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)
<u>V₂ = 1.86 m/s</u>
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(b)
For fluid flow rate we use the following equation:
Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)
<u>Q = 5.1 x 10⁻⁴ m³/s</u>
