Question

002 10.0 points A liquid of density 1268 kg/m3 flows with speed 1.31 m/s into a pipe of diameter 0.25 m The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 4.7 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.5 atm Pi 1.31 m/s iP) 1.5 atm 0.25 m 0.05 m Applying Bernoulli's principle, what is the pressure Pı at the entrance end of the pipe? Assume the viscosity of the fluid is negligible and the fluid is incompressible The acceleration of gravity is 9.8 m/s and

Answer 1

Given that, a liquid of density (rho) = 1268 kg/m^3 speed (v_1) = 1.31 m/s a pipe of diameter (d_1) = 0.25 m The diameter of the pipe decreases (d_2) to = 0.05 m The exit end of the pipe is lower than the entrance of the pipe (h_1) = 4.7 m The pressure at the exit of the pipe is = 1.5 atm The acceleration of gravity is = 9.8 m/s^2 P_atm = 1.013 times 10^5 partial differential = p_2 To find the pressure p_1 at the entrance end of the pipe: first we can calculating the v_2, A_1 V_1 = A_2 V_2 pi/4 times d^2_1 times v_1 = pi/4 times d^2_2 times v_2 The given values are substitute in above equation. \r\n pi/4(0.25)^2 times v_1 = pi/4 times (0.05)^2 V_2 pi/4 (0.25)^2 times 1.31 = pi/4 times (0.05)^2 V_2 0.785 times 0.0625 times 1.31 = 0.785 times 2.5 times 10^-3 times v_2 0.06427 = 1.9625 times 10^-3 times v_2