University:
Kentucky State University
Course:
COS 232| Programming I
Academic year:

2019
Views:

468

Pages:

21
Author:

marthamanayay9qr15

12.9 →+ ↑+ ∑F x ∑F y = 0 : P cos 21.5  − M cos 37  − M cos 80  = 0 0.93042 P − 0.9723M = 0 (1) = 0 : P sin 21.5  + M sin 37  − M sin 80  = 0 0.3665 P − 0.383M = 0 Use any method to solve equations (1) and (2): cos 21.5  A=   sin 21.5 − (cos 37  + cos 80  )  (sin 37  − sin 80  )   0 B=   0 P   M  Apply Ax = B where x =  Use Matlab or calculator for results P = 314 lb M = 300 lb ( 2) 12.10 Mass balances can be written for each reactor as 0 = Qin c A,in − Qin c A,1 − k1V1c A,1 0 = Qin cB ,1 + k1V1c A,1 0 = Qin c A,1 + Q32c A,3 − (Qin + Q32 )c A,2 − k 2V2c A,2 0 = Qin cB,1 + Q32cB,3 − (Qin + Q32 )cB,2 + k2V2c A,2 0 = (Qin + Q32 )c A,2 + Q43c A,4 − (Qin + Q43 )c A,3 − k3V3c A,3 0 = (Qin + Q32 )cB ,2 + Q43cB ,4 − (Qin + Q43 )cB ,3 + k3V3c A,3 0 = (Qin + Q43 )c A,3 − (Qin + Q43 )c A, 4 − k 4V4c A, 4 0 = (Qin + Q43 )cB ,3 − (Qin + Q43 )cB,4 + k 4V4c A,4 Collecting terms, the system can be expresses in matrix form as [A]{C} = {B} where 0 0 0 0 0 0 0 11.25 0 0 0 0 0 0 − 1.25 10 − 10 0 22 . 5 0 − 5 0 0 0  0 − 10 − 7 . 5 15 0 − 5 0 0  [A] = 0 − 15 0 68 0 −3 0  0 0 0 − 15 − 50 18 0 − 3  0 0 0 0 − 13 0 15.5 0   0 0 0 0 0 − 13 − 2.5 13   0 [B]T = [10 0 0 0 0 0 0 0 0 0] [C]T = [cA,1 cB,1 cA,2 cB,2 cA,3 cB,3 cA,4 cB,4] The system can be solved for [C]T = [0.889 0.111 0.416 0.584 0.095 0.905 0.080 0.920]. 1 0.8 B 0.6 A 0.4 0.2 0 0 1 2 3 4 12.11 Assuming a unit flow for Q1, the simultaneous equations can be written in matrix form as − 1 0 0 1 0  0 Q 2 1 0 0 0   2  0 0 −1 2 1 0  Q3  0 0 0 0 − 1 3  Q4  = 0 1 0 0 0 0  Q5  1  1 −1 −1 0 0  Q  0 0 0 1 − 1 − 1  6  0 Q7  These equations can be solved to give [Q]T = [0.7321 0.2679 0.1964 0.0714 0.0536 0.0178]. 12.20 Find the unit vectors:  1iˆ − 2 ˆj − 4 kˆ A 2 2 2  1 + 2 +4   = 0.218iˆ − 0.436 ˆj − 0.873kˆ    2iˆ + 1 ˆj − 4kˆ B 2 2 2  1 + 2 +4   = 0.436iˆ + 0.218 ˆj − 0.873kˆ   Sum moments about the origin: ∑ M ox = 50 (2) − 0.436 B(4) − 0.218 A(4) = 0 ∑ M oy = 0.436 A(4) − 0.218B(4) = 0 Solve for A & B using equations 9.10 and 9.11: In the form a11 x1 + a12 x 2 = b1 a 21 x 2 + a22 x2 = b2 − 0.872 A + −1.744 B = −100 1.744 A + −0.872B = 0 Plug into equations 9.10 and 9.11: x1 = a 22 b1 − a12 b2 