The three reservoir problem is a classic hydraulics problem that appears in nearly every textbook
on civil engineering hydraulics.
It also frequently appears on both of the licensing examinations to become a Professional
Engineer.
The problem is typically presented with three sources of water – or three reservoirs – at
three different elevations.
All three reservoirs are hydraulically connected to a single point – in this case Point D
– which resides at a specific elevation.
In other words, there is some pressure head at D, which is the result of the hydraulic
grade line from, or to, each of the reservoirs.
Why on earth would someone ever install a system that has three different sources of
water, supplying three different amounts of energy to a system, is beyond me.
Such a situation is difficult to operate and maintain, and it doesn’t really matter because
these kinds of problems are so popular that even though they do not represent the typical
reality we deal with as practicing engineers they are popular with exam writers.
Approaching the 3 Reservoir problem we use conservation of mass and apply this to point
D. In other words the summation of the flowrates
coming into, or flowing out of, point D must sum to ‘0’.
We also use conservation of energy.
Applying conservation of energy here means that the elevation of reservoir A minus the
headloss in pipe 1 And the elevation of reservoir B minus the
headloss in pipe 2 And the elevation of reservoir C plus the
headloss in pipe 3 should all be equal to the elevation plus the pressure head at point
D. A typical analysis scenario usually involves
given all of the pipe data – the diameters, lengths, and roughness of the
pipelines – as well as giving the elevations of the reservoirs
and point D, the task would be to find the flow rates from
reservoir A, B, and C, as well as the pressure head gradient at point D.
The approach to this particular problem is to pick a value for the pressure head at point
D such that the summation of the flow into, and out of, point D is ‘0’.
In other words, once you have a guess for the pressure at D, then you can solve for
the headlosses getting to or from D for each of the water sources.
Once you know the headloss, then you can use Darcy-Weisbach or Hazen-Williams to calculate
the flowrate.
In other words the solution is found through iteration.
Therefore, to help speed up the iterations it is best to calculate the pipe characteristics
(in other words the ‘r’ term from the pipes in parallel presentation) in advance.
Another scenario is the flowrate from one of the reservoirs is known along with the
pipe data and elevations as well.
Here the task would be to find the flowrate to or from either of the other reservoirs.
Because you are given one flowrate, you can find the pressure head at D.
Once you know the pressure head at D, then you can solve for the headloss to or from
the other source in question.
Knowing the headloss from the source to D will allow you to calculate the flowrate and
then you are done – no iteration necessary.
There are, of course, other scenarios, but the key to approaching these problems is understanding
conservation of mass – in other words, the summation of flows at D is ‘0’ – and
conservation of energy dictates that the headloss from each reservoir to D results in the same
pressure gradient at point D.
The rest is just Algebra.