Question

Water is flowing through a horizontal pipe with an inside diameter of 8.00cm. The pipe narrows to an inside diameter of 3.75cm. Assume the water is an ideal incompressible fluid. What is the velocity of the fluid initially, v_1 (when It is flowing in the 8.00cm diameter section) if the velocity of the water in the narrow part of the pipe is measured at V_2 = 27m/s? If the initial pressure is known to be 670 kPa what is the pressure of the fluid in the narrow part of the pipe?

Answer 1

using equation of continuity

V1A1= V2A2

V1 $\pi$ ( 4)^2 = 27 $\pi$ (1.875)^2

V1= 5.93 m/s apprx

b) use bernoulli's

P1 + 0.5 $\rho$ v1^2 = p2 + 0.5 $\rho$ ​v2^2

670 x 10^3 + 0.5 ( 1x 10^3) ( 5.93)^2 = P2 +  0.5 ( 1x 10^3) ( 27 )^2

670000 + 17582.45 = P2 + 364500

P2= 323 kPa apoprx