87.2 = = 22.94 N a11a 22 − a12 a 21 3.80192 x2 = a11b2 − a 21b1 174.4 = = 45.87 N a11a 22 − a12 a 21 3.80192 12.21  1iˆ + 6 ˆj − 4kˆ T  2 2 2  1 +6 +4   = 0.1374iˆ + 0.824 ˆj − 0.549 kˆ   ∑ M y = −5(1) + −0.549T (1) = 0 T = 9.107 kN ∴ Tx = 1.251 kN , T y = 7.50 kN , Tz = −5 kN ∑ M x = −5(3) + −7.5( 4) + −5(3) + B z (3) = 0 ∑ M z = 7.5(3) + 1.251(3) + B x (3) = 0 B x = −3.751 kN ∑ Fz = −5 + −5 + Az + 20 = 0 Az = −10 kN ∑ Fx = Ax + −3.751 + 1.251 = 0 ∑ Fy = 7.50 + Ay = 0 12.22 This problem was solved using Matlab. A = [1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 3/5 0 0 0 0 0 0 -1 0 0 -4/5 0 0 0 0 0 0 0 -1 0 0 0 0 3/5 0 0 0 0 0 0 0 -1 0 -4/5 0 0 0 0 0 -1 -3/5 0 1 0 0 0 0 0 0 0 4/5 1 0 0 0 0 0 0 0 0 0 0 -1 -3/5 0 0 0 0 0 0 0 0 0 4/5 0 0 1]; b = [0 0 –54 0 0 24 0 0 0 0]; x=inv(A)*b x = B z = 20 kN Ax = 2.5 kN Ay = −7.5 kN 24.0000 -36.0000 54.0000 -30.0000 24.0000 36.0000 -60.0000 -54.0000 -24.0000 48.0000 Therefore, in kN AB = 24 DE = 36 BC = −36 CE = −60 AD = 54 Ax = −54 BD = −30 Ay = −24 CD = 24 Ey = 48 12.27 This problem can be solved directly on a calculator capable of doing matrix operations or on Matlab. a=[60 -40 0 -40 150 -100 0 -100 130]; b=[200 0 230]; x=inv(a)*b x = 7.7901 6.6851 6.9116 Therefore, I1 = 7.79 A I2 = 6.69 A I3 = 6.91 A 12.28 This problem can be solved directly on a calculator capable of doing matrix operations or on Matlab. a=[17 -8 -3 -2 6 -3 -1 -4 13]; b=[480 0 0]; x=inv(a)*b x = 37.3585 16.4151 7.9245 Therefore, V1 = 37.4 V V2 = 16.42 V V3 = 7.92 V 12.29 This problem can be solved directly on a calculator capable of doing matrix operations or on Matlab. a=[6 0 -4 1 0 8 -8 -1 -4 -8 18 0 -1 1 0 0]; b=[0 -20 0 10]; x=inv(a)*b x = -7.7778 2.2222 -0.7407 43.7037 Therefore, I1 = -7.77 A I2 = 2.22 A I3 = -.741 A Vs = 43.7 V 12.30 This problem can be solved directly on a calculator capable of doing matrix operations or on Matlab. a=[55 0 -25 0 37 -4 -25 -4 29]; b=[-200 -250 100]; x=inv(a)*b x = -4.1103 -6.8695 -1.0426 Therefore, I1 = -4.11 A I3 = -6.87 A I4 = -1.043 A 12.37 %massspring34.m k1=10; k2=40; k3=40; k4=10; m1=1; m2=1; m3=1; km=[(1/m1)*(k2+k1), -(k2/m1),0; -(k2/m2), (1/m2)*(k2+k3), -(k3/m2); 0, -(k3/m3),(1/m3)*(k3+k4)]; X=[0.05;0.04;0.03]; kmx=km*X kmx = 0.9000 0.0000 -0.1000 Therefore, x1 = -0.9, x2 = 0 , and x3 = 0.1 m/s2.