Question

Problem 4 A steel cylinder at 80°F (room temperature) of length Lo 12" Is placed in a commercially available liquid nitrogen bath -320°F. The thermal expansion coefficient as a function of temperature is given by the following data: Temperature Thermal expansion T OF) coefficient (OF) 2.76x10 3.83 x10 4.72 x10 5.43 x10 6.00 x10 6.47 x10 -320 -240 160 -80 80 When the cylinder is placed in the liquid Nitrogen bath, there is a reduction in the length of the cylinder, which can be approximated as: S0 +320 Determine AL when the cylinder is cooled from 80°F to -320°F using the trapezoidal method using 5 segments

can someone show me the solution to this problem? thank you

Answer 1

"As per trapezoidal Rule integral^x_0 + xh_x_0 f(x) dx = h/2 [y_0 + y_x + 2 (y_1 + y_2 + ellipsis y_n-1)] taking each segment of width h = 80 integral^80_320 alpha dT = h/2 [y_0 + y_5 + 2 times (y_1 + y_2 + y_3 + y_4)] = 80/2 times [2.76 + 6.47 + 2 times (3.83 + 4.72 + 5.43 + 65.00)] times 10^-3 integral^80_-320 alpha dT = 1967.6 times 10^-3 \r\n so delta L = L_0 times integral^80_-320 alpha dT = L_0 times 1967.6 times 10^-3 as L_0 = 12"" so delta L = 12 times 1.9676 times 10^-3 = 0.0236112 inch."