Question 6 chap 9 sect 7 part 1 of 1 10 points Note Patm 10

Question 6, chap 9, sect 7. part 1 of 1 10 points Note: Patm = 101300 Pa/atm. The viscos- ity of the fluid is negligible and the fluid is incompressible.. A liquid of density 1219 kg/m3 flows with speed 2.21 m/s into a pipe of diameter 0.27 m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 7.51 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 atm. The acceleration of gravity is 9.8 m/s2 . P1 2.21 m/s 0.27 m P2 1atm v2 0.05 m 7.51m 5.2 m/s 3.2 cm v2 2.3 cm Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe? Answer in units of Pa.

Solution

GIven V₁=2.21 m/s and d=0.27 m ,

d₂=0.05 m ,V₂=?

h₁-h₂=7.51 m

From Continuity equation

A₁V₁ = A₂V₂

(pi/4)*0.27²*2.21 =(pi/4)*0.05²*V₂

V₂=64.44 m/s

From Bernoulli\'s equation

P₁+(1/2)pV₁²+pgh₁ =P₂+(1/2)pV₂²+pgh₂

P₁=P₂+(1/2)p(V₂²-V₁²)-pg(h₁-h₂)

P₁=101300+(1/2)*1219*(64.44²-2.21²)-1219*9.8*7.51

P₁=2.54*10⁶ Pa