Question

A horizontal pipe line carries water in a streamline flow. At a point along the tube where the cross-sectional area is ${10}^{-2}$ $m^2$, the water velocity is $2$ $m{s}^{-1}$ and the pressure is $8000$ Pa. The pressure of water at another point where the cross-sectional area is $0.5\times {10}^{-2}{m}^2$ is?

Answer 1

$P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)$

$\frac{1}{2}\rho (v_2^2-v_1^2)$

$\frac{1}{2}\rho v_1^2(\frac{v_2^2}{v_1^2}-1)$

$\frac{1}{2}\times 1000\times (\frac{100}{25}-1)$

$P_2-8000=1500$

$P_2=9500$ Pa