Laminar and Turbulent developing flow with/without heat transfer over a flat plate Introduction The purpose of the project was to use the FLOLAB software to model the laminar and turbulent flow over a flat plate. The results of the model under different flow conditions were compared with the theoretical predictions. Detailed analysis of the data obtained was done to characterize the flow parameters like Nusselt Number, Skin Friction Coefficient, shear layer thickness, velocity and temperature. Effect of change in geometry and initial flow conditions were also studied. Theory Boundary Layer Definition 1. Boundary layer thickness (δ ): defined as the distance away from the surface where the local velocity reaches to 99% of the free-stream velocity, that is u(y=δ )=.99U. Somewhat an easy to understand but arbitrary definition.
* 2. Displacement thickness ( δ ): Since the viscous force slows down the boundary layer flow, as a result, certain amount of the mass has been displaced (ejected) by the presence of the boundary layer (to satisfy the mass conservation requirement). Imagine that if we displace the uniform flow away from the solid surface by an amount, such that the flow rate with the uniform velocity will be the same as the flow rate being displaced by the presence of the boundary layer. The essence of the boundary layer approimation is that the shear layer is thin, δ <<. This is true if Re>>1.In the boundary layer u i.e. the velocity in the direction scales with L and v,the velocity in the y direction scales with δ.therefore u>>v.also the u u v v following approimations like << and << are made to simplify the Navier y y Stokes equations to the corresponding boundary layer forms. u u + =... continuity y u u u + v y = U e du d e 2 u + ν... momentum Steady, incompressible, 2-D 2 y laminar flow. ρc p T ( u 2 T T + v ) = k 2 y y u + µ ( ) y 2... energy du 1 dp The term U e e = = for flow over a flat plate as U e = constant. d ρ d U e Blasius found a celebrated solution for Re>>1 with a similarity variable η = y. 2ν Using this we have ' u = U f ( η) v = e νu e ( ηf ' f ) 2 Substituting these variables into the -momentum equation of the boundary is going to give us the following ODE as a function of η. f ''' + ff '' = Assuming no slip conditions we have u(,)=v(,)= and The free stream merge condition u(, )=U e
These convert to f f ' ' () = f () = ( ) = 1 Solving the Blasius Equation we obtain the following results:- Parameters Eact from Blasius θ.664 Re * δ 1.721 Re δ 5 99% Re C Re.664 f D C Re 1.328 Definitions: - For Laminar, where τ w u = µ y w C f = 2 2τ w / ρu e 1 CD ( L) = C f d = L L 2C f ( L) w 1/ 2 / k( Tw Te ).332 Re Nu = q = Pr 1/ 3 For turbulence, C f =.455 / ln Nu =.296 Re 2 (.6 Re 4 / 5 Pr Flow Conditions: - LAMINAR FLOW: - WATER: - 1/ 3 )...6 < Pr < 6
Re Heat Transfer Length of Plate 1 5 T w =4 K,T in =3 K 1 1 5 T w =3 K,T in =3 K 1 1 5 T w =4 K,T in =3 K 5 5*1 5 T w =4 K,T in =3 K 1 5*1 5 T w =3 K,T in =3 K 1 15 T w =4 K,T in =3 K 1 15 T w =3 K,T in =3 K 1 1 4 T w =4 K,T in =3 K 1 1 4 T w =3 K,T in =3 K 1 AIR : - Re Heat Transfer Length of Plate 1 5 T w =4 K,T in =3 K 1 1 5 T w =3 K,T in =3 K 1 1 5 T w =4 K,T in =3 K 5 5*1 5 T w =4 K,T in =3 K 1 5*1 5 T w =3 K,T in =3 K 1 1 T w =4 K,T in =3 K 1 1 T w =3 K,T in =3 K 1 1 6 T w =4 K,T in =3 K 1 1 6 T w =3 K,T in =3 K 1 Re = 1 6 falls in the transition zone. To compare the laminar and the turbulent models,we ran this case under both the conditions.
