Water pressure calculation from overhead tank to treatment plant

Question

I have an overhead water tank which is at a height of $38'$ from the ground level. I have a water treatment plant at ground level to which we need to provide a water line from the overhead tank. Requirement for the water treatment plant is $560 \; gpm$. The water Treatment plant is located $90'$ (horizontally) away from overhead tank. My questions are

1. How can we calculate the pressure in the line?
2. What diameter of line do we need to transfer water from the tank to the plant without using a pump?

Answer 1

Let us denote the British to metric conversion factors from $ft$ to $m$ and $gpm$ to $\frac{m^3}{s}$ by $ft\_m$ and $gpm\_m^3ps$ respectively. Let us denote the height $h=38'\cdot ft\_m$, the flow rate $q=560\;gpm\cdot gpm\_m^3ps$, the horizontal length $l=90'\cdot ft\_m$, the density of water in metric units $\rho\approx1000\;\frac{kg}{m^3}$, the acceleration due to gravity $g=9.81\frac{m^2}{s}$, the cross sectional area in $m^2$ of the pipe $A=\pi r^2$ where the radius of the connecting pipe in $m$ is denoted by $r$, the velocity of water in $\frac{m}{s}$ in the connecting pipe $v$, the height in $m$ of a point denoted $i\in\mathbb{N}$ in the connecting pipe $h_i$ and the pressure of water in $\frac{m}{s}$ at a point denoted $i\in\mathbb{N}$ in the connecting pipe $P_i$. Let $i=1$ and $i=2$ denote the points within the pipe located at the bottom of the tank and the top of the plant respectively.

Let $\bar{h}$ be the depth of the tank. From fluid statics we have that $P_1=\rho g \bar{h}$ and $v_1=0$. From the geometry we have that $h_1=h$ and $h_2=0$ and from kinematics we have that $v_2=\frac{q}{A}$. Applying Bernoulli's equation we have that $P_1+\frac{1}{2}\rho v_1^2+\rho g h=P_i+\frac{1}{2}\rho v_i^2+\rho g h_i$ and $P_1+\frac{1}{2}\rho v_1^2+\rho g h=P_2+\frac{1}{2}\rho v_2^2+\rho g h_2$ so that $P_1+\frac{1}{2}\rho v_1^2+\rho g h=P_2+\frac{1}{2}\rho v_2^2$.

1. Therefore, $P_i=P_1 + \rho g h - \frac{1}{2}\rho v_i^2 -\rho g h_i$ at any point $i$ along the pipe lying between the points $1$ and $2$
2. The pipe can be sized on considering that $A=\frac{q}{v_2}$ and that $v_2=\sqrt{2\frac{P_1}{\rho}+2 g h - 2\frac{P_2}{\rho}}$ where $P_2$ is the measured pressure within the tank (equal to atmospheric pressure if the pipe opens into the plant which is open to air and not full).

In order to account for the pipe loss (due to friction among other causes) in the absence of knowledge of loss constants, the sum on the left hand side in the Bernoulli's equations in this answer can be reduced by a nominal percentage, say $10\%$ for instance.

Answer 2

Another way to look at this could be in terms of energy transfer. The gravitational potential energy of the water in the tank, by virtue of it being above ground level, that is transferred to kinetic energy when the water comes out of the pipe into the plant at ground level. So, mgh = 1/2 m (v squared), v squared = 2gh. This will give you the velocity of the water at ground level. Then, to deliver the required volume, at this v, you can calculate the required cross-sectional area and diameter of the pipe. As previously raised, we do not know how much energy will be dissipated by the flow through the pipe due to friction, which also raises the water temperature a little, by say <4 C. So, the pipe needs to have its theoretical diameter increased, x-sectional area increased by say 10% +.You could fit a devise at the end of the pipe to adjust the x-sectional area to get the flow you want. Hope this helps.