1. Unit Volumes and Unit Weights:
1 cubic foot of water weighs 62.4 pounds
1 cubic foot of water equals 7.5 gallons
1 gallon of water weighs 8.34 pounds
2. Unit Pressure:
Pressures are usually expressed in pounds per square foot (psf)
or pounds per square inch (psi). This is known as unit
pressure.
A container which has dimensions of 1' x 1'
x 1'(the same as 12" x 12" x 12")
and is filled with water, weighs 62.4 pounds. The total
pressure on the bottom is therefore 62.4 lbs., and the unit pressure
is 62.4 psf.
Also, the unit pressure is:
From this illustration we see that a column of water 1' high
causes a pressure of 0.43 psi. How high must a column
of water stand to cause a pressure of 1 psi?
Example:
Find the pressure at the bottom of a standpipe in which water
stands 60 feet deep.
In the following diagram the unit pressure at
each valve will be the same.
The unit pressure at the valve in each case is 26 psi. In
the case of liquid at rest, the unit pressure, p, is determined
solely by the height of the liquid surface, h, and the unit weight
of the liquid, w (lbs./cu.ft.) or p = wh
Care must be taken to use (w and h) in the proper units: if
(w) is in lbs./cu.ft., (h) must be in feet.
Pressures are sometimes measured in terms of feet of head
which means the same thing as height to the water level.
For example, the head on the pipes from either of the tanks
is 60 feet.
3. Gage and Absolute Pressure:
The dials of gages are designed so that the readings
indicate pressures caused by the head of water.
Air has weight and therefore exerts a pressure. This pressure
is usually taken to be 14.7 psi (atmospheric pressures).
The absolute pressure at the bottom of the tank
is actually:
Absolute pressure = gage + atmospheric
=
26 psi + 14.7 psi
=
40.7 psi
4. Transmission of Pressure:
The principle of transmission of pressure illustrated above is that
the unit pressure at any point in a fluid at rest is transmitted
equally in all directions.
If P_{1} = 100 lbs. and A_{1} = 10 sq.
inches
Let (p) = pressure in pounds per square inch (psi)
So an exerted pressure, P_{1}, of 100 lbs. at A_{1}
results in a pressure, P_{2}, of 1,000 lbs. All unit pressures
are 10 psi. The pressure at any point in the vessel can be
figured in the same manner.
The hydraulic gage valve operates on the principle outlined above.
Water in Motion
Pressure and head in relation to bodies of water at rest have been
considered. This is called static pressure and static head.
When water is in motion, such as from the mechanical force exerted
by a pump in producing motion, the pressure created is called dynamic
pressure. There are factors other than elevation which affect
pressures when water is in motion.
Suppose a pump at a source of supply will pump water to a storage
tank with an overflow elevation 100 feet above the centerline of
the pump. A pressure gauge placed at the outlet of the pump
would show a pressure reading of 43 psi when the tank was full.
This would represent a backpressure against the pump. To
pump water, the pump would have to exert a force greater than the
pressure exerted by this 00 foot column of water. We
would say this pump would operate against a static head of 100 feet.
If the source of water for this pump came from a reservoir ten feet
above the centerline of the pump, there would be a positive pressure
on the suction side of the pump of ten feet (or 4.3 psi).
This is called suction head and would serve to relieve some of the
work of the pump, and it would only have to exert a force against
a net static head of 90 feet.(100 ft.  10 ft.).
If the source of water came from a well where the ground water stood
at an elevation ten feet below the centerline of the pump,
this would require an increased force to be exerted as a result
of the need to lift the water ten feet to the pump. This is
called suction lift and the head against which the pump must