Results Laminar Case Velocity Analysis: Fluid: Water Vinlet velocity=.8954 m/sec,re=1.25.2 X velocity at inlet X velocity at X=.2 m X velocity at X=.4 m X velocity at X=.6 m X velocity at X=.8 m X velocity at X=.9 m X velocity at X=outlet.15 Y.1 Shear stress and hence Cf decreases as X increases.the value of Delta increases..5.5.1.15 X velocity
Re=15,Vinlet=.1345 m/s.6.5.4 X velocity at inlet X velocity at X=.2 X velocity at X=.4 X velocity at X=.6 X velocity at X=.8 X velocity at X=.9 X velocity at outlet Y.3.2.1.5.1.15.2 X velocity Observations: - Shear stress and hence C f decreases as increases, this is because as Reynolds # increases with, C f goes down. - δ becomes thicker as increases. - As increases we observe that the velocity eceeds the inlet velocity. It shows a problem in Flowlab s approach in solving the velocity profile equations. - In the case of the low Re # (15), δ is much thinner than a higher Re #
Fluid: Air Re=1^5 velocity=1.34m/s y direction.2.18.16.14.12.1.8.6.4.2 -.4.1.6 1.1 1.6 velocity inlet =.2 =.4 =.6 =.9 =.8 outlet Re=1^2 velocity=.134m/s y direction.5.45.4.35.3.25.2.15.1.5.5.1.15.2 velocity inlet =.2 =.4 =.6 =.8 =.9 outlet Observations: - Same as in the case of water - For Re=1,i.e under creep flow the velocity profiles does not obey the Boundary layer theory any more.
Analysis of the Coefficient of Skin Friction: Fluid: Air/Water skin friction coefficient Water.1.9.8.7 Cf Re=1 Cf Re=15 Cf Re=5 Cf Re=1.6 Cf.5.4.3.2.1.2.4.6.8 1 1.2 X
Skin friction coefficient Water.7.6 Cf.5.4.3.2 Re=1^2 Re=1^5 Re=5*1^5 Re=1^6.1.2.4.6.8 1 1.2 direction Observations - With the increase of the value of C f decreases as Re increases. C f 1/ Re - As the value of inlet velocity is decreased so that the Re decreases the skin friction coefficient goes up. - For low Re i.e at 15 or 1 the C f does not follow the relation C f Re =.664 as predicted by the Boundary layer solution by Blasius. Cf Water at Re=1^5 Cf.3.25.2.15.1.5 Cf Re=1.2.4.6.8 1 1.2 X Cf.3.25.2.15.1.5 Cf Air at Re=1^5 Re=1^5.2.4.6.8 1 1.2
Observation - For two different fluids water and air with varying viscosity and density if the Re is maintained constant at a value say 1^5 then C f gives the same profile at different as C f depends on Re only. This shows that Re drives the problem in incompressible fluid flow. Comparison of Flowlab C f values with Theoretical (water).9 W all skin friction,re=1,vinlet=.8954 m/.18 Re=15,Vinlet=.1345 m /s Wall skin friction coefficient.8.7.6.5.4.3.2.1 wall skin friction blausius prediction.16.14.12.1.8.6.4.2 s k in fric tio n coefficient Predicted skin friction.2.4.6.8 1.5 1 1.5 X X Observations - C f matches with the theoretical value quite closely for Re=1^5.But at lower values of Re say 15 the boundary layer theory has got large discrepancies. So the theoretical values do not match the flolab generated results
Comparison of Flolab C f values with Theoretical values(air) Cf Re=1^5.9.8.7.6 Cf.5.4.3.2.1.2.4.6.8 1 1.2 direction flowlab theoretical Cf Re=1^2.7.6.5 Cf.4.3.2 flowlab theoretical.1.2.4.6.8 1 1.2 direction Observations: - At low Reynolds number the boundary layer theory breaks down,so the value of C f does not match with that predicted by B.L. The deviations are appreciable.
Nusselt Number with/without heat transfer (water) Nu,various Reynolds number,with heat transfer 5 45 4 Nu 35 3 25 Re=1 Re=1 Re=5 Re=15 2 15 1 5.2.4.6.8 1 1.2 X Observations 1/ 2 3 - Nu = q k( T T ) =.332 Re Pr 1/ for laminar case. Here for a constant Pr as Re w / w e increases with,the nusselt number increases as well as Re
Nusselt number without heat transfer 35 3 Nusselt 25 2 15 Re=1 Re=15 1 5.2.4.6.8 1 1.2 X Observations - With T wall =T e the profile doesn t obey the Polhaussen Pr 1/3 law.
