Answer:
Explanation:
Given
Pressure drop [tex]\Delta P=650\ kPa[/tex]
inlet diameter [tex]d_1=31\ cm[/tex]
Outlet diameter [tex]d_2=19\ cm[/tex]
density of water [tex]\rho=10^3\ kg/m^3[/tex]
Suppose [tex]v_1[/tex] and [tex]v_2[/tex] be the inlet and outlet velocity
According to continuity equation
[tex]A_1v_1=A_2v_2[/tex]
where A=cross-section of Pipe
[tex]A=\frac{\pi d^2}{4}[/tex]
thus [tex]d_1^2v_1=d_2^2v_2[/tex]
[tex]v_2=v_1\times (\frac{d_1}{d_2})^2[/tex]
[tex]v_2=v_1\times (\frac{31}{19})^2[/tex]
Also from Bernoulli's Equation
[tex]\Delta P=\frac{1}{2}\rho (v_2^2-v_1^2)[/tex]
[tex]650\times 10^3=\frac{1}{2}\times 10^3\times (v_1^2(\frac{31}{19})^4-v_1^2)[/tex]
[tex]v_1=\sqrt{106.79}[/tex]
[tex]v_1=10.33\ m/s[/tex]
