Answer:
Explanation:
Given
Diameter at section 1 is [tex]d_1=7.5\ cm[/tex]
diameter at section 2 is [tex]d_2=4.5\ cm[/tex]
Pressure [tex]P_1=31\ kPa[/tex]
[tex]P_2=25\ kPa[/tex]
[tex]A_1=\frac{\pi d_1^2}{4}=44.184\ cm^2[/tex]
[tex]A_2=\frac{\pi d_2^2}{4}=15.906\ cm^2[/tex]
Applying Bernoulli's Equation we get
[tex]P_1+\rho v_1^2+\rho gh_1=P_2+\rho v_2^2+\rho gh_2[/tex]
since there is no change in height therefore [tex]h_1=h_2[/tex] from continuity equation we can write as
[tex]A_1v_1=A_2v_2[/tex]
[tex]v_1=\frac{A_2}{A_1}v_2[/tex]
[tex]P_1-P_2=\frac{\rho }{2}\left [ v_2^2-v_1^2\right ][/tex]
[tex]P_1-P_2=\frac{\rho }{2}\left [ v_2^2-(\frac{A_2}{A_1})^2v_1^2\right ][/tex]
[tex]\frac{A_2}{A_1}=0.359[/tex]
[tex](31-25)\times 10^3=\frac{10^3}{2}\left [ v_2^2-(0.3599)^2v_2^2\right ][/tex]
[tex]6\times 2=v_2^2=\left [ 1-0.359^2\right ][/tex]
[tex]v_2=3.713\ m/s[/tex]
[tex]Q=A_2v_2[/tex]
[tex]Q=15.906\times 10^{-4}\times 3.173[/tex]
[tex]Q=59\times 10^{-4}\ m^3/s[/tex]