Comparison of Nu with theoretical values(water) W all nusselt number,re=1,tw=3k,tinlet=4k 25 2 Nusselt number 15 1 5 W all nusselt number Nusselt number predicted by polhaussen.2.4.6.8 1 X(m) Observations - The value from Flolab closely matches the theoretical value as calculated by 1/ 2 1/ 3 Polhaussen. Nu = q k( T T ) =.332 Re Pr w / w e
Nusselt no Re=1 without heat transfer 35 Nusselt Number 3 25 2 15 The mismatch is due to the fact that at Tw=Te,the theoretical epression for nusselt number.332*(re)^.5*(pr)^.3333 is not valid plate-wall predicted 1 5.2.4.6.8 1 1.2 Y
Nusselt Number for Air Nu number with heat transfer 35 3 Nu number 25 2 15 1 Re=1^2 Re=1^5 Re=5*1^5 Re=1^6 5.2.4.6.8 1 1.2 direction Nu number no heat transfer 5 4 Nu 3 2 1 Re=1^2 Re=1^5 Re=5*1^5 Re=1^6.2.4.6.8 1 1.2-1 direction
25 2 Nu at Re=1^5 with heat transfer Water 12 1 Nu number at Re=1^5 with heat transfer for air Nu 15 1 Re=1 Nu 8 6 4 Re=1^5 flowlab 5 2.5 1 1.5 X.2.4.6.8 1 1.2 direction Observations - Nu varies as Pr 1/3.Pr of air is appro=.72 and Pr for Water=6.So the Nu water /Nu air =2.2.From the chart it is obvious that The flolab results do agree with these for a constant Re=1^5
Temperature Profiles for Water.15 Temperature profiles Re=1^5 with heat transfer.14.13.12.11 Y.1.9.8.7.6 Tem perature at different X locations Tem perature at X=.2 m Tem perature at X=.4 m Tem perature at X=.6 m Tem perature at X=.8 m Tem perature at X=.9 m Tem perature at X=1 m.5.4.3.2.1 29 3 31 32 33 34 35 36 37 38 39 4 41 tem perature Observations: - The profiles closely matches the profile as found by polhaussen - The temperature is almost constant when T w =T in
Re=1^5 without heat transfer 3.1 T 3 299.999 299.998 299.997 299.996 299.995 299.994 inlet =.2 =.4 =.6 =.8 =.9 outlet 299.993.1.2.3.4.5.6 y direction T profile at Re=15 w ith heat transfer 41 39 T 37 35 33 31 T at X= T at X=.2 T at X=.4 T at X=.6 T at X=.8 T at X=.9 T at X=1 29 27 25.1.2.3.4.5.6.7.8.9.1 Y
Effect of length change of the plate (water) W all friction,re=1.9.8.7.6.5 W all friction at L=5 m wall friction at L=1 m Cf.4.3.2.1.2.4.6.8 1 1.2 X Nusselt number at Re=1^5 25 2 nusselt number for L=5 m Nusselt with L=1 m Nusselt No 15 1 5 1 2 3 4 5 X/L
Observations - if the length of the plate and inlet velocity is changed such that Re is constant then Nu and C f does not change at all as evident from the above graphs Temperature Profiles for Air Temperature Re=1^2 41 Temperature 39 37 35 33 31 29 27 inlet =.2 =.4 =.6 =.8 =.9 outlet 25.1.2.3.4.5 y direction Re=1^5 Temperature 3.1 Temperature 3 299.999 299.998 299.997 299.996 299.995 299.994 inlet =.2 =.4 =.6 =.8 =.9 outlet 299.993.1.2.3.4.5.6 y direction Observations: - The profiles closely matches the profile as found by polhaussen - The temperature is almost constant when T w =T in
Re=1 6 is a typical case of transition, although the laminar model shows large inaccuracies for Re=5*1 5.So we have used both the laminar as well as the turbulent model to run a case where the fluid is air for Re=1 5. Cf Re=1^6.7.6.5 Cf.4.3.2 laminar flow turbulent flow.1.5 1 direction Nu number Re=1^6 Nu 18 16 14 12 1 8 6 4 2.2.4.6.8 1 1.2 X direction lamina flowr turbulent flow
Observations: - Turbulent model predicts a much higher Nu and C f than the laminar counterpart. We believe that at 1 6 the turbulent model must be the better one to solve the flow problem. Sample velocity contour as generated by FLOLAB at different X locations
