# Problem: Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 4.80x10-2 m2; at point 3, where the water is discharged, it is 1.60x10-2 m2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.What is the gauge pressure pgauge at point 2?

###### FREE Expert Solution

Applying Bernoulli's principle to points 2 and 3:

$\overline{){{\mathbf{P}}}_{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho }}{{\mathbf{v}}_{\mathbf{2}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{h}}}_{{\mathbf{2}}}{\mathbf{\rho g}}{\mathbf{=}}{{\mathbf{P}}}_{{\mathbf{0}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho }}{{\mathbf{v}}_{\mathbf{3}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{h}}}_{{\mathbf{3}}}{\mathbf{\rho g}}}$

We're looking for P2.

We can solve for P2 as follows:

$\begin{array}{rcl}{\mathbf{P}}_{\mathbf{2}}\mathbf{-}{\mathbf{P}}_{\mathbf{0}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\mathbf{\rho }{{\mathbf{v}}_{\mathbf{3}}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\rho }{{\mathbf{v}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}{\mathbf{\rho gh}}_{\mathbf{3}}\mathbf{-}{\mathbf{\rho gh}}_{\mathbf{2}}\end{array}$

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###### Problem Details

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 4.80x10-2 m2; at point 3, where the water is discharged, it is 1.60x10-2 m2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.

What is the gauge pressure pgauge at point 2?