A liquid of density 1397 kg/m3 flows with speed 1.94 m/s into a pipe of diameter 0.13 m.The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 8.04 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.2 atm. What is the velocity v2 of the liquid flowing out of the exit 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/s2 and
P atm= 1.013 ×10^5 Pa. Answer in units of m/s.
Thank you in advance! I have been working on this for hours
P atm= 1.013 ×10^5 Pa. Answer in units of m/s.
Thank you in advance! I have been working on this for hours
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Surely if the flow rate is 1.94m/sec in the 0.13m. dia. pipe, all the rest of the information is superfluous? The fluid is incompressible!
1.94 x (0.13^2)/(0.05^2) = V out of 13.1144m/sec.
1.94 x (0.13^2)/(0.05^2) = V out of 13.1144m/sec.