Chapter 1: Water At Rest and In Motion Six Principles of Fluid Pressure 1) Fluid pressure is perpendicular to any surface on which it acts. 2) Beneath the surface of a liquid at rest, the pressure is the same in all directions (upward, sideward, downward). 3) Pressure applied to a confined fluid from without is transmitted equally in all directions. 4) The pressure of a liquid in an open vessel is proportional to its depth. 5) The pressure of a liquid in an open vessel is proportional to the density of the liquid. 6) Liquid pressure at the bottom of a vessel is unaffected by the size and shape of the vessel as long as the height of water remains the same. Pressure – Height – Density Relationship Formulas a. Pressure (P) = .434 X Height or P = .434H b. Height / Head (H) = 2.31 X Pressure or H = 2.31P Work Problems: Using the formulas above, solve the following: a. Find the pressure at the bottom of a standpipe filled with water 100 feet high. P = .434H P = .434 (100) P = 43.4 psi * The pressure in this formula is often referred to as back pressure (BP) in pumping operations. This back7pressure may be encountered during high7rise operations, while using dry standpipes, or pumping up or down hills. BP = .434H b. The static pressure in a fire hose connected to a standpipe is 150 psi. How high will that static pressure raise the water in the standpipe? H = 2.31P H = 2.31 (150) H = 346.5 ft Back Pressure 1) Multi7Story Buildings The average height per story is 10712 feet BP = .434H Therefore, BP per story is .434 (12) or 5.2 psi per story As a rule of thumb, 5 psi per story above the first floor is used for calculating BP in Chapter 1: Water At Rest and In Motion http://home.honolulu.hawaii.edu/~jkemmler/chapter1.htm 1 of 2 20/10/2011 20:05
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Chapter 1: Water At Rest and In Motion
Six Principles of Fluid Pressure
1) Fluid pressure is perpendicular to any surface on which it acts.
2) Beneath the surface of a liquid at rest, the pressure is the same in all directions
(upward, sideward, downward).
3) Pressure applied to a confined fluid from without is transmitted equally in all
directions.
4) The pressure of a liquid in an open vessel is proportional to its depth.
5) The pressure of a liquid in an open vessel is proportional to the density of the liquid.
6) Liquid pressure at the bottom of a vessel is unaffected by the size and shape of the
vessel as long as the height of water remains the same.
Pressure – Height – Density Relationship
Formulas
a. Pressure (P) = .434 X Height or P = .434H
b. Height / Head (H) = 2.31 X Pressure or H = 2.31P
Work Problems: Using the formulas above, solve the following:
a. Find the pressure at the bottom of a standpipe filled with water 100
feet high.
P = .434H
P = .434 (100)
P = 43.4 psi
* The pressure in this formula is often referred to as back pressure (BP)
in pumping operations. This back7pressure may be encountered during
high7rise operations, while using dry standpipes, or pumping up or down
hills. BP = .434H
b. The static pressure in a fire hose connected to a standpipe is 150 psi.
How high will that static pressure raise the water in the standpipe?
H = 2.31P
H = 2.31 (150)
H = 346.5 ft
Back Pressure
1) Multi7Story Buildings
The average height per story is 10712 feet
BP = .434H
Therefore, BP per story is .434 (12) or 5.2 psi per story
As a rule of thumb, 5 psi per story above the first floor is used for calculating BP in
Chapter 1: Water At Rest and In Motion http://home.honolulu.hawaii.edu/~jkemmler/chapter1.htm
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high7rise buildings.
Work Problems: Using the rule of thumb, find the BP for the following:
a. Fire on the 10th
floor level of a 207story office building
BP = 5 X 9
BP = 45 psi
note: the fire is only 9 floors above ground level (this can be tricky)
b. Fire on the roof top of a 207story office building
BP = 5 X 20
BP = 100 psi
note: in this case, the fire is actually 20 floors above the ground floor
because it is on the roof and not the floor level. (tricky too)
2) Uphill vs. Downhill
a. Uphill:
When pumping on a grade, either uphill or downhill, pump operators must take into
consideration the pressure loss or gain caused by BP.
When pumping uphill, the pump has to work harder to get the water to the desired
location because gravity is acting on the water and holding it back. The pump
pressure must be increased to overcome the back pressure.
