Question

Water flows through a horizontal tube of diameter 3.0 cm that is joined to a second horizontal tube of diameter 1.0 cm. The pressure difference between the tubes is 7.5 kPa.1. Find the speed of flow in the first tube.

Answer 1

Answer:

v₁ = 0.433 m/s

Explanation:

given,

diameter of first tube, d₁ = 3 cm

diameter of second tube, d₂ = 1 cm

Pressure difference= 7.5 kPa

using continuity equation

     A₁ v₁ = A₂ v₂

         $v_1=\dfrac{\pi r_2^2}{\pi r^2_1}v_2$

         $v_1=\dfrac{0.5^2}{1.5^2}v_2$

            v₂ = 9 v₁

Applying Bernoulli's equation

 $\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

 $7.5 \times 10^3= \dfrac{1}{2}\times 1000\times ((9v_1)^2-(v_1)^2)$

        80 v₁² = 15

        v₁² =0.1875

        v₁ = 0.433 m/s