Respuesta :
Answer:[tex]Q=0.5612 m^3/s[/tex]
Explanation:
Given
diameter of pipe([tex]d_1[/tex])=6 cm
diameter of pipe([tex]d_2[/tex])=4 cm
[tex]P_1=32 kPa[/tex]
[tex]P_2=24 kPa[/tex]
[tex]A_1=\frac{\pi }{4}6^2=9\pi cm^2[/tex]
[tex]A_2=\frac{\pi }{4}4^2=4\pi cm^2[/tex]
[tex]v_1=\frac{Q}{A_1}[/tex]
Applying bernoulli's equation
[tex]\frac{P_1}{\rho g}+\frac{v^2_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v^2_2}{2g}+z_2[/tex]
[tex]\frac{P_1}{\rho g}+\frac{\frac{Q^2}{A_1^2}}{2g}+z_1=\frac{P_2}{\rho g}+\frac{\frac{Q^2}{A_2^2}}{2g}+z_2[/tex]
since [tex]z_1=z_2[/tex]
[tex]\frac{32\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_1^2g}=\frac{24\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_2^2g}[/tex]
[tex]Q^2=\frac{8\times 2\times 81\pi ^2\times 16\pi ^2\times 10^{-4}}{65\pi ^2}[/tex]
[tex]Q^2=3149.3722\times 10^{-4} [/tex]
[tex]Q=\sqrt{3149.3722\times 10^{-4}}[/tex]
[tex]Q=0.5612 m^3/s[/tex]
