PS7_solution

Problem 3d the expression for the critical pressure

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Unformatted text preview: the critical pressure coefficient is given by ( [ * + ) The values for at different Mach numbers is summarized in table-3. Also listed are the peak negative Pressure Coefficient values obtained from the three compressibility rules. (Please see the MATLAB code at the end of section 3d) Table-3 M 0.40 0.60 0.70 0.75 Cpmin_PG -0.4508 -0.5165 -0.5786 -0.6247 -3.662 -1.294 -0.779 -0.591 Cpmin_KT -0.4595 -0.5446 -0.6308 -0.6986 Cpmin_L -0.4699 -0.5900 -0.7399 -0.8868 For M=0.4 and M=0.6, we find the peak values for the peak pressure coefficient indicated from figures23, to be far off from the value determined from the critical pressure coefficient formula. For M=0.7 and M=0.75, we find the peak value of the pressure coefficient from plots 4-5 to be closer to that of . At M=0.7, is higher in magnitude (0.779) when compared to the minimum Cp values obtained from the three compressibility rules; at M=0.75, is lower in magnitude (0.591). This trend indicates that for some Mach number between 0.7 and 0.75, matches the value from the compressibility rules, i.e., the critical Mach number is located between 0.7 and 0.75. PROBLEM 3e Note: For this problem, the non-linear equations for Mcr were solved using fzero in MATLAB (please refer the MATLAB code listed at the end of this section). The critical Mach number can be determined by equating the following expressions for the critical pressure coefficient: ( * [ + ) Prandtl Glauert √ is the minimum incompressible pressure coefficient. The MATLAB program was used to obtain this value from the potential flow data provided (naca0012_0_cp.dat). This value was determined to be To find Mcr from the Prandtl Glauert rule, we equate, E3.1 to E3.2 and solve for Mcr. This value turns out t...
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This homework help was uploaded on 03/31/2014 for the course AAE 334 taught by Professor Collicott during the Spring '09 term at Purdue.

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