Question

) Oil (with sg = 0.987) flows through a pipe of diameter 2 inches at a rate of 2 gal/min at a pressure of 80 psi. Downstream the oil flows through a reducer to a pipe diameter of 0.75 inches. a. What is the change in flow rate? b. What is the fluid velocity in both pipes (inches/sec)? c. What is the change in the fluid pressure change?

Answer 1

a) Change in Flow Rate = 0

b) υ₁ = 2.45 in/s

υ₂ = 17.46 in/s

c) ΔP = 2.4 psi

Step-by-step explanation:

a)

Assuming the steady state flow, the volume flow rate of the oil must remain constant. Therefore,

Change in Flow Rate = 0

and,

V₁ = V₂ = (2 gal/min)(231 in³/gal)(1 min/60 sec) = 7.7 in³/s

b)

Since,

V = Aυ

υ = V/A

V = Volume Flow Rate

A = Cross-Sectional Area

υ = Velocity of Fluid

For Inlet Pipe:

A = A₁ = π(1 in)² = 3.141 in²

Therefore,

υ₁ = V₁/A₁

υ₁ = (7.7 in³/s)/(3.141 in²)

υ₁ = 2.45 in/s

For Outlet Pipe:

A = A₂ = π(0.375 in)² = 0.441 in²

Therefore,

υ₂ = V₂/A₂

υ₂ = (7.7 in³/s)/(0.441 in²)

υ₂ = 17.46 in/s

c)

Using Bernoulli's Equation, between inlet and outlet:

P₁ + ρυ₁²/2 + ρgh₁ = P₂ + ρυ₂²/2 + ρgh₂

since, h₁ = h₂

Therefore,

P₁ + ρυ₁²/2 = P₂ + ρυ₂²/2

P₂ = P₁ + ρυ₁²/2 - ρυ₂²/2

P₂ = P₁ + ρ/2(υ₁² - υ₂²)

where,

P₁ = Inlet Pressure = 80 psi

ρ = density of fluid = (sg)(ρ of water) = 0.987(62.4 lb/ft³) = 61.588 lb/ft³

ρ = (61.588 lb/ft³)(1 ft/12 in)³ = 0.0356 lb/ft³

Therefore,

P₂ = 80 psi + [(0.0161 kg/in³)/2][(2.45 in/s)² - (17.46 in/s)²]

P₂ = 77.6 psi

Therefore, the change in pressure is found to be:

ΔP = P₁ - P₂ = 80 psi - 77.6 psi

ΔP = 2.4 psi