2. Simplified Method to Detail the BL Profile… P M V Subbarao Professor Mechanical Engineering Department I I T Delhi Flat Plate Boundary Layer Solutions

3. Blasius Boundary Layer Equations

4. Expectation of possible SIMILARITY Variable For a solution obeying similarity in the velocity profile we must have where f 1 is a universal function, independent of x (position along the plate). Since we have reason to believe that We can rewrite any such similarity form as Note that  is a dimensionless variable.

5. Approximate Analytical Method -1 Polhausen Approximate Solution

6. Polhausen Boundary Layer Profile

7. Approximate Analytical Method -2 Integral Method

8. The partial integration of the second term of the left side of above equation gives: Using continuity equation :

10. But Does the Problem Admit A Similarity Solution? Maybe, maybe not, you never know until you try. The problem is: This problem can be reduced with the stream function (u =  /  y, v = -  /  x) to: Note that the stream function satisfies continuity identically.

11. 10 Solution By The Method Of Guessing We want our stream function to give us a velocity u =  /  y satisfying the similarity form so that So we could start off by guessing where F is another similarity function. Will it work ???? Lets see…..

12. 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.

13. Second Guess learnt from the first 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(  ).

14. 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:

15. 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:

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

17. Summary of Conversion

18. 17 BL ODE

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

20. The Blasius Equation The Blasius equation with the above boundary conditions exhibits a boundary value problem. However, 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

22.  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

23. 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''

24. 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''

25. 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''

26. 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''

27. 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''

28. 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