Example: A fire engine is pumping water uphill through a hoseline that is
80ft above the firetruck.
BP = .434H
.434 (80)
34.7 psi
The pump operator will have to increase the pump pressure by 34.7 psi to make up
the difference in back pressure.
b. Downhill
When pumping downhill, the pump does not have to work as hard because gravity
is acting on the water, helping to move it through the fire hose. This means that the
BP gained will be in addition to the pump pressure reading. You won’t see the
pressure increase on your pump gauge, but the hoseman will feel it on the hoseline.
Example: A fire engine is pumping water downhill through a hoseline that
is 60ft below the firetruck.
BP = .434H
.434 (60)
26.04 psi
The pump operator will have to decrease the pump pressure by 26.04 psi
to negate the pressure increase caused by back pressure.
3) Types of Pressure
a. Static Pressure 7 Pressure of water at rest
b. Flow Pressure 7 Pressure of water flowing from nozzle
c. Residual Pressure – Pressure remaining in water main or inlet side of
fire pump after water is flowing
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Chapter 2: Velocity and Discharge
Drafting Operations
1) Theory
Drafting is a way in which a fire pump uses atmospheric pressure to draw water into
the fire pump from a static water source (see figure 2.1). Atmospheric pressure, at sea
level, is 14.7 psi. Fire pumps are capable of expelling its air through use of a priming
pump. If all the air within a pump is displaced and a good seal is maintained (no leaks
or loose fittings), the fire pump can create a vacuum'like atmosphere within the fire
pump. Once this vacuum'like atmosphere is obtained, the atmospheric pressure outside
the pump will be greater than the pressure within the fire pump, and water will be
forced into the pump through the drafting hose. This means that using the formula H =
2.31P, we can see how atmospheric pressure will force water into the fire pump up to
33.9 feet high (H = 2.31 X 14.7 psi).
Figure 2.1 – Drafting Operations
The theoretical lift of 33.9 feet is nearly impossible to obtain. Since the fire pump
cannot produce an absolute vacuum (0 psi), and there is friction loss in the drafting
hose, a practical lift of 22'25 feet is more realistic.
2) Factors Affecting Priming Operations (trying to obtain a vacuum)
a. Loose hose connections, loose covers, open gates, open valves, too long or too
small suction hose
b. Defective priming pump
c. Depth of water source
note: 1250 gpm pumpers should only take 30 seconds to prime
1500 gpm pumpers should only take 45 seconds to prime
3) Factors Affecting Lift
a. Altitude
b. Weather
c. Water temperature
d. Too long or too small suction hose
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Velocity
1) Formulas
For velocity, you will be given either the height of the water or the pressure. Therefore,
two formulas for velocity will be discussed.
a. Velocity = 8 X ÖHeight of water or V = 8ÖH
b. Velocity = 12.1 X ÖPressure of nozzle or V = 12.1ÖP
Work Problems:
What is the velocity of water flow is the nozzle pressure is 60 psi?
V = 12.1 ÖP
V = 12.1 Ö60
V = 12.1 (7.75)
V = 93.78 fps
What is the velocity of water flow from a water tank 50 ft high?
V = 8 ÖH
V = 8 Ö50
V = 8 (7.07)
V = 56.56 fps
Flow Velocities
The velocity of water varies inversely with the cross section of the hoseline and nozzle tip.
What does this mean??
With the same nozzle pressure:
Changing to a smaller nozzle tip will increase nozzle velocity / pressure
Changing to a larger nozzle tip will decrease nozzle velocity / pressure
The inverse relationship between velocity and nozzle size simply means that when one
increases, the other decreases and vice versa.
Nozzle Discharge – Gallons per Minute (GPM)
1) Discharge Formulas
a. With Nozzle
Discharge (GPM) = 30 X Diameter of Nozzle2 X ÖNozzle Pressure
GPM = 30d2ÖP
Work Problem: How many GPMs are flowing through a 2 ½” hoseline with a
1 1/8” nozzle tip with a nozzle pressure of 50 psi?
