Question

A liquid of density 1.13 × 103 kg/m3 flows steadily through a pipe of varying diameter and height. At location 1 along the pipe the flow speed is 9.77 m/s and the pipe diameter is 11.3 cm. At location 2 the pipe diameter is 14.5 cm. At location 1 the pipe is 8.43 m higher than it is at location 2. Ignoring viscosity, calculate the difference between the fluid pressure at location 2 and the fluid pressure at location 1.

Answer 1

Given

Density of the liquid is rho = 1.13*10^3 kg/m^3

At location 1 ,

diameter d1 = 0.113 m ,flow speed v1 = 9.77 m/s , pressure P1, height h1= 8.34+h2

At location 2 ,

diameter d2 = 0.145 m ,flow speed v2 = ? m/s , pressure P2, height h2

From continuity equation A1*v1 = A2*v2

pi*(0.113/2)^2*9.77 = pi*(0.145/2)^2*v2

v2 = 5.933 m/s

From Bernouli's equation  

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

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

(P2-P1) = 0.5*rho(v1^2-v2^2)+rho*g(8.34+h2-h2)

(P2-P1) = 0.5*1.13*10^3(9.77^2-5.933^2)+1.13*10^3(8.34)

(P2-P1) = 43466.812 Pa