Question

A liquid of density 1343 kg / m 3 flows with speed 2 . 21 m / s into a pipe of diameter 0 . 21 m . The diameter of the pipe decreases to 0 . 05 m at its exit end. The exit end of the pipe is 4 . 41 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 . 2 atm Applying Bernoulli’s principle, what is the pressure P 1 at the entrance 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 / s ^2 and P atm = 1 . 013 × 10 ^5 Pa. Answer in units of Pa.

Answer 1

Given that

p = 1343kg/m^3

v1 = 2.21m/s

p2 = 1.2atm

h1-h2 = 4.41m

d1 = 0.21m

d2 = 0.05m

g = 9.8m/s^2

p1 = ?

v2 = ?

according to equation of continuity

A1*v1=A2*v2

where A1 and A2 the cross sectional areas of initial and exit pipes

v2= A1*v1/A2 (A1/A2 = d1^2/d2^2)

v2= d1^2*v1/d2^2

v2=0.21^2*2.21/0.05^2

v2= 38.98m/s

calculation of p1 according to Bernoulli's principle

p1+pgh1+0.5pv1^2 = p2+pgh2+0.5pv2^2

p1=p2+0.5p(v2^2 - v1^2)+pg(h2-h1)

= 1.2atm+((0.5)(1343)((38.98^2)-(2.21^2))+(1343)(9.8)(-4.41))/(1.013*10^5)

=1.2atm+9.467atm

=10.667atm((1.013*10^5)/1atm)

=1080567 pa