To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.
Our values are given as
[tex]d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m[/tex]
[tex]d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m[/tex]
From the continuity equations in pipes we have to
[tex]A_1V_1 = A_2 V_2[/tex]
Where,
[tex]A_{1,2}[/tex] = Cross sectional Area at each section
[tex]V_{1,2}[/tex] = Flow Velocity at each section
Then replacing we have,
[tex](\pi r_1^2) v_1 = (\pi r_2^2) v_2[/tex]
[tex](1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2[/tex]
[tex]v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1[/tex]
From Bernoulli equation we have that the change in the pressure is
[tex]\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)[/tex]
[tex]7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)[/tex]
[tex]7300= 8919.01 v_1^2[/tex]
[tex]v_1 = 0.9m/s[/tex]
Therefore the speed of flow in the first tube is 0.9m/s
