Question

Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is $2m/s$ is $2m$ of $Hg$. What will be the pressure in the portion where velocity is $4m/s$

Answer 1

According to Bernoulli,s principle

$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1$ $=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

$\Rightarrow$ $2mHg+\frac{1}{2}\rho (2{)}^2$ $=P_2+\frac{1}{2}\rho (4{)}^2$

$\Rightarrow P_2=\frac{1}{2}\times {10}^3(2^2-4^2)+2mHg$

$\Rightarrow P_2=-6000+2Hg$

$\Rightarrow -\frac{6000}{101325}\left(0.76\right)mHg+2mHg$

$\Rightarrow P_2=1.955$ m of Hg