Question

A 6cm diameter horizontal pipe gradually narrows to 4cm.

Answer 1

Answer:$Q=0.5612 m^3/s$

Explanation:

Given

diameter of pipe($d_1$)=6 cm

diameter of pipe($d_2$)=4 cm

$P_1=32 kPa$

$P_2=24 kPa$

$A_1=\frac{\pi }{4}6^2=9\pi cm^2$

$A_2=\frac{\pi }{4}4^2=4\pi cm^2$

$v_1=\frac{Q}{A_1}$

Applying bernoulli's equation

$\frac{P_1}{\rho g}+\frac{v^2_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v^2_2}{2g}+z_2$

$\frac{P_1}{\rho g}+\frac{\frac{Q^2}{A_1^2}}{2g}+z_1=\frac{P_2}{\rho g}+\frac{\frac{Q^2}{A_2^2}}{2g}+z_2$

since $z_1=z_2$

$\frac{32\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_1^2g}=\frac{24\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_2^2g}$

$Q^2=\frac{8\times 2\times 81\pi ^2\times 16\pi ^2\times 10^{-4}}{65\pi ^2}$

$Q^2=3149.3722\times 10^{-4}$

$Q=\sqrt{3149.3722\times 10^{-4}}$

$Q=0.5612 m^3/s$