Question

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.

Answer 1

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;

                               $V_{2} = \frac{A_{1}V_{1}}{A_{2}}$

the pipe is widened to twice its original radius,

                                r₂ = 2r₁          

then the cross-sectional area A₂ = 4A₁

                           ⇒  $V_{2}= \frac{A_{1}V_{1}}{4A_{1}}$

                                  $V_{2}= \frac{V_{1}}{4}$

This implies that the water speed will drop by a factor of  $\frac{1}{4}$ because of the increase the pipe cross-sectional area.  

 The Bernoulli Equation;

     Energy per unit volume before = Energy per unit volume after    

        p₁ + $\frac{1}{2}$ρV₁²  + ρgh₁ = p₂ + $\frac{1}{2}$ρV₂²  + ρgh₂  

Total pressure is constant and $P_{T}$ = P = $\frac{1}{2}$ρV₂²ρV²  

        p₁ + $\frac{1}{2}$ρV₁²  = p₂ + $\frac{1}{2}$ρV₂²

Making p₂ the subject of the equation above;

        p₂ = p₁ + $\frac{1}{2}$ρV₁² - $\frac{1}{2}$ρV₂²

But $V_{2} = \frac{V_{1}}{4}$ so,

        p₂ = p₁ + $\frac{1}{2}$ρV₁² - $\frac{1}{2}$ρ$\frac{V_{1}^{2}}{4^{2}}$      

       p₂ = 3.0 x 10⁴ + ($\frac{1}{2}$ × 1000 × 12²) - ( $\frac{1}{2}$ × 1000 × 12²/4² )

      P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³    

       P₂ = 9.79 × 10⁴Pa      

      P₂ = 9.8 × 10⁴Pa