Answer:
9.8 × 10⁴Pa
Explanation:
Given:
Velocity V₁ = 12m/s
Pressure P₁ = 3 × 10⁴ Pa
From continuity equation we have
ρA₁V₁ = ρA₂V₂
A₁V₁ = A₂V₂
making V₂ the subject of the equation;
[tex]V_{2} = \frac{A_{1}V_{1}}{A_{2}}[/tex]
the pipe is widened to twice its original radius,
r₂ = 2r₁
then the cross-sectional area A₂ = 4A₁
⇒ [tex]V_{2}= \frac{A_{1}V_{1}}{4A_{1}}[/tex]
[tex]V_{2}= \frac{V_{1}}{4}[/tex]
This implies that the water speed will drop by a factor of [tex]\frac{1}{4}[/tex] because of the increase the pipe cross-sectional area.
The Bernoulli Equation;
Energy per unit volume before = Energy per unit volume after
p₁ + [tex]\frac{1}{2}[/tex]ρV₁² + ρgh₁ = p₂ + [tex]\frac{1}{2}[/tex]ρV₂² + ρgh₂
Total pressure is constant and [tex]P_{T}[/tex] = P = [tex]\frac{1}{2}[/tex]ρV₂²ρV²
p₁ + [tex]\frac{1}{2}[/tex]ρV₁² = p₂ + [tex]\frac{1}{2}[/tex]ρV₂²
Making p₂ the subject of the equation above;
p₂ = p₁ + [tex]\frac{1}{2}[/tex]ρV₁² - [tex]\frac{1}{2}[/tex]ρV₂²
But [tex]V_{2} = \frac{V_{1}}{4}[/tex] so,
p₂ = p₁ + [tex]\frac{1}{2}[/tex]ρV₁² - [tex]\frac{1}{2}[/tex]ρ[tex]\frac{V_{1}^{2}}{4^{2}}[/tex]
p₂ = 3.0 x 10⁴ + ([tex]\frac{1}{2}[/tex] × 1000 × 12²) - ( [tex]\frac{1}{2}[/tex] × 1000 × 12²/4² )
P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³
P₂ = 9.79 × 10⁴Pa
P₂ = 9.8 × 10⁴Pa
