Page 117 - Water Loss Control

Data Collection, Formatting, and Management 99

8.2.7 Balancing Pressures
It is equally as important to balance pressures in a water system when attempting to
identify losses, as the system pressure plays a large part in water loss especially leakage
as discussed later in this manual.
Usually, when we want to balance pressures we work out hydraulic grade lines
(HGLs). Hydraulic grade is a sum of the ground level plus the static pressure at that
particular point and the lines are the chosen points connected up.
Most often in water loss control situations we will need to know
• The supply or inlet pressure
• Average zone pressure
• Critical point pressure
• Minimum service pressure required
• The number of hours the system is pressurized in cases where there is
intermittent supply

8.2.8 Balancing Levels
In systems with large storage capacity, it is also important to include the various tank or
reservoir levels in the water balance, as the change in volume over time may represent
significant flow and could be mistaken for loss.
In systems with a small amount of storage capacity, this is not so important, however,
this should not be overlooked. It is always better to overanalyze than underanalyze!

8.2.9 Putting Data into a Common Format
Putting data into a common format is extremely important. Metric and U.S. Customary
units should never be mixed and even when using one or the other it is still a good idea
to think about the method of data recording, which
has taken place in the field and the required report-
ing units. Don’t mix incompatible units.
If working in metric, for example, it is much
easier to work in cubic meters per hour flow if you measure your velocity in meters per
second and calculate your pipe effective area in square meters. The result will always
be automatically in meters and then it is just a simple case of deciding the time units.
For example, we measure a velocity of 2 m/s in a pipe, which has a diameter of 400 mm
(400 mm is actually 0.400 m). To calculate the area we would use our formula Pi × R , which
2
2
in this case would be 3.142 × 0.2 × 0.2. The answer would be 0.12568 m . We have a velocity
3
of 2 m/s so we would multiply this figure by two, which would give us 0.25136 m /s.
Now we must decide on a unit of time. Usually when working in the field with cubic
meters we would use cubic meters per hour of flow. There are 60 seconds in a minute and
60 minutes in an hour so we would multiply our flow of 0.25136 m /s by 3600, which
3
3
3
would give us 904.896 m /hr. We know that there are a 1000 L in 1 m so we could also say
that we have a flow of 904,896 L/hr. This number is quite large and if added to other large
numbers could lead to mistakes. If we wanted to express our flow in liters we would most
likely use liters per second. If that were our desired final number we would have taken
our figure above of 0.25136 and multiplied by 1000 to take our flow units from cubic
meters to liters. We would not need to multiply anything else as our original number was