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.

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Netta00 Netta00 · Answer 1 · 2019-10-09T13:27:52+02:00

Answer:

[tex]{P_2}-P_1=49.99\ KPa[/tex]

Explanation:

[tex]v_1=9.43\ m/s[/tex]

[tex]d_1=11.7\ cm/s[/tex]

[tex]d_2=17.5\ cm/s[/tex]

From continuity equation

[tex]A_1v_1=A_2v_2[/tex]

[tex]v_1d_1^2=v_2d_2^2[/tex]

[tex]v_2=\dfrac{9.43\times 11.7^2}{17.5^2}[/tex]

[tex]v_2=4.21\ m/s[/tex]

[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2[/tex]

[tex]{P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g[/tex]

[tex]\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1[/tex]

[tex]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[/tex]

[tex]{P_2}-P_1=49.99\ KPa[/tex]