GPM = 30d2ÖP
GPM = 30 (1.125)2 Ö50 (nozzle size converted into a decimal)
GPM = 30 (1.27) (7.07)
GPM = 269.37
b. Without Nozzle (open butt)
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To find the discharge pressure of a hoseline without a nozzle, simply use 90% of
the original discharge formula. The hose diameter will substitute as the nozzle
size in this case.
GPM (open butt) = 90% X 30 X Hose Diameter2 X ÖPressure
GPM (open butt) = 27d2 ÖP
Work Problem: How many GPMs are flowing through a 2 ½” hoseline
without a nozzle attached to it at 50 psi?
GPM (open butt) = 27d2 ÖP
GPM = 27 (2.5)2 Ö50
GPM = 27 (6.25) (7.07)
GPM = 1193.06
Nozzle Reaction
1) Formula
Nozzle Reaction (NR) = 1.57 X Nozzle Diameter2 X Nozzle Pressure
NR = 1.57d2
P
note: In theory, the nozzle reaction will always be greater than the actual
nozzle reaction felt by the firefighter because:
a. The hoseline is in contact with the ground, and this absorbs
some of the nozzle reaction.
b. Bends in the hoseline as it is laid out will help to absorb some
of the nozzle reaction.
Work Problem: What is the nozzle reaction of a 2 ½” hoseline with a
1 1/8” nozzle tip flowing 50 psi?
NR = 1.57 d2 P
NR = 1.57 (1.125)2 (50) (nozzle size converted to a decimal)
NR = 1.57 (1.27) (50)
NR = 99.70 lbs
2) Safety Factors to Consider
a. Handling Hose Lines
i. Bends near the nozzle tend to straighten out. The hoseline
should be straight at least 10 feet back of the nozzle
ii. Nozzle reaction from fog streams is less than straight streams
iii. Open and close nozzle slowly because:
1. Initial nozzle reaction is greater than the nozzle reaction when
water is flowing
2. Sudden closing of nozzle sends pressure surges backwards. This
is called a “water hammer”. A water hammer can break the
hoseline, fire pump, and/or water main.
iv. When using handlines on ladders, the nozzle reaction could
cause the ladder to lift'off and fall away from the building. To help
avoid this dangerous situation from occurring firefighters should:
1. Fasten the ladder to the window sill
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2. Set the base of the ladder further away from the building
b. Ladder Truck Operations
i. If the hose should burst, the ladder / boom will whip violently
ii. Always try and shoot the fire stream in'line with the ladder.
Never turn the nozzle more than 15 degrees from the center of the
ladder.
Work Problem: What is the nozzle reaction of a ladder pipe
operation flowing 80 psi from a 2” nozzle tip?
NR = 1.57 d2
P
NR = 1.57 (2)2
(80)
NR = 1.57 (4) (80)
NR = 502.4 lbs (force)
Friction Loss
1) Effect of Flow Pattern
a. Laminar Flow – Low Flow Velocities
b. Turbulent Flow – High Flow Velocities
i. Friction loss in hose affected by:
1. Inner lining of hose
2. Age of hose
3. Thickness of hose lining
4. Type of hose jacked weave (will it expand or not)
2) Friction Loss in Hoses
a. Friction loss varies with quality of hose
b. Friction loss varies directly with length of hose line (the longer the hoseline, the
greater the friction loss).
c. Friction loss varies approximately as the square of the velocity of flow (the
faster the flow velocity, the greater the friction loss)
Example:
If the flow velocity is doubled – Friction loss is 4 times
greater (2)2 = 4
If the flow velocity is tripled – Friction loss is 9 times
greater (3)2 = 9
d. For a given velocity, friction loss varies inversely as the fifth power of the hose
diameter (the bigger the hose diameter, the less the friction loss)
Example:
If the hose size was doubled from 1 ½” to 3”
We see how 1 ½” X 2 = 3”
Inverting the 2 we get ½
Now we take the ½ and multiply it to the fifth power
½ 5
= ½ X ½ X ½ X ½ X ½ = 1/32
We can now conclude that if the hose size is doubled, the
new friction loss is only 1/32 as much as the original figure.
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e. For a given velocity of flow, friction loss is nearly independent of pressure. In
other words, he velocity of flow, and not the pressure, is the determining factor
in friction loss.
