Calculating PSI of water tower system with intermediate tank

I'm stumped.

I'm trying to design a system with an elevated water tank that feeds into an intermediate closed tank at ground level and then on to an irrigation pipe.

The reason for the elevated water tank is increased PSI. The reason for the intermediate ground level tank is increased water holding capacity without the added cost of an insanely heavy duty water tower.

My thinking is that if the intermediate tank is closed, the PSI at the end of the pipe (X), should be computed from the difference between the bottom of the tower tank and the end of the pipe.

Is that right?

Yes, as long as the intermediate tank is fully closed, the pressure will be measured from the top tank.

Agreed.  And it should be calculated from the top of the water level in the tower tank (just to be overly detail oriented)

Thanks guys! And I don't mean to be ungrateful, but are you sure on that?

Reason being, I'm thinking of the elevated tank being 275 gallons, and the ground level tank being some 2,500 or 3,000 gallons.

Where does the PSI come from? That is, how much of the PSI is generated by the weight of the 275 gallons pushing down on the other 2,500 ?

Or imagine it was only a 1 gallon container at the top of that water tower, seems magical that the PSI would be the same as if it were a lot larger mass up there.

I'm sure, but maybe I'm the only one.  Just pretend the lower tank is not there at all.  Your one gallon tank way up high would give you 23 psi (per the math in your graphic).  All the water in the line between the gallon and the spigot at the bottom is at that same pressure despite its lower elevation.  If that line happened to have a fat part in it that held an extra 3000 gallons, it would still transmit that pressure from above.  This is assuming that the water lines are sized big enough to not restrict the flow significantly.

The trick is maintaining that gallon way up high.  If you use two gallons of water, now you've lost the head pressure of that gallon and the pressure would drop to wherever the highest bit of water is in the system.

Municipal water towers have widely varying amounts of water in them over the course of the week.  Because of their shape, the water level only changes by a few feet.  The pressure for the city only varies by a bit due to the height change in the tank, not the massive amount of added water up in the tank.

The reason the size of the intermediate tank doesn't matter is that we're calculating pressure, which is force per square inch (psi), not total force. Pressure always comes from the total height of water, regardless of how much water there is (be it a barometer or the Oroville dam).  Try not to think of it as 275 gallons vs 2,500 gallons, but rather the pipe itself.  

Let's say you have a 2" pipe coming out the tower and 10psi of pressure in a 200" round tank. The total force in the pipe is 3.14 in^2 (pi*r^2) * 10 lbs per in^2 = 31.4 lbs. However, the total force in the tower is 314.15 in^2 * 10 lbs per in^2 = 3,141 lbs! That's why your tower needs to be so strong, but the pipe can still handle the pressure — it's a smaller cross section, and the pressure results in a much smaller force.

All that being said, with this setup your practical pressure will be governed by the intermediate tank's water level since it is likely your water tower will empty and water will be coming solely from the intermediate tank. If you go with this setup, I'd recommend installing a water pump at the intermediate tank so you can maintain pressure at all times, but turn the pump off and save energy when the water tower is empty.

thanks all! I think I get it now.

I'm drawing the water off the top of the system as it were, and not the bottom.

If that makes any sense.