A liquid of density 1208kg/m^3 flows with speed 2.07m/s into a pipe of diameter

Question

A liquid of density 1208kg/m^3 flows with speed 2.07m/s into a pipe of diameter 0.16m. The diameter of the pipe decreases to 0.05m at its exit end. The exit end of the pipe is 5.68m lower than the entrance to the pipe, and the pressure at the exit of the pipe is 1.3atm. What is the velocity v2 of the liquid flowing out of the exit end of the pipe and applying Bernoullis's principle what is the pressure P1 in units of Pa at the entrance of the pipe? Assume the viscosity of the fluid is negligible and the fluid is incompressible. The acceleration of gravity is 9.8m/s^2 and Patm=1.013*10^5 Pa. Answer in units of m/s for velocity v2.

Explanation / Answer

by equation of continuity
A1v1 = A2v2

(0.08)2 x 2.07 = (0.025)2 xv2

v2 = 21.1968m/s

now applying bernoulli theorem

P1 + (1/2)v12 + gh1 = P2 + (1/2)v22 + gh2

taking gh2 = 0

P1 + (1/2)(1208)(2.07)2 + (1208)(9.8)(5.68) = 1.3 x 105 + (1/2)1208 x 21.19682

P1 = 3.315 x 105 pa

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