Question

A liquid of density 1394 kg/m3 flows with speed 2.48 m/s into a pipe of diameter 0.21 m . The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 5.43 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.5 atm. Applying Bernoulli’s principle, what is the pressure P1 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 2 and Patm = 1.013 × 105 Pa. Answer in units of Pa

Answer 1

by bernoulli's equation

pressure1 + density * g * height1 + 0.5 * density * velocity1 = pressure2 + density * g * height2 + 0.5 * density * velocity2

by continuity equation

area1 * velocity1 = area2 * velocity2

pi * (0.21 / 2)^2 * 2.48 = pi * (0.05 / 2)^2 * v2

v2 = 43.7472 m/s

pressure2 = 1.5 atm or 151988 Pa

pressure1 + density * g * (height1 - height2) + 0.5 * density * velocity1 = pressure2 + 0.5 * density * velocity2

pressure1 + 1394 * 9.8 * 5.43 + 0.5 * 1394 * 2.48^2 = 151988 + 0.5 * 1394 * 43.7472^2

pressure1 = 1407451.65816 Pa