Pressure at the outlet of a pipe

Question

I am revising for a fluid dynamics exams and I am struggling with the following question:

Consider a pipe that consists of a horizontal segment with a varying cross-sectional area that is aligned along the $x-axis$ and a downward segment of height, $h$, that is parallel to the $z -axis$. The two segments are joined together by a smooth bend.

Water flowing smoothly within the pipe experiences a gravitational force $g = −gk$ where $g$ is the gravitational acceleration and $\mathbf k$ is the unit normal pointing in the positive $z -direction$, with $z = 0$ along the horizontal segment of the pipe. The cross-sectional area at the inlet is given by $A_1$ while the area at the outlet is $A_2$ .

If the pressure at the inlet of the horizontal section is at atmospheric pressure given by $p_a$ , and the inlet velocity is $V_1$ , find the pressure at the outlet of the pipe.

I think I'm supposed to use Bernoulli's equations but I'm not sure which form and I struggle to interpret it so an explanation would be great.

Thanks!

Answer 1

Very simply , Let velocity at end be $v_2$ and Area of cross section be $A_2$

$$v_1\times A_1=v_2 \times A_2\tag{mass conservation in fluids}$$

Now, we know that $P+h\rho g+ \frac{1}{2}\rho v^2$ is constant,

$$P_a+h\rho g +\frac{1}{2}\rho v_1^2=P+\frac{1}{2}\rho\frac {(v_1A_1)^2}{(A_2)^2}$$

Therefore, answer is : $$P_a+h\rho g + \frac{1}{2}\rho \frac{v_1^2(A_2^2-A_1^2)}{A_2^2}$$