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This is the problem and I am trying to find the velocity of water in the pipe.

I made my first point at the free surface of the water and my datum to be the very bottom. Part b of the question asked to find the velocity of the water in the pipe. I set my second point to be at the pipe end up top at 4 meters. I know that p1=50 kN/m^3 and V1=0, p2=0 and therefore I can find the velocity of the water in the pipe. However in the solutions for this part, they assumed the first point to be at the end of the pipe and the second point to be at the maximum height of the water. These points make sense to me, but the answer that I get in the solutions is not the same as mine when I apply Bernouli the way I explained. I am not sure if there is something I am missing but help on this problem would be great.

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$p_2$ is at equilibrium of pressure, therefore $p_2 = p_{atm} = 1bar$.

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Let point $1$ be the fluid surface of the tank and point $2$ the outlet of the vertical pipe. Now apply Bernoulli's Principle along that flowline:

$$\frac12 v_1^2+gz_1+\frac{p_1}{\rho}=\frac12 v_2^2+gz_2+\frac{p_2}{\rho}$$

Because the tank is large (and wide), $v_1\approx 0$ because the continuity equation tells us:

$$A_1v_1=A_2v_2$$

where the $A$s are the resp. cross sections and since as $A_1 \gg A_2$, then $v_1\approx 0$.

The first equation then rewrites as:

$$v_2=\sqrt{ 2\Big(g(z_1-z_2)+\frac{p_1-p_2}{\rho}\Big)}$$

Everything in that expression is known. $p_2$ is simply the atmospheric pressure ($101325\ \mathrm{Pa}$)

To detemine the final height $h$, apply Conservation of Energy to a mass element $\mathrm{d}m$:

$$\frac12 \mathrm{d}m v_2^2=\mathrm{d}m g \Delta z$$

So:

$$\Delta z=\frac{1}{2g}v_2^2$$

Where $\Delta z$ is with respect to $z_2$: $h=z_2+\Delta z$.

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