Question

don’t us excel please

Problem 3. (30 points) From inviscid potential flow analysis, as a uniform stream approaches a cylinder of radius R along the symmetry line AB in the figure, the velocity has only one component: u(x)U for 0<xs-R where U, is the stream velocity far from the cylinder. a. Using EXCEL, plot u(x) along AB from -0.4<x<-0.05 when R-0.05 m and U,-4 m/s. Label axes with variables and dimensions and add a title to the plot. Since the solution is analytical, it should be plotted using a smooth curve and no symbols. Include enough points in your spreadsheet to resolve the curve. Du along AB as a function of U, R and x. b. Find the acceleration a, c. Using EXCEL, plotfrom-0.4<x<-0.05 when R=0.05 m and U. 4 m/s. Label Dt axes with variables and dimensions and add a title to the plot. Since the solution is analytical, it should be plotted usinga smooth curve and no symbols. Include enough points in your spreadsheet to resolve the curve. d. Find the maximum flow deceleration along AB and its location when R-0.05 m and U4 m/s. (Partial answer: Xmin-0.0646 m) U R B

Answer 1

u(x) -0.4 3.9375 4.5 -0.39 3.934254 4 -0.38 3.930748 3.5 -0.37 3.926954 -0.36 3.92284 Chart Area 2.5 -0.35 3.918367 -0.34 3.913495 velocity 2 Series1 -0.33 3.908173 1.5 -0.32 3.902344 -0.31 3.895942 0.5 -0.3 3.888889 -0.29 3.881094 -0.5 -0.2 -0.1 -0.4 -0.3 3.872449 -0.28 -0.5 distance -0.27 3.862826 -0.26 3.852071 -0.25 3.84 -0.24 3.826389

Given that equation. s-R for b) DuL Dt Du u du du Dt Du udu Dt Jui du dz 27 Du RY DE 27 (1C Du - DE l

du/dt X -0.4 15.75788 -0.39 15.74469 18 -0.38 15.73046 -0.37 15.71508 16 -0.36 15.69842 14 -0.35 15.68033 -0.34 15.66063 12 -0.33 15.63914 -0.32 15.61562 10 -0.31 15.58981 Acceleratio Series1 n -0.3 15.56139 -0.29 15.53 6 -0.28 15.49522 -0.27 15.45652 -0.26 15.41329 -0.25 15.3648 -0.24 15.31015 -0.4 -0.3 distance -0.5 -0.2 -0.1 -0.23 15.24824 -2 -0.22 15.17773

S (d) Du (ft) D 2.PT 2n For Maximum Coa) minimum dari

RY R4 224 - 27 2x4 2x4 R4 4 3 2. raking common of RY 2 -3 3 - 2 -. O 3eY -4 -aV. The above eayuahon weiFten as i 4- 3R 0 GR=005 4-3(o.05o G boling e above Equatibn Can get We 0 ool874 4 Sub st , Valueg în equaion (A) RY =-4 2.7 1 +(D05JV 2X4 (4) (o.05) 15. m/s 1=-4