Question

An oil refinery uses a Venturi tube to measure the flow rate of gasoline. The density of the gasoline is

Answer 1

Answer:

(a) V₂ = 1.86 m/s

(b) Q = 5.1 x 10⁻⁴ m³/s

Explanation:

(a)

The formula derived for Venturi tube is as follows:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa

ρ = Density of Gasoline = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

Therefore,

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²   ------------------- equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

Therefore,

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

using this value in equation (1):

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

(b)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)

Q = 5.1 x 10⁻⁴ m³/s