Question 13, chap 115, sect 10 part 1 of 1 10 points Note: Patm 101300 Pa/atm. T

Question

Question 13, chap 115, sect 10 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 1356 kg/m3 flows with speed 2.94 m/s into a pipe of diameter 0.19 m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 6.66 m lower than the entrance of the pipe, and the pressure 1.2 atm at the exit of the pipe is The acceleration of gravity is 9.8 m/s2 Pi 1.2 atm 2.94 m/s 6.66 m 0.19 m D2 0.05 m Applying Bernoulli's principle, what is the pressure Pi at the entrance end of the pipe? Answer in units of Pa.

Explanation / Answer

Using Bernoullies principle

P1 + (rho*g*h1)+(0.5*rho*v1^2) = P2 + (rho*g*h2)+(0.5*rho*v2^2)

P1+((1356*9.8*6.66)+(0.5*1356*2.94^2) = (1.2*1.013*10^5)+(1356*9.8*0)+(0.5*1356*v2^2)

using equation of continuty

A1*v1 = A2*v2
pi*r1^2*v1 = pi*r2^2*v2

(0.19/2)^2*2.94 = (0.05/2)^2*v2

v2 = 42.45 m/s

then

P1+((1356*9.8*6.66)+(0.5*1356*2.94^2) = (1.2*1.013*10^5)+(1356*9.8*0)+(0.5*1356*v2^2)

P1+94363.76 = 121560+(0.5*1356*42.45^2)

P2 = 1.25*10^6 Pa

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