3) Formulas (more details in Chapter 5)
a. Friction Loss = 2 X Q2 + Q
FL = 2Q2 + Q Q = GPM ̧100
note: This formula is only for 2 ½” diameter hoselines.
The friction loss figure represents friction loss per 100 ft of 2 ½” hose.
i.e. If the hoseline is 600 ft, the friction loss figure must be
multiplied by 6 to get the total friction loss in that hoseline.
Work Problem: Find the friction loss in a 2 ½” hoseline 100 ft in length
flowing 300 gpm.
FL = 2Q2 + Q Q = GPM ¸ 100
FL = 2 (3)2 + 3 Q = 300 ¸ 100
FL = 2 (9) + 3 Q = 3
FL = 21 # per 100 ft length
FL = 21# (since there is only 100 ft of hoseline)
c. Engine Pressure = Friction Loss + Nozzle Pressure
EP = FL + NP
Work Problem: An engine is pumping through 600 ft of 2 ½” hoseline to a
nozzle that is flowing 200 gpm at 100 psi nozzle pressure. Find the engine
pressure.
EP = FL + NP FL = 2Q2
+ Q Q = GPM ¸ 100
EP = FL + 100 FL = 2(2)2 + 2 Q = 200 ¸ 100
EP = 60 + 100 FL = 2(4) +2 Q = 2
EP = 160 psi FL = 10 per 100ft hoseline
FL = 10 X 6 (# of hundreds of feet of hose)
FL = 60
d. Engine Pressure = Friction Loss + Nozzle Pressure + Back Pressure
EP = FL + NP + BP
Work Problem: An engine is pumping through 300 ft of 2 ½” hoseline
to a nozzle that is flowing 200 gpm at 100 psi nozzle
pressure. The nozzle is on a hill that is 60 ft higher
than the fire pump. Find the engine pressure.
EP = FL + NP + BP
FL = 2Q2 + Q Q = GPM ¸ 100 BP = .434H
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FL = 2 (2)2 + 2 Q = 200 ¸ 100 BP = .434 (60)
FL = 2 (4) + 2 Q = 2 BP = 26.04
FL = 10 per 100 ft hoseline
FL = 10 X 3 ( # of hundreds of feet of hose)
FL = 30
EP = FL + NP + BP
EP = 30 + 100 + 26
EP = 156 psi
note: If the nozzle was below the fire pump (downhill or in a
basement), the BP would have to be subtracted from the EP.
Remember that the gravity would cause the pressure to
increase, thus giving the firefighter on the hoseline too much
pressure.
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Chapter 3: Water Distribution System
General Remarks
1) Public Water Systems are designed to perform two functions.
a. Provide water for domestic, commercial and industrial use
b. Provide water for fire protection
i. Combination systems use same water source for both functions
ii. Separate systems use “potable” water for domestic use and use
brackish, salt or treated sewage water for firefighting.
2) Fire hydrants must be installed and maintained in accordance to standards set forth by
the American Water Works Association.
3) The term “Fire Plug” originated from the old days when pipes were made from
hollowed out wood and buried underground. These pipes were gravity fed from water
sources located in the area. If there were a need for water, the fire department would
expose the pipe by digging up the ground. Once exposed, a hole was drilled into the
wooden pipe. Water would then flow out of the hole by means of gravity. When the
fire was out, a tapered wooden plug was inserted into the hole and the pipe would be
re,buried.
Fire Hydrants
1) Dry Barrel Hydrants (see figure 3.1)
a. Dry barrel hydrants do not have water stored in the fire hydrant itself. Instead,
the water is stored in the piping below the hydrant. When the operating stem is
opened, water will begin to fill the hydrant. This type of hydrant is used in
areas where freezing can occur. If the water is stored in the hydrant, it could
freeze and that hydrant would be useless in a fire situation. Because the water
in the piping below the hydrant is constantly moving, it usually does not freeze.
b. Main Parts:
i. Dry Barrel
ii. Footpiece
iii. Bonnet
iv. Operating Stem
v. Main Valve
vi. Drain
2) Wet Barrel Hydrants (see figure 3.2)
a. Unlike the dry barrel hydrant, the wet barrel hydrant has water in the hydrant
right up to the discharge outlet. These hydrants should not be used in areas
where freezing could occur. These are the types of hydrants commonly used in
Hawai`i. However, there are some dry barrel hydrants in use today in Hawai`i.
b. Wet barrel hydrants have fewer parts than dry barrel hydrants.
c. Each outlet has an independent valve on a threaded stem with
operating nut on opposite side of barrel.
