# What will be the pressure rise due to a freely falling water from a Tank from a 20 m height to a partially filled pipe?

My question is that what would be the rise in pressure due to the fall of water from a certain height to a partially filled pipe?

To be more clear, an overhead tank is connected with a pipe and the pipe is filled with water in the bottom reach where there is saddle in the alignment. The depth of fall from the top of the pipe to the filled up portion is about 20m. Let the water that flows in the pipe from the over head tank is q litres/second and let the dia. of the pipe is d mm. The flow of water in this 20 m reach is initially be a free fall just like a water fall ie. it does not follow the pipe flow characteristics in this reach.

Once, the pipe is fully filled with water, the pressure would be 20 m of water column. But in this case, ie. of free fall what would be the pressure (or pressure rise)?

• Welcome to Physics! This question is right at the boundary between conceptual homework problems that we like and the do-my-work problems that we discourage. I look forward to answers that address the conceptual part of the question.
– rob
May 1 '18 at 20:15

I'm not gonna give a complete answer, but some hints how to arrive at the result. As you may know, the force $F$ exerted on the water in the pipe is $F=\frac{dp}{dt}$. This means that the force acting on the water surface in the pipe is the infinitesimal change in the momentum in an infinitesimal amount of time. The pressure on the water becomes $\frac{F}{{\frac1 4}\pi d^2}$ (the force divided by the area enclosed by the pipe's circumference). Let's assume all the momentum is conversed in pressure (exerted on the surface and thus on the bottom of the pipe).
To calculate $\frac{dp}{dt}$ we can calculate the change in momentum of $q$ liter water (weighing about $q(kg)$) in one second (which is the same as $\frac{dp}{dt}$ because of the constant stream) and considering the $q(kg)$ as a point mass when reaching the surface (of course the distance of $20(m)$ is in reality diminished if $q(l)$ has fallen on the water in the pipe, but in this way we can calculate $\frac{dp}{dt}$).
We can use the equation $s=\frac{1}{2}10t^2$ to calculate the time it takes if $s=20(m)$, after which we can use $v=10t$ to calculate the velocity of the $q(kg)$, and thus the momentum change in one second, which is the same as $\frac{dp}{dt}$.