operate would be increased to 100 feet.
The maximum lift which can be accomplished by suction is 33.9 feet.
This is the theoretical maximum, determined and limited by atmospheric
pressure, and is not attainable in field operations. It takes
just as much energy for a pump to lift water 10 feet by suction
as it does to force water 10 feet in elevation by pressure.
Fluid Flow Calculations
The amount of water passing a given section of a pipe line in a
unit time is called the discharge, Q.
If: A = cross sectional
area of pipe (square feet)
V
= velocity of water in pipe (feet per second)
Then: Q = AV = cubic feet of water
per second
Find the quantity of water passing through a 6" diameter pipe
if the velocity is 5 feet per second:
The crosssectional area, A, of a pipe is
In water works practice it is customary to express discharge in
gallons per minute (gpm) or in million gallons per day (mgd).
450 gpm x 60 min. = 27,000 gal./hr.
2,700 gph x 24 hrs. = 648,000 gal./day
=
0.65 mgd
The effect of reducing a pipe diameter is to increase the velocity.
For example, if the 6" pipe just mentioned discharges into
a 4" pipe, what will be the velocity in the 4" pipe?
Since the discharge through each pipe is the same,
(too high for water works practice).
Loss of Pressure
If only static head was involved in pumping operations, any force
exerted by the pump which was greater than the static head (or back
pressure), would result in water being pumped. But as
soon as water is in motion, there are other factors to be
considered. Most important of these factors is friction loss.
There is always friction in any moving body of water. In pressure
pipe lines, friction is caused by the water rubbing on the inside
surface of the pipe. Friction results in loss of pressure.
Friction loss is usually measured in "feet per 1,000 feet of pipe"
and may be easily converted to pressure loss in pounds per square
inch.
Pumps are designed to operate under specific head conditions.
In addition to the static head, all friction losses and minor
losses should be computed in order to determine the total
head against which the pump will operate. The pump will then
be specifically designed or selected from a standard design to provide
the desired capacity for the conditions under which it will operate.
The total pressure provided at the discharge side of the pump represents
the discharge pressure of the discharge head.
When water flows through a pipe, the water molecules rub against
each other and against the wall of the pipe. This is called
friction and it requires pressure to overcome it. The amount
of pressure required (or feet of head) to overcome the friction
cannot be used for any other purpose and it is therefore called
lost pressure or lost head.
Factors Influencing Head
Loss in Pipes
Friction loss in a pipe depends on the length of the line since
the head loss varies directly to the pipe length. The rate
of flow and the condition of the interior surface of the pipe
also influence head loss.
Type of pipe, pipe coating (if any), age of pipe, and the smoothness
or roughness of the interior surface of the pipe affects the friction
loss. A smooth interior surface would result in very little
friction loss at a reasonable velocity, whereas a very rough
surface would develop a considerable loss through friction, at the
same velocity.
The rate of flow is referred to as velocity, which is the speed
of the water. Velocity is measured in feet per second. Friction
loss increases as the velocity through the pipe increases.
If the flow through a pipe is doubled, the friction loss in the
pipe will be increased by almost four times.
The diameter of the pipe also affects the head loss, as the diameter
determines the area of wall in contact with flowing water.
Also, for a given discharge the diameter determines the velocity
of the water.
Increased friction may also be due to any of the following factors:

