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+clc
+// Example 4.8.py
+// A uniform supersonic stream with M1 = 1.5, p1 = 1700 lb/ft^2, and T1 = 460.0 R
+// encounters an expansion corner which deflects the stream by and angle theta_2
+// = 20 degrees. Calculate M2, p2, T2, po2, To2, and the angles the forward and
+// rearward Mach lines make with respect to the upstream flow direction.
+
+
+// Variable declaration
+M1 = 1.5          // upstream mach number
+p1 = 1700.0       // upstream pressure (in lb/ft^2)
+T1 = 460.0        // upstream temperature (in R)
+theta_2 = 20.0    // deflection (in degrees)
+
+
+// Calculations 
+// subscript 2 means after the expansion fan
+
+// from Table A5 for M1 = 1.5
+v1 = 11.91        // (in degrees)
+mu1 = 41.81       // (in degrees)
+
+v2 = v1 + theta_2
+
+// from Table A5, for v2 = 31.91
+M2 = 2.207        // Mach behind the expansion fan
+mu2 = 26.95       // (in degrees)
+
+// from Table A1 for M1 = 1.5
+po1_by_p1 = 3.671 // po1/p1
+To1_by_T1 = 1.45  // To1/T1
+
+// from Table A1 for M2 = 2.207
+po2_by_p2 = 10.81 // po2/p2
+To2_by_T2 = 1.974 // To2/T2
+
+p2 = 1/po2_by_p2 * po1_by_p1 * p1 // p2 (in lb/ft^2) = p2/po2 * po2/po1 * po1/p1 * p1 and po2 = po1
+T2 = 1/To2_by_T2 * To1_by_T1 * T1 // T2 (in R) = T2/To2 * To2/To1 * To1/T1 * T1 and To2 = To1
+
+
+angle_forward  = mu1              // angle of forward ray (in degrees)
+angle_rearward = mu2 - theta_2    // angle of backward ray (in degrees)
+
+po2 = po1_by_p1 * p1        // po2 (in lb/ft^2) = po1/p1 * p1
+To2  = To1_by_T1 * T1        // To2 (in R) = To1/T1 * T1
+ po1 = po1_by_p1 * p1        // po2 (in lb/ft^2) = po1/p1 * p1
+ To1 = To1_by_T1 * T1        // To2 (in R) = To1/T1 * T1
+
+// Result 
+printf("\n M2 = %.3f", M2)
+
+printf("\n p2 = %.2f lb/ft^2", p2)
+
+printf("\n T2 = %.2f deg R", T2)
+
+printf("\n po2 = %.2f lb/ft^2", po2)
+
+printf("\n To2 = %.2f deg R", To2)
+
+printf("\n Angle forward = %.2f degrees", angle_forward)
+
+printf("\n Angle rearward = %.2f degrees", angle_rearward)
+