Question

At a certain point in a pipeline the velocity is $1m/s$ and gauge pressure is $3\times {10}^{5}N/m^2$. Find the gauge pressure at a second point in the line $20m$ lower than the first, if the cross-section at the second point is half that at the first. The liquid in the pipe is water.

Answer 1

Given,

$g=9.8m/s^2$

$\rho ={10}^{3}kg/m^3$

$h_1=hm$

$h_2=(h-20)m$

$P_1=3\times {10}^{5}m/s$

$P_2=?$

$A_1=A{m}^2$

$A_2=\frac{A}{2}{m}^2$

$v_1=1m/s$

$v_2=?$

From the equation of continuity,

$A_1{v}_1=A_2{v}_2$

$A\left(1\right)=\frac{A}{2}{v}_2$

$v_2=2m/s$

From the Bernoulli's equation,

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

$(P_1-P_2)+\rho g(h_1-h_2)=\frac{1}{2}\rho (v_2^2-v_1^2)$

$(P_1-P_2)+\rho g(h-h+20)=\frac{1}{2}\rho (4-1)$

$(P_1-P_2)+20\rho g=\frac{3}{2}\times {10}^3$

$(P_1-P_2)+20\times {10}^3\times 9.8=1.5\times {10}^3$

$P_1-P_2=-194.5\times {10}^3$

$P_2=P_1+194.5\times {10}^3$

$P_2=300\times {10}^3+194.5\times {10}^3$

$P_2=4.945\times {10}^{5}Pa$