Sedimentation: mud, silt or sand.

Obstruction of the pipe due to debris:
sticks, boards, stones, tools and other things that may have
gotten into the pipe during construction.

Partly closed valves.

Accumulation of air at summits.

Mineral deposits and slime growths on walls of
pipe.

Tuberculation.
There are other minor losses such as losses resulting from changes
in size of pipe or direction of flow, meter losses, losses through
check valves or undersize gate valves, pump friction, etc.
The amount of friction developed is the criterion by which the size
of pipe and the amount of power required for pumping are determined.
When a given amount of water is to be transported, the total amount
of friction developed depends on the diameter and length of pipe
and the condition of its interior.
In determining the proper size of a new water pipe to be installed
in any water distribution system, it is necessary that the main
be large enough to deliver the expected demand at adequate pressure.
The probable maximum demand must be estimated and should be based
on use in comparable areas. Fire demands may be exceeded by
lawn sprinkling demands in residential areas. Allowance should
be made for probable future increases in demand. It may be
more economical to anticipate future demand rather than to replace
the main with a larger one at some future date.
Mains which serve fire hydrants should not be less than 6 inches
in diameter. Mains of smaller size will not be recognized by fire
insurance underwriters. Many large cities do not install any mains
smaller than 8 inches in size. An 8 inch main costs little
more than a 6 inch main to install, since trenching and other costs
remain the same. Yet an 8 inch main has more than twice the
capacity of a 6 inch main, under the same head. It is much
more desirable to oversize, rather than to undersize, when installing
new mains.
When mains are not of sufficient size to meet peak demands in an
area, pressures drop rapidly and inadequate service results.
Conditions of this nature have been rather prevalent in many cities
during the past few years because of increasing demands resulting
from improved standards of living, rapid population growth, new
appliances, air conditioning and other factors.
The American Water Works Association recommends a normal static
pressure of 60 to 75 psi through the system. A minimum pressure
of 45 psi is desirable in a residential area and 75 psi would be
desirable in a commercial or industrial area.
Calculation of Head Loss
There are several formulas for computing the carrying capacities
of water mains. The one most commonly used in the water works
industry is the HazenWilliams Formula. A factor called "C"
represents the coefficient of friction. It is a measure
of the roughness of the interior surface of pipe. A "C" factor
of 140 would indicate a very smooth pipe, whereas a "C" factor of
70 would indicate a very rough pipe. The smoother the
surface of the pipe wall, the larger the value of "C" and, consequently,
the greater the carrying capacity. To save time, these formulas
can be converted into chart form from which head loss values can
be read directly. A good example to show differences in head loss
is a flow of 200 gallons per minute in the following:

2 inch pipe  690 feet per 1000 feet

3 inch pipe  96 feet per 1000 feet

6 inch pipe  3.31 feet per 1000 feet

8 inch pipe  .82 feet per 1000 feet

10 inch pipe  .28 feet per 1000 feet

12 inch pipe  .11 feet per 1000 feet
Compare the head loss due to friction between the 2 inch and 12
inch pipe.
Problem:
A ground storage tank is to be constructed
on a hill with an overflow elevation of 586 feet. An
8" line is to be used to serve the town approximately 20,000
feet away from the tank. The elevation of the town is
approximately 380 feet. Find:
What fire flow can be expected in the center of
town 20,000 feet away while maintaining a 20 psi pressure
residual?
With a fire flow of 700 gpm what will be the head
loss in the 20,000 feet of pipe?
With a fire flow of 700 gpm what will the pressure
residual be at the hydrant in the center of town 10,000 feet
from the tank?
Solution:
a.
First find the difference in elevation
586'  380' = 206'.
Determine the feet of head necessary to provide a
20 psi pressure residual. (2.31' of water is equal to 1 pound
of pressure, therefore, 20 x 2.31 = 46.2')
Therefore, 46.2' of head is necessary.
Subtract this figure from the elevation difference
as this head must be constant and maintained at all times.
206'  46.2' = 159.8'. This 159.8'
of head may be used up in friction losses or 7.92' per 1000' of
pipe may be used. (Using 159' at Head).
Referring to the friction loss chart, in an 8" pipe
with a head loss of 7.92' per 1000' of pipe length, slightly less
than 700 gpm fire flow can be expected. (Because of fitting loss
and changes in the pipe with time).
From the solution in b), 158.4' of head is
lost due to friction. To find the available head and resulting
pressure remaining, subtract 158.4' from the total elevation difference
(head) as found in a) above or 206'. Therefore, 206'
 158.4' = 47.6'. Since 2.31' of head equals
1 psi, divide 47.6' by 2.31' to result in a residual pressure of
20.6 psi.
b.
Using the table with a 700 gpm fire flow, the friction
loss per 1000' of pipe is 7.92/1000'.
The correct multiplier for b.1) is 20,000 of
pipe per 1000' of pipe, or 20. Therefore, 7.92' x 20
results in 158.4' of frictional head loss in 20,000 feet of pipe.
c.
At 10,000' the pressure drop is 79.2'. The
residual pressure is 206'  79.2' = 126.8'
Diagram for Lost Pressure or Lost Head.
The pressure lines shown in the above sketch are called hydraulic
gradients. The elevation of any point on the hydraulic
gradient represents the height to which water would rise due to
pressure in the line at a point immediately below the given point
on the gradient.
The discharge pressure of a high service pump equals the elevation
through which the water is pumped plus pipe friction.