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Figure 3.1 – Dry Barrel Hydrant
Figure 3.2 – Wet Barrel Hydrant
3) Other Information
d. Hydrant Outlets
i. Every hydrant must have at least two outlets.
1. One pumper suction hose outlet (usually 4”)
2. One regular hose outlet (usually 2 ½”)
ii. Outlets must not be less than 18 inches from the ground level.
iii. Outlets must have a cap and chain.
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e. Hydrant Spacing
i. Hydrants should not be spaced more than 250 ft apart in
commercial / industrial areas and should have a minimum flow of 1000
gpm.
ii. Hydrants should not be spaced more than 350 ft apart in
residential areas and should have a minimum flow of 1000 gpm.
iii. Hydrants should not be spaced more than 700 ft apart in rural
areas and should have a minimum flow of 1000 gpm.
f. Branch Connection
i. The minimum size water main supplying fire hydrants in
Honolulu is 8 inches. Main sizes smaller than 6 inches are not suitable
for providing fire protection.
ii. Each wet barrel hydrant has its own gate valve. This gate valve
is located somewhere near the hydrant, and its location is indicated on
the fire hydrant. In case the hydrant should be damaged (if a car knocks
one over) the gate valve can be used to stop the flow of water to that
one hydrant without interrupting the flow to other hydrants on that same
water main.
4) Estimating Available Flow From Fire Hydrants
a. In order to estimate the amount of flow we have available in a hydrant, we
must first find the percentage of drop in pressure between static and residual
pressure.
i. Open hydrant with suction hose connected to fire truck. At this
time take note as to what your intake (suction) pressure gauge is
reading. This is the hydrant’s static pressure.
ii. Open a discharge gate for one hoseline. Again look at your
suction gauge. The pressure will be lower because some of the static
pressure will have been used for the first firefighting line. This is the
hydrant’s residual pressure.
iii. Subtract the residual pressure from the static pressure and
convert that number into a percentage. This is the percentage of drop in
pressure between static and residual.
Work Problem: A hydrant is connected to your fire truck with no
firefighting lines flowing. Your suction gauge reads 60
psi. After opening one firefighting line, your suction
gauge now reads 55 psi. What is the percentage of drop in
pressure?
Static Pressure = 60 psi
Residual Pressure = 55 psi
Pressure Difference = 5 psi
To find the drop in a %, simply divide the difference in
pressure by the static pressure.
5 psi ¸ 60 psi = .083 or 8.3%
b. Applying percentage of drop in pressure to practical situations.
i. Once the percentage of drop between static and residual
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pressure is found, that number can be used to help estimate the number
of additional hoselines the hydrant can supply. The estimates are based
on hoselines of the same diameter utilizing nozzles of the same diameter
also. The following chart shows the general rule of thumb regarding
additional hoselines:
10% or less 3 more hoselines
11,15% 2 more hoselines
16,25% 1 more hoseline
more than 25% no more hoselines
Chapter 3: Water Distribution System http://home.honolulu.hawaii.edu/~jkemmler/chapter3.htm
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Chapter 4: Fire Service Pumps
Introduction
1) Three Basic Types of Fire Pumps
a. Piston Type Fire Pump
b. Rotary Type Fire Pump
c. Centrifugal Type Fire Pump
2) Pump Mounting on Apparatus
a. Mid"ship (middle) – 2 ways
i. Between road transmission and rear axle in line with drive shaft
(most common)
ii. Ahead of clutch and transmission with flywheel and power take
off. This type allows for direct engine power to pump transmission
connection. It allows for driving and pumping simultaneously.
b. Front Mounting
i. From front of engine crankshaft connected to pump
transmission. This type also allows for driving and pumping
simultaneously.
3) Pump Ratings
a. Standard pumper capacity ratings start from 500 gpm and increase in 250 gpm
increments (NFPA 19 Specification) up to 2000 gpm.