Question

A horizontal pipe of diameter 0.768 m has a smooth constriction to a section of diameter 0.4608 m. The density of oil flowing in the pipe is 821 kg/m3 . If the pressure in the pipe is 7120 N/m2 and in the constricted section is 5340 N/m2 , what is the rate at which oil is flowing? Answer in units of m3/s.

Answer 1

As you know the equation of continuity i.e. volumetric liquid flow rate is constant for any cross section as the liquid we consider is incompressible, i.e AV = const. , A is the area of the cross section,

π(0.768)²*V₁ / 4 = π(0.4608)²*v₂ / 4 ;   V₁ & V₂ are the velocities of water in pipe and constriction respectively,

V₁/V₂ = (0.4608 / 0.768)²

V₁/V₂ = 0.36 (i)

Hence the volumetric flow rate given for the pipe of diameter 0.768m will be equal to the volumetric flow rate for the constriction of diameter of 0.4608m.

Now using Bernoulli's equation for the two cross-section :

P + 0.5 ρ ν² + ρ gh = Const

as the height for both the cross sections is same therefore equation reduces to:

P + 0.5 ρ ν² =Const

So now,

7120 + 0.5* 821*V₁² = 5340 + 0.5*821*V₂² ;    V₁ & V₂ are the velocities of water in pipe and constriction respectively,

0.5*821*(V₂² - V₁²) = 1780

(V₂² - V₁²)= 4.336 (ii)

Solving (i) & (ii),

We get,

V₂² - (0.36V₂)² = 4.336

(1-(0.36)2)V₂² = 4.336

0.8704 V₂² = 4.336

V₂² = 4.98

V₂ = 2.23

Hence,

Volumetric flow rate: = π(0.4608)²*v₂ / 4 = π(0.4608)²* 2.23/ 4 = 0.372 m³/s