2. Nature of Zero Pressure Gradient BL Flows…… P M V Subbarao Professor Mechanical Engineering Department I I T Delhi Solutions for Flat Plate Boundary Layer Equations

3. The Philosophy of Similarity Belief For a solution obeying similarity in the velocity profile we must have where f is a universal function, independent of x (position along the plate). We can rewrite any such similarity form as Note that  is a dimensionless variable. Since we have a reason to believe that

4. Are these Equations Eligible for Similarity Solution Maybe, maybe not, you never know, Try to convert these equations into an eligible form. Introduce Stream function,  (x,y). Note that the stream function satisfies continuity identically.

5. Discovery of Similarity Variable by The Method Of Guessing We want our stream function to give us a velocity u =  /  y satisfying the similarity variable so that So we could start-off by guessing where F is a similarity function. Will it work ???? Lets see….. Along with a similarity function for velocity profile

6. Is it a right Guess ? If we assume not OK OK then we obtain This form does not satisfy the condition that u/U is a function of  alone. If F is a function of  alone then its first derivative F’(  ) is also a function of  alone. But note the extra function in x via the term (U x) -1/2 ! So our first guess failed because of the term (U x) -1/2. Use this guess to generate a better guess.

7. Second Guess learnt from the first Guess This time we assume Now remembering that x and y are independent of each other and recalling the evaluation of  /  y, Thus the first guess helped in guessing the correct function this new form of  that satisfies similarity in velocity! But this does not mean that we are done. We have to solve for the function F(  ).

8. The Basic of Similarity Variable Our goal is to reduce the partial differential equation using  and . To do this we will need the following basic derivatives:

9. The next steps involve tedious differential calculus, to evaluate the terms in the BL Stream Function equation. Conversion of Third Order PD into OD The third order Partial derivative is:

10. we now work out the two second order derivatives: Conversion of Second Order PDs into ODs

11. Summary of Conversion

12. ODE governing Flat Plate BL

13. The boundary conditions are BOUNDARY CONDITIONS But we already showed that

14. The Blasius Equation The Blasius equation with the above boundary conditions exhibits a boundary value problem. However, the location of one boundary is unknown, though boundary condition is known. However, using an iterative method, it can be converted into an initial value problem. Assuming a certain initial value for F  =0 , Blasius equation can be solved using Runge-Kutta or Predictor- Corrector methods

16.  FF'F'' 0000.33206 0.10.001660.033210.33205 0.20.006640.066410.33199 0.30.014940.09960.33181 0.40.026560.132770.33147 0.50.041490.165890.33091 0.60.059740.198940.33008 0.70.081280.231890.32892 0.80.106110.264710.32739 0.90.134210.297360.32544 10.165570.329780.32301 1.10.200160.361940.32007 1.20.237950.393780.31659 1.30.278910.425240.31253 1.40.322980.456270.30787 1.50.370140.486790.30258 1.60.420320.516760.29667 1.70.473470.546110.29011 1.80.529520.574760.28293 1.90.58840.602670.27514 2 0.650030.629770.26675 Runge’s Numerical Reults

17. 2.20.78120.681320.24835 2.30.850560.705660.23843 2.40.92230.728990.22809 2.50.996320.751270.21741 2.61.072510.772460.20646 2.71.150770.792550.19529 2.81.230990.811520.18401 2.91.313040.829350.17267 31.396820.846050.16136 3.11.482210.861620.15016 3.21.569110.876090.13913 3.31.657390.889460.12835 3.41.746960.901770.11788 3.51.837710.913050.10777 3.61.929540.923340.09809 3.72.022350.932680.08886 3.82.116040.941120.08013 3.92.210540.948720.07191  FF'F''

18. 42.305760.955520.06423 4.12.401620.961590.0571 4.22.498060.966960.05052 4.32.5950.971710.04448 4.42.692380.975880.03897 4.52.790150.979520.03398 4.62.888270.982690.02948 4.72.986680.985430.02546 4.83.085340.987790.02187 4.93.184220.989820.0187 53.28330.991550.01591 5.13.382530.993010.01347 5.23.481890.994250.01134 5.33.581370.995290.00951 5.43.680940.996160.00793 5.53.78060.996880.00658  FF'F''

19. 5.63.880320.997480.00543 5.73.980090.997980.00446 5.84.079910.998380.00365 5.94.179760.998710.00297 64.279650.998980.0024 6.14.379560.999190.00193 6.24.479490.999370.00155 6.34.579430.999510.00124 6.44.679390.999620.00098 6.54.779350.99970.00077 6.64.879330.999770.00061 6.74.979310.999830.00048 6.85.079290.999870.00037 6.95.179280.99990.00029 75.279270.999930.00022  FF'F''

20. 7.15.379270.999950.00017 7.25.479260.999960.00013 7.35.579260.999979.8E-05 7.45.679260.999987.4E-05 7.55.779250.999995.5E-05 7.65.879250.999994.1E-05 7.75.9792513.1E-05 7.86.0792512.3E-05 7.96.1792511.7E-05 86.2792511.2E-05 8.16.3792518.9E-06 8.26.4792516.5E-06 8.36.5792514.7E-06 8.46.6792513.4E-06 8.56.7792512.4E-06 8.66.8792511.7E-06 8.76.9792511.2E-06 8.87.0792518.5E-07  FF'F''

21. 8.97.179261.000015.9E-07 97.279261.000014.1E-07 9.17.379261.000012.9E-07 9.27.479261.000012E-07 9.37.579261.000011.4E-07 9.47.679261.000019.3E-08 9.57.779261.000016.3E-08 9.67.879261.000014.3E-08 9.77.979261.000012.9E-08 9.88.079261.000011.9E-08 9.98.179261.000011.3E-08 108.279261.000018.5E-09  FF'F''

22. Recall that the nominal boundary thickness  is defined such that u = 0.99 U when y = . By interpolating on the table, it is seen that u/U = F’ = 0.99 when  = 4.91. Since u = 0.99 U when  = 4.91 and  = y[U/( x)] 1/2, it follows that the relation for nominal boundary layer thickness is NOMINAL BOUNDARY LAYER THICKNESS 4.62.888270.982690.02948 4.72.986680.985430.02546 4.83.085340.987790.02187 4.93.184220.989820.0187 53.28330.991550.01591 5.13.382530.993010.01347  FF'F''

23. Let the flat plate have length L and width b out of the page: DRAG FORCE ON THE FLAT PLATE L b The shear stress  o (drag force per unit area) acting on one side of the plate is given as Since the flow is assumed to be uniform out of the page, the total drag force F D acting on the plate is given as The term  u/  y =  2  /  y 2 is given from as

24. The shear stress  o (x) on the flat plate is then given as But from the table F  (0) = 0.332, so that boundary shear stress is given as Thus the boundary shear stress varies as x -1/2. A sample case is illustrated on the next slide for the case U = 10 m/s, = 1x10 -6 m 2 /s, L = 10 m and  = 1000 kg/m 3 (water).  FF'F'' 0000.33206 0.10.001660.033210.33205 0.20.006640.066410.33199 0.30.014940.09960.33181 0.40.026560.132770.33147 0.50.041490.165890.33091

25. Sample distribution of shear stress  o (x) on a flat plate: Variation of Local Shear Stress along the Length Note that  o =  at x = 0. Does this mean that the drag force F D is also infinite? U = 0.04 m/s L = 0.1 m = 1.5x10 -5 m 2 /s  = 1.2 kg/m 3 (air)

26. The drag force converges to a finite value! DRAG FORCE ON THE FLAT PLATE