Contour plot from flolab Vector Plot from flolab
Streamlines as plotted by flolab
Turbulent Theory : - Approach suggested by Kestin and Persen is to assume that the streamwise velocity u in the boundary layer is correlated by inner variables u(, y) + + + yv *( ) = u f ( y ), y = * v ( ) ν These velocity assumptions when substituted directly into the streamwise boundary layer momentum equation with du e / d = u u 1 τ u + v = y ρ y The turbulent boundary layer begins at = and the integral evaluated algebraically Re = λ Gdλ G in its practical range 2<λ<4 is G(λ)=8.e.48λ.455 C f for flat plate 2 ln (.6 Re ) So from Prandtl s seventh power law yields.58 C f is not accurate. The inner variable approach can be etended readily 1/ 5 Re to variable pressure gradients.
Local Skin friction on a smooth flat plate for a turbulent flow showing several theories The transition occurs from laminar to turbulent Re=5e5.
TURBULENT FLOW CONDITIONS: - WATER Re Heat Transfer Length of Plate 2.23*1 9 T w =4 K,T in =1 K 1 2.23*1 9 T w =4 K,T in =4 K 1 5*1 6 T w =3 K,T in =3 K 1 5.586*1 8 T w =3 K,T in =4 K 1 1 7 T w =3 K,T in =1 K 1 1 7 T w =3 K,T in =4 K 1 1 7 T w =3 K,T in =4 K 5 15 T w =3 K,T in =3 K 1 AIR Re Heat Transfer Length of Plate 1 7 T w =3 K,T in =4 K 1 5*1 6 T w =3 K,T in =4 K 1 1 7 T w =3 K,T in =4 K 1 1 6 T w =3 K,T in =3 K 1 5*1 6 T w =3 K,T in =3 K 1 1 7 T w =3 K,T in =3 K 1 1 7 T w =3 K,T in =1 K 1 1 7 T w =3 K,T in =4 K 5
Results Turbulent Case Velocity Analysis: Fluid: Water Re=5*1^6,Vinlet=4.477m/s,T=3-3).1.9 y.8.7.6.5.4.3.2.1 3.5 3.7 3.9 4.1 4.3 4.5 4.7 velocity inlet.2l.4l.6l.8l.9l outlet Re=2.23*1^9,Vinlet=2m/s,T=1-4).1.9 y.8.7.6.5.4.3.2.1 19 2 velocity inlet.2l.4l.6l.8l.9l outlet
Observations: - Shear stress and hence C f decreases as increases, this is because as Reynolds # increases with, C f goes down. - δ becomes thicker as increases. - As increases we observe that the velocity eceeds the inlet velocity. It shows a problem in Flowlab s approach in solving the velocity profile equations. - The difference with the laminar case is that the shear layer in turbulence is much thinner than its laminar counterpart. Air: Re= 1^7, V inlet= 134.2 m/s, T= 3-4.1.9 Y.8.7.6.5.4.3.2 Inlet =.2 =.4 =.6 =.8 =.9 Outlet.1 9 95 1 15 11 115 12 125 13 135 14 Velocity Re= 5*1^6, Vinlet=67.14286, T=3-4.4.35 Y.3.25.2.15.1 Inlet =.2 =.4 =.6 =.8 =.9 Outlet.5 3 35 4 45 5 55 6 65 7 Velocity
Analysis of the Coefficient of Skin Friction: Fluid: Air/Water Cf vs Reynolds.5 Cf.4.3.2.1 5*1^6 1^7 5.5*e^8 2.23*e^9.2.4.6.8 1 1.2 Cf Vs Reynolds Cf.7.6.5.4.3.2.1 Re 5*1^6 Re 1^6 Re 1^7.2.4.6.8 1 1.2 X
Observations - With the increase of the value of C f decreases as Re increases..455 - C f ln 2 (.6 Re ) - As the value of inlet velocity is decreased so that the Re decreases the skin friction coefficient goes up. Re=1^7,Vinlet=8.954m/s water.45.4.35.3 Cf.25.2 flowlab theoretical.15.1.5.2.4.6.8 1 1.2 Re= 1^7, Vinlet=134.28 m/s air Cf.45.4.35.3.25.2.15.1.5.2.4.6.8 1 1.2 X Flowlab Prandtl
Observations: As the Re is the same the values of C f remains unaltered immaterial of the fluid. Comparison of Flowlab C f values with Theoretical (water) Re=2.2*1^9,Vinlet=2m/s,water.2.15 Cf.1.5 flowlab theoretical.2.4.6.8 1 1.2 Re= 1^7, Vinlet=134.28 m/s, air Cf.45.4.35.3.25.2.15.1.5.2.4.6.8 1 1.2 X Flowlab Prandtl Observations - C f matches with the theoretical value quite closely for all Re values.
