Question

A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.43 m/s and the pipe diameter ????1 is 11.7 cm . At Location 2, the pipe diameter ????2 is 17.5 cm . At Location 1, the pipe is Δy=9.95 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.

Answer 1

Answer:

${P_2}-P_1=49.99\ KPa$

Explanation:

$v_1=9.43\ m/s$

$d_1=11.7\ cm/s$

$d_2=17.5\ cm/s$

From continuity equation

$A_1v_1=A_2v_2$

$v_1d_1^2=v_2d_2^2$

$v_2=\dfrac{9.43\times 11.7^2}{17.5^2}$

$v_2=4.21\ m/s$

$\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2$

${P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g$

$\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1$

$1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1$

${P_2}-P_1=49.99\ KPa$