is given by eqation as, therefoere, C D F D 1 2 ρ A U2 F D 1 2 ρau2 x C D Where C D from Blasis s soltion, C D C D.297 F D 1 2 ρau2 x C D x 1.24 x.6 x.5 x 52 x.297 F D N.

15 Note. If no velocity profile is given in the nmerical problem bt bondary layer is laminar, then Blasis s soltion is sed. A plate of 6 mm length and 4 mm wide is immersed in a flid of sp.gr..9 and kinematic viscosity (υ) 1-4 m 2 /s. The flid is moving with a velocity of 6 m/s. Determine (i) bondary layer thickness (ii) shear stress at the end of the plate, and (iii) drag force on one side of the plate. Given: As no velocity profile is given in the above problem, hence Blasis s soltion will be sed. length of plate, Width of plate, Sp.gr.of flid, S.9 L 6 mm.6 m b 4 mm.4 m Density, ρ.9 x 1 9 kg/m 3 Velocity of flid U 6 m/s Kinematic viscosity υ 1-4 m 2 /s Soltion: U x L Reynold nmber, R el υ 6 x x 14. As R el is less than 5 x 1 5, hence bondary layer is laminar over the entire length of the plate. (i) Thickness of bondary layer at the end of the plate from Blasis s soltion is 4.91 x R e x where x.6m and R ex 3.6x x x m 15.5 mm (ii) Shear stress at the end of the plate is τ.33 ρu x 9 x N/m x 1 4 (iii) Drag force (F D ) on one side of the plate is given by F D 1 2 ρau2 x C D Where from Blasis s soltion is C D x F D 1 2 ρau2 x C D 1 2 x 9 x.6 x.4 x 62 x.699 [ A Lxb.6x. 4] F D N

4 Exact laminar boundary layer solutions

4 Exact laminar boundary layer solutions 4 Eact laminar bondary layer soltions 4.1 Bondary layer on a flat plate (Blasis 1908 In Sec. 3, we derived the bondary layer eqations for 2D incompressible flow of constant viscosity past a weakly crved