Question

At a processing plant, olive oil of density 875 kg/m³ flows in a horizontal section of hose that constricts from a diameter of 3.20 cm to a diameter of 1.25 cm. Assume steady, ideal flow.

(a) What is the volume flow rate if the change in pressure between the two sections of hose is 4.95 kPa?

Your response differs from the correct answer by more than 10%. Double check your calculations. m³/s

(b) What is the volume flow rate if the change in pressure between the two sections of hose is 14.0 kPa?

Answer 1

let
rho = 875 kg/m^3
d1 = 3.20 cm
d2 = 1.25 cm

a) use Bernoulli's equation,

use continuity equation,

A1*v1 = A2*v2

(pi*d1^2/4)*v1 = (pi*d2^2/4)*v2

v2 = v1*(d1/d2)^2

= v1*(3.2/1.25)^2

= 6.55*v1

now use Bernoullis equation

P1 + (1/2)*rho*v1^2 = P2 + (1/2)*rho*v2^2

P1 - P2 = (1/2)*rho*(v2^2 - v1^2)

= (1/2)*rho*((6.55*v1)^2 - v1^2)

= 41.9*(1/2)*rho*v1^2

v1^2 = 2*(P1 - P2)/(41.9*rho)

v1 = sqrt(2*(P1 - P2)/(41.9*rho) )

= sqrt(2*4.95*10^3/(41.9*875 ) )

= 0.5196 m/s

volume flow rate, dV/dt = A1*v1

= (pi*d1^2/4)*v1

= (pi*0.032^2/4)*0.5196

= 4.18*10^-4 m^3/s <<<<<<<<<-----------Answer

b)

v1 = sqrt(2*(P1 - P2)/(41.9*rho) )

= sqrt(2*14*10^3/(41.9*875 ) )

= 0.8739 m/s

volume flow rate, dV/dt = A1*v1

= (pi*d1^2/4)*v1

= (pi*0.032^2/4)*0.8739

= 7.03*10^-4 m^3/s <<<<<<<<<-----------Answer