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diff --git a/3765/CH4/EX4.9/Ex4_9.sce b/3765/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..c4a730371 --- /dev/null +++ b/3765/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,63 @@ +clc +// Example 4.9.py +// Consider the arrangement shows in fig. 4.29. A 15 degree half angle diamond +// wedge airfoil is in supersonic flow at zero angle of attack. A pitot tube is +// inserted into the flow at the location shown in fig 4.29. The pressure measured +// by the Pitot tube is 2.596 atm. At point a on the backface, the pressure is 0.1 +// atm. Calculate the freestream Mach number M1. + +// + +// Variable declaration +theta = 15.0 // wedge angle/deflection (in degrees) +po4 = 2.596 // measured pressure (in atm) +p3 = 0.1 // pressure at point a (in atm) + +// Calculations + +po4_by_p3 = po4/p3 + +// from Table A 2 for po4/p3 = 25.96 +M3 = 4.45 +v3 = 71.27 +v2 = v3 - 2*theta + +// from Table A 5, for v2 = 41.27 degrees +M2 = 2.6 +// Mn2 = M2*sin((beta1-theta)*%pi/180) @equation 1 + +// Guessing + +// Guess 1 +M1 = 4.0 // Guess for freestream number +beta1 = 27.0 // from fig 4.5 (in degrees) +Mn1 = M1*sin(beta1*%pi/180) // mach number normal to shock + +// from Table A2 for Mn1 = 1.816 +Mn2 = 0.612 +// but Mn2 from equation 1 is 0.54 + +// Guess 2 +M1 = 4.5 // Guess for freestream number +beta1 = 25.5 // from fig 4.5 (in degrees) +Mn1 = M1*sin(beta1*%pi/180) // mach number normal to shock + +// from Table A2 for Mn1 = 1.937 +Mn2 = 0.588 +// but Mn2 from equation 1 is 0.47 + +// Guess 3 +M1 = 3.5 // Guess for freestream number +beta1 = 29.2 // from fig 4.5 (in degrees) +Mn1 = M1*sin(beta1*%pi/180) // mach number normal to shock + +// from Table A2 for Mn1 = 1.71 +Mn2 = 0.638 +// but Mn2 from equation 1 is 0.638 + + + + +// Result +printf("\n Freestream mach number is %.1f", M1) + |