Nusselt Number with/without heat transfer (water) Nu vs Re(with heat transfer) Nu 6 55 5 45 4 35 3 25 2 15 1 5.2.4.6.8 1 1.2 1^7 5.586e^8 2.233e^9 Nu vs Re(without heat transfer) Nu 5 45 4 35 3 25 2 15 1 5.2.4.6.8 1 1.2 1^7 5.583e^8 2.23e9
Observations: 4 / 5 3 Nu =.296 Re Pr 1/...6 < Pr < 6 for turbulent case. Here for a constant Pr as Re increases with, the nusselt number increases as well as Re 4/5.Without heat transfer, the empirical law of Nu is not appropriate. Re=2.2*1^9,Vinlet=2m/s,water without heat transfer 5 45 4 35 Nu 3 25 2 flowlab theoretical 15 1 5.2.4.6.8 1 1.2 Re=1^7,Vinlet=8.954m/s,water with heat transfer 45 4 35 3 Nu 25 2 Flowlab Theory 15 1 5.2.4.6.8 1 1.2
Observations: 4 / 5 3 Nu =.296 Re Pr 1/...6 < Pr < 6 for turbulent case. For without heat transfer case the deviation from the empirical law is much more substantial than the case with heat transfer. Fluid Air: Nu Vs Re Temp with heat transfer 14 12 1 Nu 8 6 4 2 Re 1^6 Re 5*1^6 Re 1^7.2.4.6.8 1 1.2 X Re= 1^7, Vinlet= 134.28m/s, with heat tranfer Nu 14 12 1 8 6 4 2.2.4.6.8 1 1.2 X Flowlab Theory
Temperature Profiles for Water Re=2.233*1^9,Vinlet=2m/s,T=4-1 y.1.9.8.7.6.5.4.3.2.1 395 397 399 41 43 45 47 49 Temp inlet.2l.4l.6l.8l.9l outlet Re=1^7, Vinlet=8.954m/s,with heat transfer y.1.9.8.7.6.5.4.3.2.1 56 57 58 59 6 61 62 63 64 65 temp inlet.2l.4l.6l.8l.9l outlet
Observations: - - The thickness of the thermal boundary layer in the case of turbulence reduces as compared to the laminar case. Fluid Air Re= 5*1^6, Vinlet= 67.14, T= 3-4.1.9 Y.8.7.6.5.4.3.2.1 29 3 31 32 33 34 35 Temp inlet =.2*l =.4*l =.6*l =.8*l =.9*l outlet Observations: - - The thickness of the thermal boundary layer in the case of turbulence reduces as compared to the laminar case.
Conclusions: - - The Flowlab shows fairly good agreement in all respects as compared to the theoretical values obtained by the boundary layer theory. - However it is not possible to get δ *, δ,θ directly from the flowlab results. - The overshoot in the velocity profiles above the inlet velocity also couldn t be eplained. Y Boundary layer solution. Full solution of the Navier Stokes equation. The Flowlab solves the full Navier Stokes Equation