Question

Water is flowing smoothly through a closed-pipe system. At one point the speed of the water is $3.0m/s$,while at another point $1.0m$ higher the speed is $4.0m/s$. If the pressure is $20kPa$ at the lower point, what is the pressure at the upper point? What would the pressure at the upper point be if the water were to stop flowing and the pressure at the lower point were $18kPa$ ?

Answer 1

(i) $p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
$(20\times {10}^3)+\frac{1}{2}\times {10}^3\times {\left(3\right)}^2+0=p_2+\frac{1}{2}\times {10}^3\times {\left(4\right)}^2+{10}^3\times 10\times 1$
$\therefore$ $p_2=6.5\times {10}^3$ N/m${}^2$=$6.5$ kPa
(ii) Again applying the same equation, we have:
$(18\times {10}^3)+0+0=p_2+0+\left({10}^3\right)\left(10\right)\left(1\right)$
$\Rightarrow$ $p_2=8\times {10}^3$ N/m${}^2$
$=8$ kPa