PEARSON ETEXT ENGINEERING MECH & STATS
PEARSON ETEXT ENGINEERING MECH & STATS
15th Edition
ISBN: 9780137514724
Author: HIBBELER
Publisher: PEARSON
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A boiler with 80% efficiency produces steam at 40bar and 500 C at a rate of 1.128kg/s. The temperature of the feed water is raised from 25 C to 125 C in the economizer and the ambient air is drawn to the boiler at a rate of 2.70 kg/s at 16 C. The flue gases leave the chimney at rate of 3 kg/s at 150 C with specific heat of 1.01 kJ/kg.K. The dryness fraction of steam collected in the steam drum is 0.95. 1- Determine the heat value of the fuel. 2- The equivalence evaporation. 3- Draw the heat balance sheet.
A rotating shaft is made of 42 mm by 4 mm thick cold-drawn round steel tubing and has a 6 mm diameter hole drilled transversely through it. The shaft is subjected to a pulsating torque fluctuating from 20 to 160 Nm and a completely reversed bending moment of 200 Nm. The steel tubing has a minimum strength of Sut = 410 MPa (60 ksi). The static stress-concentration factor for the hole is 2.4 for bending and 1.9 for torsion. The maximum operating temperature is 400˚C and a reliability of 99.9% is to be assumed. Find the factor of safety for infinite life using the modified Goodman failure criterion.
I need help with a MATLAB code. This code just keeps running and does not give me any plots. I even reduced the tolerance from 1e-9 to 1e-6. Can you help me fix this? Please make sure your solution runs. % Initial Conditions rev = 0:0.001:2;   g1 = deg2rad(1); g2 = deg2rad(3); g3 = deg2rad(6); g4 = deg2rad(30); g0 = deg2rad(0);   Z0 = 0; w0 = [0; Z0*cos(g0); -Z0*sin(g0)];   Z1 = 5; w1 = [0; Z1*cos(g1); -Z1*sin(g1)];   Z2 = 11; w2 = [0; Z2*cos(g2); -Z2*sin(g2)];   [v3, psi3, eta3] = Nut_angle(Z2, g2, w2);   plot(v3, psi3)     function dwedt = K_DDE(~, w_en)   % Extracting the initial condtions to a variable % Extracting the initial condtions to a variable w = w_en(1:3); e = w_en(4:7); Z = w_en(8);     I = 0.060214; J = 0.015707; x = (J/I) - 1; y = Z - 1; s = Z;   % Kinematic Differential Equations dedt = zeros(4,1); dedt(1) = pi*(e(3)*(s-w(2)-1) + e(2)*w(3) + e(4)*w(1)); dedt(2) = pi*(e(4)*(w(2)-1-s) + e(3)*w(1) - e(1)*w(3)); dedt(3) = pi*(-e(1)*(s-w(2)-1) - e(2)*w(1) + e(4)*w(3));…
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